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bull Preacutesenter la simplification 3D vers 2D 1D
bull Rappeler des eacuteleacutements manquants dans le modegravele simplifieacute
bull Preacutesenter le triangle de vitesses
bull Deacutecrire le triangle normal et les triangles speacuteciaux
bull Preacutesenter les coefficients de charge de deacutebit et le degreacute dereacuteaction
Dans le machines axiales le fluide entre et sort avec une vitesseprincipalement axiale
Dans ces dispositifs le deacutebit est important mais le gain (oulrsquoinverse) en pression est faible Pour cette raison ont utiliseplusieurs eacutetages pour traiter le gain (ou chute) en pression
Chaque eacutetage comprend un rotor (mobile) et un stator (fixe)
Dans un compresseur le fluide est drsquoabord acceacuteleacutereacute dans lerotor ougrave il y a un eacutechange drsquoeacutenergie avec le fluide
Dans le stator on modifie la forme drsquoeacutenergie lrsquoeacutenergie cineacutetiquereacutesiduelle du fluide est transformeacutee en pression
Dans une turbine le fluide agrave haute eacutenergie est drsquoabord etacceacuteleacutereacute et guideacute dans le stator pour lrsquoajuster aux exigencesgeacuteomeacutetriques est cineacutematiques du rotorPar la suite lrsquoeacutenergie cineacutetique de lrsquoeacutecoulement est transmise aurotor en mecircme temps qursquoun changement de direction a lieu
Ces procegraves sont reacutepeacuteteacutes pour des eacutetages conseacutecutifs
Agrave cause du mouvement du rotor par rapport au stator deux typesde vitesses relatives et absolues seront preacutesentes
C2x
C2
C2uC2m
C1uC1m
C1x
Machine axiale
➊
➋
Machine radiale
119963119963
120637120637119955119955
Plan 119903119903 minus 120579120579119911119911 = 119888119888119888119888119888119888119888119888119888119888
Carter
119955119955 Moyeu
Plan 119903119903 minus 119911119911
Plan 119903119903 minus 119911119911120579120579 = 119888119888119888119888119888119888119888119888119888119888
119955119955
Rayon moyen
Vues drsquoun compresseur axial
Plan 119911119911 minus 120579120579119903119903 = 119888119888119888119888119888119888119888119888119888119888
119880119880 = 119903119903120596120596
Moyeu
Carter
Rm
X
CmCx
C
Ω
119888119888 119907119907119907119907119888119888119888119888119888119888119888119888119888119888 119889119889119888119888 119897119897primeeacute119888119888119888119888119888119888119897119897119888119888119888119888119888119888119888119888119888119888
intrados
extrados surface S1
surface S2 Rayon moyen r
Lrsquoeacutetude est reacutealiseacute sur la surface 1198781198781
119932119932 = 119955119955120654120654
120654120654
θ
x
Zone aubeacutee
surface S1
119932119932 = 119955119955120654120654
Rotor Stator
Rotor-Stator
RotorStator
Dans une turbomachinelrsquoensemble stator-rotor estappeleacute eacutetage
W2
C3
U
W3
U
C1W3
U
C2
1
2
3
W vitesse relative de lrsquoeacutecoulement
U vitesse peacuteripheacuterique du rotor
C vitesse absolue de lrsquoeacutecoulement
W2
U
C2
C3W3
U
C1W1
URotor
Stator
➀
➁
➂
C
W
U
Stator Rotor
U2
Stator
C2
U3
W3
Rotor
Stator
Rotor
U2
W2
Rotor
Stator
C2
U3
W3C3
Stator
Rotor
Effets 3DEffets visqueuxDistribution de pressionTraicircneacutee PortanceEacutecoulement instationnaireInteacuteraction fuide structureTurbulence
Extrados
Intrados
L
D
Profil isoleacuteGrille drsquoaubes
Pour analyser lrsquoeacutechange drsquoeacutenergie entre le fluide et le rotor drsquouneturbomachine on utilise la variation de la quantiteacute demouvement (changement de direction etou de vitesse)
Lrsquoeacutetude classique unidimensionnelle utilise le concept de trianglede vitesse Ce modegravele est appliqueacute agrave lrsquoentreacutee et agrave la sortie ducanal inter-aube (grille drsquoaubes)
Dans celui-ci on retrouve la vitesse absolue exprimeacutee dans unrepegravere lieacute aux parties fixes (le stator) accompagneacutee de lavitesse relative lieacutee aux partie tournantes ( le rotor)
Approche classique
Lrsquoapplication du triangle de vitesse permet drsquoestimer
bull lrsquoangle drsquoincidence agrave lrsquoentreacutee du rotor
bull lrsquoangle au bord de fuite du rotor
bull la deacuteflection de lrsquoeacutecoulement dans le stator
bull le travail produit ou consommeacute par la roue
Approche classique
C vitesse absolue de lrsquoeacutecoulement
W vitesse relative de lrsquoeacutecoulement
U vitesse peacuteripheacuterique du rotor
Cu Cm Cx composante tangentielle radiale et axiale de lavitesse absolue du fluide
Wu Wm Wx composante tangentielle radiale et axiale de lavitesse relative du fluide
α lrsquoangle des vitesses absolues mesureacutees par rapport agrave la directionaxiale
β lrsquoangle des vitesses relatives mesureacutees par rapport agrave la directionaxiale
La forme des pales du rotor deacutepend des angles 120631120631(des vitesses relatives dans le repegravere du rotor)
La forme des pales du stator deacutepend des angles 120630120630(des vitesses absolues dans le repegravere fixe)
Si lrsquoon veut garder la vitesse axiale 119940119940119961119961 = 119940119940119940119940119940119940119940119940119940119940 lrsquoeacutequation deconservation de la masse 119950 = 120646120646120488120488119940119940119961119961 = 119940119940119940119940119940119940119940119940119940119940 impose 120646120646120488120488 = 119940119940119940119940119940119940119940119940119940119940
Compresseur
Cu
Cm
Cu
Plan normal
Plan meacuteridional
BABF
W2
C2
U
U
W3C3
u
Les vecteurs vitesse (vitesse absolue 119940119940 pour le stator etrelative 119960119960 pour le rotor) sont consideacutereacutes tangents agrave laligne de squelette de lrsquoaube
β2
β3W3
W2
Pour les machines axiales (u= uent = usort ) le trianglede vitesse donne lieu agrave des eacutequations scalaires tellesque
cu= wu+u
cx= wxcx= = wx c
U
w
wu
cu
Pour les machines agrave eacutetages multiples on suppose que lavitesse axiale demeure constante Crsquoest agrave dire que119940119940120784120784119961119961= 119940119940120785120785119961119961= 119940119940119961119961= 119940119940119940119940119940119940119940119940119940119940
Cette condition jumeleacutee agrave 119932119932120784120784= 119932119932120785120785= 119932119932 megravene agrave lrsquohypothegravese drsquoune cineacutematique reacutepeacutetitive En pratique
Les triangles de vitesse agrave lrsquoentreacutee et agrave la sortie drsquouneacutetage peuvent ecirctre superposeacutes
Nous commenccedilons par la repreacutesentation de triangles de vitessespour le cas drsquoune turbineNous rappelons que par hypothegravese le rayon ne varie pas dans leplan inter-aube et que la vitesse axiale demeure constante dansle sens axialCes conditions permettent drsquoeffectuer la superposition destriangles de vitesse au bord drsquoattaque et bord de fuite du rotoravec la vitesse 119932119932 commune
c1w1
u
c2
u2
w2
➊
➋
➌ c3w3
u
Turbine
C2
W2 C3
W3
Stator Rotor
C2u C3u
W3u
α2β2 β3
α3
∆Cu= ∆Wu
U
W2u
119940119940119961119961119880119880 = 119888119888119888119888119888119888119888119888119888119888119888119888119909119909 = 119888119888119888119888119888119888119888119888119888119888
1198881198883 = 1198881198881
1205721205723 = 1205721205721
Cineacutematique reacutepeacutetitive
➁ ➂
➁U➀
Turbine
119914119914120784120784
119914119914120785120785
119934119934120785120785
119932119932119934119934120784120784
119932119932
119914119914120783120783120630120630120783120783
120630120630120784120784120631120631120784120784
120630120630120785120785
120631120631120785120785
Stator Rotor
➀ ➁ ➂
Le cas du compresseur est similaire agrave celui drsquoune turbineLe rayon ne varie pas (lrsquoeacutetude se deacuteroule dans le plan inter-aube) et la vitesse axiale demeure constante dans le sens axialComme pour la turbine nous pouvons superposer les triangles devitesse au bord drsquoattaque et bord de fuite du rotor
W2 C2
C3W3
U
C1W1
U
U
Compresseur
➊
➋
➌
α2 β1
C2 W1
C1W2
U
β2α1
W2U C1U
CX
( )2 1 = minuse u uW U C C
Lrsquoopeacuteration est vectorielle
C2U
119914119914120784120784119958119958 minus 119914119914120783120783119958119958 =- 119934119934120784120784119958119958 minus119934119934120783120783119958119958
➀
➁
➁
Compresseur
119914119914120784120784119914119914120785120785
119934119934120783120783
119932119932119914119914120783120783120630120630120783120783 120630120630120784120784
120631120631120784120784120630120630120785120785
Rotor Stator
119934119934120784120784120631120631120783120783
119932119932
➀ ➁ ➂
Trois quantiteacutes fondamentales sont utiliseacutees dans laconception des turbomachines
bull le coefficient de charge ψ
bull le coefficient de deacutebit Φ
bull le degreacute de reacuteaction R
Approche classique
Parfois ψ est consideacutereacute positif pour les turbines et neacutegatifpour les compresseurs 119882119882119890119890 = frasl119882 119888
Coefficient de charge
2Ψ = eWU
Coefficient de charge
2e
UW
=ψ 02 01 02 2
h h hU Uminus ∆
= = 02
s shU
η ∆=
=
eWWm
120578120578119904119904 =∆ℎ0∆ℎ0119904119904
Turbine
Coefficient de deacutebit
xcU
Φ =
xCU
Φ =ρRT
pρ
= xC RTRTU p
=RT m RT
U Ap
xCU
Φ =119888 = ρ119862119862119909119909119860119860
11988811987711987711987911987901199011199010119860119860
= 119872119872 120574120574 1 +120574120574 minus 1
21198721198722
minus 120574120574+1)2(120574120574minus1
Afin de geacuteneacuteraliser le triangle de vitesses drsquoune machineaxiale on effectue une mise agrave lrsquoeacutechelle en divisant chaquecocircteacute par la vitesse tangentielle U
Cette opeacuteration permettra de retrouver directement lescoefficients de charge ψ et de deacutebit Φ
Par la suite un nouveau paramegravetre R qursquoon appelle degreacutede reacuteaction sera introduit pour compleacuteter la caracteacuterisationde lrsquoaubage
Approche classique
α2
β2 α3
β3C2 W3
W2
c3u - c2u= w3u - w2u
W3uW2u
U
CX CX
Lrsquoopeacuteration est vectorielle
C3
120511120511 = frasl119934119934119940119940 119932119932120784120784 = frasl119914119914120785120785119958119958 minus 119914119914120784120784119958119958 119932119932
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
Turbine
3
2
3 3
2 2
12
12
12
12
u
u
u u
u u
c RUc RU
w c RU Uw c RU U
Ψ= minus +
Ψ= minus +
Ψ= minus = +
Ψ= minus = minus
2 22 23 2
2 22 23 2
1 12 2
2 2
c cR RU U
w wR RU U
Ψ Ψ = Φ + minus + = Φ + minus +
Ψ Ψ = Φ + + = Φ + minus
3 2
3 2
1 2 1 2
2 2
R Ratan atan
R Ratan atan
minus + Ψ minus + Ψ = = Φ Φ
Ψ + Ψ minus = = Φ Φ
α α
β β
α2β1
C2U
W1U
C1UW2U β2 α1
Compresseur
ΦC2U
C3UW3U
W2U
ψ
1
ψ
ΦC2UC3U
W3U
W2U
1
Il faut deacutefinir un troisiegraveme paramegravetre pour diffeacuterencier ces deux triangles de vitesses ayant le mecircme 120509120509 et le mecircme 120537120537
Les coefficients 120509120509 et 120537120537 ne suffisent pas pour deacutefinir de maniegravereunivoque le triangle de vitesses
Le degreacute de reacuteaction est un paramegravetre adimensionnel quiexprime la variation drsquoenthalpie subie par le fluide dans le rotorpar rapport agrave la chute totale drsquoenthalpie dans lrsquoeacutetage
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
Variation d enthalpie dans le rotorRVariation d enthalpie dans l eacutetage
=
❸❶
❷
Rotor
Stator
R=0 machine drsquoaction (Δhrotor=0 Δheacutetage=Δhstator)
R=1 machine agrave reacuteaction (Δhstator=0 Δheacutetage=Δhrotor)
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
R indique la reacutepartition du saut drsquoenthalpie (de pression) statique entre le stator et le rotor
Le degreacute de reacuteaction R deacutefini en fonction de variablesthermodynamiques (lrsquoenthalpie) peut aussi srsquoexprimer en fonctionde variables cineacutematiques preacutesentes dans le triangle de vitessesNotamment en fonction de la vitesse peacuteripheacuterique U et descomposantes des vitesses absolues et relatives
La vitesse absolue agrave lrsquoentreacutee du stator est eacutegale agrave celle agrave la sortie du rotor (c1= c3 eacutetage reacutepeacutetitif )
On neacuteglige les pertes et le transfert de chaleur dans le stator
1 2 3
ℎ3 minus ℎ1= ℎ3 + frasl11988811988832 2 minus ℎ1 minus frasl11988811988812 2 = ℎ03 minus ℎ01 = 120549120549ℎ0
ℎ03 minus ℎ01 = ℎ03 minus ℎ02 = 120549120549ℎ )0(3minus2
ℎ03 minus ℎ02 = ℎ3 minus ℎ1
119945119945120782120782120783120783 = 119945119945120782120782120784120784
frasl11988811988832 2 = frasl11988811988812 2
3 2
3 1
rotor
eacutetage
h h hRh h h
∆∆
minus= =
minus
2 20 3 2 2 3
0 3 2
2 2( )
( )
h c c h
∆∆
minus
minus
+ minus=
ℎ03 ℎ02
(ℎ01= ℎ02 1198881198881= 1198881198883)
120549120549ℎ )0(3minus2 = (ℎ3 + frasl11988811988832 2) minus (ℎ2 + frasl11988811988822 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
0(3 2) 3 2( )∆ minus = minusu uh U c c
= 1 minusfrasl11988811988832 2 minus frasl11988811988822 2120549120549ℎ )0(3minus2
119877119877 =120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
120549120549ℎ )0(3minus2
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
3 212+
= minus u uc cRU
119888119888119909119909 = 1198881198882119909119909 = 1198881198883119909119909 = 119888119888119888119888119888119888119888119888119888119888
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 11988811988821199061199061198881198882 = 1198881198881199091199092+1198881198881199061199062
119877119877 = 1 minus11988811988831199061199062 minus 11988811988821199061199062
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
Indirectement lrsquoeacutequation pour R deacutecrite en fonction de composantes cineacutematiques eacutetablit un lien entre les angles drsquoentreacutee et de sortie
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
= 1 minus1198881198883119906119906 + 1198881198882119906119906
2119880119880
2232
2 32 2wwh h+ = +
2 23 2
2 3 2 2w wh hminus = minus
2 2 2 23 3 2 2
2 3 2 2u x u xw w w wh h + +
minus = minus
Machine axiale U2 = U3
La composante axiale de la vitesse demeure constante w3x = w 2x
( eacutequation justifieacutee plus tard lors de la rothalpie)
2 23 2
2 3 2 2u uw wh hminus = minus
3 2 3 22 3
( )( )2
u u u uw w w wh h minus +minus =
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
Eacutetage reacutepeacutetitif c3 = c1
Si les pertes dans le stator sont neacutegligeacutees
c2u ndash c3u= w2u - w3u
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
3 2 3 23 2
( )( )2
u u u uw w w wh h minus +minus = minus
c2u ndash c3u= w2u - w3u
119877119877 = minus1199081199083119906119906 + 1199081199082119906119906
2119880119880
2 2 2 22 3 2 3 2 3( ) 2 ( ) 2h h u u w wminus = minus minus minus
2 2 2 2 2 23 2 3 2 3 2
2 2 2e
I II III
u u c c w wW minus minus minus= + minus
2 2 2 22 3 2 3
2 2 2 2 2 22 3 2 3 2 3
( ) ( )( ) ( ) ( )
minus minus minus=
minus + minus minus minusu u w wR
c c u u w w
3 2= =U u u
Expliqueacutee + tard
2 23 2
03 02 3 2 2ec cW h h h h minus
= minus = minus +
119877119877 =ℎ2 minus ℎ3ℎ1 minus ℎ3
=ℎ2 minus ℎ3ℎ02 minus ℎ03
Degreacute de reacuteaction
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
( )3u 2uw wR2U+
= minus
3 212
u uc cRU+
= minus
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
Puisque le degreacute de reacuteaction R peut aussi srsquoexprimer enfonction de variables cineacutematiques on peut chercher sarepreacutesentation dans le triangle adimensionnel de vitesses
Turbine
α2
β2 α3
β3
C2 W3
C3W2
C3uC2u
U
C2xC3x
W2u
c2u ndash c3u= w2u - w3u
2Ψ minusR
frasl119934119934120784120784119958119958 119932119932 = frasl1198621198622119906119906 119880119880 minus frasl119880119880 119880119880Ψ = frasl1198621198622119906119906 minus 1198621198623119906119906 119880119880
119877119877 = 1 minus (1198621198623119906119906 + 1198621198622119906119906 frasl) 2119880119880Φ = frasl1198621198622119909119909 119880119880 = frasl1198621198623119909119909 119880119880
= (119914119914120784120784119958119958 minus 119914119914120785120785119958119958 frasl) 2119880119880 minus 1 + (119914119914120784120784119958119958 + 119914119914120785120785119958119958 frasl) 2119880119880
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α2
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
Les repreacutesentations des triangles de vitesses adimensionnellessont tregraves riches en information et contiennent plusieurs formules
Par exemple pour un compresseur on deacuteduit aiseacutement que
Avec 120595120595 = 1198821198821198901198901198801198802
Φ = 119888119888119909119909119880119880
et que pour 119888119888119901119901 = 119888119888119888119888119888119888119888119888119888119888 119882119882119890119890 = 119888119888119901119901(11987911987902minus11987911987901)on trouve
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
(11987911987902minus11987911987901) =119880119880119888119888119909119909119888119888119901119901
(tan1205731205731 minus tan1205731205732)
Alors si lrsquoon veut obtenir une plus grande variation detempeacuterature 11987911987902 minus 11987911987901 indirectement un plus grand rapport(1199011199010211990111990101) on peut augmenter
bull la vitesse peacuteripheacuterique 119880119880 (la vitesse de rotation)bull la vitesse axiale 119888119888119909119909 (le deacutebit massique)bull lrsquoangle de deacuteflection de la pale(tan1205731205731 minus tan1205731205732)
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
Dans le machines axiales le fluide entre et sort avec une vitesseprincipalement axiale
Dans ces dispositifs le deacutebit est important mais le gain (oulrsquoinverse) en pression est faible Pour cette raison ont utiliseplusieurs eacutetages pour traiter le gain (ou chute) en pression
Chaque eacutetage comprend un rotor (mobile) et un stator (fixe)
Dans un compresseur le fluide est drsquoabord acceacuteleacutereacute dans lerotor ougrave il y a un eacutechange drsquoeacutenergie avec le fluide
Dans le stator on modifie la forme drsquoeacutenergie lrsquoeacutenergie cineacutetiquereacutesiduelle du fluide est transformeacutee en pression
Dans une turbine le fluide agrave haute eacutenergie est drsquoabord etacceacuteleacutereacute et guideacute dans le stator pour lrsquoajuster aux exigencesgeacuteomeacutetriques est cineacutematiques du rotorPar la suite lrsquoeacutenergie cineacutetique de lrsquoeacutecoulement est transmise aurotor en mecircme temps qursquoun changement de direction a lieu
Ces procegraves sont reacutepeacuteteacutes pour des eacutetages conseacutecutifs
Agrave cause du mouvement du rotor par rapport au stator deux typesde vitesses relatives et absolues seront preacutesentes
C2x
C2
C2uC2m
C1uC1m
C1x
Machine axiale
➊
➋
Machine radiale
119963119963
120637120637119955119955
Plan 119903119903 minus 120579120579119911119911 = 119888119888119888119888119888119888119888119888119888119888
Carter
119955119955 Moyeu
Plan 119903119903 minus 119911119911
Plan 119903119903 minus 119911119911120579120579 = 119888119888119888119888119888119888119888119888119888119888
119955119955
Rayon moyen
Vues drsquoun compresseur axial
Plan 119911119911 minus 120579120579119903119903 = 119888119888119888119888119888119888119888119888119888119888
119880119880 = 119903119903120596120596
Moyeu
Carter
Rm
X
CmCx
C
Ω
119888119888 119907119907119907119907119888119888119888119888119888119888119888119888119888119888 119889119889119888119888 119897119897primeeacute119888119888119888119888119888119888119897119897119888119888119888119888119888119888119888119888119888119888
intrados
extrados surface S1
surface S2 Rayon moyen r
Lrsquoeacutetude est reacutealiseacute sur la surface 1198781198781
119932119932 = 119955119955120654120654
120654120654
θ
x
Zone aubeacutee
surface S1
119932119932 = 119955119955120654120654
Rotor Stator
Rotor-Stator
RotorStator
Dans une turbomachinelrsquoensemble stator-rotor estappeleacute eacutetage
W2
C3
U
W3
U
C1W3
U
C2
1
2
3
W vitesse relative de lrsquoeacutecoulement
U vitesse peacuteripheacuterique du rotor
C vitesse absolue de lrsquoeacutecoulement
W2
U
C2
C3W3
U
C1W1
URotor
Stator
➀
➁
➂
C
W
U
Stator Rotor
U2
Stator
C2
U3
W3
Rotor
Stator
Rotor
U2
W2
Rotor
Stator
C2
U3
W3C3
Stator
Rotor
Effets 3DEffets visqueuxDistribution de pressionTraicircneacutee PortanceEacutecoulement instationnaireInteacuteraction fuide structureTurbulence
Extrados
Intrados
L
D
Profil isoleacuteGrille drsquoaubes
Pour analyser lrsquoeacutechange drsquoeacutenergie entre le fluide et le rotor drsquouneturbomachine on utilise la variation de la quantiteacute demouvement (changement de direction etou de vitesse)
Lrsquoeacutetude classique unidimensionnelle utilise le concept de trianglede vitesse Ce modegravele est appliqueacute agrave lrsquoentreacutee et agrave la sortie ducanal inter-aube (grille drsquoaubes)
Dans celui-ci on retrouve la vitesse absolue exprimeacutee dans unrepegravere lieacute aux parties fixes (le stator) accompagneacutee de lavitesse relative lieacutee aux partie tournantes ( le rotor)
Approche classique
Lrsquoapplication du triangle de vitesse permet drsquoestimer
bull lrsquoangle drsquoincidence agrave lrsquoentreacutee du rotor
bull lrsquoangle au bord de fuite du rotor
bull la deacuteflection de lrsquoeacutecoulement dans le stator
bull le travail produit ou consommeacute par la roue
Approche classique
C vitesse absolue de lrsquoeacutecoulement
W vitesse relative de lrsquoeacutecoulement
U vitesse peacuteripheacuterique du rotor
Cu Cm Cx composante tangentielle radiale et axiale de lavitesse absolue du fluide
Wu Wm Wx composante tangentielle radiale et axiale de lavitesse relative du fluide
α lrsquoangle des vitesses absolues mesureacutees par rapport agrave la directionaxiale
β lrsquoangle des vitesses relatives mesureacutees par rapport agrave la directionaxiale
La forme des pales du rotor deacutepend des angles 120631120631(des vitesses relatives dans le repegravere du rotor)
La forme des pales du stator deacutepend des angles 120630120630(des vitesses absolues dans le repegravere fixe)
Si lrsquoon veut garder la vitesse axiale 119940119940119961119961 = 119940119940119940119940119940119940119940119940119940119940 lrsquoeacutequation deconservation de la masse 119950 = 120646120646120488120488119940119940119961119961 = 119940119940119940119940119940119940119940119940119940119940 impose 120646120646120488120488 = 119940119940119940119940119940119940119940119940119940119940
Compresseur
Cu
Cm
Cu
Plan normal
Plan meacuteridional
BABF
W2
C2
U
U
W3C3
u
Les vecteurs vitesse (vitesse absolue 119940119940 pour le stator etrelative 119960119960 pour le rotor) sont consideacutereacutes tangents agrave laligne de squelette de lrsquoaube
β2
β3W3
W2
Pour les machines axiales (u= uent = usort ) le trianglede vitesse donne lieu agrave des eacutequations scalaires tellesque
cu= wu+u
cx= wxcx= = wx c
U
w
wu
cu
Pour les machines agrave eacutetages multiples on suppose que lavitesse axiale demeure constante Crsquoest agrave dire que119940119940120784120784119961119961= 119940119940120785120785119961119961= 119940119940119961119961= 119940119940119940119940119940119940119940119940119940119940
Cette condition jumeleacutee agrave 119932119932120784120784= 119932119932120785120785= 119932119932 megravene agrave lrsquohypothegravese drsquoune cineacutematique reacutepeacutetitive En pratique
Les triangles de vitesse agrave lrsquoentreacutee et agrave la sortie drsquouneacutetage peuvent ecirctre superposeacutes
Nous commenccedilons par la repreacutesentation de triangles de vitessespour le cas drsquoune turbineNous rappelons que par hypothegravese le rayon ne varie pas dans leplan inter-aube et que la vitesse axiale demeure constante dansle sens axialCes conditions permettent drsquoeffectuer la superposition destriangles de vitesse au bord drsquoattaque et bord de fuite du rotoravec la vitesse 119932119932 commune
c1w1
u
c2
u2
w2
➊
➋
➌ c3w3
u
Turbine
C2
W2 C3
W3
Stator Rotor
C2u C3u
W3u
α2β2 β3
α3
∆Cu= ∆Wu
U
W2u
119940119940119961119961119880119880 = 119888119888119888119888119888119888119888119888119888119888119888119888119909119909 = 119888119888119888119888119888119888119888119888119888119888
1198881198883 = 1198881198881
1205721205723 = 1205721205721
Cineacutematique reacutepeacutetitive
➁ ➂
➁U➀
Turbine
119914119914120784120784
119914119914120785120785
119934119934120785120785
119932119932119934119934120784120784
119932119932
119914119914120783120783120630120630120783120783
120630120630120784120784120631120631120784120784
120630120630120785120785
120631120631120785120785
Stator Rotor
➀ ➁ ➂
Le cas du compresseur est similaire agrave celui drsquoune turbineLe rayon ne varie pas (lrsquoeacutetude se deacuteroule dans le plan inter-aube) et la vitesse axiale demeure constante dans le sens axialComme pour la turbine nous pouvons superposer les triangles devitesse au bord drsquoattaque et bord de fuite du rotor
W2 C2
C3W3
U
C1W1
U
U
Compresseur
➊
➋
➌
α2 β1
C2 W1
C1W2
U
β2α1
W2U C1U
CX
( )2 1 = minuse u uW U C C
Lrsquoopeacuteration est vectorielle
C2U
119914119914120784120784119958119958 minus 119914119914120783120783119958119958 =- 119934119934120784120784119958119958 minus119934119934120783120783119958119958
➀
➁
➁
Compresseur
119914119914120784120784119914119914120785120785
119934119934120783120783
119932119932119914119914120783120783120630120630120783120783 120630120630120784120784
120631120631120784120784120630120630120785120785
Rotor Stator
119934119934120784120784120631120631120783120783
119932119932
➀ ➁ ➂
Trois quantiteacutes fondamentales sont utiliseacutees dans laconception des turbomachines
bull le coefficient de charge ψ
bull le coefficient de deacutebit Φ
bull le degreacute de reacuteaction R
Approche classique
Parfois ψ est consideacutereacute positif pour les turbines et neacutegatifpour les compresseurs 119882119882119890119890 = frasl119882 119888
Coefficient de charge
2Ψ = eWU
Coefficient de charge
2e
UW
=ψ 02 01 02 2
h h hU Uminus ∆
= = 02
s shU
η ∆=
=
eWWm
120578120578119904119904 =∆ℎ0∆ℎ0119904119904
Turbine
Coefficient de deacutebit
xcU
Φ =
xCU
Φ =ρRT
pρ
= xC RTRTU p
=RT m RT
U Ap
xCU
Φ =119888 = ρ119862119862119909119909119860119860
11988811987711987711987911987901199011199010119860119860
= 119872119872 120574120574 1 +120574120574 minus 1
21198721198722
minus 120574120574+1)2(120574120574minus1
Afin de geacuteneacuteraliser le triangle de vitesses drsquoune machineaxiale on effectue une mise agrave lrsquoeacutechelle en divisant chaquecocircteacute par la vitesse tangentielle U
Cette opeacuteration permettra de retrouver directement lescoefficients de charge ψ et de deacutebit Φ
Par la suite un nouveau paramegravetre R qursquoon appelle degreacutede reacuteaction sera introduit pour compleacuteter la caracteacuterisationde lrsquoaubage
Approche classique
α2
β2 α3
β3C2 W3
W2
c3u - c2u= w3u - w2u
W3uW2u
U
CX CX
Lrsquoopeacuteration est vectorielle
C3
120511120511 = frasl119934119934119940119940 119932119932120784120784 = frasl119914119914120785120785119958119958 minus 119914119914120784120784119958119958 119932119932
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
Turbine
3
2
3 3
2 2
12
12
12
12
u
u
u u
u u
c RUc RU
w c RU Uw c RU U
Ψ= minus +
Ψ= minus +
Ψ= minus = +
Ψ= minus = minus
2 22 23 2
2 22 23 2
1 12 2
2 2
c cR RU U
w wR RU U
Ψ Ψ = Φ + minus + = Φ + minus +
Ψ Ψ = Φ + + = Φ + minus
3 2
3 2
1 2 1 2
2 2
R Ratan atan
R Ratan atan
minus + Ψ minus + Ψ = = Φ Φ
Ψ + Ψ minus = = Φ Φ
α α
β β
α2β1
C2U
W1U
C1UW2U β2 α1
Compresseur
ΦC2U
C3UW3U
W2U
ψ
1
ψ
ΦC2UC3U
W3U
W2U
1
Il faut deacutefinir un troisiegraveme paramegravetre pour diffeacuterencier ces deux triangles de vitesses ayant le mecircme 120509120509 et le mecircme 120537120537
Les coefficients 120509120509 et 120537120537 ne suffisent pas pour deacutefinir de maniegravereunivoque le triangle de vitesses
Le degreacute de reacuteaction est un paramegravetre adimensionnel quiexprime la variation drsquoenthalpie subie par le fluide dans le rotorpar rapport agrave la chute totale drsquoenthalpie dans lrsquoeacutetage
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
Variation d enthalpie dans le rotorRVariation d enthalpie dans l eacutetage
=
❸❶
❷
Rotor
Stator
R=0 machine drsquoaction (Δhrotor=0 Δheacutetage=Δhstator)
R=1 machine agrave reacuteaction (Δhstator=0 Δheacutetage=Δhrotor)
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
R indique la reacutepartition du saut drsquoenthalpie (de pression) statique entre le stator et le rotor
Le degreacute de reacuteaction R deacutefini en fonction de variablesthermodynamiques (lrsquoenthalpie) peut aussi srsquoexprimer en fonctionde variables cineacutematiques preacutesentes dans le triangle de vitessesNotamment en fonction de la vitesse peacuteripheacuterique U et descomposantes des vitesses absolues et relatives
La vitesse absolue agrave lrsquoentreacutee du stator est eacutegale agrave celle agrave la sortie du rotor (c1= c3 eacutetage reacutepeacutetitif )
On neacuteglige les pertes et le transfert de chaleur dans le stator
1 2 3
ℎ3 minus ℎ1= ℎ3 + frasl11988811988832 2 minus ℎ1 minus frasl11988811988812 2 = ℎ03 minus ℎ01 = 120549120549ℎ0
ℎ03 minus ℎ01 = ℎ03 minus ℎ02 = 120549120549ℎ )0(3minus2
ℎ03 minus ℎ02 = ℎ3 minus ℎ1
119945119945120782120782120783120783 = 119945119945120782120782120784120784
frasl11988811988832 2 = frasl11988811988812 2
3 2
3 1
rotor
eacutetage
h h hRh h h
∆∆
minus= =
minus
2 20 3 2 2 3
0 3 2
2 2( )
( )
h c c h
∆∆
minus
minus
+ minus=
ℎ03 ℎ02
(ℎ01= ℎ02 1198881198881= 1198881198883)
120549120549ℎ )0(3minus2 = (ℎ3 + frasl11988811988832 2) minus (ℎ2 + frasl11988811988822 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
0(3 2) 3 2( )∆ minus = minusu uh U c c
= 1 minusfrasl11988811988832 2 minus frasl11988811988822 2120549120549ℎ )0(3minus2
119877119877 =120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
120549120549ℎ )0(3minus2
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
3 212+
= minus u uc cRU
119888119888119909119909 = 1198881198882119909119909 = 1198881198883119909119909 = 119888119888119888119888119888119888119888119888119888119888
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 11988811988821199061199061198881198882 = 1198881198881199091199092+1198881198881199061199062
119877119877 = 1 minus11988811988831199061199062 minus 11988811988821199061199062
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
Indirectement lrsquoeacutequation pour R deacutecrite en fonction de composantes cineacutematiques eacutetablit un lien entre les angles drsquoentreacutee et de sortie
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
= 1 minus1198881198883119906119906 + 1198881198882119906119906
2119880119880
2232
2 32 2wwh h+ = +
2 23 2
2 3 2 2w wh hminus = minus
2 2 2 23 3 2 2
2 3 2 2u x u xw w w wh h + +
minus = minus
Machine axiale U2 = U3
La composante axiale de la vitesse demeure constante w3x = w 2x
( eacutequation justifieacutee plus tard lors de la rothalpie)
2 23 2
2 3 2 2u uw wh hminus = minus
3 2 3 22 3
( )( )2
u u u uw w w wh h minus +minus =
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
Eacutetage reacutepeacutetitif c3 = c1
Si les pertes dans le stator sont neacutegligeacutees
c2u ndash c3u= w2u - w3u
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
3 2 3 23 2
( )( )2
u u u uw w w wh h minus +minus = minus
c2u ndash c3u= w2u - w3u
119877119877 = minus1199081199083119906119906 + 1199081199082119906119906
2119880119880
2 2 2 22 3 2 3 2 3( ) 2 ( ) 2h h u u w wminus = minus minus minus
2 2 2 2 2 23 2 3 2 3 2
2 2 2e
I II III
u u c c w wW minus minus minus= + minus
2 2 2 22 3 2 3
2 2 2 2 2 22 3 2 3 2 3
( ) ( )( ) ( ) ( )
minus minus minus=
minus + minus minus minusu u w wR
c c u u w w
3 2= =U u u
Expliqueacutee + tard
2 23 2
03 02 3 2 2ec cW h h h h minus
= minus = minus +
119877119877 =ℎ2 minus ℎ3ℎ1 minus ℎ3
=ℎ2 minus ℎ3ℎ02 minus ℎ03
Degreacute de reacuteaction
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
( )3u 2uw wR2U+
= minus
3 212
u uc cRU+
= minus
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
Puisque le degreacute de reacuteaction R peut aussi srsquoexprimer enfonction de variables cineacutematiques on peut chercher sarepreacutesentation dans le triangle adimensionnel de vitesses
Turbine
α2
β2 α3
β3
C2 W3
C3W2
C3uC2u
U
C2xC3x
W2u
c2u ndash c3u= w2u - w3u
2Ψ minusR
frasl119934119934120784120784119958119958 119932119932 = frasl1198621198622119906119906 119880119880 minus frasl119880119880 119880119880Ψ = frasl1198621198622119906119906 minus 1198621198623119906119906 119880119880
119877119877 = 1 minus (1198621198623119906119906 + 1198621198622119906119906 frasl) 2119880119880Φ = frasl1198621198622119909119909 119880119880 = frasl1198621198623119909119909 119880119880
= (119914119914120784120784119958119958 minus 119914119914120785120785119958119958 frasl) 2119880119880 minus 1 + (119914119914120784120784119958119958 + 119914119914120785120785119958119958 frasl) 2119880119880
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α2
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
Les repreacutesentations des triangles de vitesses adimensionnellessont tregraves riches en information et contiennent plusieurs formules
Par exemple pour un compresseur on deacuteduit aiseacutement que
Avec 120595120595 = 1198821198821198901198901198801198802
Φ = 119888119888119909119909119880119880
et que pour 119888119888119901119901 = 119888119888119888119888119888119888119888119888119888119888 119882119882119890119890 = 119888119888119901119901(11987911987902minus11987911987901)on trouve
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
(11987911987902minus11987911987901) =119880119880119888119888119909119909119888119888119901119901
(tan1205731205731 minus tan1205731205732)
Alors si lrsquoon veut obtenir une plus grande variation detempeacuterature 11987911987902 minus 11987911987901 indirectement un plus grand rapport(1199011199010211990111990101) on peut augmenter
bull la vitesse peacuteripheacuterique 119880119880 (la vitesse de rotation)bull la vitesse axiale 119888119888119909119909 (le deacutebit massique)bull lrsquoangle de deacuteflection de la pale(tan1205731205731 minus tan1205731205732)
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
Dans une turbine le fluide agrave haute eacutenergie est drsquoabord etacceacuteleacutereacute et guideacute dans le stator pour lrsquoajuster aux exigencesgeacuteomeacutetriques est cineacutematiques du rotorPar la suite lrsquoeacutenergie cineacutetique de lrsquoeacutecoulement est transmise aurotor en mecircme temps qursquoun changement de direction a lieu
Ces procegraves sont reacutepeacuteteacutes pour des eacutetages conseacutecutifs
Agrave cause du mouvement du rotor par rapport au stator deux typesde vitesses relatives et absolues seront preacutesentes
C2x
C2
C2uC2m
C1uC1m
C1x
Machine axiale
➊
➋
Machine radiale
119963119963
120637120637119955119955
Plan 119903119903 minus 120579120579119911119911 = 119888119888119888119888119888119888119888119888119888119888
Carter
119955119955 Moyeu
Plan 119903119903 minus 119911119911
Plan 119903119903 minus 119911119911120579120579 = 119888119888119888119888119888119888119888119888119888119888
119955119955
Rayon moyen
Vues drsquoun compresseur axial
Plan 119911119911 minus 120579120579119903119903 = 119888119888119888119888119888119888119888119888119888119888
119880119880 = 119903119903120596120596
Moyeu
Carter
Rm
X
CmCx
C
Ω
119888119888 119907119907119907119907119888119888119888119888119888119888119888119888119888119888 119889119889119888119888 119897119897primeeacute119888119888119888119888119888119888119897119897119888119888119888119888119888119888119888119888119888119888
intrados
extrados surface S1
surface S2 Rayon moyen r
Lrsquoeacutetude est reacutealiseacute sur la surface 1198781198781
119932119932 = 119955119955120654120654
120654120654
θ
x
Zone aubeacutee
surface S1
119932119932 = 119955119955120654120654
Rotor Stator
Rotor-Stator
RotorStator
Dans une turbomachinelrsquoensemble stator-rotor estappeleacute eacutetage
W2
C3
U
W3
U
C1W3
U
C2
1
2
3
W vitesse relative de lrsquoeacutecoulement
U vitesse peacuteripheacuterique du rotor
C vitesse absolue de lrsquoeacutecoulement
W2
U
C2
C3W3
U
C1W1
URotor
Stator
➀
➁
➂
C
W
U
Stator Rotor
U2
Stator
C2
U3
W3
Rotor
Stator
Rotor
U2
W2
Rotor
Stator
C2
U3
W3C3
Stator
Rotor
Effets 3DEffets visqueuxDistribution de pressionTraicircneacutee PortanceEacutecoulement instationnaireInteacuteraction fuide structureTurbulence
Extrados
Intrados
L
D
Profil isoleacuteGrille drsquoaubes
Pour analyser lrsquoeacutechange drsquoeacutenergie entre le fluide et le rotor drsquouneturbomachine on utilise la variation de la quantiteacute demouvement (changement de direction etou de vitesse)
Lrsquoeacutetude classique unidimensionnelle utilise le concept de trianglede vitesse Ce modegravele est appliqueacute agrave lrsquoentreacutee et agrave la sortie ducanal inter-aube (grille drsquoaubes)
Dans celui-ci on retrouve la vitesse absolue exprimeacutee dans unrepegravere lieacute aux parties fixes (le stator) accompagneacutee de lavitesse relative lieacutee aux partie tournantes ( le rotor)
Approche classique
Lrsquoapplication du triangle de vitesse permet drsquoestimer
bull lrsquoangle drsquoincidence agrave lrsquoentreacutee du rotor
bull lrsquoangle au bord de fuite du rotor
bull la deacuteflection de lrsquoeacutecoulement dans le stator
bull le travail produit ou consommeacute par la roue
Approche classique
C vitesse absolue de lrsquoeacutecoulement
W vitesse relative de lrsquoeacutecoulement
U vitesse peacuteripheacuterique du rotor
Cu Cm Cx composante tangentielle radiale et axiale de lavitesse absolue du fluide
Wu Wm Wx composante tangentielle radiale et axiale de lavitesse relative du fluide
α lrsquoangle des vitesses absolues mesureacutees par rapport agrave la directionaxiale
β lrsquoangle des vitesses relatives mesureacutees par rapport agrave la directionaxiale
La forme des pales du rotor deacutepend des angles 120631120631(des vitesses relatives dans le repegravere du rotor)
La forme des pales du stator deacutepend des angles 120630120630(des vitesses absolues dans le repegravere fixe)
Si lrsquoon veut garder la vitesse axiale 119940119940119961119961 = 119940119940119940119940119940119940119940119940119940119940 lrsquoeacutequation deconservation de la masse 119950 = 120646120646120488120488119940119940119961119961 = 119940119940119940119940119940119940119940119940119940119940 impose 120646120646120488120488 = 119940119940119940119940119940119940119940119940119940119940
Compresseur
Cu
Cm
Cu
Plan normal
Plan meacuteridional
BABF
W2
C2
U
U
W3C3
u
Les vecteurs vitesse (vitesse absolue 119940119940 pour le stator etrelative 119960119960 pour le rotor) sont consideacutereacutes tangents agrave laligne de squelette de lrsquoaube
β2
β3W3
W2
Pour les machines axiales (u= uent = usort ) le trianglede vitesse donne lieu agrave des eacutequations scalaires tellesque
cu= wu+u
cx= wxcx= = wx c
U
w
wu
cu
Pour les machines agrave eacutetages multiples on suppose que lavitesse axiale demeure constante Crsquoest agrave dire que119940119940120784120784119961119961= 119940119940120785120785119961119961= 119940119940119961119961= 119940119940119940119940119940119940119940119940119940119940
Cette condition jumeleacutee agrave 119932119932120784120784= 119932119932120785120785= 119932119932 megravene agrave lrsquohypothegravese drsquoune cineacutematique reacutepeacutetitive En pratique
Les triangles de vitesse agrave lrsquoentreacutee et agrave la sortie drsquouneacutetage peuvent ecirctre superposeacutes
Nous commenccedilons par la repreacutesentation de triangles de vitessespour le cas drsquoune turbineNous rappelons que par hypothegravese le rayon ne varie pas dans leplan inter-aube et que la vitesse axiale demeure constante dansle sens axialCes conditions permettent drsquoeffectuer la superposition destriangles de vitesse au bord drsquoattaque et bord de fuite du rotoravec la vitesse 119932119932 commune
c1w1
u
c2
u2
w2
➊
➋
➌ c3w3
u
Turbine
C2
W2 C3
W3
Stator Rotor
C2u C3u
W3u
α2β2 β3
α3
∆Cu= ∆Wu
U
W2u
119940119940119961119961119880119880 = 119888119888119888119888119888119888119888119888119888119888119888119888119909119909 = 119888119888119888119888119888119888119888119888119888119888
1198881198883 = 1198881198881
1205721205723 = 1205721205721
Cineacutematique reacutepeacutetitive
➁ ➂
➁U➀
Turbine
119914119914120784120784
119914119914120785120785
119934119934120785120785
119932119932119934119934120784120784
119932119932
119914119914120783120783120630120630120783120783
120630120630120784120784120631120631120784120784
120630120630120785120785
120631120631120785120785
Stator Rotor
➀ ➁ ➂
Le cas du compresseur est similaire agrave celui drsquoune turbineLe rayon ne varie pas (lrsquoeacutetude se deacuteroule dans le plan inter-aube) et la vitesse axiale demeure constante dans le sens axialComme pour la turbine nous pouvons superposer les triangles devitesse au bord drsquoattaque et bord de fuite du rotor
W2 C2
C3W3
U
C1W1
U
U
Compresseur
➊
➋
➌
α2 β1
C2 W1
C1W2
U
β2α1
W2U C1U
CX
( )2 1 = minuse u uW U C C
Lrsquoopeacuteration est vectorielle
C2U
119914119914120784120784119958119958 minus 119914119914120783120783119958119958 =- 119934119934120784120784119958119958 minus119934119934120783120783119958119958
➀
➁
➁
Compresseur
119914119914120784120784119914119914120785120785
119934119934120783120783
119932119932119914119914120783120783120630120630120783120783 120630120630120784120784
120631120631120784120784120630120630120785120785
Rotor Stator
119934119934120784120784120631120631120783120783
119932119932
➀ ➁ ➂
Trois quantiteacutes fondamentales sont utiliseacutees dans laconception des turbomachines
bull le coefficient de charge ψ
bull le coefficient de deacutebit Φ
bull le degreacute de reacuteaction R
Approche classique
Parfois ψ est consideacutereacute positif pour les turbines et neacutegatifpour les compresseurs 119882119882119890119890 = frasl119882 119888
Coefficient de charge
2Ψ = eWU
Coefficient de charge
2e
UW
=ψ 02 01 02 2
h h hU Uminus ∆
= = 02
s shU
η ∆=
=
eWWm
120578120578119904119904 =∆ℎ0∆ℎ0119904119904
Turbine
Coefficient de deacutebit
xcU
Φ =
xCU
Φ =ρRT
pρ
= xC RTRTU p
=RT m RT
U Ap
xCU
Φ =119888 = ρ119862119862119909119909119860119860
11988811987711987711987911987901199011199010119860119860
= 119872119872 120574120574 1 +120574120574 minus 1
21198721198722
minus 120574120574+1)2(120574120574minus1
Afin de geacuteneacuteraliser le triangle de vitesses drsquoune machineaxiale on effectue une mise agrave lrsquoeacutechelle en divisant chaquecocircteacute par la vitesse tangentielle U
Cette opeacuteration permettra de retrouver directement lescoefficients de charge ψ et de deacutebit Φ
Par la suite un nouveau paramegravetre R qursquoon appelle degreacutede reacuteaction sera introduit pour compleacuteter la caracteacuterisationde lrsquoaubage
Approche classique
α2
β2 α3
β3C2 W3
W2
c3u - c2u= w3u - w2u
W3uW2u
U
CX CX
Lrsquoopeacuteration est vectorielle
C3
120511120511 = frasl119934119934119940119940 119932119932120784120784 = frasl119914119914120785120785119958119958 minus 119914119914120784120784119958119958 119932119932
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
Turbine
3
2
3 3
2 2
12
12
12
12
u
u
u u
u u
c RUc RU
w c RU Uw c RU U
Ψ= minus +
Ψ= minus +
Ψ= minus = +
Ψ= minus = minus
2 22 23 2
2 22 23 2
1 12 2
2 2
c cR RU U
w wR RU U
Ψ Ψ = Φ + minus + = Φ + minus +
Ψ Ψ = Φ + + = Φ + minus
3 2
3 2
1 2 1 2
2 2
R Ratan atan
R Ratan atan
minus + Ψ minus + Ψ = = Φ Φ
Ψ + Ψ minus = = Φ Φ
α α
β β
α2β1
C2U
W1U
C1UW2U β2 α1
Compresseur
ΦC2U
C3UW3U
W2U
ψ
1
ψ
ΦC2UC3U
W3U
W2U
1
Il faut deacutefinir un troisiegraveme paramegravetre pour diffeacuterencier ces deux triangles de vitesses ayant le mecircme 120509120509 et le mecircme 120537120537
Les coefficients 120509120509 et 120537120537 ne suffisent pas pour deacutefinir de maniegravereunivoque le triangle de vitesses
Le degreacute de reacuteaction est un paramegravetre adimensionnel quiexprime la variation drsquoenthalpie subie par le fluide dans le rotorpar rapport agrave la chute totale drsquoenthalpie dans lrsquoeacutetage
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
Variation d enthalpie dans le rotorRVariation d enthalpie dans l eacutetage
=
❸❶
❷
Rotor
Stator
R=0 machine drsquoaction (Δhrotor=0 Δheacutetage=Δhstator)
R=1 machine agrave reacuteaction (Δhstator=0 Δheacutetage=Δhrotor)
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
R indique la reacutepartition du saut drsquoenthalpie (de pression) statique entre le stator et le rotor
Le degreacute de reacuteaction R deacutefini en fonction de variablesthermodynamiques (lrsquoenthalpie) peut aussi srsquoexprimer en fonctionde variables cineacutematiques preacutesentes dans le triangle de vitessesNotamment en fonction de la vitesse peacuteripheacuterique U et descomposantes des vitesses absolues et relatives
La vitesse absolue agrave lrsquoentreacutee du stator est eacutegale agrave celle agrave la sortie du rotor (c1= c3 eacutetage reacutepeacutetitif )
On neacuteglige les pertes et le transfert de chaleur dans le stator
1 2 3
ℎ3 minus ℎ1= ℎ3 + frasl11988811988832 2 minus ℎ1 minus frasl11988811988812 2 = ℎ03 minus ℎ01 = 120549120549ℎ0
ℎ03 minus ℎ01 = ℎ03 minus ℎ02 = 120549120549ℎ )0(3minus2
ℎ03 minus ℎ02 = ℎ3 minus ℎ1
119945119945120782120782120783120783 = 119945119945120782120782120784120784
frasl11988811988832 2 = frasl11988811988812 2
3 2
3 1
rotor
eacutetage
h h hRh h h
∆∆
minus= =
minus
2 20 3 2 2 3
0 3 2
2 2( )
( )
h c c h
∆∆
minus
minus
+ minus=
ℎ03 ℎ02
(ℎ01= ℎ02 1198881198881= 1198881198883)
120549120549ℎ )0(3minus2 = (ℎ3 + frasl11988811988832 2) minus (ℎ2 + frasl11988811988822 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
0(3 2) 3 2( )∆ minus = minusu uh U c c
= 1 minusfrasl11988811988832 2 minus frasl11988811988822 2120549120549ℎ )0(3minus2
119877119877 =120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
120549120549ℎ )0(3minus2
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
3 212+
= minus u uc cRU
119888119888119909119909 = 1198881198882119909119909 = 1198881198883119909119909 = 119888119888119888119888119888119888119888119888119888119888
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 11988811988821199061199061198881198882 = 1198881198881199091199092+1198881198881199061199062
119877119877 = 1 minus11988811988831199061199062 minus 11988811988821199061199062
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
Indirectement lrsquoeacutequation pour R deacutecrite en fonction de composantes cineacutematiques eacutetablit un lien entre les angles drsquoentreacutee et de sortie
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
= 1 minus1198881198883119906119906 + 1198881198882119906119906
2119880119880
2232
2 32 2wwh h+ = +
2 23 2
2 3 2 2w wh hminus = minus
2 2 2 23 3 2 2
2 3 2 2u x u xw w w wh h + +
minus = minus
Machine axiale U2 = U3
La composante axiale de la vitesse demeure constante w3x = w 2x
( eacutequation justifieacutee plus tard lors de la rothalpie)
2 23 2
2 3 2 2u uw wh hminus = minus
3 2 3 22 3
( )( )2
u u u uw w w wh h minus +minus =
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
Eacutetage reacutepeacutetitif c3 = c1
Si les pertes dans le stator sont neacutegligeacutees
c2u ndash c3u= w2u - w3u
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
3 2 3 23 2
( )( )2
u u u uw w w wh h minus +minus = minus
c2u ndash c3u= w2u - w3u
119877119877 = minus1199081199083119906119906 + 1199081199082119906119906
2119880119880
2 2 2 22 3 2 3 2 3( ) 2 ( ) 2h h u u w wminus = minus minus minus
2 2 2 2 2 23 2 3 2 3 2
2 2 2e
I II III
u u c c w wW minus minus minus= + minus
2 2 2 22 3 2 3
2 2 2 2 2 22 3 2 3 2 3
( ) ( )( ) ( ) ( )
minus minus minus=
minus + minus minus minusu u w wR
c c u u w w
3 2= =U u u
Expliqueacutee + tard
2 23 2
03 02 3 2 2ec cW h h h h minus
= minus = minus +
119877119877 =ℎ2 minus ℎ3ℎ1 minus ℎ3
=ℎ2 minus ℎ3ℎ02 minus ℎ03
Degreacute de reacuteaction
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
( )3u 2uw wR2U+
= minus
3 212
u uc cRU+
= minus
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
Puisque le degreacute de reacuteaction R peut aussi srsquoexprimer enfonction de variables cineacutematiques on peut chercher sarepreacutesentation dans le triangle adimensionnel de vitesses
Turbine
α2
β2 α3
β3
C2 W3
C3W2
C3uC2u
U
C2xC3x
W2u
c2u ndash c3u= w2u - w3u
2Ψ minusR
frasl119934119934120784120784119958119958 119932119932 = frasl1198621198622119906119906 119880119880 minus frasl119880119880 119880119880Ψ = frasl1198621198622119906119906 minus 1198621198623119906119906 119880119880
119877119877 = 1 minus (1198621198623119906119906 + 1198621198622119906119906 frasl) 2119880119880Φ = frasl1198621198622119909119909 119880119880 = frasl1198621198623119909119909 119880119880
= (119914119914120784120784119958119958 minus 119914119914120785120785119958119958 frasl) 2119880119880 minus 1 + (119914119914120784120784119958119958 + 119914119914120785120785119958119958 frasl) 2119880119880
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α2
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
Les repreacutesentations des triangles de vitesses adimensionnellessont tregraves riches en information et contiennent plusieurs formules
Par exemple pour un compresseur on deacuteduit aiseacutement que
Avec 120595120595 = 1198821198821198901198901198801198802
Φ = 119888119888119909119909119880119880
et que pour 119888119888119901119901 = 119888119888119888119888119888119888119888119888119888119888 119882119882119890119890 = 119888119888119901119901(11987911987902minus11987911987901)on trouve
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
(11987911987902minus11987911987901) =119880119880119888119888119909119909119888119888119901119901
(tan1205731205731 minus tan1205731205732)
Alors si lrsquoon veut obtenir une plus grande variation detempeacuterature 11987911987902 minus 11987911987901 indirectement un plus grand rapport(1199011199010211990111990101) on peut augmenter
bull la vitesse peacuteripheacuterique 119880119880 (la vitesse de rotation)bull la vitesse axiale 119888119888119909119909 (le deacutebit massique)bull lrsquoangle de deacuteflection de la pale(tan1205731205731 minus tan1205731205732)
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
C2x
C2
C2uC2m
C1uC1m
C1x
Machine axiale
➊
➋
Machine radiale
119963119963
120637120637119955119955
Plan 119903119903 minus 120579120579119911119911 = 119888119888119888119888119888119888119888119888119888119888
Carter
119955119955 Moyeu
Plan 119903119903 minus 119911119911
Plan 119903119903 minus 119911119911120579120579 = 119888119888119888119888119888119888119888119888119888119888
119955119955
Rayon moyen
Vues drsquoun compresseur axial
Plan 119911119911 minus 120579120579119903119903 = 119888119888119888119888119888119888119888119888119888119888
119880119880 = 119903119903120596120596
Moyeu
Carter
Rm
X
CmCx
C
Ω
119888119888 119907119907119907119907119888119888119888119888119888119888119888119888119888119888 119889119889119888119888 119897119897primeeacute119888119888119888119888119888119888119897119897119888119888119888119888119888119888119888119888119888119888
intrados
extrados surface S1
surface S2 Rayon moyen r
Lrsquoeacutetude est reacutealiseacute sur la surface 1198781198781
119932119932 = 119955119955120654120654
120654120654
θ
x
Zone aubeacutee
surface S1
119932119932 = 119955119955120654120654
Rotor Stator
Rotor-Stator
RotorStator
Dans une turbomachinelrsquoensemble stator-rotor estappeleacute eacutetage
W2
C3
U
W3
U
C1W3
U
C2
1
2
3
W vitesse relative de lrsquoeacutecoulement
U vitesse peacuteripheacuterique du rotor
C vitesse absolue de lrsquoeacutecoulement
W2
U
C2
C3W3
U
C1W1
URotor
Stator
➀
➁
➂
C
W
U
Stator Rotor
U2
Stator
C2
U3
W3
Rotor
Stator
Rotor
U2
W2
Rotor
Stator
C2
U3
W3C3
Stator
Rotor
Effets 3DEffets visqueuxDistribution de pressionTraicircneacutee PortanceEacutecoulement instationnaireInteacuteraction fuide structureTurbulence
Extrados
Intrados
L
D
Profil isoleacuteGrille drsquoaubes
Pour analyser lrsquoeacutechange drsquoeacutenergie entre le fluide et le rotor drsquouneturbomachine on utilise la variation de la quantiteacute demouvement (changement de direction etou de vitesse)
Lrsquoeacutetude classique unidimensionnelle utilise le concept de trianglede vitesse Ce modegravele est appliqueacute agrave lrsquoentreacutee et agrave la sortie ducanal inter-aube (grille drsquoaubes)
Dans celui-ci on retrouve la vitesse absolue exprimeacutee dans unrepegravere lieacute aux parties fixes (le stator) accompagneacutee de lavitesse relative lieacutee aux partie tournantes ( le rotor)
Approche classique
Lrsquoapplication du triangle de vitesse permet drsquoestimer
bull lrsquoangle drsquoincidence agrave lrsquoentreacutee du rotor
bull lrsquoangle au bord de fuite du rotor
bull la deacuteflection de lrsquoeacutecoulement dans le stator
bull le travail produit ou consommeacute par la roue
Approche classique
C vitesse absolue de lrsquoeacutecoulement
W vitesse relative de lrsquoeacutecoulement
U vitesse peacuteripheacuterique du rotor
Cu Cm Cx composante tangentielle radiale et axiale de lavitesse absolue du fluide
Wu Wm Wx composante tangentielle radiale et axiale de lavitesse relative du fluide
α lrsquoangle des vitesses absolues mesureacutees par rapport agrave la directionaxiale
β lrsquoangle des vitesses relatives mesureacutees par rapport agrave la directionaxiale
La forme des pales du rotor deacutepend des angles 120631120631(des vitesses relatives dans le repegravere du rotor)
La forme des pales du stator deacutepend des angles 120630120630(des vitesses absolues dans le repegravere fixe)
Si lrsquoon veut garder la vitesse axiale 119940119940119961119961 = 119940119940119940119940119940119940119940119940119940119940 lrsquoeacutequation deconservation de la masse 119950 = 120646120646120488120488119940119940119961119961 = 119940119940119940119940119940119940119940119940119940119940 impose 120646120646120488120488 = 119940119940119940119940119940119940119940119940119940119940
Compresseur
Cu
Cm
Cu
Plan normal
Plan meacuteridional
BABF
W2
C2
U
U
W3C3
u
Les vecteurs vitesse (vitesse absolue 119940119940 pour le stator etrelative 119960119960 pour le rotor) sont consideacutereacutes tangents agrave laligne de squelette de lrsquoaube
β2
β3W3
W2
Pour les machines axiales (u= uent = usort ) le trianglede vitesse donne lieu agrave des eacutequations scalaires tellesque
cu= wu+u
cx= wxcx= = wx c
U
w
wu
cu
Pour les machines agrave eacutetages multiples on suppose que lavitesse axiale demeure constante Crsquoest agrave dire que119940119940120784120784119961119961= 119940119940120785120785119961119961= 119940119940119961119961= 119940119940119940119940119940119940119940119940119940119940
Cette condition jumeleacutee agrave 119932119932120784120784= 119932119932120785120785= 119932119932 megravene agrave lrsquohypothegravese drsquoune cineacutematique reacutepeacutetitive En pratique
Les triangles de vitesse agrave lrsquoentreacutee et agrave la sortie drsquouneacutetage peuvent ecirctre superposeacutes
Nous commenccedilons par la repreacutesentation de triangles de vitessespour le cas drsquoune turbineNous rappelons que par hypothegravese le rayon ne varie pas dans leplan inter-aube et que la vitesse axiale demeure constante dansle sens axialCes conditions permettent drsquoeffectuer la superposition destriangles de vitesse au bord drsquoattaque et bord de fuite du rotoravec la vitesse 119932119932 commune
c1w1
u
c2
u2
w2
➊
➋
➌ c3w3
u
Turbine
C2
W2 C3
W3
Stator Rotor
C2u C3u
W3u
α2β2 β3
α3
∆Cu= ∆Wu
U
W2u
119940119940119961119961119880119880 = 119888119888119888119888119888119888119888119888119888119888119888119888119909119909 = 119888119888119888119888119888119888119888119888119888119888
1198881198883 = 1198881198881
1205721205723 = 1205721205721
Cineacutematique reacutepeacutetitive
➁ ➂
➁U➀
Turbine
119914119914120784120784
119914119914120785120785
119934119934120785120785
119932119932119934119934120784120784
119932119932
119914119914120783120783120630120630120783120783
120630120630120784120784120631120631120784120784
120630120630120785120785
120631120631120785120785
Stator Rotor
➀ ➁ ➂
Le cas du compresseur est similaire agrave celui drsquoune turbineLe rayon ne varie pas (lrsquoeacutetude se deacuteroule dans le plan inter-aube) et la vitesse axiale demeure constante dans le sens axialComme pour la turbine nous pouvons superposer les triangles devitesse au bord drsquoattaque et bord de fuite du rotor
W2 C2
C3W3
U
C1W1
U
U
Compresseur
➊
➋
➌
α2 β1
C2 W1
C1W2
U
β2α1
W2U C1U
CX
( )2 1 = minuse u uW U C C
Lrsquoopeacuteration est vectorielle
C2U
119914119914120784120784119958119958 minus 119914119914120783120783119958119958 =- 119934119934120784120784119958119958 minus119934119934120783120783119958119958
➀
➁
➁
Compresseur
119914119914120784120784119914119914120785120785
119934119934120783120783
119932119932119914119914120783120783120630120630120783120783 120630120630120784120784
120631120631120784120784120630120630120785120785
Rotor Stator
119934119934120784120784120631120631120783120783
119932119932
➀ ➁ ➂
Trois quantiteacutes fondamentales sont utiliseacutees dans laconception des turbomachines
bull le coefficient de charge ψ
bull le coefficient de deacutebit Φ
bull le degreacute de reacuteaction R
Approche classique
Parfois ψ est consideacutereacute positif pour les turbines et neacutegatifpour les compresseurs 119882119882119890119890 = frasl119882 119888
Coefficient de charge
2Ψ = eWU
Coefficient de charge
2e
UW
=ψ 02 01 02 2
h h hU Uminus ∆
= = 02
s shU
η ∆=
=
eWWm
120578120578119904119904 =∆ℎ0∆ℎ0119904119904
Turbine
Coefficient de deacutebit
xcU
Φ =
xCU
Φ =ρRT
pρ
= xC RTRTU p
=RT m RT
U Ap
xCU
Φ =119888 = ρ119862119862119909119909119860119860
11988811987711987711987911987901199011199010119860119860
= 119872119872 120574120574 1 +120574120574 minus 1
21198721198722
minus 120574120574+1)2(120574120574minus1
Afin de geacuteneacuteraliser le triangle de vitesses drsquoune machineaxiale on effectue une mise agrave lrsquoeacutechelle en divisant chaquecocircteacute par la vitesse tangentielle U
Cette opeacuteration permettra de retrouver directement lescoefficients de charge ψ et de deacutebit Φ
Par la suite un nouveau paramegravetre R qursquoon appelle degreacutede reacuteaction sera introduit pour compleacuteter la caracteacuterisationde lrsquoaubage
Approche classique
α2
β2 α3
β3C2 W3
W2
c3u - c2u= w3u - w2u
W3uW2u
U
CX CX
Lrsquoopeacuteration est vectorielle
C3
120511120511 = frasl119934119934119940119940 119932119932120784120784 = frasl119914119914120785120785119958119958 minus 119914119914120784120784119958119958 119932119932
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
Turbine
3
2
3 3
2 2
12
12
12
12
u
u
u u
u u
c RUc RU
w c RU Uw c RU U
Ψ= minus +
Ψ= minus +
Ψ= minus = +
Ψ= minus = minus
2 22 23 2
2 22 23 2
1 12 2
2 2
c cR RU U
w wR RU U
Ψ Ψ = Φ + minus + = Φ + minus +
Ψ Ψ = Φ + + = Φ + minus
3 2
3 2
1 2 1 2
2 2
R Ratan atan
R Ratan atan
minus + Ψ minus + Ψ = = Φ Φ
Ψ + Ψ minus = = Φ Φ
α α
β β
α2β1
C2U
W1U
C1UW2U β2 α1
Compresseur
ΦC2U
C3UW3U
W2U
ψ
1
ψ
ΦC2UC3U
W3U
W2U
1
Il faut deacutefinir un troisiegraveme paramegravetre pour diffeacuterencier ces deux triangles de vitesses ayant le mecircme 120509120509 et le mecircme 120537120537
Les coefficients 120509120509 et 120537120537 ne suffisent pas pour deacutefinir de maniegravereunivoque le triangle de vitesses
Le degreacute de reacuteaction est un paramegravetre adimensionnel quiexprime la variation drsquoenthalpie subie par le fluide dans le rotorpar rapport agrave la chute totale drsquoenthalpie dans lrsquoeacutetage
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
Variation d enthalpie dans le rotorRVariation d enthalpie dans l eacutetage
=
❸❶
❷
Rotor
Stator
R=0 machine drsquoaction (Δhrotor=0 Δheacutetage=Δhstator)
R=1 machine agrave reacuteaction (Δhstator=0 Δheacutetage=Δhrotor)
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
R indique la reacutepartition du saut drsquoenthalpie (de pression) statique entre le stator et le rotor
Le degreacute de reacuteaction R deacutefini en fonction de variablesthermodynamiques (lrsquoenthalpie) peut aussi srsquoexprimer en fonctionde variables cineacutematiques preacutesentes dans le triangle de vitessesNotamment en fonction de la vitesse peacuteripheacuterique U et descomposantes des vitesses absolues et relatives
La vitesse absolue agrave lrsquoentreacutee du stator est eacutegale agrave celle agrave la sortie du rotor (c1= c3 eacutetage reacutepeacutetitif )
On neacuteglige les pertes et le transfert de chaleur dans le stator
1 2 3
ℎ3 minus ℎ1= ℎ3 + frasl11988811988832 2 minus ℎ1 minus frasl11988811988812 2 = ℎ03 minus ℎ01 = 120549120549ℎ0
ℎ03 minus ℎ01 = ℎ03 minus ℎ02 = 120549120549ℎ )0(3minus2
ℎ03 minus ℎ02 = ℎ3 minus ℎ1
119945119945120782120782120783120783 = 119945119945120782120782120784120784
frasl11988811988832 2 = frasl11988811988812 2
3 2
3 1
rotor
eacutetage
h h hRh h h
∆∆
minus= =
minus
2 20 3 2 2 3
0 3 2
2 2( )
( )
h c c h
∆∆
minus
minus
+ minus=
ℎ03 ℎ02
(ℎ01= ℎ02 1198881198881= 1198881198883)
120549120549ℎ )0(3minus2 = (ℎ3 + frasl11988811988832 2) minus (ℎ2 + frasl11988811988822 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
0(3 2) 3 2( )∆ minus = minusu uh U c c
= 1 minusfrasl11988811988832 2 minus frasl11988811988822 2120549120549ℎ )0(3minus2
119877119877 =120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
120549120549ℎ )0(3minus2
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
3 212+
= minus u uc cRU
119888119888119909119909 = 1198881198882119909119909 = 1198881198883119909119909 = 119888119888119888119888119888119888119888119888119888119888
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 11988811988821199061199061198881198882 = 1198881198881199091199092+1198881198881199061199062
119877119877 = 1 minus11988811988831199061199062 minus 11988811988821199061199062
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
Indirectement lrsquoeacutequation pour R deacutecrite en fonction de composantes cineacutematiques eacutetablit un lien entre les angles drsquoentreacutee et de sortie
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
= 1 minus1198881198883119906119906 + 1198881198882119906119906
2119880119880
2232
2 32 2wwh h+ = +
2 23 2
2 3 2 2w wh hminus = minus
2 2 2 23 3 2 2
2 3 2 2u x u xw w w wh h + +
minus = minus
Machine axiale U2 = U3
La composante axiale de la vitesse demeure constante w3x = w 2x
( eacutequation justifieacutee plus tard lors de la rothalpie)
2 23 2
2 3 2 2u uw wh hminus = minus
3 2 3 22 3
( )( )2
u u u uw w w wh h minus +minus =
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
Eacutetage reacutepeacutetitif c3 = c1
Si les pertes dans le stator sont neacutegligeacutees
c2u ndash c3u= w2u - w3u
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
3 2 3 23 2
( )( )2
u u u uw w w wh h minus +minus = minus
c2u ndash c3u= w2u - w3u
119877119877 = minus1199081199083119906119906 + 1199081199082119906119906
2119880119880
2 2 2 22 3 2 3 2 3( ) 2 ( ) 2h h u u w wminus = minus minus minus
2 2 2 2 2 23 2 3 2 3 2
2 2 2e
I II III
u u c c w wW minus minus minus= + minus
2 2 2 22 3 2 3
2 2 2 2 2 22 3 2 3 2 3
( ) ( )( ) ( ) ( )
minus minus minus=
minus + minus minus minusu u w wR
c c u u w w
3 2= =U u u
Expliqueacutee + tard
2 23 2
03 02 3 2 2ec cW h h h h minus
= minus = minus +
119877119877 =ℎ2 minus ℎ3ℎ1 minus ℎ3
=ℎ2 minus ℎ3ℎ02 minus ℎ03
Degreacute de reacuteaction
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
( )3u 2uw wR2U+
= minus
3 212
u uc cRU+
= minus
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
Puisque le degreacute de reacuteaction R peut aussi srsquoexprimer enfonction de variables cineacutematiques on peut chercher sarepreacutesentation dans le triangle adimensionnel de vitesses
Turbine
α2
β2 α3
β3
C2 W3
C3W2
C3uC2u
U
C2xC3x
W2u
c2u ndash c3u= w2u - w3u
2Ψ minusR
frasl119934119934120784120784119958119958 119932119932 = frasl1198621198622119906119906 119880119880 minus frasl119880119880 119880119880Ψ = frasl1198621198622119906119906 minus 1198621198623119906119906 119880119880
119877119877 = 1 minus (1198621198623119906119906 + 1198621198622119906119906 frasl) 2119880119880Φ = frasl1198621198622119909119909 119880119880 = frasl1198621198623119909119909 119880119880
= (119914119914120784120784119958119958 minus 119914119914120785120785119958119958 frasl) 2119880119880 minus 1 + (119914119914120784120784119958119958 + 119914119914120785120785119958119958 frasl) 2119880119880
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α2
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
Les repreacutesentations des triangles de vitesses adimensionnellessont tregraves riches en information et contiennent plusieurs formules
Par exemple pour un compresseur on deacuteduit aiseacutement que
Avec 120595120595 = 1198821198821198901198901198801198802
Φ = 119888119888119909119909119880119880
et que pour 119888119888119901119901 = 119888119888119888119888119888119888119888119888119888119888 119882119882119890119890 = 119888119888119901119901(11987911987902minus11987911987901)on trouve
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
(11987911987902minus11987911987901) =119880119880119888119888119909119909119888119888119901119901
(tan1205731205731 minus tan1205731205732)
Alors si lrsquoon veut obtenir une plus grande variation detempeacuterature 11987911987902 minus 11987911987901 indirectement un plus grand rapport(1199011199010211990111990101) on peut augmenter
bull la vitesse peacuteripheacuterique 119880119880 (la vitesse de rotation)bull la vitesse axiale 119888119888119909119909 (le deacutebit massique)bull lrsquoangle de deacuteflection de la pale(tan1205731205731 minus tan1205731205732)
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
119963119963
120637120637119955119955
Plan 119903119903 minus 120579120579119911119911 = 119888119888119888119888119888119888119888119888119888119888
Carter
119955119955 Moyeu
Plan 119903119903 minus 119911119911
Plan 119903119903 minus 119911119911120579120579 = 119888119888119888119888119888119888119888119888119888119888
119955119955
Rayon moyen
Vues drsquoun compresseur axial
Plan 119911119911 minus 120579120579119903119903 = 119888119888119888119888119888119888119888119888119888119888
119880119880 = 119903119903120596120596
Moyeu
Carter
Rm
X
CmCx
C
Ω
119888119888 119907119907119907119907119888119888119888119888119888119888119888119888119888119888 119889119889119888119888 119897119897primeeacute119888119888119888119888119888119888119897119897119888119888119888119888119888119888119888119888119888119888
intrados
extrados surface S1
surface S2 Rayon moyen r
Lrsquoeacutetude est reacutealiseacute sur la surface 1198781198781
119932119932 = 119955119955120654120654
120654120654
θ
x
Zone aubeacutee
surface S1
119932119932 = 119955119955120654120654
Rotor Stator
Rotor-Stator
RotorStator
Dans une turbomachinelrsquoensemble stator-rotor estappeleacute eacutetage
W2
C3
U
W3
U
C1W3
U
C2
1
2
3
W vitesse relative de lrsquoeacutecoulement
U vitesse peacuteripheacuterique du rotor
C vitesse absolue de lrsquoeacutecoulement
W2
U
C2
C3W3
U
C1W1
URotor
Stator
➀
➁
➂
C
W
U
Stator Rotor
U2
Stator
C2
U3
W3
Rotor
Stator
Rotor
U2
W2
Rotor
Stator
C2
U3
W3C3
Stator
Rotor
Effets 3DEffets visqueuxDistribution de pressionTraicircneacutee PortanceEacutecoulement instationnaireInteacuteraction fuide structureTurbulence
Extrados
Intrados
L
D
Profil isoleacuteGrille drsquoaubes
Pour analyser lrsquoeacutechange drsquoeacutenergie entre le fluide et le rotor drsquouneturbomachine on utilise la variation de la quantiteacute demouvement (changement de direction etou de vitesse)
Lrsquoeacutetude classique unidimensionnelle utilise le concept de trianglede vitesse Ce modegravele est appliqueacute agrave lrsquoentreacutee et agrave la sortie ducanal inter-aube (grille drsquoaubes)
Dans celui-ci on retrouve la vitesse absolue exprimeacutee dans unrepegravere lieacute aux parties fixes (le stator) accompagneacutee de lavitesse relative lieacutee aux partie tournantes ( le rotor)
Approche classique
Lrsquoapplication du triangle de vitesse permet drsquoestimer
bull lrsquoangle drsquoincidence agrave lrsquoentreacutee du rotor
bull lrsquoangle au bord de fuite du rotor
bull la deacuteflection de lrsquoeacutecoulement dans le stator
bull le travail produit ou consommeacute par la roue
Approche classique
C vitesse absolue de lrsquoeacutecoulement
W vitesse relative de lrsquoeacutecoulement
U vitesse peacuteripheacuterique du rotor
Cu Cm Cx composante tangentielle radiale et axiale de lavitesse absolue du fluide
Wu Wm Wx composante tangentielle radiale et axiale de lavitesse relative du fluide
α lrsquoangle des vitesses absolues mesureacutees par rapport agrave la directionaxiale
β lrsquoangle des vitesses relatives mesureacutees par rapport agrave la directionaxiale
La forme des pales du rotor deacutepend des angles 120631120631(des vitesses relatives dans le repegravere du rotor)
La forme des pales du stator deacutepend des angles 120630120630(des vitesses absolues dans le repegravere fixe)
Si lrsquoon veut garder la vitesse axiale 119940119940119961119961 = 119940119940119940119940119940119940119940119940119940119940 lrsquoeacutequation deconservation de la masse 119950 = 120646120646120488120488119940119940119961119961 = 119940119940119940119940119940119940119940119940119940119940 impose 120646120646120488120488 = 119940119940119940119940119940119940119940119940119940119940
Compresseur
Cu
Cm
Cu
Plan normal
Plan meacuteridional
BABF
W2
C2
U
U
W3C3
u
Les vecteurs vitesse (vitesse absolue 119940119940 pour le stator etrelative 119960119960 pour le rotor) sont consideacutereacutes tangents agrave laligne de squelette de lrsquoaube
β2
β3W3
W2
Pour les machines axiales (u= uent = usort ) le trianglede vitesse donne lieu agrave des eacutequations scalaires tellesque
cu= wu+u
cx= wxcx= = wx c
U
w
wu
cu
Pour les machines agrave eacutetages multiples on suppose que lavitesse axiale demeure constante Crsquoest agrave dire que119940119940120784120784119961119961= 119940119940120785120785119961119961= 119940119940119961119961= 119940119940119940119940119940119940119940119940119940119940
Cette condition jumeleacutee agrave 119932119932120784120784= 119932119932120785120785= 119932119932 megravene agrave lrsquohypothegravese drsquoune cineacutematique reacutepeacutetitive En pratique
Les triangles de vitesse agrave lrsquoentreacutee et agrave la sortie drsquouneacutetage peuvent ecirctre superposeacutes
Nous commenccedilons par la repreacutesentation de triangles de vitessespour le cas drsquoune turbineNous rappelons que par hypothegravese le rayon ne varie pas dans leplan inter-aube et que la vitesse axiale demeure constante dansle sens axialCes conditions permettent drsquoeffectuer la superposition destriangles de vitesse au bord drsquoattaque et bord de fuite du rotoravec la vitesse 119932119932 commune
c1w1
u
c2
u2
w2
➊
➋
➌ c3w3
u
Turbine
C2
W2 C3
W3
Stator Rotor
C2u C3u
W3u
α2β2 β3
α3
∆Cu= ∆Wu
U
W2u
119940119940119961119961119880119880 = 119888119888119888119888119888119888119888119888119888119888119888119888119909119909 = 119888119888119888119888119888119888119888119888119888119888
1198881198883 = 1198881198881
1205721205723 = 1205721205721
Cineacutematique reacutepeacutetitive
➁ ➂
➁U➀
Turbine
119914119914120784120784
119914119914120785120785
119934119934120785120785
119932119932119934119934120784120784
119932119932
119914119914120783120783120630120630120783120783
120630120630120784120784120631120631120784120784
120630120630120785120785
120631120631120785120785
Stator Rotor
➀ ➁ ➂
Le cas du compresseur est similaire agrave celui drsquoune turbineLe rayon ne varie pas (lrsquoeacutetude se deacuteroule dans le plan inter-aube) et la vitesse axiale demeure constante dans le sens axialComme pour la turbine nous pouvons superposer les triangles devitesse au bord drsquoattaque et bord de fuite du rotor
W2 C2
C3W3
U
C1W1
U
U
Compresseur
➊
➋
➌
α2 β1
C2 W1
C1W2
U
β2α1
W2U C1U
CX
( )2 1 = minuse u uW U C C
Lrsquoopeacuteration est vectorielle
C2U
119914119914120784120784119958119958 minus 119914119914120783120783119958119958 =- 119934119934120784120784119958119958 minus119934119934120783120783119958119958
➀
➁
➁
Compresseur
119914119914120784120784119914119914120785120785
119934119934120783120783
119932119932119914119914120783120783120630120630120783120783 120630120630120784120784
120631120631120784120784120630120630120785120785
Rotor Stator
119934119934120784120784120631120631120783120783
119932119932
➀ ➁ ➂
Trois quantiteacutes fondamentales sont utiliseacutees dans laconception des turbomachines
bull le coefficient de charge ψ
bull le coefficient de deacutebit Φ
bull le degreacute de reacuteaction R
Approche classique
Parfois ψ est consideacutereacute positif pour les turbines et neacutegatifpour les compresseurs 119882119882119890119890 = frasl119882 119888
Coefficient de charge
2Ψ = eWU
Coefficient de charge
2e
UW
=ψ 02 01 02 2
h h hU Uminus ∆
= = 02
s shU
η ∆=
=
eWWm
120578120578119904119904 =∆ℎ0∆ℎ0119904119904
Turbine
Coefficient de deacutebit
xcU
Φ =
xCU
Φ =ρRT
pρ
= xC RTRTU p
=RT m RT
U Ap
xCU
Φ =119888 = ρ119862119862119909119909119860119860
11988811987711987711987911987901199011199010119860119860
= 119872119872 120574120574 1 +120574120574 minus 1
21198721198722
minus 120574120574+1)2(120574120574minus1
Afin de geacuteneacuteraliser le triangle de vitesses drsquoune machineaxiale on effectue une mise agrave lrsquoeacutechelle en divisant chaquecocircteacute par la vitesse tangentielle U
Cette opeacuteration permettra de retrouver directement lescoefficients de charge ψ et de deacutebit Φ
Par la suite un nouveau paramegravetre R qursquoon appelle degreacutede reacuteaction sera introduit pour compleacuteter la caracteacuterisationde lrsquoaubage
Approche classique
α2
β2 α3
β3C2 W3
W2
c3u - c2u= w3u - w2u
W3uW2u
U
CX CX
Lrsquoopeacuteration est vectorielle
C3
120511120511 = frasl119934119934119940119940 119932119932120784120784 = frasl119914119914120785120785119958119958 minus 119914119914120784120784119958119958 119932119932
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
Turbine
3
2
3 3
2 2
12
12
12
12
u
u
u u
u u
c RUc RU
w c RU Uw c RU U
Ψ= minus +
Ψ= minus +
Ψ= minus = +
Ψ= minus = minus
2 22 23 2
2 22 23 2
1 12 2
2 2
c cR RU U
w wR RU U
Ψ Ψ = Φ + minus + = Φ + minus +
Ψ Ψ = Φ + + = Φ + minus
3 2
3 2
1 2 1 2
2 2
R Ratan atan
R Ratan atan
minus + Ψ minus + Ψ = = Φ Φ
Ψ + Ψ minus = = Φ Φ
α α
β β
α2β1
C2U
W1U
C1UW2U β2 α1
Compresseur
ΦC2U
C3UW3U
W2U
ψ
1
ψ
ΦC2UC3U
W3U
W2U
1
Il faut deacutefinir un troisiegraveme paramegravetre pour diffeacuterencier ces deux triangles de vitesses ayant le mecircme 120509120509 et le mecircme 120537120537
Les coefficients 120509120509 et 120537120537 ne suffisent pas pour deacutefinir de maniegravereunivoque le triangle de vitesses
Le degreacute de reacuteaction est un paramegravetre adimensionnel quiexprime la variation drsquoenthalpie subie par le fluide dans le rotorpar rapport agrave la chute totale drsquoenthalpie dans lrsquoeacutetage
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
Variation d enthalpie dans le rotorRVariation d enthalpie dans l eacutetage
=
❸❶
❷
Rotor
Stator
R=0 machine drsquoaction (Δhrotor=0 Δheacutetage=Δhstator)
R=1 machine agrave reacuteaction (Δhstator=0 Δheacutetage=Δhrotor)
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
R indique la reacutepartition du saut drsquoenthalpie (de pression) statique entre le stator et le rotor
Le degreacute de reacuteaction R deacutefini en fonction de variablesthermodynamiques (lrsquoenthalpie) peut aussi srsquoexprimer en fonctionde variables cineacutematiques preacutesentes dans le triangle de vitessesNotamment en fonction de la vitesse peacuteripheacuterique U et descomposantes des vitesses absolues et relatives
La vitesse absolue agrave lrsquoentreacutee du stator est eacutegale agrave celle agrave la sortie du rotor (c1= c3 eacutetage reacutepeacutetitif )
On neacuteglige les pertes et le transfert de chaleur dans le stator
1 2 3
ℎ3 minus ℎ1= ℎ3 + frasl11988811988832 2 minus ℎ1 minus frasl11988811988812 2 = ℎ03 minus ℎ01 = 120549120549ℎ0
ℎ03 minus ℎ01 = ℎ03 minus ℎ02 = 120549120549ℎ )0(3minus2
ℎ03 minus ℎ02 = ℎ3 minus ℎ1
119945119945120782120782120783120783 = 119945119945120782120782120784120784
frasl11988811988832 2 = frasl11988811988812 2
3 2
3 1
rotor
eacutetage
h h hRh h h
∆∆
minus= =
minus
2 20 3 2 2 3
0 3 2
2 2( )
( )
h c c h
∆∆
minus
minus
+ minus=
ℎ03 ℎ02
(ℎ01= ℎ02 1198881198881= 1198881198883)
120549120549ℎ )0(3minus2 = (ℎ3 + frasl11988811988832 2) minus (ℎ2 + frasl11988811988822 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
0(3 2) 3 2( )∆ minus = minusu uh U c c
= 1 minusfrasl11988811988832 2 minus frasl11988811988822 2120549120549ℎ )0(3minus2
119877119877 =120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
120549120549ℎ )0(3minus2
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
3 212+
= minus u uc cRU
119888119888119909119909 = 1198881198882119909119909 = 1198881198883119909119909 = 119888119888119888119888119888119888119888119888119888119888
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 11988811988821199061199061198881198882 = 1198881198881199091199092+1198881198881199061199062
119877119877 = 1 minus11988811988831199061199062 minus 11988811988821199061199062
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
Indirectement lrsquoeacutequation pour R deacutecrite en fonction de composantes cineacutematiques eacutetablit un lien entre les angles drsquoentreacutee et de sortie
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
= 1 minus1198881198883119906119906 + 1198881198882119906119906
2119880119880
2232
2 32 2wwh h+ = +
2 23 2
2 3 2 2w wh hminus = minus
2 2 2 23 3 2 2
2 3 2 2u x u xw w w wh h + +
minus = minus
Machine axiale U2 = U3
La composante axiale de la vitesse demeure constante w3x = w 2x
( eacutequation justifieacutee plus tard lors de la rothalpie)
2 23 2
2 3 2 2u uw wh hminus = minus
3 2 3 22 3
( )( )2
u u u uw w w wh h minus +minus =
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
Eacutetage reacutepeacutetitif c3 = c1
Si les pertes dans le stator sont neacutegligeacutees
c2u ndash c3u= w2u - w3u
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
3 2 3 23 2
( )( )2
u u u uw w w wh h minus +minus = minus
c2u ndash c3u= w2u - w3u
119877119877 = minus1199081199083119906119906 + 1199081199082119906119906
2119880119880
2 2 2 22 3 2 3 2 3( ) 2 ( ) 2h h u u w wminus = minus minus minus
2 2 2 2 2 23 2 3 2 3 2
2 2 2e
I II III
u u c c w wW minus minus minus= + minus
2 2 2 22 3 2 3
2 2 2 2 2 22 3 2 3 2 3
( ) ( )( ) ( ) ( )
minus minus minus=
minus + minus minus minusu u w wR
c c u u w w
3 2= =U u u
Expliqueacutee + tard
2 23 2
03 02 3 2 2ec cW h h h h minus
= minus = minus +
119877119877 =ℎ2 minus ℎ3ℎ1 minus ℎ3
=ℎ2 minus ℎ3ℎ02 minus ℎ03
Degreacute de reacuteaction
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
( )3u 2uw wR2U+
= minus
3 212
u uc cRU+
= minus
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
Puisque le degreacute de reacuteaction R peut aussi srsquoexprimer enfonction de variables cineacutematiques on peut chercher sarepreacutesentation dans le triangle adimensionnel de vitesses
Turbine
α2
β2 α3
β3
C2 W3
C3W2
C3uC2u
U
C2xC3x
W2u
c2u ndash c3u= w2u - w3u
2Ψ minusR
frasl119934119934120784120784119958119958 119932119932 = frasl1198621198622119906119906 119880119880 minus frasl119880119880 119880119880Ψ = frasl1198621198622119906119906 minus 1198621198623119906119906 119880119880
119877119877 = 1 minus (1198621198623119906119906 + 1198621198622119906119906 frasl) 2119880119880Φ = frasl1198621198622119909119909 119880119880 = frasl1198621198623119909119909 119880119880
= (119914119914120784120784119958119958 minus 119914119914120785120785119958119958 frasl) 2119880119880 minus 1 + (119914119914120784120784119958119958 + 119914119914120785120785119958119958 frasl) 2119880119880
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α2
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
Les repreacutesentations des triangles de vitesses adimensionnellessont tregraves riches en information et contiennent plusieurs formules
Par exemple pour un compresseur on deacuteduit aiseacutement que
Avec 120595120595 = 1198821198821198901198901198801198802
Φ = 119888119888119909119909119880119880
et que pour 119888119888119901119901 = 119888119888119888119888119888119888119888119888119888119888 119882119882119890119890 = 119888119888119901119901(11987911987902minus11987911987901)on trouve
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
(11987911987902minus11987911987901) =119880119880119888119888119909119909119888119888119901119901
(tan1205731205731 minus tan1205731205732)
Alors si lrsquoon veut obtenir une plus grande variation detempeacuterature 11987911987902 minus 11987911987901 indirectement un plus grand rapport(1199011199010211990111990101) on peut augmenter
bull la vitesse peacuteripheacuterique 119880119880 (la vitesse de rotation)bull la vitesse axiale 119888119888119909119909 (le deacutebit massique)bull lrsquoangle de deacuteflection de la pale(tan1205731205731 minus tan1205731205732)
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
Plan 119903119903 minus 120579120579119911119911 = 119888119888119888119888119888119888119888119888119888119888
Carter
119955119955 Moyeu
Plan 119903119903 minus 119911119911
Plan 119903119903 minus 119911119911120579120579 = 119888119888119888119888119888119888119888119888119888119888
119955119955
Rayon moyen
Vues drsquoun compresseur axial
Plan 119911119911 minus 120579120579119903119903 = 119888119888119888119888119888119888119888119888119888119888
119880119880 = 119903119903120596120596
Moyeu
Carter
Rm
X
CmCx
C
Ω
119888119888 119907119907119907119907119888119888119888119888119888119888119888119888119888119888 119889119889119888119888 119897119897primeeacute119888119888119888119888119888119888119897119897119888119888119888119888119888119888119888119888119888119888
intrados
extrados surface S1
surface S2 Rayon moyen r
Lrsquoeacutetude est reacutealiseacute sur la surface 1198781198781
119932119932 = 119955119955120654120654
120654120654
θ
x
Zone aubeacutee
surface S1
119932119932 = 119955119955120654120654
Rotor Stator
Rotor-Stator
RotorStator
Dans une turbomachinelrsquoensemble stator-rotor estappeleacute eacutetage
W2
C3
U
W3
U
C1W3
U
C2
1
2
3
W vitesse relative de lrsquoeacutecoulement
U vitesse peacuteripheacuterique du rotor
C vitesse absolue de lrsquoeacutecoulement
W2
U
C2
C3W3
U
C1W1
URotor
Stator
➀
➁
➂
C
W
U
Stator Rotor
U2
Stator
C2
U3
W3
Rotor
Stator
Rotor
U2
W2
Rotor
Stator
C2
U3
W3C3
Stator
Rotor
Effets 3DEffets visqueuxDistribution de pressionTraicircneacutee PortanceEacutecoulement instationnaireInteacuteraction fuide structureTurbulence
Extrados
Intrados
L
D
Profil isoleacuteGrille drsquoaubes
Pour analyser lrsquoeacutechange drsquoeacutenergie entre le fluide et le rotor drsquouneturbomachine on utilise la variation de la quantiteacute demouvement (changement de direction etou de vitesse)
Lrsquoeacutetude classique unidimensionnelle utilise le concept de trianglede vitesse Ce modegravele est appliqueacute agrave lrsquoentreacutee et agrave la sortie ducanal inter-aube (grille drsquoaubes)
Dans celui-ci on retrouve la vitesse absolue exprimeacutee dans unrepegravere lieacute aux parties fixes (le stator) accompagneacutee de lavitesse relative lieacutee aux partie tournantes ( le rotor)
Approche classique
Lrsquoapplication du triangle de vitesse permet drsquoestimer
bull lrsquoangle drsquoincidence agrave lrsquoentreacutee du rotor
bull lrsquoangle au bord de fuite du rotor
bull la deacuteflection de lrsquoeacutecoulement dans le stator
bull le travail produit ou consommeacute par la roue
Approche classique
C vitesse absolue de lrsquoeacutecoulement
W vitesse relative de lrsquoeacutecoulement
U vitesse peacuteripheacuterique du rotor
Cu Cm Cx composante tangentielle radiale et axiale de lavitesse absolue du fluide
Wu Wm Wx composante tangentielle radiale et axiale de lavitesse relative du fluide
α lrsquoangle des vitesses absolues mesureacutees par rapport agrave la directionaxiale
β lrsquoangle des vitesses relatives mesureacutees par rapport agrave la directionaxiale
La forme des pales du rotor deacutepend des angles 120631120631(des vitesses relatives dans le repegravere du rotor)
La forme des pales du stator deacutepend des angles 120630120630(des vitesses absolues dans le repegravere fixe)
Si lrsquoon veut garder la vitesse axiale 119940119940119961119961 = 119940119940119940119940119940119940119940119940119940119940 lrsquoeacutequation deconservation de la masse 119950 = 120646120646120488120488119940119940119961119961 = 119940119940119940119940119940119940119940119940119940119940 impose 120646120646120488120488 = 119940119940119940119940119940119940119940119940119940119940
Compresseur
Cu
Cm
Cu
Plan normal
Plan meacuteridional
BABF
W2
C2
U
U
W3C3
u
Les vecteurs vitesse (vitesse absolue 119940119940 pour le stator etrelative 119960119960 pour le rotor) sont consideacutereacutes tangents agrave laligne de squelette de lrsquoaube
β2
β3W3
W2
Pour les machines axiales (u= uent = usort ) le trianglede vitesse donne lieu agrave des eacutequations scalaires tellesque
cu= wu+u
cx= wxcx= = wx c
U
w
wu
cu
Pour les machines agrave eacutetages multiples on suppose que lavitesse axiale demeure constante Crsquoest agrave dire que119940119940120784120784119961119961= 119940119940120785120785119961119961= 119940119940119961119961= 119940119940119940119940119940119940119940119940119940119940
Cette condition jumeleacutee agrave 119932119932120784120784= 119932119932120785120785= 119932119932 megravene agrave lrsquohypothegravese drsquoune cineacutematique reacutepeacutetitive En pratique
Les triangles de vitesse agrave lrsquoentreacutee et agrave la sortie drsquouneacutetage peuvent ecirctre superposeacutes
Nous commenccedilons par la repreacutesentation de triangles de vitessespour le cas drsquoune turbineNous rappelons que par hypothegravese le rayon ne varie pas dans leplan inter-aube et que la vitesse axiale demeure constante dansle sens axialCes conditions permettent drsquoeffectuer la superposition destriangles de vitesse au bord drsquoattaque et bord de fuite du rotoravec la vitesse 119932119932 commune
c1w1
u
c2
u2
w2
➊
➋
➌ c3w3
u
Turbine
C2
W2 C3
W3
Stator Rotor
C2u C3u
W3u
α2β2 β3
α3
∆Cu= ∆Wu
U
W2u
119940119940119961119961119880119880 = 119888119888119888119888119888119888119888119888119888119888119888119888119909119909 = 119888119888119888119888119888119888119888119888119888119888
1198881198883 = 1198881198881
1205721205723 = 1205721205721
Cineacutematique reacutepeacutetitive
➁ ➂
➁U➀
Turbine
119914119914120784120784
119914119914120785120785
119934119934120785120785
119932119932119934119934120784120784
119932119932
119914119914120783120783120630120630120783120783
120630120630120784120784120631120631120784120784
120630120630120785120785
120631120631120785120785
Stator Rotor
➀ ➁ ➂
Le cas du compresseur est similaire agrave celui drsquoune turbineLe rayon ne varie pas (lrsquoeacutetude se deacuteroule dans le plan inter-aube) et la vitesse axiale demeure constante dans le sens axialComme pour la turbine nous pouvons superposer les triangles devitesse au bord drsquoattaque et bord de fuite du rotor
W2 C2
C3W3
U
C1W1
U
U
Compresseur
➊
➋
➌
α2 β1
C2 W1
C1W2
U
β2α1
W2U C1U
CX
( )2 1 = minuse u uW U C C
Lrsquoopeacuteration est vectorielle
C2U
119914119914120784120784119958119958 minus 119914119914120783120783119958119958 =- 119934119934120784120784119958119958 minus119934119934120783120783119958119958
➀
➁
➁
Compresseur
119914119914120784120784119914119914120785120785
119934119934120783120783
119932119932119914119914120783120783120630120630120783120783 120630120630120784120784
120631120631120784120784120630120630120785120785
Rotor Stator
119934119934120784120784120631120631120783120783
119932119932
➀ ➁ ➂
Trois quantiteacutes fondamentales sont utiliseacutees dans laconception des turbomachines
bull le coefficient de charge ψ
bull le coefficient de deacutebit Φ
bull le degreacute de reacuteaction R
Approche classique
Parfois ψ est consideacutereacute positif pour les turbines et neacutegatifpour les compresseurs 119882119882119890119890 = frasl119882 119888
Coefficient de charge
2Ψ = eWU
Coefficient de charge
2e
UW
=ψ 02 01 02 2
h h hU Uminus ∆
= = 02
s shU
η ∆=
=
eWWm
120578120578119904119904 =∆ℎ0∆ℎ0119904119904
Turbine
Coefficient de deacutebit
xcU
Φ =
xCU
Φ =ρRT
pρ
= xC RTRTU p
=RT m RT
U Ap
xCU
Φ =119888 = ρ119862119862119909119909119860119860
11988811987711987711987911987901199011199010119860119860
= 119872119872 120574120574 1 +120574120574 minus 1
21198721198722
minus 120574120574+1)2(120574120574minus1
Afin de geacuteneacuteraliser le triangle de vitesses drsquoune machineaxiale on effectue une mise agrave lrsquoeacutechelle en divisant chaquecocircteacute par la vitesse tangentielle U
Cette opeacuteration permettra de retrouver directement lescoefficients de charge ψ et de deacutebit Φ
Par la suite un nouveau paramegravetre R qursquoon appelle degreacutede reacuteaction sera introduit pour compleacuteter la caracteacuterisationde lrsquoaubage
Approche classique
α2
β2 α3
β3C2 W3
W2
c3u - c2u= w3u - w2u
W3uW2u
U
CX CX
Lrsquoopeacuteration est vectorielle
C3
120511120511 = frasl119934119934119940119940 119932119932120784120784 = frasl119914119914120785120785119958119958 minus 119914119914120784120784119958119958 119932119932
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
Turbine
3
2
3 3
2 2
12
12
12
12
u
u
u u
u u
c RUc RU
w c RU Uw c RU U
Ψ= minus +
Ψ= minus +
Ψ= minus = +
Ψ= minus = minus
2 22 23 2
2 22 23 2
1 12 2
2 2
c cR RU U
w wR RU U
Ψ Ψ = Φ + minus + = Φ + minus +
Ψ Ψ = Φ + + = Φ + minus
3 2
3 2
1 2 1 2
2 2
R Ratan atan
R Ratan atan
minus + Ψ minus + Ψ = = Φ Φ
Ψ + Ψ minus = = Φ Φ
α α
β β
α2β1
C2U
W1U
C1UW2U β2 α1
Compresseur
ΦC2U
C3UW3U
W2U
ψ
1
ψ
ΦC2UC3U
W3U
W2U
1
Il faut deacutefinir un troisiegraveme paramegravetre pour diffeacuterencier ces deux triangles de vitesses ayant le mecircme 120509120509 et le mecircme 120537120537
Les coefficients 120509120509 et 120537120537 ne suffisent pas pour deacutefinir de maniegravereunivoque le triangle de vitesses
Le degreacute de reacuteaction est un paramegravetre adimensionnel quiexprime la variation drsquoenthalpie subie par le fluide dans le rotorpar rapport agrave la chute totale drsquoenthalpie dans lrsquoeacutetage
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
Variation d enthalpie dans le rotorRVariation d enthalpie dans l eacutetage
=
❸❶
❷
Rotor
Stator
R=0 machine drsquoaction (Δhrotor=0 Δheacutetage=Δhstator)
R=1 machine agrave reacuteaction (Δhstator=0 Δheacutetage=Δhrotor)
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
R indique la reacutepartition du saut drsquoenthalpie (de pression) statique entre le stator et le rotor
Le degreacute de reacuteaction R deacutefini en fonction de variablesthermodynamiques (lrsquoenthalpie) peut aussi srsquoexprimer en fonctionde variables cineacutematiques preacutesentes dans le triangle de vitessesNotamment en fonction de la vitesse peacuteripheacuterique U et descomposantes des vitesses absolues et relatives
La vitesse absolue agrave lrsquoentreacutee du stator est eacutegale agrave celle agrave la sortie du rotor (c1= c3 eacutetage reacutepeacutetitif )
On neacuteglige les pertes et le transfert de chaleur dans le stator
1 2 3
ℎ3 minus ℎ1= ℎ3 + frasl11988811988832 2 minus ℎ1 minus frasl11988811988812 2 = ℎ03 minus ℎ01 = 120549120549ℎ0
ℎ03 minus ℎ01 = ℎ03 minus ℎ02 = 120549120549ℎ )0(3minus2
ℎ03 minus ℎ02 = ℎ3 minus ℎ1
119945119945120782120782120783120783 = 119945119945120782120782120784120784
frasl11988811988832 2 = frasl11988811988812 2
3 2
3 1
rotor
eacutetage
h h hRh h h
∆∆
minus= =
minus
2 20 3 2 2 3
0 3 2
2 2( )
( )
h c c h
∆∆
minus
minus
+ minus=
ℎ03 ℎ02
(ℎ01= ℎ02 1198881198881= 1198881198883)
120549120549ℎ )0(3minus2 = (ℎ3 + frasl11988811988832 2) minus (ℎ2 + frasl11988811988822 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
0(3 2) 3 2( )∆ minus = minusu uh U c c
= 1 minusfrasl11988811988832 2 minus frasl11988811988822 2120549120549ℎ )0(3minus2
119877119877 =120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
120549120549ℎ )0(3minus2
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
3 212+
= minus u uc cRU
119888119888119909119909 = 1198881198882119909119909 = 1198881198883119909119909 = 119888119888119888119888119888119888119888119888119888119888
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 11988811988821199061199061198881198882 = 1198881198881199091199092+1198881198881199061199062
119877119877 = 1 minus11988811988831199061199062 minus 11988811988821199061199062
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
Indirectement lrsquoeacutequation pour R deacutecrite en fonction de composantes cineacutematiques eacutetablit un lien entre les angles drsquoentreacutee et de sortie
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
= 1 minus1198881198883119906119906 + 1198881198882119906119906
2119880119880
2232
2 32 2wwh h+ = +
2 23 2
2 3 2 2w wh hminus = minus
2 2 2 23 3 2 2
2 3 2 2u x u xw w w wh h + +
minus = minus
Machine axiale U2 = U3
La composante axiale de la vitesse demeure constante w3x = w 2x
( eacutequation justifieacutee plus tard lors de la rothalpie)
2 23 2
2 3 2 2u uw wh hminus = minus
3 2 3 22 3
( )( )2
u u u uw w w wh h minus +minus =
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
Eacutetage reacutepeacutetitif c3 = c1
Si les pertes dans le stator sont neacutegligeacutees
c2u ndash c3u= w2u - w3u
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
3 2 3 23 2
( )( )2
u u u uw w w wh h minus +minus = minus
c2u ndash c3u= w2u - w3u
119877119877 = minus1199081199083119906119906 + 1199081199082119906119906
2119880119880
2 2 2 22 3 2 3 2 3( ) 2 ( ) 2h h u u w wminus = minus minus minus
2 2 2 2 2 23 2 3 2 3 2
2 2 2e
I II III
u u c c w wW minus minus minus= + minus
2 2 2 22 3 2 3
2 2 2 2 2 22 3 2 3 2 3
( ) ( )( ) ( ) ( )
minus minus minus=
minus + minus minus minusu u w wR
c c u u w w
3 2= =U u u
Expliqueacutee + tard
2 23 2
03 02 3 2 2ec cW h h h h minus
= minus = minus +
119877119877 =ℎ2 minus ℎ3ℎ1 minus ℎ3
=ℎ2 minus ℎ3ℎ02 minus ℎ03
Degreacute de reacuteaction
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
( )3u 2uw wR2U+
= minus
3 212
u uc cRU+
= minus
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
Puisque le degreacute de reacuteaction R peut aussi srsquoexprimer enfonction de variables cineacutematiques on peut chercher sarepreacutesentation dans le triangle adimensionnel de vitesses
Turbine
α2
β2 α3
β3
C2 W3
C3W2
C3uC2u
U
C2xC3x
W2u
c2u ndash c3u= w2u - w3u
2Ψ minusR
frasl119934119934120784120784119958119958 119932119932 = frasl1198621198622119906119906 119880119880 minus frasl119880119880 119880119880Ψ = frasl1198621198622119906119906 minus 1198621198623119906119906 119880119880
119877119877 = 1 minus (1198621198623119906119906 + 1198621198622119906119906 frasl) 2119880119880Φ = frasl1198621198622119909119909 119880119880 = frasl1198621198623119909119909 119880119880
= (119914119914120784120784119958119958 minus 119914119914120785120785119958119958 frasl) 2119880119880 minus 1 + (119914119914120784120784119958119958 + 119914119914120785120785119958119958 frasl) 2119880119880
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α2
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
Les repreacutesentations des triangles de vitesses adimensionnellessont tregraves riches en information et contiennent plusieurs formules
Par exemple pour un compresseur on deacuteduit aiseacutement que
Avec 120595120595 = 1198821198821198901198901198801198802
Φ = 119888119888119909119909119880119880
et que pour 119888119888119901119901 = 119888119888119888119888119888119888119888119888119888119888 119882119882119890119890 = 119888119888119901119901(11987911987902minus11987911987901)on trouve
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
(11987911987902minus11987911987901) =119880119880119888119888119909119909119888119888119901119901
(tan1205731205731 minus tan1205731205732)
Alors si lrsquoon veut obtenir une plus grande variation detempeacuterature 11987911987902 minus 11987911987901 indirectement un plus grand rapport(1199011199010211990111990101) on peut augmenter
bull la vitesse peacuteripheacuterique 119880119880 (la vitesse de rotation)bull la vitesse axiale 119888119888119909119909 (le deacutebit massique)bull lrsquoangle de deacuteflection de la pale(tan1205731205731 minus tan1205731205732)
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
Moyeu
Carter
Rm
X
CmCx
C
Ω
119888119888 119907119907119907119907119888119888119888119888119888119888119888119888119888119888 119889119889119888119888 119897119897primeeacute119888119888119888119888119888119888119897119897119888119888119888119888119888119888119888119888119888119888
intrados
extrados surface S1
surface S2 Rayon moyen r
Lrsquoeacutetude est reacutealiseacute sur la surface 1198781198781
119932119932 = 119955119955120654120654
120654120654
θ
x
Zone aubeacutee
surface S1
119932119932 = 119955119955120654120654
Rotor Stator
Rotor-Stator
RotorStator
Dans une turbomachinelrsquoensemble stator-rotor estappeleacute eacutetage
W2
C3
U
W3
U
C1W3
U
C2
1
2
3
W vitesse relative de lrsquoeacutecoulement
U vitesse peacuteripheacuterique du rotor
C vitesse absolue de lrsquoeacutecoulement
W2
U
C2
C3W3
U
C1W1
URotor
Stator
➀
➁
➂
C
W
U
Stator Rotor
U2
Stator
C2
U3
W3
Rotor
Stator
Rotor
U2
W2
Rotor
Stator
C2
U3
W3C3
Stator
Rotor
Effets 3DEffets visqueuxDistribution de pressionTraicircneacutee PortanceEacutecoulement instationnaireInteacuteraction fuide structureTurbulence
Extrados
Intrados
L
D
Profil isoleacuteGrille drsquoaubes
Pour analyser lrsquoeacutechange drsquoeacutenergie entre le fluide et le rotor drsquouneturbomachine on utilise la variation de la quantiteacute demouvement (changement de direction etou de vitesse)
Lrsquoeacutetude classique unidimensionnelle utilise le concept de trianglede vitesse Ce modegravele est appliqueacute agrave lrsquoentreacutee et agrave la sortie ducanal inter-aube (grille drsquoaubes)
Dans celui-ci on retrouve la vitesse absolue exprimeacutee dans unrepegravere lieacute aux parties fixes (le stator) accompagneacutee de lavitesse relative lieacutee aux partie tournantes ( le rotor)
Approche classique
Lrsquoapplication du triangle de vitesse permet drsquoestimer
bull lrsquoangle drsquoincidence agrave lrsquoentreacutee du rotor
bull lrsquoangle au bord de fuite du rotor
bull la deacuteflection de lrsquoeacutecoulement dans le stator
bull le travail produit ou consommeacute par la roue
Approche classique
C vitesse absolue de lrsquoeacutecoulement
W vitesse relative de lrsquoeacutecoulement
U vitesse peacuteripheacuterique du rotor
Cu Cm Cx composante tangentielle radiale et axiale de lavitesse absolue du fluide
Wu Wm Wx composante tangentielle radiale et axiale de lavitesse relative du fluide
α lrsquoangle des vitesses absolues mesureacutees par rapport agrave la directionaxiale
β lrsquoangle des vitesses relatives mesureacutees par rapport agrave la directionaxiale
La forme des pales du rotor deacutepend des angles 120631120631(des vitesses relatives dans le repegravere du rotor)
La forme des pales du stator deacutepend des angles 120630120630(des vitesses absolues dans le repegravere fixe)
Si lrsquoon veut garder la vitesse axiale 119940119940119961119961 = 119940119940119940119940119940119940119940119940119940119940 lrsquoeacutequation deconservation de la masse 119950 = 120646120646120488120488119940119940119961119961 = 119940119940119940119940119940119940119940119940119940119940 impose 120646120646120488120488 = 119940119940119940119940119940119940119940119940119940119940
Compresseur
Cu
Cm
Cu
Plan normal
Plan meacuteridional
BABF
W2
C2
U
U
W3C3
u
Les vecteurs vitesse (vitesse absolue 119940119940 pour le stator etrelative 119960119960 pour le rotor) sont consideacutereacutes tangents agrave laligne de squelette de lrsquoaube
β2
β3W3
W2
Pour les machines axiales (u= uent = usort ) le trianglede vitesse donne lieu agrave des eacutequations scalaires tellesque
cu= wu+u
cx= wxcx= = wx c
U
w
wu
cu
Pour les machines agrave eacutetages multiples on suppose que lavitesse axiale demeure constante Crsquoest agrave dire que119940119940120784120784119961119961= 119940119940120785120785119961119961= 119940119940119961119961= 119940119940119940119940119940119940119940119940119940119940
Cette condition jumeleacutee agrave 119932119932120784120784= 119932119932120785120785= 119932119932 megravene agrave lrsquohypothegravese drsquoune cineacutematique reacutepeacutetitive En pratique
Les triangles de vitesse agrave lrsquoentreacutee et agrave la sortie drsquouneacutetage peuvent ecirctre superposeacutes
Nous commenccedilons par la repreacutesentation de triangles de vitessespour le cas drsquoune turbineNous rappelons que par hypothegravese le rayon ne varie pas dans leplan inter-aube et que la vitesse axiale demeure constante dansle sens axialCes conditions permettent drsquoeffectuer la superposition destriangles de vitesse au bord drsquoattaque et bord de fuite du rotoravec la vitesse 119932119932 commune
c1w1
u
c2
u2
w2
➊
➋
➌ c3w3
u
Turbine
C2
W2 C3
W3
Stator Rotor
C2u C3u
W3u
α2β2 β3
α3
∆Cu= ∆Wu
U
W2u
119940119940119961119961119880119880 = 119888119888119888119888119888119888119888119888119888119888119888119888119909119909 = 119888119888119888119888119888119888119888119888119888119888
1198881198883 = 1198881198881
1205721205723 = 1205721205721
Cineacutematique reacutepeacutetitive
➁ ➂
➁U➀
Turbine
119914119914120784120784
119914119914120785120785
119934119934120785120785
119932119932119934119934120784120784
119932119932
119914119914120783120783120630120630120783120783
120630120630120784120784120631120631120784120784
120630120630120785120785
120631120631120785120785
Stator Rotor
➀ ➁ ➂
Le cas du compresseur est similaire agrave celui drsquoune turbineLe rayon ne varie pas (lrsquoeacutetude se deacuteroule dans le plan inter-aube) et la vitesse axiale demeure constante dans le sens axialComme pour la turbine nous pouvons superposer les triangles devitesse au bord drsquoattaque et bord de fuite du rotor
W2 C2
C3W3
U
C1W1
U
U
Compresseur
➊
➋
➌
α2 β1
C2 W1
C1W2
U
β2α1
W2U C1U
CX
( )2 1 = minuse u uW U C C
Lrsquoopeacuteration est vectorielle
C2U
119914119914120784120784119958119958 minus 119914119914120783120783119958119958 =- 119934119934120784120784119958119958 minus119934119934120783120783119958119958
➀
➁
➁
Compresseur
119914119914120784120784119914119914120785120785
119934119934120783120783
119932119932119914119914120783120783120630120630120783120783 120630120630120784120784
120631120631120784120784120630120630120785120785
Rotor Stator
119934119934120784120784120631120631120783120783
119932119932
➀ ➁ ➂
Trois quantiteacutes fondamentales sont utiliseacutees dans laconception des turbomachines
bull le coefficient de charge ψ
bull le coefficient de deacutebit Φ
bull le degreacute de reacuteaction R
Approche classique
Parfois ψ est consideacutereacute positif pour les turbines et neacutegatifpour les compresseurs 119882119882119890119890 = frasl119882 119888
Coefficient de charge
2Ψ = eWU
Coefficient de charge
2e
UW
=ψ 02 01 02 2
h h hU Uminus ∆
= = 02
s shU
η ∆=
=
eWWm
120578120578119904119904 =∆ℎ0∆ℎ0119904119904
Turbine
Coefficient de deacutebit
xcU
Φ =
xCU
Φ =ρRT
pρ
= xC RTRTU p
=RT m RT
U Ap
xCU
Φ =119888 = ρ119862119862119909119909119860119860
11988811987711987711987911987901199011199010119860119860
= 119872119872 120574120574 1 +120574120574 minus 1
21198721198722
minus 120574120574+1)2(120574120574minus1
Afin de geacuteneacuteraliser le triangle de vitesses drsquoune machineaxiale on effectue une mise agrave lrsquoeacutechelle en divisant chaquecocircteacute par la vitesse tangentielle U
Cette opeacuteration permettra de retrouver directement lescoefficients de charge ψ et de deacutebit Φ
Par la suite un nouveau paramegravetre R qursquoon appelle degreacutede reacuteaction sera introduit pour compleacuteter la caracteacuterisationde lrsquoaubage
Approche classique
α2
β2 α3
β3C2 W3
W2
c3u - c2u= w3u - w2u
W3uW2u
U
CX CX
Lrsquoopeacuteration est vectorielle
C3
120511120511 = frasl119934119934119940119940 119932119932120784120784 = frasl119914119914120785120785119958119958 minus 119914119914120784120784119958119958 119932119932
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
Turbine
3
2
3 3
2 2
12
12
12
12
u
u
u u
u u
c RUc RU
w c RU Uw c RU U
Ψ= minus +
Ψ= minus +
Ψ= minus = +
Ψ= minus = minus
2 22 23 2
2 22 23 2
1 12 2
2 2
c cR RU U
w wR RU U
Ψ Ψ = Φ + minus + = Φ + minus +
Ψ Ψ = Φ + + = Φ + minus
3 2
3 2
1 2 1 2
2 2
R Ratan atan
R Ratan atan
minus + Ψ minus + Ψ = = Φ Φ
Ψ + Ψ minus = = Φ Φ
α α
β β
α2β1
C2U
W1U
C1UW2U β2 α1
Compresseur
ΦC2U
C3UW3U
W2U
ψ
1
ψ
ΦC2UC3U
W3U
W2U
1
Il faut deacutefinir un troisiegraveme paramegravetre pour diffeacuterencier ces deux triangles de vitesses ayant le mecircme 120509120509 et le mecircme 120537120537
Les coefficients 120509120509 et 120537120537 ne suffisent pas pour deacutefinir de maniegravereunivoque le triangle de vitesses
Le degreacute de reacuteaction est un paramegravetre adimensionnel quiexprime la variation drsquoenthalpie subie par le fluide dans le rotorpar rapport agrave la chute totale drsquoenthalpie dans lrsquoeacutetage
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
Variation d enthalpie dans le rotorRVariation d enthalpie dans l eacutetage
=
❸❶
❷
Rotor
Stator
R=0 machine drsquoaction (Δhrotor=0 Δheacutetage=Δhstator)
R=1 machine agrave reacuteaction (Δhstator=0 Δheacutetage=Δhrotor)
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
R indique la reacutepartition du saut drsquoenthalpie (de pression) statique entre le stator et le rotor
Le degreacute de reacuteaction R deacutefini en fonction de variablesthermodynamiques (lrsquoenthalpie) peut aussi srsquoexprimer en fonctionde variables cineacutematiques preacutesentes dans le triangle de vitessesNotamment en fonction de la vitesse peacuteripheacuterique U et descomposantes des vitesses absolues et relatives
La vitesse absolue agrave lrsquoentreacutee du stator est eacutegale agrave celle agrave la sortie du rotor (c1= c3 eacutetage reacutepeacutetitif )
On neacuteglige les pertes et le transfert de chaleur dans le stator
1 2 3
ℎ3 minus ℎ1= ℎ3 + frasl11988811988832 2 minus ℎ1 minus frasl11988811988812 2 = ℎ03 minus ℎ01 = 120549120549ℎ0
ℎ03 minus ℎ01 = ℎ03 minus ℎ02 = 120549120549ℎ )0(3minus2
ℎ03 minus ℎ02 = ℎ3 minus ℎ1
119945119945120782120782120783120783 = 119945119945120782120782120784120784
frasl11988811988832 2 = frasl11988811988812 2
3 2
3 1
rotor
eacutetage
h h hRh h h
∆∆
minus= =
minus
2 20 3 2 2 3
0 3 2
2 2( )
( )
h c c h
∆∆
minus
minus
+ minus=
ℎ03 ℎ02
(ℎ01= ℎ02 1198881198881= 1198881198883)
120549120549ℎ )0(3minus2 = (ℎ3 + frasl11988811988832 2) minus (ℎ2 + frasl11988811988822 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
0(3 2) 3 2( )∆ minus = minusu uh U c c
= 1 minusfrasl11988811988832 2 minus frasl11988811988822 2120549120549ℎ )0(3minus2
119877119877 =120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
120549120549ℎ )0(3minus2
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
3 212+
= minus u uc cRU
119888119888119909119909 = 1198881198882119909119909 = 1198881198883119909119909 = 119888119888119888119888119888119888119888119888119888119888
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 11988811988821199061199061198881198882 = 1198881198881199091199092+1198881198881199061199062
119877119877 = 1 minus11988811988831199061199062 minus 11988811988821199061199062
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
Indirectement lrsquoeacutequation pour R deacutecrite en fonction de composantes cineacutematiques eacutetablit un lien entre les angles drsquoentreacutee et de sortie
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
= 1 minus1198881198883119906119906 + 1198881198882119906119906
2119880119880
2232
2 32 2wwh h+ = +
2 23 2
2 3 2 2w wh hminus = minus
2 2 2 23 3 2 2
2 3 2 2u x u xw w w wh h + +
minus = minus
Machine axiale U2 = U3
La composante axiale de la vitesse demeure constante w3x = w 2x
( eacutequation justifieacutee plus tard lors de la rothalpie)
2 23 2
2 3 2 2u uw wh hminus = minus
3 2 3 22 3
( )( )2
u u u uw w w wh h minus +minus =
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
Eacutetage reacutepeacutetitif c3 = c1
Si les pertes dans le stator sont neacutegligeacutees
c2u ndash c3u= w2u - w3u
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
3 2 3 23 2
( )( )2
u u u uw w w wh h minus +minus = minus
c2u ndash c3u= w2u - w3u
119877119877 = minus1199081199083119906119906 + 1199081199082119906119906
2119880119880
2 2 2 22 3 2 3 2 3( ) 2 ( ) 2h h u u w wminus = minus minus minus
2 2 2 2 2 23 2 3 2 3 2
2 2 2e
I II III
u u c c w wW minus minus minus= + minus
2 2 2 22 3 2 3
2 2 2 2 2 22 3 2 3 2 3
( ) ( )( ) ( ) ( )
minus minus minus=
minus + minus minus minusu u w wR
c c u u w w
3 2= =U u u
Expliqueacutee + tard
2 23 2
03 02 3 2 2ec cW h h h h minus
= minus = minus +
119877119877 =ℎ2 minus ℎ3ℎ1 minus ℎ3
=ℎ2 minus ℎ3ℎ02 minus ℎ03
Degreacute de reacuteaction
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
( )3u 2uw wR2U+
= minus
3 212
u uc cRU+
= minus
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
Puisque le degreacute de reacuteaction R peut aussi srsquoexprimer enfonction de variables cineacutematiques on peut chercher sarepreacutesentation dans le triangle adimensionnel de vitesses
Turbine
α2
β2 α3
β3
C2 W3
C3W2
C3uC2u
U
C2xC3x
W2u
c2u ndash c3u= w2u - w3u
2Ψ minusR
frasl119934119934120784120784119958119958 119932119932 = frasl1198621198622119906119906 119880119880 minus frasl119880119880 119880119880Ψ = frasl1198621198622119906119906 minus 1198621198623119906119906 119880119880
119877119877 = 1 minus (1198621198623119906119906 + 1198621198622119906119906 frasl) 2119880119880Φ = frasl1198621198622119909119909 119880119880 = frasl1198621198623119909119909 119880119880
= (119914119914120784120784119958119958 minus 119914119914120785120785119958119958 frasl) 2119880119880 minus 1 + (119914119914120784120784119958119958 + 119914119914120785120785119958119958 frasl) 2119880119880
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α2
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
Les repreacutesentations des triangles de vitesses adimensionnellessont tregraves riches en information et contiennent plusieurs formules
Par exemple pour un compresseur on deacuteduit aiseacutement que
Avec 120595120595 = 1198821198821198901198901198801198802
Φ = 119888119888119909119909119880119880
et que pour 119888119888119901119901 = 119888119888119888119888119888119888119888119888119888119888 119882119882119890119890 = 119888119888119901119901(11987911987902minus11987911987901)on trouve
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
(11987911987902minus11987911987901) =119880119880119888119888119909119909119888119888119901119901
(tan1205731205731 minus tan1205731205732)
Alors si lrsquoon veut obtenir une plus grande variation detempeacuterature 11987911987902 minus 11987911987901 indirectement un plus grand rapport(1199011199010211990111990101) on peut augmenter
bull la vitesse peacuteripheacuterique 119880119880 (la vitesse de rotation)bull la vitesse axiale 119888119888119909119909 (le deacutebit massique)bull lrsquoangle de deacuteflection de la pale(tan1205731205731 minus tan1205731205732)
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
intrados
extrados surface S1
surface S2 Rayon moyen r
Lrsquoeacutetude est reacutealiseacute sur la surface 1198781198781
119932119932 = 119955119955120654120654
120654120654
θ
x
Zone aubeacutee
surface S1
119932119932 = 119955119955120654120654
Rotor Stator
Rotor-Stator
RotorStator
Dans une turbomachinelrsquoensemble stator-rotor estappeleacute eacutetage
W2
C3
U
W3
U
C1W3
U
C2
1
2
3
W vitesse relative de lrsquoeacutecoulement
U vitesse peacuteripheacuterique du rotor
C vitesse absolue de lrsquoeacutecoulement
W2
U
C2
C3W3
U
C1W1
URotor
Stator
➀
➁
➂
C
W
U
Stator Rotor
U2
Stator
C2
U3
W3
Rotor
Stator
Rotor
U2
W2
Rotor
Stator
C2
U3
W3C3
Stator
Rotor
Effets 3DEffets visqueuxDistribution de pressionTraicircneacutee PortanceEacutecoulement instationnaireInteacuteraction fuide structureTurbulence
Extrados
Intrados
L
D
Profil isoleacuteGrille drsquoaubes
Pour analyser lrsquoeacutechange drsquoeacutenergie entre le fluide et le rotor drsquouneturbomachine on utilise la variation de la quantiteacute demouvement (changement de direction etou de vitesse)
Lrsquoeacutetude classique unidimensionnelle utilise le concept de trianglede vitesse Ce modegravele est appliqueacute agrave lrsquoentreacutee et agrave la sortie ducanal inter-aube (grille drsquoaubes)
Dans celui-ci on retrouve la vitesse absolue exprimeacutee dans unrepegravere lieacute aux parties fixes (le stator) accompagneacutee de lavitesse relative lieacutee aux partie tournantes ( le rotor)
Approche classique
Lrsquoapplication du triangle de vitesse permet drsquoestimer
bull lrsquoangle drsquoincidence agrave lrsquoentreacutee du rotor
bull lrsquoangle au bord de fuite du rotor
bull la deacuteflection de lrsquoeacutecoulement dans le stator
bull le travail produit ou consommeacute par la roue
Approche classique
C vitesse absolue de lrsquoeacutecoulement
W vitesse relative de lrsquoeacutecoulement
U vitesse peacuteripheacuterique du rotor
Cu Cm Cx composante tangentielle radiale et axiale de lavitesse absolue du fluide
Wu Wm Wx composante tangentielle radiale et axiale de lavitesse relative du fluide
α lrsquoangle des vitesses absolues mesureacutees par rapport agrave la directionaxiale
β lrsquoangle des vitesses relatives mesureacutees par rapport agrave la directionaxiale
La forme des pales du rotor deacutepend des angles 120631120631(des vitesses relatives dans le repegravere du rotor)
La forme des pales du stator deacutepend des angles 120630120630(des vitesses absolues dans le repegravere fixe)
Si lrsquoon veut garder la vitesse axiale 119940119940119961119961 = 119940119940119940119940119940119940119940119940119940119940 lrsquoeacutequation deconservation de la masse 119950 = 120646120646120488120488119940119940119961119961 = 119940119940119940119940119940119940119940119940119940119940 impose 120646120646120488120488 = 119940119940119940119940119940119940119940119940119940119940
Compresseur
Cu
Cm
Cu
Plan normal
Plan meacuteridional
BABF
W2
C2
U
U
W3C3
u
Les vecteurs vitesse (vitesse absolue 119940119940 pour le stator etrelative 119960119960 pour le rotor) sont consideacutereacutes tangents agrave laligne de squelette de lrsquoaube
β2
β3W3
W2
Pour les machines axiales (u= uent = usort ) le trianglede vitesse donne lieu agrave des eacutequations scalaires tellesque
cu= wu+u
cx= wxcx= = wx c
U
w
wu
cu
Pour les machines agrave eacutetages multiples on suppose que lavitesse axiale demeure constante Crsquoest agrave dire que119940119940120784120784119961119961= 119940119940120785120785119961119961= 119940119940119961119961= 119940119940119940119940119940119940119940119940119940119940
Cette condition jumeleacutee agrave 119932119932120784120784= 119932119932120785120785= 119932119932 megravene agrave lrsquohypothegravese drsquoune cineacutematique reacutepeacutetitive En pratique
Les triangles de vitesse agrave lrsquoentreacutee et agrave la sortie drsquouneacutetage peuvent ecirctre superposeacutes
Nous commenccedilons par la repreacutesentation de triangles de vitessespour le cas drsquoune turbineNous rappelons que par hypothegravese le rayon ne varie pas dans leplan inter-aube et que la vitesse axiale demeure constante dansle sens axialCes conditions permettent drsquoeffectuer la superposition destriangles de vitesse au bord drsquoattaque et bord de fuite du rotoravec la vitesse 119932119932 commune
c1w1
u
c2
u2
w2
➊
➋
➌ c3w3
u
Turbine
C2
W2 C3
W3
Stator Rotor
C2u C3u
W3u
α2β2 β3
α3
∆Cu= ∆Wu
U
W2u
119940119940119961119961119880119880 = 119888119888119888119888119888119888119888119888119888119888119888119888119909119909 = 119888119888119888119888119888119888119888119888119888119888
1198881198883 = 1198881198881
1205721205723 = 1205721205721
Cineacutematique reacutepeacutetitive
➁ ➂
➁U➀
Turbine
119914119914120784120784
119914119914120785120785
119934119934120785120785
119932119932119934119934120784120784
119932119932
119914119914120783120783120630120630120783120783
120630120630120784120784120631120631120784120784
120630120630120785120785
120631120631120785120785
Stator Rotor
➀ ➁ ➂
Le cas du compresseur est similaire agrave celui drsquoune turbineLe rayon ne varie pas (lrsquoeacutetude se deacuteroule dans le plan inter-aube) et la vitesse axiale demeure constante dans le sens axialComme pour la turbine nous pouvons superposer les triangles devitesse au bord drsquoattaque et bord de fuite du rotor
W2 C2
C3W3
U
C1W1
U
U
Compresseur
➊
➋
➌
α2 β1
C2 W1
C1W2
U
β2α1
W2U C1U
CX
( )2 1 = minuse u uW U C C
Lrsquoopeacuteration est vectorielle
C2U
119914119914120784120784119958119958 minus 119914119914120783120783119958119958 =- 119934119934120784120784119958119958 minus119934119934120783120783119958119958
➀
➁
➁
Compresseur
119914119914120784120784119914119914120785120785
119934119934120783120783
119932119932119914119914120783120783120630120630120783120783 120630120630120784120784
120631120631120784120784120630120630120785120785
Rotor Stator
119934119934120784120784120631120631120783120783
119932119932
➀ ➁ ➂
Trois quantiteacutes fondamentales sont utiliseacutees dans laconception des turbomachines
bull le coefficient de charge ψ
bull le coefficient de deacutebit Φ
bull le degreacute de reacuteaction R
Approche classique
Parfois ψ est consideacutereacute positif pour les turbines et neacutegatifpour les compresseurs 119882119882119890119890 = frasl119882 119888
Coefficient de charge
2Ψ = eWU
Coefficient de charge
2e
UW
=ψ 02 01 02 2
h h hU Uminus ∆
= = 02
s shU
η ∆=
=
eWWm
120578120578119904119904 =∆ℎ0∆ℎ0119904119904
Turbine
Coefficient de deacutebit
xcU
Φ =
xCU
Φ =ρRT
pρ
= xC RTRTU p
=RT m RT
U Ap
xCU
Φ =119888 = ρ119862119862119909119909119860119860
11988811987711987711987911987901199011199010119860119860
= 119872119872 120574120574 1 +120574120574 minus 1
21198721198722
minus 120574120574+1)2(120574120574minus1
Afin de geacuteneacuteraliser le triangle de vitesses drsquoune machineaxiale on effectue une mise agrave lrsquoeacutechelle en divisant chaquecocircteacute par la vitesse tangentielle U
Cette opeacuteration permettra de retrouver directement lescoefficients de charge ψ et de deacutebit Φ
Par la suite un nouveau paramegravetre R qursquoon appelle degreacutede reacuteaction sera introduit pour compleacuteter la caracteacuterisationde lrsquoaubage
Approche classique
α2
β2 α3
β3C2 W3
W2
c3u - c2u= w3u - w2u
W3uW2u
U
CX CX
Lrsquoopeacuteration est vectorielle
C3
120511120511 = frasl119934119934119940119940 119932119932120784120784 = frasl119914119914120785120785119958119958 minus 119914119914120784120784119958119958 119932119932
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
Turbine
3
2
3 3
2 2
12
12
12
12
u
u
u u
u u
c RUc RU
w c RU Uw c RU U
Ψ= minus +
Ψ= minus +
Ψ= minus = +
Ψ= minus = minus
2 22 23 2
2 22 23 2
1 12 2
2 2
c cR RU U
w wR RU U
Ψ Ψ = Φ + minus + = Φ + minus +
Ψ Ψ = Φ + + = Φ + minus
3 2
3 2
1 2 1 2
2 2
R Ratan atan
R Ratan atan
minus + Ψ minus + Ψ = = Φ Φ
Ψ + Ψ minus = = Φ Φ
α α
β β
α2β1
C2U
W1U
C1UW2U β2 α1
Compresseur
ΦC2U
C3UW3U
W2U
ψ
1
ψ
ΦC2UC3U
W3U
W2U
1
Il faut deacutefinir un troisiegraveme paramegravetre pour diffeacuterencier ces deux triangles de vitesses ayant le mecircme 120509120509 et le mecircme 120537120537
Les coefficients 120509120509 et 120537120537 ne suffisent pas pour deacutefinir de maniegravereunivoque le triangle de vitesses
Le degreacute de reacuteaction est un paramegravetre adimensionnel quiexprime la variation drsquoenthalpie subie par le fluide dans le rotorpar rapport agrave la chute totale drsquoenthalpie dans lrsquoeacutetage
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
Variation d enthalpie dans le rotorRVariation d enthalpie dans l eacutetage
=
❸❶
❷
Rotor
Stator
R=0 machine drsquoaction (Δhrotor=0 Δheacutetage=Δhstator)
R=1 machine agrave reacuteaction (Δhstator=0 Δheacutetage=Δhrotor)
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
R indique la reacutepartition du saut drsquoenthalpie (de pression) statique entre le stator et le rotor
Le degreacute de reacuteaction R deacutefini en fonction de variablesthermodynamiques (lrsquoenthalpie) peut aussi srsquoexprimer en fonctionde variables cineacutematiques preacutesentes dans le triangle de vitessesNotamment en fonction de la vitesse peacuteripheacuterique U et descomposantes des vitesses absolues et relatives
La vitesse absolue agrave lrsquoentreacutee du stator est eacutegale agrave celle agrave la sortie du rotor (c1= c3 eacutetage reacutepeacutetitif )
On neacuteglige les pertes et le transfert de chaleur dans le stator
1 2 3
ℎ3 minus ℎ1= ℎ3 + frasl11988811988832 2 minus ℎ1 minus frasl11988811988812 2 = ℎ03 minus ℎ01 = 120549120549ℎ0
ℎ03 minus ℎ01 = ℎ03 minus ℎ02 = 120549120549ℎ )0(3minus2
ℎ03 minus ℎ02 = ℎ3 minus ℎ1
119945119945120782120782120783120783 = 119945119945120782120782120784120784
frasl11988811988832 2 = frasl11988811988812 2
3 2
3 1
rotor
eacutetage
h h hRh h h
∆∆
minus= =
minus
2 20 3 2 2 3
0 3 2
2 2( )
( )
h c c h
∆∆
minus
minus
+ minus=
ℎ03 ℎ02
(ℎ01= ℎ02 1198881198881= 1198881198883)
120549120549ℎ )0(3minus2 = (ℎ3 + frasl11988811988832 2) minus (ℎ2 + frasl11988811988822 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
0(3 2) 3 2( )∆ minus = minusu uh U c c
= 1 minusfrasl11988811988832 2 minus frasl11988811988822 2120549120549ℎ )0(3minus2
119877119877 =120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
120549120549ℎ )0(3minus2
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
3 212+
= minus u uc cRU
119888119888119909119909 = 1198881198882119909119909 = 1198881198883119909119909 = 119888119888119888119888119888119888119888119888119888119888
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 11988811988821199061199061198881198882 = 1198881198881199091199092+1198881198881199061199062
119877119877 = 1 minus11988811988831199061199062 minus 11988811988821199061199062
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
Indirectement lrsquoeacutequation pour R deacutecrite en fonction de composantes cineacutematiques eacutetablit un lien entre les angles drsquoentreacutee et de sortie
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
= 1 minus1198881198883119906119906 + 1198881198882119906119906
2119880119880
2232
2 32 2wwh h+ = +
2 23 2
2 3 2 2w wh hminus = minus
2 2 2 23 3 2 2
2 3 2 2u x u xw w w wh h + +
minus = minus
Machine axiale U2 = U3
La composante axiale de la vitesse demeure constante w3x = w 2x
( eacutequation justifieacutee plus tard lors de la rothalpie)
2 23 2
2 3 2 2u uw wh hminus = minus
3 2 3 22 3
( )( )2
u u u uw w w wh h minus +minus =
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
Eacutetage reacutepeacutetitif c3 = c1
Si les pertes dans le stator sont neacutegligeacutees
c2u ndash c3u= w2u - w3u
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
3 2 3 23 2
( )( )2
u u u uw w w wh h minus +minus = minus
c2u ndash c3u= w2u - w3u
119877119877 = minus1199081199083119906119906 + 1199081199082119906119906
2119880119880
2 2 2 22 3 2 3 2 3( ) 2 ( ) 2h h u u w wminus = minus minus minus
2 2 2 2 2 23 2 3 2 3 2
2 2 2e
I II III
u u c c w wW minus minus minus= + minus
2 2 2 22 3 2 3
2 2 2 2 2 22 3 2 3 2 3
( ) ( )( ) ( ) ( )
minus minus minus=
minus + minus minus minusu u w wR
c c u u w w
3 2= =U u u
Expliqueacutee + tard
2 23 2
03 02 3 2 2ec cW h h h h minus
= minus = minus +
119877119877 =ℎ2 minus ℎ3ℎ1 minus ℎ3
=ℎ2 minus ℎ3ℎ02 minus ℎ03
Degreacute de reacuteaction
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
( )3u 2uw wR2U+
= minus
3 212
u uc cRU+
= minus
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
Puisque le degreacute de reacuteaction R peut aussi srsquoexprimer enfonction de variables cineacutematiques on peut chercher sarepreacutesentation dans le triangle adimensionnel de vitesses
Turbine
α2
β2 α3
β3
C2 W3
C3W2
C3uC2u
U
C2xC3x
W2u
c2u ndash c3u= w2u - w3u
2Ψ minusR
frasl119934119934120784120784119958119958 119932119932 = frasl1198621198622119906119906 119880119880 minus frasl119880119880 119880119880Ψ = frasl1198621198622119906119906 minus 1198621198623119906119906 119880119880
119877119877 = 1 minus (1198621198623119906119906 + 1198621198622119906119906 frasl) 2119880119880Φ = frasl1198621198622119909119909 119880119880 = frasl1198621198623119909119909 119880119880
= (119914119914120784120784119958119958 minus 119914119914120785120785119958119958 frasl) 2119880119880 minus 1 + (119914119914120784120784119958119958 + 119914119914120785120785119958119958 frasl) 2119880119880
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α2
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
Les repreacutesentations des triangles de vitesses adimensionnellessont tregraves riches en information et contiennent plusieurs formules
Par exemple pour un compresseur on deacuteduit aiseacutement que
Avec 120595120595 = 1198821198821198901198901198801198802
Φ = 119888119888119909119909119880119880
et que pour 119888119888119901119901 = 119888119888119888119888119888119888119888119888119888119888 119882119882119890119890 = 119888119888119901119901(11987911987902minus11987911987901)on trouve
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
(11987911987902minus11987911987901) =119880119880119888119888119909119909119888119888119901119901
(tan1205731205731 minus tan1205731205732)
Alors si lrsquoon veut obtenir une plus grande variation detempeacuterature 11987911987902 minus 11987911987901 indirectement un plus grand rapport(1199011199010211990111990101) on peut augmenter
bull la vitesse peacuteripheacuterique 119880119880 (la vitesse de rotation)bull la vitesse axiale 119888119888119909119909 (le deacutebit massique)bull lrsquoangle de deacuteflection de la pale(tan1205731205731 minus tan1205731205732)
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
θ
x
Zone aubeacutee
surface S1
119932119932 = 119955119955120654120654
Rotor Stator
Rotor-Stator
RotorStator
Dans une turbomachinelrsquoensemble stator-rotor estappeleacute eacutetage
W2
C3
U
W3
U
C1W3
U
C2
1
2
3
W vitesse relative de lrsquoeacutecoulement
U vitesse peacuteripheacuterique du rotor
C vitesse absolue de lrsquoeacutecoulement
W2
U
C2
C3W3
U
C1W1
URotor
Stator
➀
➁
➂
C
W
U
Stator Rotor
U2
Stator
C2
U3
W3
Rotor
Stator
Rotor
U2
W2
Rotor
Stator
C2
U3
W3C3
Stator
Rotor
Effets 3DEffets visqueuxDistribution de pressionTraicircneacutee PortanceEacutecoulement instationnaireInteacuteraction fuide structureTurbulence
Extrados
Intrados
L
D
Profil isoleacuteGrille drsquoaubes
Pour analyser lrsquoeacutechange drsquoeacutenergie entre le fluide et le rotor drsquouneturbomachine on utilise la variation de la quantiteacute demouvement (changement de direction etou de vitesse)
Lrsquoeacutetude classique unidimensionnelle utilise le concept de trianglede vitesse Ce modegravele est appliqueacute agrave lrsquoentreacutee et agrave la sortie ducanal inter-aube (grille drsquoaubes)
Dans celui-ci on retrouve la vitesse absolue exprimeacutee dans unrepegravere lieacute aux parties fixes (le stator) accompagneacutee de lavitesse relative lieacutee aux partie tournantes ( le rotor)
Approche classique
Lrsquoapplication du triangle de vitesse permet drsquoestimer
bull lrsquoangle drsquoincidence agrave lrsquoentreacutee du rotor
bull lrsquoangle au bord de fuite du rotor
bull la deacuteflection de lrsquoeacutecoulement dans le stator
bull le travail produit ou consommeacute par la roue
Approche classique
C vitesse absolue de lrsquoeacutecoulement
W vitesse relative de lrsquoeacutecoulement
U vitesse peacuteripheacuterique du rotor
Cu Cm Cx composante tangentielle radiale et axiale de lavitesse absolue du fluide
Wu Wm Wx composante tangentielle radiale et axiale de lavitesse relative du fluide
α lrsquoangle des vitesses absolues mesureacutees par rapport agrave la directionaxiale
β lrsquoangle des vitesses relatives mesureacutees par rapport agrave la directionaxiale
La forme des pales du rotor deacutepend des angles 120631120631(des vitesses relatives dans le repegravere du rotor)
La forme des pales du stator deacutepend des angles 120630120630(des vitesses absolues dans le repegravere fixe)
Si lrsquoon veut garder la vitesse axiale 119940119940119961119961 = 119940119940119940119940119940119940119940119940119940119940 lrsquoeacutequation deconservation de la masse 119950 = 120646120646120488120488119940119940119961119961 = 119940119940119940119940119940119940119940119940119940119940 impose 120646120646120488120488 = 119940119940119940119940119940119940119940119940119940119940
Compresseur
Cu
Cm
Cu
Plan normal
Plan meacuteridional
BABF
W2
C2
U
U
W3C3
u
Les vecteurs vitesse (vitesse absolue 119940119940 pour le stator etrelative 119960119960 pour le rotor) sont consideacutereacutes tangents agrave laligne de squelette de lrsquoaube
β2
β3W3
W2
Pour les machines axiales (u= uent = usort ) le trianglede vitesse donne lieu agrave des eacutequations scalaires tellesque
cu= wu+u
cx= wxcx= = wx c
U
w
wu
cu
Pour les machines agrave eacutetages multiples on suppose que lavitesse axiale demeure constante Crsquoest agrave dire que119940119940120784120784119961119961= 119940119940120785120785119961119961= 119940119940119961119961= 119940119940119940119940119940119940119940119940119940119940
Cette condition jumeleacutee agrave 119932119932120784120784= 119932119932120785120785= 119932119932 megravene agrave lrsquohypothegravese drsquoune cineacutematique reacutepeacutetitive En pratique
Les triangles de vitesse agrave lrsquoentreacutee et agrave la sortie drsquouneacutetage peuvent ecirctre superposeacutes
Nous commenccedilons par la repreacutesentation de triangles de vitessespour le cas drsquoune turbineNous rappelons que par hypothegravese le rayon ne varie pas dans leplan inter-aube et que la vitesse axiale demeure constante dansle sens axialCes conditions permettent drsquoeffectuer la superposition destriangles de vitesse au bord drsquoattaque et bord de fuite du rotoravec la vitesse 119932119932 commune
c1w1
u
c2
u2
w2
➊
➋
➌ c3w3
u
Turbine
C2
W2 C3
W3
Stator Rotor
C2u C3u
W3u
α2β2 β3
α3
∆Cu= ∆Wu
U
W2u
119940119940119961119961119880119880 = 119888119888119888119888119888119888119888119888119888119888119888119888119909119909 = 119888119888119888119888119888119888119888119888119888119888
1198881198883 = 1198881198881
1205721205723 = 1205721205721
Cineacutematique reacutepeacutetitive
➁ ➂
➁U➀
Turbine
119914119914120784120784
119914119914120785120785
119934119934120785120785
119932119932119934119934120784120784
119932119932
119914119914120783120783120630120630120783120783
120630120630120784120784120631120631120784120784
120630120630120785120785
120631120631120785120785
Stator Rotor
➀ ➁ ➂
Le cas du compresseur est similaire agrave celui drsquoune turbineLe rayon ne varie pas (lrsquoeacutetude se deacuteroule dans le plan inter-aube) et la vitesse axiale demeure constante dans le sens axialComme pour la turbine nous pouvons superposer les triangles devitesse au bord drsquoattaque et bord de fuite du rotor
W2 C2
C3W3
U
C1W1
U
U
Compresseur
➊
➋
➌
α2 β1
C2 W1
C1W2
U
β2α1
W2U C1U
CX
( )2 1 = minuse u uW U C C
Lrsquoopeacuteration est vectorielle
C2U
119914119914120784120784119958119958 minus 119914119914120783120783119958119958 =- 119934119934120784120784119958119958 minus119934119934120783120783119958119958
➀
➁
➁
Compresseur
119914119914120784120784119914119914120785120785
119934119934120783120783
119932119932119914119914120783120783120630120630120783120783 120630120630120784120784
120631120631120784120784120630120630120785120785
Rotor Stator
119934119934120784120784120631120631120783120783
119932119932
➀ ➁ ➂
Trois quantiteacutes fondamentales sont utiliseacutees dans laconception des turbomachines
bull le coefficient de charge ψ
bull le coefficient de deacutebit Φ
bull le degreacute de reacuteaction R
Approche classique
Parfois ψ est consideacutereacute positif pour les turbines et neacutegatifpour les compresseurs 119882119882119890119890 = frasl119882 119888
Coefficient de charge
2Ψ = eWU
Coefficient de charge
2e
UW
=ψ 02 01 02 2
h h hU Uminus ∆
= = 02
s shU
η ∆=
=
eWWm
120578120578119904119904 =∆ℎ0∆ℎ0119904119904
Turbine
Coefficient de deacutebit
xcU
Φ =
xCU
Φ =ρRT
pρ
= xC RTRTU p
=RT m RT
U Ap
xCU
Φ =119888 = ρ119862119862119909119909119860119860
11988811987711987711987911987901199011199010119860119860
= 119872119872 120574120574 1 +120574120574 minus 1
21198721198722
minus 120574120574+1)2(120574120574minus1
Afin de geacuteneacuteraliser le triangle de vitesses drsquoune machineaxiale on effectue une mise agrave lrsquoeacutechelle en divisant chaquecocircteacute par la vitesse tangentielle U
Cette opeacuteration permettra de retrouver directement lescoefficients de charge ψ et de deacutebit Φ
Par la suite un nouveau paramegravetre R qursquoon appelle degreacutede reacuteaction sera introduit pour compleacuteter la caracteacuterisationde lrsquoaubage
Approche classique
α2
β2 α3
β3C2 W3
W2
c3u - c2u= w3u - w2u
W3uW2u
U
CX CX
Lrsquoopeacuteration est vectorielle
C3
120511120511 = frasl119934119934119940119940 119932119932120784120784 = frasl119914119914120785120785119958119958 minus 119914119914120784120784119958119958 119932119932
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
Turbine
3
2
3 3
2 2
12
12
12
12
u
u
u u
u u
c RUc RU
w c RU Uw c RU U
Ψ= minus +
Ψ= minus +
Ψ= minus = +
Ψ= minus = minus
2 22 23 2
2 22 23 2
1 12 2
2 2
c cR RU U
w wR RU U
Ψ Ψ = Φ + minus + = Φ + minus +
Ψ Ψ = Φ + + = Φ + minus
3 2
3 2
1 2 1 2
2 2
R Ratan atan
R Ratan atan
minus + Ψ minus + Ψ = = Φ Φ
Ψ + Ψ minus = = Φ Φ
α α
β β
α2β1
C2U
W1U
C1UW2U β2 α1
Compresseur
ΦC2U
C3UW3U
W2U
ψ
1
ψ
ΦC2UC3U
W3U
W2U
1
Il faut deacutefinir un troisiegraveme paramegravetre pour diffeacuterencier ces deux triangles de vitesses ayant le mecircme 120509120509 et le mecircme 120537120537
Les coefficients 120509120509 et 120537120537 ne suffisent pas pour deacutefinir de maniegravereunivoque le triangle de vitesses
Le degreacute de reacuteaction est un paramegravetre adimensionnel quiexprime la variation drsquoenthalpie subie par le fluide dans le rotorpar rapport agrave la chute totale drsquoenthalpie dans lrsquoeacutetage
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
Variation d enthalpie dans le rotorRVariation d enthalpie dans l eacutetage
=
❸❶
❷
Rotor
Stator
R=0 machine drsquoaction (Δhrotor=0 Δheacutetage=Δhstator)
R=1 machine agrave reacuteaction (Δhstator=0 Δheacutetage=Δhrotor)
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
R indique la reacutepartition du saut drsquoenthalpie (de pression) statique entre le stator et le rotor
Le degreacute de reacuteaction R deacutefini en fonction de variablesthermodynamiques (lrsquoenthalpie) peut aussi srsquoexprimer en fonctionde variables cineacutematiques preacutesentes dans le triangle de vitessesNotamment en fonction de la vitesse peacuteripheacuterique U et descomposantes des vitesses absolues et relatives
La vitesse absolue agrave lrsquoentreacutee du stator est eacutegale agrave celle agrave la sortie du rotor (c1= c3 eacutetage reacutepeacutetitif )
On neacuteglige les pertes et le transfert de chaleur dans le stator
1 2 3
ℎ3 minus ℎ1= ℎ3 + frasl11988811988832 2 minus ℎ1 minus frasl11988811988812 2 = ℎ03 minus ℎ01 = 120549120549ℎ0
ℎ03 minus ℎ01 = ℎ03 minus ℎ02 = 120549120549ℎ )0(3minus2
ℎ03 minus ℎ02 = ℎ3 minus ℎ1
119945119945120782120782120783120783 = 119945119945120782120782120784120784
frasl11988811988832 2 = frasl11988811988812 2
3 2
3 1
rotor
eacutetage
h h hRh h h
∆∆
minus= =
minus
2 20 3 2 2 3
0 3 2
2 2( )
( )
h c c h
∆∆
minus
minus
+ minus=
ℎ03 ℎ02
(ℎ01= ℎ02 1198881198881= 1198881198883)
120549120549ℎ )0(3minus2 = (ℎ3 + frasl11988811988832 2) minus (ℎ2 + frasl11988811988822 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
0(3 2) 3 2( )∆ minus = minusu uh U c c
= 1 minusfrasl11988811988832 2 minus frasl11988811988822 2120549120549ℎ )0(3minus2
119877119877 =120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
120549120549ℎ )0(3minus2
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
3 212+
= minus u uc cRU
119888119888119909119909 = 1198881198882119909119909 = 1198881198883119909119909 = 119888119888119888119888119888119888119888119888119888119888
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 11988811988821199061199061198881198882 = 1198881198881199091199092+1198881198881199061199062
119877119877 = 1 minus11988811988831199061199062 minus 11988811988821199061199062
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
Indirectement lrsquoeacutequation pour R deacutecrite en fonction de composantes cineacutematiques eacutetablit un lien entre les angles drsquoentreacutee et de sortie
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
= 1 minus1198881198883119906119906 + 1198881198882119906119906
2119880119880
2232
2 32 2wwh h+ = +
2 23 2
2 3 2 2w wh hminus = minus
2 2 2 23 3 2 2
2 3 2 2u x u xw w w wh h + +
minus = minus
Machine axiale U2 = U3
La composante axiale de la vitesse demeure constante w3x = w 2x
( eacutequation justifieacutee plus tard lors de la rothalpie)
2 23 2
2 3 2 2u uw wh hminus = minus
3 2 3 22 3
( )( )2
u u u uw w w wh h minus +minus =
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
Eacutetage reacutepeacutetitif c3 = c1
Si les pertes dans le stator sont neacutegligeacutees
c2u ndash c3u= w2u - w3u
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
3 2 3 23 2
( )( )2
u u u uw w w wh h minus +minus = minus
c2u ndash c3u= w2u - w3u
119877119877 = minus1199081199083119906119906 + 1199081199082119906119906
2119880119880
2 2 2 22 3 2 3 2 3( ) 2 ( ) 2h h u u w wminus = minus minus minus
2 2 2 2 2 23 2 3 2 3 2
2 2 2e
I II III
u u c c w wW minus minus minus= + minus
2 2 2 22 3 2 3
2 2 2 2 2 22 3 2 3 2 3
( ) ( )( ) ( ) ( )
minus minus minus=
minus + minus minus minusu u w wR
c c u u w w
3 2= =U u u
Expliqueacutee + tard
2 23 2
03 02 3 2 2ec cW h h h h minus
= minus = minus +
119877119877 =ℎ2 minus ℎ3ℎ1 minus ℎ3
=ℎ2 minus ℎ3ℎ02 minus ℎ03
Degreacute de reacuteaction
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
( )3u 2uw wR2U+
= minus
3 212
u uc cRU+
= minus
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
Puisque le degreacute de reacuteaction R peut aussi srsquoexprimer enfonction de variables cineacutematiques on peut chercher sarepreacutesentation dans le triangle adimensionnel de vitesses
Turbine
α2
β2 α3
β3
C2 W3
C3W2
C3uC2u
U
C2xC3x
W2u
c2u ndash c3u= w2u - w3u
2Ψ minusR
frasl119934119934120784120784119958119958 119932119932 = frasl1198621198622119906119906 119880119880 minus frasl119880119880 119880119880Ψ = frasl1198621198622119906119906 minus 1198621198623119906119906 119880119880
119877119877 = 1 minus (1198621198623119906119906 + 1198621198622119906119906 frasl) 2119880119880Φ = frasl1198621198622119909119909 119880119880 = frasl1198621198623119909119909 119880119880
= (119914119914120784120784119958119958 minus 119914119914120785120785119958119958 frasl) 2119880119880 minus 1 + (119914119914120784120784119958119958 + 119914119914120785120785119958119958 frasl) 2119880119880
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α2
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
Les repreacutesentations des triangles de vitesses adimensionnellessont tregraves riches en information et contiennent plusieurs formules
Par exemple pour un compresseur on deacuteduit aiseacutement que
Avec 120595120595 = 1198821198821198901198901198801198802
Φ = 119888119888119909119909119880119880
et que pour 119888119888119901119901 = 119888119888119888119888119888119888119888119888119888119888 119882119882119890119890 = 119888119888119901119901(11987911987902minus11987911987901)on trouve
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
(11987911987902minus11987911987901) =119880119880119888119888119909119909119888119888119901119901
(tan1205731205731 minus tan1205731205732)
Alors si lrsquoon veut obtenir une plus grande variation detempeacuterature 11987911987902 minus 11987911987901 indirectement un plus grand rapport(1199011199010211990111990101) on peut augmenter
bull la vitesse peacuteripheacuterique 119880119880 (la vitesse de rotation)bull la vitesse axiale 119888119888119909119909 (le deacutebit massique)bull lrsquoangle de deacuteflection de la pale(tan1205731205731 minus tan1205731205732)
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
Rotor Stator
Rotor-Stator
RotorStator
Dans une turbomachinelrsquoensemble stator-rotor estappeleacute eacutetage
W2
C3
U
W3
U
C1W3
U
C2
1
2
3
W vitesse relative de lrsquoeacutecoulement
U vitesse peacuteripheacuterique du rotor
C vitesse absolue de lrsquoeacutecoulement
W2
U
C2
C3W3
U
C1W1
URotor
Stator
➀
➁
➂
C
W
U
Stator Rotor
U2
Stator
C2
U3
W3
Rotor
Stator
Rotor
U2
W2
Rotor
Stator
C2
U3
W3C3
Stator
Rotor
Effets 3DEffets visqueuxDistribution de pressionTraicircneacutee PortanceEacutecoulement instationnaireInteacuteraction fuide structureTurbulence
Extrados
Intrados
L
D
Profil isoleacuteGrille drsquoaubes
Pour analyser lrsquoeacutechange drsquoeacutenergie entre le fluide et le rotor drsquouneturbomachine on utilise la variation de la quantiteacute demouvement (changement de direction etou de vitesse)
Lrsquoeacutetude classique unidimensionnelle utilise le concept de trianglede vitesse Ce modegravele est appliqueacute agrave lrsquoentreacutee et agrave la sortie ducanal inter-aube (grille drsquoaubes)
Dans celui-ci on retrouve la vitesse absolue exprimeacutee dans unrepegravere lieacute aux parties fixes (le stator) accompagneacutee de lavitesse relative lieacutee aux partie tournantes ( le rotor)
Approche classique
Lrsquoapplication du triangle de vitesse permet drsquoestimer
bull lrsquoangle drsquoincidence agrave lrsquoentreacutee du rotor
bull lrsquoangle au bord de fuite du rotor
bull la deacuteflection de lrsquoeacutecoulement dans le stator
bull le travail produit ou consommeacute par la roue
Approche classique
C vitesse absolue de lrsquoeacutecoulement
W vitesse relative de lrsquoeacutecoulement
U vitesse peacuteripheacuterique du rotor
Cu Cm Cx composante tangentielle radiale et axiale de lavitesse absolue du fluide
Wu Wm Wx composante tangentielle radiale et axiale de lavitesse relative du fluide
α lrsquoangle des vitesses absolues mesureacutees par rapport agrave la directionaxiale
β lrsquoangle des vitesses relatives mesureacutees par rapport agrave la directionaxiale
La forme des pales du rotor deacutepend des angles 120631120631(des vitesses relatives dans le repegravere du rotor)
La forme des pales du stator deacutepend des angles 120630120630(des vitesses absolues dans le repegravere fixe)
Si lrsquoon veut garder la vitesse axiale 119940119940119961119961 = 119940119940119940119940119940119940119940119940119940119940 lrsquoeacutequation deconservation de la masse 119950 = 120646120646120488120488119940119940119961119961 = 119940119940119940119940119940119940119940119940119940119940 impose 120646120646120488120488 = 119940119940119940119940119940119940119940119940119940119940
Compresseur
Cu
Cm
Cu
Plan normal
Plan meacuteridional
BABF
W2
C2
U
U
W3C3
u
Les vecteurs vitesse (vitesse absolue 119940119940 pour le stator etrelative 119960119960 pour le rotor) sont consideacutereacutes tangents agrave laligne de squelette de lrsquoaube
β2
β3W3
W2
Pour les machines axiales (u= uent = usort ) le trianglede vitesse donne lieu agrave des eacutequations scalaires tellesque
cu= wu+u
cx= wxcx= = wx c
U
w
wu
cu
Pour les machines agrave eacutetages multiples on suppose que lavitesse axiale demeure constante Crsquoest agrave dire que119940119940120784120784119961119961= 119940119940120785120785119961119961= 119940119940119961119961= 119940119940119940119940119940119940119940119940119940119940
Cette condition jumeleacutee agrave 119932119932120784120784= 119932119932120785120785= 119932119932 megravene agrave lrsquohypothegravese drsquoune cineacutematique reacutepeacutetitive En pratique
Les triangles de vitesse agrave lrsquoentreacutee et agrave la sortie drsquouneacutetage peuvent ecirctre superposeacutes
Nous commenccedilons par la repreacutesentation de triangles de vitessespour le cas drsquoune turbineNous rappelons que par hypothegravese le rayon ne varie pas dans leplan inter-aube et que la vitesse axiale demeure constante dansle sens axialCes conditions permettent drsquoeffectuer la superposition destriangles de vitesse au bord drsquoattaque et bord de fuite du rotoravec la vitesse 119932119932 commune
c1w1
u
c2
u2
w2
➊
➋
➌ c3w3
u
Turbine
C2
W2 C3
W3
Stator Rotor
C2u C3u
W3u
α2β2 β3
α3
∆Cu= ∆Wu
U
W2u
119940119940119961119961119880119880 = 119888119888119888119888119888119888119888119888119888119888119888119888119909119909 = 119888119888119888119888119888119888119888119888119888119888
1198881198883 = 1198881198881
1205721205723 = 1205721205721
Cineacutematique reacutepeacutetitive
➁ ➂
➁U➀
Turbine
119914119914120784120784
119914119914120785120785
119934119934120785120785
119932119932119934119934120784120784
119932119932
119914119914120783120783120630120630120783120783
120630120630120784120784120631120631120784120784
120630120630120785120785
120631120631120785120785
Stator Rotor
➀ ➁ ➂
Le cas du compresseur est similaire agrave celui drsquoune turbineLe rayon ne varie pas (lrsquoeacutetude se deacuteroule dans le plan inter-aube) et la vitesse axiale demeure constante dans le sens axialComme pour la turbine nous pouvons superposer les triangles devitesse au bord drsquoattaque et bord de fuite du rotor
W2 C2
C3W3
U
C1W1
U
U
Compresseur
➊
➋
➌
α2 β1
C2 W1
C1W2
U
β2α1
W2U C1U
CX
( )2 1 = minuse u uW U C C
Lrsquoopeacuteration est vectorielle
C2U
119914119914120784120784119958119958 minus 119914119914120783120783119958119958 =- 119934119934120784120784119958119958 minus119934119934120783120783119958119958
➀
➁
➁
Compresseur
119914119914120784120784119914119914120785120785
119934119934120783120783
119932119932119914119914120783120783120630120630120783120783 120630120630120784120784
120631120631120784120784120630120630120785120785
Rotor Stator
119934119934120784120784120631120631120783120783
119932119932
➀ ➁ ➂
Trois quantiteacutes fondamentales sont utiliseacutees dans laconception des turbomachines
bull le coefficient de charge ψ
bull le coefficient de deacutebit Φ
bull le degreacute de reacuteaction R
Approche classique
Parfois ψ est consideacutereacute positif pour les turbines et neacutegatifpour les compresseurs 119882119882119890119890 = frasl119882 119888
Coefficient de charge
2Ψ = eWU
Coefficient de charge
2e
UW
=ψ 02 01 02 2
h h hU Uminus ∆
= = 02
s shU
η ∆=
=
eWWm
120578120578119904119904 =∆ℎ0∆ℎ0119904119904
Turbine
Coefficient de deacutebit
xcU
Φ =
xCU
Φ =ρRT
pρ
= xC RTRTU p
=RT m RT
U Ap
xCU
Φ =119888 = ρ119862119862119909119909119860119860
11988811987711987711987911987901199011199010119860119860
= 119872119872 120574120574 1 +120574120574 minus 1
21198721198722
minus 120574120574+1)2(120574120574minus1
Afin de geacuteneacuteraliser le triangle de vitesses drsquoune machineaxiale on effectue une mise agrave lrsquoeacutechelle en divisant chaquecocircteacute par la vitesse tangentielle U
Cette opeacuteration permettra de retrouver directement lescoefficients de charge ψ et de deacutebit Φ
Par la suite un nouveau paramegravetre R qursquoon appelle degreacutede reacuteaction sera introduit pour compleacuteter la caracteacuterisationde lrsquoaubage
Approche classique
α2
β2 α3
β3C2 W3
W2
c3u - c2u= w3u - w2u
W3uW2u
U
CX CX
Lrsquoopeacuteration est vectorielle
C3
120511120511 = frasl119934119934119940119940 119932119932120784120784 = frasl119914119914120785120785119958119958 minus 119914119914120784120784119958119958 119932119932
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
Turbine
3
2
3 3
2 2
12
12
12
12
u
u
u u
u u
c RUc RU
w c RU Uw c RU U
Ψ= minus +
Ψ= minus +
Ψ= minus = +
Ψ= minus = minus
2 22 23 2
2 22 23 2
1 12 2
2 2
c cR RU U
w wR RU U
Ψ Ψ = Φ + minus + = Φ + minus +
Ψ Ψ = Φ + + = Φ + minus
3 2
3 2
1 2 1 2
2 2
R Ratan atan
R Ratan atan
minus + Ψ minus + Ψ = = Φ Φ
Ψ + Ψ minus = = Φ Φ
α α
β β
α2β1
C2U
W1U
C1UW2U β2 α1
Compresseur
ΦC2U
C3UW3U
W2U
ψ
1
ψ
ΦC2UC3U
W3U
W2U
1
Il faut deacutefinir un troisiegraveme paramegravetre pour diffeacuterencier ces deux triangles de vitesses ayant le mecircme 120509120509 et le mecircme 120537120537
Les coefficients 120509120509 et 120537120537 ne suffisent pas pour deacutefinir de maniegravereunivoque le triangle de vitesses
Le degreacute de reacuteaction est un paramegravetre adimensionnel quiexprime la variation drsquoenthalpie subie par le fluide dans le rotorpar rapport agrave la chute totale drsquoenthalpie dans lrsquoeacutetage
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
Variation d enthalpie dans le rotorRVariation d enthalpie dans l eacutetage
=
❸❶
❷
Rotor
Stator
R=0 machine drsquoaction (Δhrotor=0 Δheacutetage=Δhstator)
R=1 machine agrave reacuteaction (Δhstator=0 Δheacutetage=Δhrotor)
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
R indique la reacutepartition du saut drsquoenthalpie (de pression) statique entre le stator et le rotor
Le degreacute de reacuteaction R deacutefini en fonction de variablesthermodynamiques (lrsquoenthalpie) peut aussi srsquoexprimer en fonctionde variables cineacutematiques preacutesentes dans le triangle de vitessesNotamment en fonction de la vitesse peacuteripheacuterique U et descomposantes des vitesses absolues et relatives
La vitesse absolue agrave lrsquoentreacutee du stator est eacutegale agrave celle agrave la sortie du rotor (c1= c3 eacutetage reacutepeacutetitif )
On neacuteglige les pertes et le transfert de chaleur dans le stator
1 2 3
ℎ3 minus ℎ1= ℎ3 + frasl11988811988832 2 minus ℎ1 minus frasl11988811988812 2 = ℎ03 minus ℎ01 = 120549120549ℎ0
ℎ03 minus ℎ01 = ℎ03 minus ℎ02 = 120549120549ℎ )0(3minus2
ℎ03 minus ℎ02 = ℎ3 minus ℎ1
119945119945120782120782120783120783 = 119945119945120782120782120784120784
frasl11988811988832 2 = frasl11988811988812 2
3 2
3 1
rotor
eacutetage
h h hRh h h
∆∆
minus= =
minus
2 20 3 2 2 3
0 3 2
2 2( )
( )
h c c h
∆∆
minus
minus
+ minus=
ℎ03 ℎ02
(ℎ01= ℎ02 1198881198881= 1198881198883)
120549120549ℎ )0(3minus2 = (ℎ3 + frasl11988811988832 2) minus (ℎ2 + frasl11988811988822 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
0(3 2) 3 2( )∆ minus = minusu uh U c c
= 1 minusfrasl11988811988832 2 minus frasl11988811988822 2120549120549ℎ )0(3minus2
119877119877 =120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
120549120549ℎ )0(3minus2
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
3 212+
= minus u uc cRU
119888119888119909119909 = 1198881198882119909119909 = 1198881198883119909119909 = 119888119888119888119888119888119888119888119888119888119888
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 11988811988821199061199061198881198882 = 1198881198881199091199092+1198881198881199061199062
119877119877 = 1 minus11988811988831199061199062 minus 11988811988821199061199062
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
Indirectement lrsquoeacutequation pour R deacutecrite en fonction de composantes cineacutematiques eacutetablit un lien entre les angles drsquoentreacutee et de sortie
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
= 1 minus1198881198883119906119906 + 1198881198882119906119906
2119880119880
2232
2 32 2wwh h+ = +
2 23 2
2 3 2 2w wh hminus = minus
2 2 2 23 3 2 2
2 3 2 2u x u xw w w wh h + +
minus = minus
Machine axiale U2 = U3
La composante axiale de la vitesse demeure constante w3x = w 2x
( eacutequation justifieacutee plus tard lors de la rothalpie)
2 23 2
2 3 2 2u uw wh hminus = minus
3 2 3 22 3
( )( )2
u u u uw w w wh h minus +minus =
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
Eacutetage reacutepeacutetitif c3 = c1
Si les pertes dans le stator sont neacutegligeacutees
c2u ndash c3u= w2u - w3u
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
3 2 3 23 2
( )( )2
u u u uw w w wh h minus +minus = minus
c2u ndash c3u= w2u - w3u
119877119877 = minus1199081199083119906119906 + 1199081199082119906119906
2119880119880
2 2 2 22 3 2 3 2 3( ) 2 ( ) 2h h u u w wminus = minus minus minus
2 2 2 2 2 23 2 3 2 3 2
2 2 2e
I II III
u u c c w wW minus minus minus= + minus
2 2 2 22 3 2 3
2 2 2 2 2 22 3 2 3 2 3
( ) ( )( ) ( ) ( )
minus minus minus=
minus + minus minus minusu u w wR
c c u u w w
3 2= =U u u
Expliqueacutee + tard
2 23 2
03 02 3 2 2ec cW h h h h minus
= minus = minus +
119877119877 =ℎ2 minus ℎ3ℎ1 minus ℎ3
=ℎ2 minus ℎ3ℎ02 minus ℎ03
Degreacute de reacuteaction
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
( )3u 2uw wR2U+
= minus
3 212
u uc cRU+
= minus
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
Puisque le degreacute de reacuteaction R peut aussi srsquoexprimer enfonction de variables cineacutematiques on peut chercher sarepreacutesentation dans le triangle adimensionnel de vitesses
Turbine
α2
β2 α3
β3
C2 W3
C3W2
C3uC2u
U
C2xC3x
W2u
c2u ndash c3u= w2u - w3u
2Ψ minusR
frasl119934119934120784120784119958119958 119932119932 = frasl1198621198622119906119906 119880119880 minus frasl119880119880 119880119880Ψ = frasl1198621198622119906119906 minus 1198621198623119906119906 119880119880
119877119877 = 1 minus (1198621198623119906119906 + 1198621198622119906119906 frasl) 2119880119880Φ = frasl1198621198622119909119909 119880119880 = frasl1198621198623119909119909 119880119880
= (119914119914120784120784119958119958 minus 119914119914120785120785119958119958 frasl) 2119880119880 minus 1 + (119914119914120784120784119958119958 + 119914119914120785120785119958119958 frasl) 2119880119880
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α2
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
Les repreacutesentations des triangles de vitesses adimensionnellessont tregraves riches en information et contiennent plusieurs formules
Par exemple pour un compresseur on deacuteduit aiseacutement que
Avec 120595120595 = 1198821198821198901198901198801198802
Φ = 119888119888119909119909119880119880
et que pour 119888119888119901119901 = 119888119888119888119888119888119888119888119888119888119888 119882119882119890119890 = 119888119888119901119901(11987911987902minus11987911987901)on trouve
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
(11987911987902minus11987911987901) =119880119880119888119888119909119909119888119888119901119901
(tan1205731205731 minus tan1205731205732)
Alors si lrsquoon veut obtenir une plus grande variation detempeacuterature 11987911987902 minus 11987911987901 indirectement un plus grand rapport(1199011199010211990111990101) on peut augmenter
bull la vitesse peacuteripheacuterique 119880119880 (la vitesse de rotation)bull la vitesse axiale 119888119888119909119909 (le deacutebit massique)bull lrsquoangle de deacuteflection de la pale(tan1205731205731 minus tan1205731205732)
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
RotorStator
Dans une turbomachinelrsquoensemble stator-rotor estappeleacute eacutetage
W2
C3
U
W3
U
C1W3
U
C2
1
2
3
W vitesse relative de lrsquoeacutecoulement
U vitesse peacuteripheacuterique du rotor
C vitesse absolue de lrsquoeacutecoulement
W2
U
C2
C3W3
U
C1W1
URotor
Stator
➀
➁
➂
C
W
U
Stator Rotor
U2
Stator
C2
U3
W3
Rotor
Stator
Rotor
U2
W2
Rotor
Stator
C2
U3
W3C3
Stator
Rotor
Effets 3DEffets visqueuxDistribution de pressionTraicircneacutee PortanceEacutecoulement instationnaireInteacuteraction fuide structureTurbulence
Extrados
Intrados
L
D
Profil isoleacuteGrille drsquoaubes
Pour analyser lrsquoeacutechange drsquoeacutenergie entre le fluide et le rotor drsquouneturbomachine on utilise la variation de la quantiteacute demouvement (changement de direction etou de vitesse)
Lrsquoeacutetude classique unidimensionnelle utilise le concept de trianglede vitesse Ce modegravele est appliqueacute agrave lrsquoentreacutee et agrave la sortie ducanal inter-aube (grille drsquoaubes)
Dans celui-ci on retrouve la vitesse absolue exprimeacutee dans unrepegravere lieacute aux parties fixes (le stator) accompagneacutee de lavitesse relative lieacutee aux partie tournantes ( le rotor)
Approche classique
Lrsquoapplication du triangle de vitesse permet drsquoestimer
bull lrsquoangle drsquoincidence agrave lrsquoentreacutee du rotor
bull lrsquoangle au bord de fuite du rotor
bull la deacuteflection de lrsquoeacutecoulement dans le stator
bull le travail produit ou consommeacute par la roue
Approche classique
C vitesse absolue de lrsquoeacutecoulement
W vitesse relative de lrsquoeacutecoulement
U vitesse peacuteripheacuterique du rotor
Cu Cm Cx composante tangentielle radiale et axiale de lavitesse absolue du fluide
Wu Wm Wx composante tangentielle radiale et axiale de lavitesse relative du fluide
α lrsquoangle des vitesses absolues mesureacutees par rapport agrave la directionaxiale
β lrsquoangle des vitesses relatives mesureacutees par rapport agrave la directionaxiale
La forme des pales du rotor deacutepend des angles 120631120631(des vitesses relatives dans le repegravere du rotor)
La forme des pales du stator deacutepend des angles 120630120630(des vitesses absolues dans le repegravere fixe)
Si lrsquoon veut garder la vitesse axiale 119940119940119961119961 = 119940119940119940119940119940119940119940119940119940119940 lrsquoeacutequation deconservation de la masse 119950 = 120646120646120488120488119940119940119961119961 = 119940119940119940119940119940119940119940119940119940119940 impose 120646120646120488120488 = 119940119940119940119940119940119940119940119940119940119940
Compresseur
Cu
Cm
Cu
Plan normal
Plan meacuteridional
BABF
W2
C2
U
U
W3C3
u
Les vecteurs vitesse (vitesse absolue 119940119940 pour le stator etrelative 119960119960 pour le rotor) sont consideacutereacutes tangents agrave laligne de squelette de lrsquoaube
β2
β3W3
W2
Pour les machines axiales (u= uent = usort ) le trianglede vitesse donne lieu agrave des eacutequations scalaires tellesque
cu= wu+u
cx= wxcx= = wx c
U
w
wu
cu
Pour les machines agrave eacutetages multiples on suppose que lavitesse axiale demeure constante Crsquoest agrave dire que119940119940120784120784119961119961= 119940119940120785120785119961119961= 119940119940119961119961= 119940119940119940119940119940119940119940119940119940119940
Cette condition jumeleacutee agrave 119932119932120784120784= 119932119932120785120785= 119932119932 megravene agrave lrsquohypothegravese drsquoune cineacutematique reacutepeacutetitive En pratique
Les triangles de vitesse agrave lrsquoentreacutee et agrave la sortie drsquouneacutetage peuvent ecirctre superposeacutes
Nous commenccedilons par la repreacutesentation de triangles de vitessespour le cas drsquoune turbineNous rappelons que par hypothegravese le rayon ne varie pas dans leplan inter-aube et que la vitesse axiale demeure constante dansle sens axialCes conditions permettent drsquoeffectuer la superposition destriangles de vitesse au bord drsquoattaque et bord de fuite du rotoravec la vitesse 119932119932 commune
c1w1
u
c2
u2
w2
➊
➋
➌ c3w3
u
Turbine
C2
W2 C3
W3
Stator Rotor
C2u C3u
W3u
α2β2 β3
α3
∆Cu= ∆Wu
U
W2u
119940119940119961119961119880119880 = 119888119888119888119888119888119888119888119888119888119888119888119888119909119909 = 119888119888119888119888119888119888119888119888119888119888
1198881198883 = 1198881198881
1205721205723 = 1205721205721
Cineacutematique reacutepeacutetitive
➁ ➂
➁U➀
Turbine
119914119914120784120784
119914119914120785120785
119934119934120785120785
119932119932119934119934120784120784
119932119932
119914119914120783120783120630120630120783120783
120630120630120784120784120631120631120784120784
120630120630120785120785
120631120631120785120785
Stator Rotor
➀ ➁ ➂
Le cas du compresseur est similaire agrave celui drsquoune turbineLe rayon ne varie pas (lrsquoeacutetude se deacuteroule dans le plan inter-aube) et la vitesse axiale demeure constante dans le sens axialComme pour la turbine nous pouvons superposer les triangles devitesse au bord drsquoattaque et bord de fuite du rotor
W2 C2
C3W3
U
C1W1
U
U
Compresseur
➊
➋
➌
α2 β1
C2 W1
C1W2
U
β2α1
W2U C1U
CX
( )2 1 = minuse u uW U C C
Lrsquoopeacuteration est vectorielle
C2U
119914119914120784120784119958119958 minus 119914119914120783120783119958119958 =- 119934119934120784120784119958119958 minus119934119934120783120783119958119958
➀
➁
➁
Compresseur
119914119914120784120784119914119914120785120785
119934119934120783120783
119932119932119914119914120783120783120630120630120783120783 120630120630120784120784
120631120631120784120784120630120630120785120785
Rotor Stator
119934119934120784120784120631120631120783120783
119932119932
➀ ➁ ➂
Trois quantiteacutes fondamentales sont utiliseacutees dans laconception des turbomachines
bull le coefficient de charge ψ
bull le coefficient de deacutebit Φ
bull le degreacute de reacuteaction R
Approche classique
Parfois ψ est consideacutereacute positif pour les turbines et neacutegatifpour les compresseurs 119882119882119890119890 = frasl119882 119888
Coefficient de charge
2Ψ = eWU
Coefficient de charge
2e
UW
=ψ 02 01 02 2
h h hU Uminus ∆
= = 02
s shU
η ∆=
=
eWWm
120578120578119904119904 =∆ℎ0∆ℎ0119904119904
Turbine
Coefficient de deacutebit
xcU
Φ =
xCU
Φ =ρRT
pρ
= xC RTRTU p
=RT m RT
U Ap
xCU
Φ =119888 = ρ119862119862119909119909119860119860
11988811987711987711987911987901199011199010119860119860
= 119872119872 120574120574 1 +120574120574 minus 1
21198721198722
minus 120574120574+1)2(120574120574minus1
Afin de geacuteneacuteraliser le triangle de vitesses drsquoune machineaxiale on effectue une mise agrave lrsquoeacutechelle en divisant chaquecocircteacute par la vitesse tangentielle U
Cette opeacuteration permettra de retrouver directement lescoefficients de charge ψ et de deacutebit Φ
Par la suite un nouveau paramegravetre R qursquoon appelle degreacutede reacuteaction sera introduit pour compleacuteter la caracteacuterisationde lrsquoaubage
Approche classique
α2
β2 α3
β3C2 W3
W2
c3u - c2u= w3u - w2u
W3uW2u
U
CX CX
Lrsquoopeacuteration est vectorielle
C3
120511120511 = frasl119934119934119940119940 119932119932120784120784 = frasl119914119914120785120785119958119958 minus 119914119914120784120784119958119958 119932119932
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
Turbine
3
2
3 3
2 2
12
12
12
12
u
u
u u
u u
c RUc RU
w c RU Uw c RU U
Ψ= minus +
Ψ= minus +
Ψ= minus = +
Ψ= minus = minus
2 22 23 2
2 22 23 2
1 12 2
2 2
c cR RU U
w wR RU U
Ψ Ψ = Φ + minus + = Φ + minus +
Ψ Ψ = Φ + + = Φ + minus
3 2
3 2
1 2 1 2
2 2
R Ratan atan
R Ratan atan
minus + Ψ minus + Ψ = = Φ Φ
Ψ + Ψ minus = = Φ Φ
α α
β β
α2β1
C2U
W1U
C1UW2U β2 α1
Compresseur
ΦC2U
C3UW3U
W2U
ψ
1
ψ
ΦC2UC3U
W3U
W2U
1
Il faut deacutefinir un troisiegraveme paramegravetre pour diffeacuterencier ces deux triangles de vitesses ayant le mecircme 120509120509 et le mecircme 120537120537
Les coefficients 120509120509 et 120537120537 ne suffisent pas pour deacutefinir de maniegravereunivoque le triangle de vitesses
Le degreacute de reacuteaction est un paramegravetre adimensionnel quiexprime la variation drsquoenthalpie subie par le fluide dans le rotorpar rapport agrave la chute totale drsquoenthalpie dans lrsquoeacutetage
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
Variation d enthalpie dans le rotorRVariation d enthalpie dans l eacutetage
=
❸❶
❷
Rotor
Stator
R=0 machine drsquoaction (Δhrotor=0 Δheacutetage=Δhstator)
R=1 machine agrave reacuteaction (Δhstator=0 Δheacutetage=Δhrotor)
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
R indique la reacutepartition du saut drsquoenthalpie (de pression) statique entre le stator et le rotor
Le degreacute de reacuteaction R deacutefini en fonction de variablesthermodynamiques (lrsquoenthalpie) peut aussi srsquoexprimer en fonctionde variables cineacutematiques preacutesentes dans le triangle de vitessesNotamment en fonction de la vitesse peacuteripheacuterique U et descomposantes des vitesses absolues et relatives
La vitesse absolue agrave lrsquoentreacutee du stator est eacutegale agrave celle agrave la sortie du rotor (c1= c3 eacutetage reacutepeacutetitif )
On neacuteglige les pertes et le transfert de chaleur dans le stator
1 2 3
ℎ3 minus ℎ1= ℎ3 + frasl11988811988832 2 minus ℎ1 minus frasl11988811988812 2 = ℎ03 minus ℎ01 = 120549120549ℎ0
ℎ03 minus ℎ01 = ℎ03 minus ℎ02 = 120549120549ℎ )0(3minus2
ℎ03 minus ℎ02 = ℎ3 minus ℎ1
119945119945120782120782120783120783 = 119945119945120782120782120784120784
frasl11988811988832 2 = frasl11988811988812 2
3 2
3 1
rotor
eacutetage
h h hRh h h
∆∆
minus= =
minus
2 20 3 2 2 3
0 3 2
2 2( )
( )
h c c h
∆∆
minus
minus
+ minus=
ℎ03 ℎ02
(ℎ01= ℎ02 1198881198881= 1198881198883)
120549120549ℎ )0(3minus2 = (ℎ3 + frasl11988811988832 2) minus (ℎ2 + frasl11988811988822 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
0(3 2) 3 2( )∆ minus = minusu uh U c c
= 1 minusfrasl11988811988832 2 minus frasl11988811988822 2120549120549ℎ )0(3minus2
119877119877 =120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
120549120549ℎ )0(3minus2
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
3 212+
= minus u uc cRU
119888119888119909119909 = 1198881198882119909119909 = 1198881198883119909119909 = 119888119888119888119888119888119888119888119888119888119888
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 11988811988821199061199061198881198882 = 1198881198881199091199092+1198881198881199061199062
119877119877 = 1 minus11988811988831199061199062 minus 11988811988821199061199062
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
Indirectement lrsquoeacutequation pour R deacutecrite en fonction de composantes cineacutematiques eacutetablit un lien entre les angles drsquoentreacutee et de sortie
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
= 1 minus1198881198883119906119906 + 1198881198882119906119906
2119880119880
2232
2 32 2wwh h+ = +
2 23 2
2 3 2 2w wh hminus = minus
2 2 2 23 3 2 2
2 3 2 2u x u xw w w wh h + +
minus = minus
Machine axiale U2 = U3
La composante axiale de la vitesse demeure constante w3x = w 2x
( eacutequation justifieacutee plus tard lors de la rothalpie)
2 23 2
2 3 2 2u uw wh hminus = minus
3 2 3 22 3
( )( )2
u u u uw w w wh h minus +minus =
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
Eacutetage reacutepeacutetitif c3 = c1
Si les pertes dans le stator sont neacutegligeacutees
c2u ndash c3u= w2u - w3u
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
3 2 3 23 2
( )( )2
u u u uw w w wh h minus +minus = minus
c2u ndash c3u= w2u - w3u
119877119877 = minus1199081199083119906119906 + 1199081199082119906119906
2119880119880
2 2 2 22 3 2 3 2 3( ) 2 ( ) 2h h u u w wminus = minus minus minus
2 2 2 2 2 23 2 3 2 3 2
2 2 2e
I II III
u u c c w wW minus minus minus= + minus
2 2 2 22 3 2 3
2 2 2 2 2 22 3 2 3 2 3
( ) ( )( ) ( ) ( )
minus minus minus=
minus + minus minus minusu u w wR
c c u u w w
3 2= =U u u
Expliqueacutee + tard
2 23 2
03 02 3 2 2ec cW h h h h minus
= minus = minus +
119877119877 =ℎ2 minus ℎ3ℎ1 minus ℎ3
=ℎ2 minus ℎ3ℎ02 minus ℎ03
Degreacute de reacuteaction
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
( )3u 2uw wR2U+
= minus
3 212
u uc cRU+
= minus
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
Puisque le degreacute de reacuteaction R peut aussi srsquoexprimer enfonction de variables cineacutematiques on peut chercher sarepreacutesentation dans le triangle adimensionnel de vitesses
Turbine
α2
β2 α3
β3
C2 W3
C3W2
C3uC2u
U
C2xC3x
W2u
c2u ndash c3u= w2u - w3u
2Ψ minusR
frasl119934119934120784120784119958119958 119932119932 = frasl1198621198622119906119906 119880119880 minus frasl119880119880 119880119880Ψ = frasl1198621198622119906119906 minus 1198621198623119906119906 119880119880
119877119877 = 1 minus (1198621198623119906119906 + 1198621198622119906119906 frasl) 2119880119880Φ = frasl1198621198622119909119909 119880119880 = frasl1198621198623119909119909 119880119880
= (119914119914120784120784119958119958 minus 119914119914120785120785119958119958 frasl) 2119880119880 minus 1 + (119914119914120784120784119958119958 + 119914119914120785120785119958119958 frasl) 2119880119880
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α2
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
Les repreacutesentations des triangles de vitesses adimensionnellessont tregraves riches en information et contiennent plusieurs formules
Par exemple pour un compresseur on deacuteduit aiseacutement que
Avec 120595120595 = 1198821198821198901198901198801198802
Φ = 119888119888119909119909119880119880
et que pour 119888119888119901119901 = 119888119888119888119888119888119888119888119888119888119888 119882119882119890119890 = 119888119888119901119901(11987911987902minus11987911987901)on trouve
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
(11987911987902minus11987911987901) =119880119880119888119888119909119909119888119888119901119901
(tan1205731205731 minus tan1205731205732)
Alors si lrsquoon veut obtenir une plus grande variation detempeacuterature 11987911987902 minus 11987911987901 indirectement un plus grand rapport(1199011199010211990111990101) on peut augmenter
bull la vitesse peacuteripheacuterique 119880119880 (la vitesse de rotation)bull la vitesse axiale 119888119888119909119909 (le deacutebit massique)bull lrsquoangle de deacuteflection de la pale(tan1205731205731 minus tan1205731205732)
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
W2
C3
U
W3
U
C1W3
U
C2
1
2
3
W vitesse relative de lrsquoeacutecoulement
U vitesse peacuteripheacuterique du rotor
C vitesse absolue de lrsquoeacutecoulement
W2
U
C2
C3W3
U
C1W1
URotor
Stator
➀
➁
➂
C
W
U
Stator Rotor
U2
Stator
C2
U3
W3
Rotor
Stator
Rotor
U2
W2
Rotor
Stator
C2
U3
W3C3
Stator
Rotor
Effets 3DEffets visqueuxDistribution de pressionTraicircneacutee PortanceEacutecoulement instationnaireInteacuteraction fuide structureTurbulence
Extrados
Intrados
L
D
Profil isoleacuteGrille drsquoaubes
Pour analyser lrsquoeacutechange drsquoeacutenergie entre le fluide et le rotor drsquouneturbomachine on utilise la variation de la quantiteacute demouvement (changement de direction etou de vitesse)
Lrsquoeacutetude classique unidimensionnelle utilise le concept de trianglede vitesse Ce modegravele est appliqueacute agrave lrsquoentreacutee et agrave la sortie ducanal inter-aube (grille drsquoaubes)
Dans celui-ci on retrouve la vitesse absolue exprimeacutee dans unrepegravere lieacute aux parties fixes (le stator) accompagneacutee de lavitesse relative lieacutee aux partie tournantes ( le rotor)
Approche classique
Lrsquoapplication du triangle de vitesse permet drsquoestimer
bull lrsquoangle drsquoincidence agrave lrsquoentreacutee du rotor
bull lrsquoangle au bord de fuite du rotor
bull la deacuteflection de lrsquoeacutecoulement dans le stator
bull le travail produit ou consommeacute par la roue
Approche classique
C vitesse absolue de lrsquoeacutecoulement
W vitesse relative de lrsquoeacutecoulement
U vitesse peacuteripheacuterique du rotor
Cu Cm Cx composante tangentielle radiale et axiale de lavitesse absolue du fluide
Wu Wm Wx composante tangentielle radiale et axiale de lavitesse relative du fluide
α lrsquoangle des vitesses absolues mesureacutees par rapport agrave la directionaxiale
β lrsquoangle des vitesses relatives mesureacutees par rapport agrave la directionaxiale
La forme des pales du rotor deacutepend des angles 120631120631(des vitesses relatives dans le repegravere du rotor)
La forme des pales du stator deacutepend des angles 120630120630(des vitesses absolues dans le repegravere fixe)
Si lrsquoon veut garder la vitesse axiale 119940119940119961119961 = 119940119940119940119940119940119940119940119940119940119940 lrsquoeacutequation deconservation de la masse 119950 = 120646120646120488120488119940119940119961119961 = 119940119940119940119940119940119940119940119940119940119940 impose 120646120646120488120488 = 119940119940119940119940119940119940119940119940119940119940
Compresseur
Cu
Cm
Cu
Plan normal
Plan meacuteridional
BABF
W2
C2
U
U
W3C3
u
Les vecteurs vitesse (vitesse absolue 119940119940 pour le stator etrelative 119960119960 pour le rotor) sont consideacutereacutes tangents agrave laligne de squelette de lrsquoaube
β2
β3W3
W2
Pour les machines axiales (u= uent = usort ) le trianglede vitesse donne lieu agrave des eacutequations scalaires tellesque
cu= wu+u
cx= wxcx= = wx c
U
w
wu
cu
Pour les machines agrave eacutetages multiples on suppose que lavitesse axiale demeure constante Crsquoest agrave dire que119940119940120784120784119961119961= 119940119940120785120785119961119961= 119940119940119961119961= 119940119940119940119940119940119940119940119940119940119940
Cette condition jumeleacutee agrave 119932119932120784120784= 119932119932120785120785= 119932119932 megravene agrave lrsquohypothegravese drsquoune cineacutematique reacutepeacutetitive En pratique
Les triangles de vitesse agrave lrsquoentreacutee et agrave la sortie drsquouneacutetage peuvent ecirctre superposeacutes
Nous commenccedilons par la repreacutesentation de triangles de vitessespour le cas drsquoune turbineNous rappelons que par hypothegravese le rayon ne varie pas dans leplan inter-aube et que la vitesse axiale demeure constante dansle sens axialCes conditions permettent drsquoeffectuer la superposition destriangles de vitesse au bord drsquoattaque et bord de fuite du rotoravec la vitesse 119932119932 commune
c1w1
u
c2
u2
w2
➊
➋
➌ c3w3
u
Turbine
C2
W2 C3
W3
Stator Rotor
C2u C3u
W3u
α2β2 β3
α3
∆Cu= ∆Wu
U
W2u
119940119940119961119961119880119880 = 119888119888119888119888119888119888119888119888119888119888119888119888119909119909 = 119888119888119888119888119888119888119888119888119888119888
1198881198883 = 1198881198881
1205721205723 = 1205721205721
Cineacutematique reacutepeacutetitive
➁ ➂
➁U➀
Turbine
119914119914120784120784
119914119914120785120785
119934119934120785120785
119932119932119934119934120784120784
119932119932
119914119914120783120783120630120630120783120783
120630120630120784120784120631120631120784120784
120630120630120785120785
120631120631120785120785
Stator Rotor
➀ ➁ ➂
Le cas du compresseur est similaire agrave celui drsquoune turbineLe rayon ne varie pas (lrsquoeacutetude se deacuteroule dans le plan inter-aube) et la vitesse axiale demeure constante dans le sens axialComme pour la turbine nous pouvons superposer les triangles devitesse au bord drsquoattaque et bord de fuite du rotor
W2 C2
C3W3
U
C1W1
U
U
Compresseur
➊
➋
➌
α2 β1
C2 W1
C1W2
U
β2α1
W2U C1U
CX
( )2 1 = minuse u uW U C C
Lrsquoopeacuteration est vectorielle
C2U
119914119914120784120784119958119958 minus 119914119914120783120783119958119958 =- 119934119934120784120784119958119958 minus119934119934120783120783119958119958
➀
➁
➁
Compresseur
119914119914120784120784119914119914120785120785
119934119934120783120783
119932119932119914119914120783120783120630120630120783120783 120630120630120784120784
120631120631120784120784120630120630120785120785
Rotor Stator
119934119934120784120784120631120631120783120783
119932119932
➀ ➁ ➂
Trois quantiteacutes fondamentales sont utiliseacutees dans laconception des turbomachines
bull le coefficient de charge ψ
bull le coefficient de deacutebit Φ
bull le degreacute de reacuteaction R
Approche classique
Parfois ψ est consideacutereacute positif pour les turbines et neacutegatifpour les compresseurs 119882119882119890119890 = frasl119882 119888
Coefficient de charge
2Ψ = eWU
Coefficient de charge
2e
UW
=ψ 02 01 02 2
h h hU Uminus ∆
= = 02
s shU
η ∆=
=
eWWm
120578120578119904119904 =∆ℎ0∆ℎ0119904119904
Turbine
Coefficient de deacutebit
xcU
Φ =
xCU
Φ =ρRT
pρ
= xC RTRTU p
=RT m RT
U Ap
xCU
Φ =119888 = ρ119862119862119909119909119860119860
11988811987711987711987911987901199011199010119860119860
= 119872119872 120574120574 1 +120574120574 minus 1
21198721198722
minus 120574120574+1)2(120574120574minus1
Afin de geacuteneacuteraliser le triangle de vitesses drsquoune machineaxiale on effectue une mise agrave lrsquoeacutechelle en divisant chaquecocircteacute par la vitesse tangentielle U
Cette opeacuteration permettra de retrouver directement lescoefficients de charge ψ et de deacutebit Φ
Par la suite un nouveau paramegravetre R qursquoon appelle degreacutede reacuteaction sera introduit pour compleacuteter la caracteacuterisationde lrsquoaubage
Approche classique
α2
β2 α3
β3C2 W3
W2
c3u - c2u= w3u - w2u
W3uW2u
U
CX CX
Lrsquoopeacuteration est vectorielle
C3
120511120511 = frasl119934119934119940119940 119932119932120784120784 = frasl119914119914120785120785119958119958 minus 119914119914120784120784119958119958 119932119932
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
Turbine
3
2
3 3
2 2
12
12
12
12
u
u
u u
u u
c RUc RU
w c RU Uw c RU U
Ψ= minus +
Ψ= minus +
Ψ= minus = +
Ψ= minus = minus
2 22 23 2
2 22 23 2
1 12 2
2 2
c cR RU U
w wR RU U
Ψ Ψ = Φ + minus + = Φ + minus +
Ψ Ψ = Φ + + = Φ + minus
3 2
3 2
1 2 1 2
2 2
R Ratan atan
R Ratan atan
minus + Ψ minus + Ψ = = Φ Φ
Ψ + Ψ minus = = Φ Φ
α α
β β
α2β1
C2U
W1U
C1UW2U β2 α1
Compresseur
ΦC2U
C3UW3U
W2U
ψ
1
ψ
ΦC2UC3U
W3U
W2U
1
Il faut deacutefinir un troisiegraveme paramegravetre pour diffeacuterencier ces deux triangles de vitesses ayant le mecircme 120509120509 et le mecircme 120537120537
Les coefficients 120509120509 et 120537120537 ne suffisent pas pour deacutefinir de maniegravereunivoque le triangle de vitesses
Le degreacute de reacuteaction est un paramegravetre adimensionnel quiexprime la variation drsquoenthalpie subie par le fluide dans le rotorpar rapport agrave la chute totale drsquoenthalpie dans lrsquoeacutetage
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
Variation d enthalpie dans le rotorRVariation d enthalpie dans l eacutetage
=
❸❶
❷
Rotor
Stator
R=0 machine drsquoaction (Δhrotor=0 Δheacutetage=Δhstator)
R=1 machine agrave reacuteaction (Δhstator=0 Δheacutetage=Δhrotor)
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
R indique la reacutepartition du saut drsquoenthalpie (de pression) statique entre le stator et le rotor
Le degreacute de reacuteaction R deacutefini en fonction de variablesthermodynamiques (lrsquoenthalpie) peut aussi srsquoexprimer en fonctionde variables cineacutematiques preacutesentes dans le triangle de vitessesNotamment en fonction de la vitesse peacuteripheacuterique U et descomposantes des vitesses absolues et relatives
La vitesse absolue agrave lrsquoentreacutee du stator est eacutegale agrave celle agrave la sortie du rotor (c1= c3 eacutetage reacutepeacutetitif )
On neacuteglige les pertes et le transfert de chaleur dans le stator
1 2 3
ℎ3 minus ℎ1= ℎ3 + frasl11988811988832 2 minus ℎ1 minus frasl11988811988812 2 = ℎ03 minus ℎ01 = 120549120549ℎ0
ℎ03 minus ℎ01 = ℎ03 minus ℎ02 = 120549120549ℎ )0(3minus2
ℎ03 minus ℎ02 = ℎ3 minus ℎ1
119945119945120782120782120783120783 = 119945119945120782120782120784120784
frasl11988811988832 2 = frasl11988811988812 2
3 2
3 1
rotor
eacutetage
h h hRh h h
∆∆
minus= =
minus
2 20 3 2 2 3
0 3 2
2 2( )
( )
h c c h
∆∆
minus
minus
+ minus=
ℎ03 ℎ02
(ℎ01= ℎ02 1198881198881= 1198881198883)
120549120549ℎ )0(3minus2 = (ℎ3 + frasl11988811988832 2) minus (ℎ2 + frasl11988811988822 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
0(3 2) 3 2( )∆ minus = minusu uh U c c
= 1 minusfrasl11988811988832 2 minus frasl11988811988822 2120549120549ℎ )0(3minus2
119877119877 =120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
120549120549ℎ )0(3minus2
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
3 212+
= minus u uc cRU
119888119888119909119909 = 1198881198882119909119909 = 1198881198883119909119909 = 119888119888119888119888119888119888119888119888119888119888
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 11988811988821199061199061198881198882 = 1198881198881199091199092+1198881198881199061199062
119877119877 = 1 minus11988811988831199061199062 minus 11988811988821199061199062
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
Indirectement lrsquoeacutequation pour R deacutecrite en fonction de composantes cineacutematiques eacutetablit un lien entre les angles drsquoentreacutee et de sortie
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
= 1 minus1198881198883119906119906 + 1198881198882119906119906
2119880119880
2232
2 32 2wwh h+ = +
2 23 2
2 3 2 2w wh hminus = minus
2 2 2 23 3 2 2
2 3 2 2u x u xw w w wh h + +
minus = minus
Machine axiale U2 = U3
La composante axiale de la vitesse demeure constante w3x = w 2x
( eacutequation justifieacutee plus tard lors de la rothalpie)
2 23 2
2 3 2 2u uw wh hminus = minus
3 2 3 22 3
( )( )2
u u u uw w w wh h minus +minus =
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
Eacutetage reacutepeacutetitif c3 = c1
Si les pertes dans le stator sont neacutegligeacutees
c2u ndash c3u= w2u - w3u
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
3 2 3 23 2
( )( )2
u u u uw w w wh h minus +minus = minus
c2u ndash c3u= w2u - w3u
119877119877 = minus1199081199083119906119906 + 1199081199082119906119906
2119880119880
2 2 2 22 3 2 3 2 3( ) 2 ( ) 2h h u u w wminus = minus minus minus
2 2 2 2 2 23 2 3 2 3 2
2 2 2e
I II III
u u c c w wW minus minus minus= + minus
2 2 2 22 3 2 3
2 2 2 2 2 22 3 2 3 2 3
( ) ( )( ) ( ) ( )
minus minus minus=
minus + minus minus minusu u w wR
c c u u w w
3 2= =U u u
Expliqueacutee + tard
2 23 2
03 02 3 2 2ec cW h h h h minus
= minus = minus +
119877119877 =ℎ2 minus ℎ3ℎ1 minus ℎ3
=ℎ2 minus ℎ3ℎ02 minus ℎ03
Degreacute de reacuteaction
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
( )3u 2uw wR2U+
= minus
3 212
u uc cRU+
= minus
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
Puisque le degreacute de reacuteaction R peut aussi srsquoexprimer enfonction de variables cineacutematiques on peut chercher sarepreacutesentation dans le triangle adimensionnel de vitesses
Turbine
α2
β2 α3
β3
C2 W3
C3W2
C3uC2u
U
C2xC3x
W2u
c2u ndash c3u= w2u - w3u
2Ψ minusR
frasl119934119934120784120784119958119958 119932119932 = frasl1198621198622119906119906 119880119880 minus frasl119880119880 119880119880Ψ = frasl1198621198622119906119906 minus 1198621198623119906119906 119880119880
119877119877 = 1 minus (1198621198623119906119906 + 1198621198622119906119906 frasl) 2119880119880Φ = frasl1198621198622119909119909 119880119880 = frasl1198621198623119909119909 119880119880
= (119914119914120784120784119958119958 minus 119914119914120785120785119958119958 frasl) 2119880119880 minus 1 + (119914119914120784120784119958119958 + 119914119914120785120785119958119958 frasl) 2119880119880
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α2
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
Les repreacutesentations des triangles de vitesses adimensionnellessont tregraves riches en information et contiennent plusieurs formules
Par exemple pour un compresseur on deacuteduit aiseacutement que
Avec 120595120595 = 1198821198821198901198901198801198802
Φ = 119888119888119909119909119880119880
et que pour 119888119888119901119901 = 119888119888119888119888119888119888119888119888119888119888 119882119882119890119890 = 119888119888119901119901(11987911987902minus11987911987901)on trouve
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
(11987911987902minus11987911987901) =119880119880119888119888119909119909119888119888119901119901
(tan1205731205731 minus tan1205731205732)
Alors si lrsquoon veut obtenir une plus grande variation detempeacuterature 11987911987902 minus 11987911987901 indirectement un plus grand rapport(1199011199010211990111990101) on peut augmenter
bull la vitesse peacuteripheacuterique 119880119880 (la vitesse de rotation)bull la vitesse axiale 119888119888119909119909 (le deacutebit massique)bull lrsquoangle de deacuteflection de la pale(tan1205731205731 minus tan1205731205732)
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
Download Sound Effects - SoundDogs - Voice Box
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
W2
U
C2
C3W3
U
C1W1
URotor
Stator
➀
➁
➂
C
W
U
Stator Rotor
U2
Stator
C2
U3
W3
Rotor
Stator
Rotor
U2
W2
Rotor
Stator
C2
U3
W3C3
Stator
Rotor
Effets 3DEffets visqueuxDistribution de pressionTraicircneacutee PortanceEacutecoulement instationnaireInteacuteraction fuide structureTurbulence
Extrados
Intrados
L
D
Profil isoleacuteGrille drsquoaubes
Pour analyser lrsquoeacutechange drsquoeacutenergie entre le fluide et le rotor drsquouneturbomachine on utilise la variation de la quantiteacute demouvement (changement de direction etou de vitesse)
Lrsquoeacutetude classique unidimensionnelle utilise le concept de trianglede vitesse Ce modegravele est appliqueacute agrave lrsquoentreacutee et agrave la sortie ducanal inter-aube (grille drsquoaubes)
Dans celui-ci on retrouve la vitesse absolue exprimeacutee dans unrepegravere lieacute aux parties fixes (le stator) accompagneacutee de lavitesse relative lieacutee aux partie tournantes ( le rotor)
Approche classique
Lrsquoapplication du triangle de vitesse permet drsquoestimer
bull lrsquoangle drsquoincidence agrave lrsquoentreacutee du rotor
bull lrsquoangle au bord de fuite du rotor
bull la deacuteflection de lrsquoeacutecoulement dans le stator
bull le travail produit ou consommeacute par la roue
Approche classique
C vitesse absolue de lrsquoeacutecoulement
W vitesse relative de lrsquoeacutecoulement
U vitesse peacuteripheacuterique du rotor
Cu Cm Cx composante tangentielle radiale et axiale de lavitesse absolue du fluide
Wu Wm Wx composante tangentielle radiale et axiale de lavitesse relative du fluide
α lrsquoangle des vitesses absolues mesureacutees par rapport agrave la directionaxiale
β lrsquoangle des vitesses relatives mesureacutees par rapport agrave la directionaxiale
La forme des pales du rotor deacutepend des angles 120631120631(des vitesses relatives dans le repegravere du rotor)
La forme des pales du stator deacutepend des angles 120630120630(des vitesses absolues dans le repegravere fixe)
Si lrsquoon veut garder la vitesse axiale 119940119940119961119961 = 119940119940119940119940119940119940119940119940119940119940 lrsquoeacutequation deconservation de la masse 119950 = 120646120646120488120488119940119940119961119961 = 119940119940119940119940119940119940119940119940119940119940 impose 120646120646120488120488 = 119940119940119940119940119940119940119940119940119940119940
Compresseur
Cu
Cm
Cu
Plan normal
Plan meacuteridional
BABF
W2
C2
U
U
W3C3
u
Les vecteurs vitesse (vitesse absolue 119940119940 pour le stator etrelative 119960119960 pour le rotor) sont consideacutereacutes tangents agrave laligne de squelette de lrsquoaube
β2
β3W3
W2
Pour les machines axiales (u= uent = usort ) le trianglede vitesse donne lieu agrave des eacutequations scalaires tellesque
cu= wu+u
cx= wxcx= = wx c
U
w
wu
cu
Pour les machines agrave eacutetages multiples on suppose que lavitesse axiale demeure constante Crsquoest agrave dire que119940119940120784120784119961119961= 119940119940120785120785119961119961= 119940119940119961119961= 119940119940119940119940119940119940119940119940119940119940
Cette condition jumeleacutee agrave 119932119932120784120784= 119932119932120785120785= 119932119932 megravene agrave lrsquohypothegravese drsquoune cineacutematique reacutepeacutetitive En pratique
Les triangles de vitesse agrave lrsquoentreacutee et agrave la sortie drsquouneacutetage peuvent ecirctre superposeacutes
Nous commenccedilons par la repreacutesentation de triangles de vitessespour le cas drsquoune turbineNous rappelons que par hypothegravese le rayon ne varie pas dans leplan inter-aube et que la vitesse axiale demeure constante dansle sens axialCes conditions permettent drsquoeffectuer la superposition destriangles de vitesse au bord drsquoattaque et bord de fuite du rotoravec la vitesse 119932119932 commune
c1w1
u
c2
u2
w2
➊
➋
➌ c3w3
u
Turbine
C2
W2 C3
W3
Stator Rotor
C2u C3u
W3u
α2β2 β3
α3
∆Cu= ∆Wu
U
W2u
119940119940119961119961119880119880 = 119888119888119888119888119888119888119888119888119888119888119888119888119909119909 = 119888119888119888119888119888119888119888119888119888119888
1198881198883 = 1198881198881
1205721205723 = 1205721205721
Cineacutematique reacutepeacutetitive
➁ ➂
➁U➀
Turbine
119914119914120784120784
119914119914120785120785
119934119934120785120785
119932119932119934119934120784120784
119932119932
119914119914120783120783120630120630120783120783
120630120630120784120784120631120631120784120784
120630120630120785120785
120631120631120785120785
Stator Rotor
➀ ➁ ➂
Le cas du compresseur est similaire agrave celui drsquoune turbineLe rayon ne varie pas (lrsquoeacutetude se deacuteroule dans le plan inter-aube) et la vitesse axiale demeure constante dans le sens axialComme pour la turbine nous pouvons superposer les triangles devitesse au bord drsquoattaque et bord de fuite du rotor
W2 C2
C3W3
U
C1W1
U
U
Compresseur
➊
➋
➌
α2 β1
C2 W1
C1W2
U
β2α1
W2U C1U
CX
( )2 1 = minuse u uW U C C
Lrsquoopeacuteration est vectorielle
C2U
119914119914120784120784119958119958 minus 119914119914120783120783119958119958 =- 119934119934120784120784119958119958 minus119934119934120783120783119958119958
➀
➁
➁
Compresseur
119914119914120784120784119914119914120785120785
119934119934120783120783
119932119932119914119914120783120783120630120630120783120783 120630120630120784120784
120631120631120784120784120630120630120785120785
Rotor Stator
119934119934120784120784120631120631120783120783
119932119932
➀ ➁ ➂
Trois quantiteacutes fondamentales sont utiliseacutees dans laconception des turbomachines
bull le coefficient de charge ψ
bull le coefficient de deacutebit Φ
bull le degreacute de reacuteaction R
Approche classique
Parfois ψ est consideacutereacute positif pour les turbines et neacutegatifpour les compresseurs 119882119882119890119890 = frasl119882 119888
Coefficient de charge
2Ψ = eWU
Coefficient de charge
2e
UW
=ψ 02 01 02 2
h h hU Uminus ∆
= = 02
s shU
η ∆=
=
eWWm
120578120578119904119904 =∆ℎ0∆ℎ0119904119904
Turbine
Coefficient de deacutebit
xcU
Φ =
xCU
Φ =ρRT
pρ
= xC RTRTU p
=RT m RT
U Ap
xCU
Φ =119888 = ρ119862119862119909119909119860119860
11988811987711987711987911987901199011199010119860119860
= 119872119872 120574120574 1 +120574120574 minus 1
21198721198722
minus 120574120574+1)2(120574120574minus1
Afin de geacuteneacuteraliser le triangle de vitesses drsquoune machineaxiale on effectue une mise agrave lrsquoeacutechelle en divisant chaquecocircteacute par la vitesse tangentielle U
Cette opeacuteration permettra de retrouver directement lescoefficients de charge ψ et de deacutebit Φ
Par la suite un nouveau paramegravetre R qursquoon appelle degreacutede reacuteaction sera introduit pour compleacuteter la caracteacuterisationde lrsquoaubage
Approche classique
α2
β2 α3
β3C2 W3
W2
c3u - c2u= w3u - w2u
W3uW2u
U
CX CX
Lrsquoopeacuteration est vectorielle
C3
120511120511 = frasl119934119934119940119940 119932119932120784120784 = frasl119914119914120785120785119958119958 minus 119914119914120784120784119958119958 119932119932
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
Turbine
3
2
3 3
2 2
12
12
12
12
u
u
u u
u u
c RUc RU
w c RU Uw c RU U
Ψ= minus +
Ψ= minus +
Ψ= minus = +
Ψ= minus = minus
2 22 23 2
2 22 23 2
1 12 2
2 2
c cR RU U
w wR RU U
Ψ Ψ = Φ + minus + = Φ + minus +
Ψ Ψ = Φ + + = Φ + minus
3 2
3 2
1 2 1 2
2 2
R Ratan atan
R Ratan atan
minus + Ψ minus + Ψ = = Φ Φ
Ψ + Ψ minus = = Φ Φ
α α
β β
α2β1
C2U
W1U
C1UW2U β2 α1
Compresseur
ΦC2U
C3UW3U
W2U
ψ
1
ψ
ΦC2UC3U
W3U
W2U
1
Il faut deacutefinir un troisiegraveme paramegravetre pour diffeacuterencier ces deux triangles de vitesses ayant le mecircme 120509120509 et le mecircme 120537120537
Les coefficients 120509120509 et 120537120537 ne suffisent pas pour deacutefinir de maniegravereunivoque le triangle de vitesses
Le degreacute de reacuteaction est un paramegravetre adimensionnel quiexprime la variation drsquoenthalpie subie par le fluide dans le rotorpar rapport agrave la chute totale drsquoenthalpie dans lrsquoeacutetage
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
Variation d enthalpie dans le rotorRVariation d enthalpie dans l eacutetage
=
❸❶
❷
Rotor
Stator
R=0 machine drsquoaction (Δhrotor=0 Δheacutetage=Δhstator)
R=1 machine agrave reacuteaction (Δhstator=0 Δheacutetage=Δhrotor)
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
R indique la reacutepartition du saut drsquoenthalpie (de pression) statique entre le stator et le rotor
Le degreacute de reacuteaction R deacutefini en fonction de variablesthermodynamiques (lrsquoenthalpie) peut aussi srsquoexprimer en fonctionde variables cineacutematiques preacutesentes dans le triangle de vitessesNotamment en fonction de la vitesse peacuteripheacuterique U et descomposantes des vitesses absolues et relatives
La vitesse absolue agrave lrsquoentreacutee du stator est eacutegale agrave celle agrave la sortie du rotor (c1= c3 eacutetage reacutepeacutetitif )
On neacuteglige les pertes et le transfert de chaleur dans le stator
1 2 3
ℎ3 minus ℎ1= ℎ3 + frasl11988811988832 2 minus ℎ1 minus frasl11988811988812 2 = ℎ03 minus ℎ01 = 120549120549ℎ0
ℎ03 minus ℎ01 = ℎ03 minus ℎ02 = 120549120549ℎ )0(3minus2
ℎ03 minus ℎ02 = ℎ3 minus ℎ1
119945119945120782120782120783120783 = 119945119945120782120782120784120784
frasl11988811988832 2 = frasl11988811988812 2
3 2
3 1
rotor
eacutetage
h h hRh h h
∆∆
minus= =
minus
2 20 3 2 2 3
0 3 2
2 2( )
( )
h c c h
∆∆
minus
minus
+ minus=
ℎ03 ℎ02
(ℎ01= ℎ02 1198881198881= 1198881198883)
120549120549ℎ )0(3minus2 = (ℎ3 + frasl11988811988832 2) minus (ℎ2 + frasl11988811988822 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
0(3 2) 3 2( )∆ minus = minusu uh U c c
= 1 minusfrasl11988811988832 2 minus frasl11988811988822 2120549120549ℎ )0(3minus2
119877119877 =120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
120549120549ℎ )0(3minus2
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
3 212+
= minus u uc cRU
119888119888119909119909 = 1198881198882119909119909 = 1198881198883119909119909 = 119888119888119888119888119888119888119888119888119888119888
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 11988811988821199061199061198881198882 = 1198881198881199091199092+1198881198881199061199062
119877119877 = 1 minus11988811988831199061199062 minus 11988811988821199061199062
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
Indirectement lrsquoeacutequation pour R deacutecrite en fonction de composantes cineacutematiques eacutetablit un lien entre les angles drsquoentreacutee et de sortie
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
= 1 minus1198881198883119906119906 + 1198881198882119906119906
2119880119880
2232
2 32 2wwh h+ = +
2 23 2
2 3 2 2w wh hminus = minus
2 2 2 23 3 2 2
2 3 2 2u x u xw w w wh h + +
minus = minus
Machine axiale U2 = U3
La composante axiale de la vitesse demeure constante w3x = w 2x
( eacutequation justifieacutee plus tard lors de la rothalpie)
2 23 2
2 3 2 2u uw wh hminus = minus
3 2 3 22 3
( )( )2
u u u uw w w wh h minus +minus =
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
Eacutetage reacutepeacutetitif c3 = c1
Si les pertes dans le stator sont neacutegligeacutees
c2u ndash c3u= w2u - w3u
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
3 2 3 23 2
( )( )2
u u u uw w w wh h minus +minus = minus
c2u ndash c3u= w2u - w3u
119877119877 = minus1199081199083119906119906 + 1199081199082119906119906
2119880119880
2 2 2 22 3 2 3 2 3( ) 2 ( ) 2h h u u w wminus = minus minus minus
2 2 2 2 2 23 2 3 2 3 2
2 2 2e
I II III
u u c c w wW minus minus minus= + minus
2 2 2 22 3 2 3
2 2 2 2 2 22 3 2 3 2 3
( ) ( )( ) ( ) ( )
minus minus minus=
minus + minus minus minusu u w wR
c c u u w w
3 2= =U u u
Expliqueacutee + tard
2 23 2
03 02 3 2 2ec cW h h h h minus
= minus = minus +
119877119877 =ℎ2 minus ℎ3ℎ1 minus ℎ3
=ℎ2 minus ℎ3ℎ02 minus ℎ03
Degreacute de reacuteaction
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
( )3u 2uw wR2U+
= minus
3 212
u uc cRU+
= minus
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
Puisque le degreacute de reacuteaction R peut aussi srsquoexprimer enfonction de variables cineacutematiques on peut chercher sarepreacutesentation dans le triangle adimensionnel de vitesses
Turbine
α2
β2 α3
β3
C2 W3
C3W2
C3uC2u
U
C2xC3x
W2u
c2u ndash c3u= w2u - w3u
2Ψ minusR
frasl119934119934120784120784119958119958 119932119932 = frasl1198621198622119906119906 119880119880 minus frasl119880119880 119880119880Ψ = frasl1198621198622119906119906 minus 1198621198623119906119906 119880119880
119877119877 = 1 minus (1198621198623119906119906 + 1198621198622119906119906 frasl) 2119880119880Φ = frasl1198621198622119909119909 119880119880 = frasl1198621198623119909119909 119880119880
= (119914119914120784120784119958119958 minus 119914119914120785120785119958119958 frasl) 2119880119880 minus 1 + (119914119914120784120784119958119958 + 119914119914120785120785119958119958 frasl) 2119880119880
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α2
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
Les repreacutesentations des triangles de vitesses adimensionnellessont tregraves riches en information et contiennent plusieurs formules
Par exemple pour un compresseur on deacuteduit aiseacutement que
Avec 120595120595 = 1198821198821198901198901198801198802
Φ = 119888119888119909119909119880119880
et que pour 119888119888119901119901 = 119888119888119888119888119888119888119888119888119888119888 119882119882119890119890 = 119888119888119901119901(11987911987902minus11987911987901)on trouve
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
(11987911987902minus11987911987901) =119880119880119888119888119909119909119888119888119901119901
(tan1205731205731 minus tan1205731205732)
Alors si lrsquoon veut obtenir une plus grande variation detempeacuterature 11987911987902 minus 11987911987901 indirectement un plus grand rapport(1199011199010211990111990101) on peut augmenter
bull la vitesse peacuteripheacuterique 119880119880 (la vitesse de rotation)bull la vitesse axiale 119888119888119909119909 (le deacutebit massique)bull lrsquoangle de deacuteflection de la pale(tan1205731205731 minus tan1205731205732)
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
Download Sound Effects - SoundDogs - Voice Box
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
C
W
U
Stator Rotor
U2
Stator
C2
U3
W3
Rotor
Stator
Rotor
U2
W2
Rotor
Stator
C2
U3
W3C3
Stator
Rotor
Effets 3DEffets visqueuxDistribution de pressionTraicircneacutee PortanceEacutecoulement instationnaireInteacuteraction fuide structureTurbulence
Extrados
Intrados
L
D
Profil isoleacuteGrille drsquoaubes
Pour analyser lrsquoeacutechange drsquoeacutenergie entre le fluide et le rotor drsquouneturbomachine on utilise la variation de la quantiteacute demouvement (changement de direction etou de vitesse)
Lrsquoeacutetude classique unidimensionnelle utilise le concept de trianglede vitesse Ce modegravele est appliqueacute agrave lrsquoentreacutee et agrave la sortie ducanal inter-aube (grille drsquoaubes)
Dans celui-ci on retrouve la vitesse absolue exprimeacutee dans unrepegravere lieacute aux parties fixes (le stator) accompagneacutee de lavitesse relative lieacutee aux partie tournantes ( le rotor)
Approche classique
Lrsquoapplication du triangle de vitesse permet drsquoestimer
bull lrsquoangle drsquoincidence agrave lrsquoentreacutee du rotor
bull lrsquoangle au bord de fuite du rotor
bull la deacuteflection de lrsquoeacutecoulement dans le stator
bull le travail produit ou consommeacute par la roue
Approche classique
C vitesse absolue de lrsquoeacutecoulement
W vitesse relative de lrsquoeacutecoulement
U vitesse peacuteripheacuterique du rotor
Cu Cm Cx composante tangentielle radiale et axiale de lavitesse absolue du fluide
Wu Wm Wx composante tangentielle radiale et axiale de lavitesse relative du fluide
α lrsquoangle des vitesses absolues mesureacutees par rapport agrave la directionaxiale
β lrsquoangle des vitesses relatives mesureacutees par rapport agrave la directionaxiale
La forme des pales du rotor deacutepend des angles 120631120631(des vitesses relatives dans le repegravere du rotor)
La forme des pales du stator deacutepend des angles 120630120630(des vitesses absolues dans le repegravere fixe)
Si lrsquoon veut garder la vitesse axiale 119940119940119961119961 = 119940119940119940119940119940119940119940119940119940119940 lrsquoeacutequation deconservation de la masse 119950 = 120646120646120488120488119940119940119961119961 = 119940119940119940119940119940119940119940119940119940119940 impose 120646120646120488120488 = 119940119940119940119940119940119940119940119940119940119940
Compresseur
Cu
Cm
Cu
Plan normal
Plan meacuteridional
BABF
W2
C2
U
U
W3C3
u
Les vecteurs vitesse (vitesse absolue 119940119940 pour le stator etrelative 119960119960 pour le rotor) sont consideacutereacutes tangents agrave laligne de squelette de lrsquoaube
β2
β3W3
W2
Pour les machines axiales (u= uent = usort ) le trianglede vitesse donne lieu agrave des eacutequations scalaires tellesque
cu= wu+u
cx= wxcx= = wx c
U
w
wu
cu
Pour les machines agrave eacutetages multiples on suppose que lavitesse axiale demeure constante Crsquoest agrave dire que119940119940120784120784119961119961= 119940119940120785120785119961119961= 119940119940119961119961= 119940119940119940119940119940119940119940119940119940119940
Cette condition jumeleacutee agrave 119932119932120784120784= 119932119932120785120785= 119932119932 megravene agrave lrsquohypothegravese drsquoune cineacutematique reacutepeacutetitive En pratique
Les triangles de vitesse agrave lrsquoentreacutee et agrave la sortie drsquouneacutetage peuvent ecirctre superposeacutes
Nous commenccedilons par la repreacutesentation de triangles de vitessespour le cas drsquoune turbineNous rappelons que par hypothegravese le rayon ne varie pas dans leplan inter-aube et que la vitesse axiale demeure constante dansle sens axialCes conditions permettent drsquoeffectuer la superposition destriangles de vitesse au bord drsquoattaque et bord de fuite du rotoravec la vitesse 119932119932 commune
c1w1
u
c2
u2
w2
➊
➋
➌ c3w3
u
Turbine
C2
W2 C3
W3
Stator Rotor
C2u C3u
W3u
α2β2 β3
α3
∆Cu= ∆Wu
U
W2u
119940119940119961119961119880119880 = 119888119888119888119888119888119888119888119888119888119888119888119888119909119909 = 119888119888119888119888119888119888119888119888119888119888
1198881198883 = 1198881198881
1205721205723 = 1205721205721
Cineacutematique reacutepeacutetitive
➁ ➂
➁U➀
Turbine
119914119914120784120784
119914119914120785120785
119934119934120785120785
119932119932119934119934120784120784
119932119932
119914119914120783120783120630120630120783120783
120630120630120784120784120631120631120784120784
120630120630120785120785
120631120631120785120785
Stator Rotor
➀ ➁ ➂
Le cas du compresseur est similaire agrave celui drsquoune turbineLe rayon ne varie pas (lrsquoeacutetude se deacuteroule dans le plan inter-aube) et la vitesse axiale demeure constante dans le sens axialComme pour la turbine nous pouvons superposer les triangles devitesse au bord drsquoattaque et bord de fuite du rotor
W2 C2
C3W3
U
C1W1
U
U
Compresseur
➊
➋
➌
α2 β1
C2 W1
C1W2
U
β2α1
W2U C1U
CX
( )2 1 = minuse u uW U C C
Lrsquoopeacuteration est vectorielle
C2U
119914119914120784120784119958119958 minus 119914119914120783120783119958119958 =- 119934119934120784120784119958119958 minus119934119934120783120783119958119958
➀
➁
➁
Compresseur
119914119914120784120784119914119914120785120785
119934119934120783120783
119932119932119914119914120783120783120630120630120783120783 120630120630120784120784
120631120631120784120784120630120630120785120785
Rotor Stator
119934119934120784120784120631120631120783120783
119932119932
➀ ➁ ➂
Trois quantiteacutes fondamentales sont utiliseacutees dans laconception des turbomachines
bull le coefficient de charge ψ
bull le coefficient de deacutebit Φ
bull le degreacute de reacuteaction R
Approche classique
Parfois ψ est consideacutereacute positif pour les turbines et neacutegatifpour les compresseurs 119882119882119890119890 = frasl119882 119888
Coefficient de charge
2Ψ = eWU
Coefficient de charge
2e
UW
=ψ 02 01 02 2
h h hU Uminus ∆
= = 02
s shU
η ∆=
=
eWWm
120578120578119904119904 =∆ℎ0∆ℎ0119904119904
Turbine
Coefficient de deacutebit
xcU
Φ =
xCU
Φ =ρRT
pρ
= xC RTRTU p
=RT m RT
U Ap
xCU
Φ =119888 = ρ119862119862119909119909119860119860
11988811987711987711987911987901199011199010119860119860
= 119872119872 120574120574 1 +120574120574 minus 1
21198721198722
minus 120574120574+1)2(120574120574minus1
Afin de geacuteneacuteraliser le triangle de vitesses drsquoune machineaxiale on effectue une mise agrave lrsquoeacutechelle en divisant chaquecocircteacute par la vitesse tangentielle U
Cette opeacuteration permettra de retrouver directement lescoefficients de charge ψ et de deacutebit Φ
Par la suite un nouveau paramegravetre R qursquoon appelle degreacutede reacuteaction sera introduit pour compleacuteter la caracteacuterisationde lrsquoaubage
Approche classique
α2
β2 α3
β3C2 W3
W2
c3u - c2u= w3u - w2u
W3uW2u
U
CX CX
Lrsquoopeacuteration est vectorielle
C3
120511120511 = frasl119934119934119940119940 119932119932120784120784 = frasl119914119914120785120785119958119958 minus 119914119914120784120784119958119958 119932119932
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
Turbine
3
2
3 3
2 2
12
12
12
12
u
u
u u
u u
c RUc RU
w c RU Uw c RU U
Ψ= minus +
Ψ= minus +
Ψ= minus = +
Ψ= minus = minus
2 22 23 2
2 22 23 2
1 12 2
2 2
c cR RU U
w wR RU U
Ψ Ψ = Φ + minus + = Φ + minus +
Ψ Ψ = Φ + + = Φ + minus
3 2
3 2
1 2 1 2
2 2
R Ratan atan
R Ratan atan
minus + Ψ minus + Ψ = = Φ Φ
Ψ + Ψ minus = = Φ Φ
α α
β β
α2β1
C2U
W1U
C1UW2U β2 α1
Compresseur
ΦC2U
C3UW3U
W2U
ψ
1
ψ
ΦC2UC3U
W3U
W2U
1
Il faut deacutefinir un troisiegraveme paramegravetre pour diffeacuterencier ces deux triangles de vitesses ayant le mecircme 120509120509 et le mecircme 120537120537
Les coefficients 120509120509 et 120537120537 ne suffisent pas pour deacutefinir de maniegravereunivoque le triangle de vitesses
Le degreacute de reacuteaction est un paramegravetre adimensionnel quiexprime la variation drsquoenthalpie subie par le fluide dans le rotorpar rapport agrave la chute totale drsquoenthalpie dans lrsquoeacutetage
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
Variation d enthalpie dans le rotorRVariation d enthalpie dans l eacutetage
=
❸❶
❷
Rotor
Stator
R=0 machine drsquoaction (Δhrotor=0 Δheacutetage=Δhstator)
R=1 machine agrave reacuteaction (Δhstator=0 Δheacutetage=Δhrotor)
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
R indique la reacutepartition du saut drsquoenthalpie (de pression) statique entre le stator et le rotor
Le degreacute de reacuteaction R deacutefini en fonction de variablesthermodynamiques (lrsquoenthalpie) peut aussi srsquoexprimer en fonctionde variables cineacutematiques preacutesentes dans le triangle de vitessesNotamment en fonction de la vitesse peacuteripheacuterique U et descomposantes des vitesses absolues et relatives
La vitesse absolue agrave lrsquoentreacutee du stator est eacutegale agrave celle agrave la sortie du rotor (c1= c3 eacutetage reacutepeacutetitif )
On neacuteglige les pertes et le transfert de chaleur dans le stator
1 2 3
ℎ3 minus ℎ1= ℎ3 + frasl11988811988832 2 minus ℎ1 minus frasl11988811988812 2 = ℎ03 minus ℎ01 = 120549120549ℎ0
ℎ03 minus ℎ01 = ℎ03 minus ℎ02 = 120549120549ℎ )0(3minus2
ℎ03 minus ℎ02 = ℎ3 minus ℎ1
119945119945120782120782120783120783 = 119945119945120782120782120784120784
frasl11988811988832 2 = frasl11988811988812 2
3 2
3 1
rotor
eacutetage
h h hRh h h
∆∆
minus= =
minus
2 20 3 2 2 3
0 3 2
2 2( )
( )
h c c h
∆∆
minus
minus
+ minus=
ℎ03 ℎ02
(ℎ01= ℎ02 1198881198881= 1198881198883)
120549120549ℎ )0(3minus2 = (ℎ3 + frasl11988811988832 2) minus (ℎ2 + frasl11988811988822 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
0(3 2) 3 2( )∆ minus = minusu uh U c c
= 1 minusfrasl11988811988832 2 minus frasl11988811988822 2120549120549ℎ )0(3minus2
119877119877 =120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
120549120549ℎ )0(3minus2
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
3 212+
= minus u uc cRU
119888119888119909119909 = 1198881198882119909119909 = 1198881198883119909119909 = 119888119888119888119888119888119888119888119888119888119888
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 11988811988821199061199061198881198882 = 1198881198881199091199092+1198881198881199061199062
119877119877 = 1 minus11988811988831199061199062 minus 11988811988821199061199062
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
Indirectement lrsquoeacutequation pour R deacutecrite en fonction de composantes cineacutematiques eacutetablit un lien entre les angles drsquoentreacutee et de sortie
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
= 1 minus1198881198883119906119906 + 1198881198882119906119906
2119880119880
2232
2 32 2wwh h+ = +
2 23 2
2 3 2 2w wh hminus = minus
2 2 2 23 3 2 2
2 3 2 2u x u xw w w wh h + +
minus = minus
Machine axiale U2 = U3
La composante axiale de la vitesse demeure constante w3x = w 2x
( eacutequation justifieacutee plus tard lors de la rothalpie)
2 23 2
2 3 2 2u uw wh hminus = minus
3 2 3 22 3
( )( )2
u u u uw w w wh h minus +minus =
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
Eacutetage reacutepeacutetitif c3 = c1
Si les pertes dans le stator sont neacutegligeacutees
c2u ndash c3u= w2u - w3u
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
3 2 3 23 2
( )( )2
u u u uw w w wh h minus +minus = minus
c2u ndash c3u= w2u - w3u
119877119877 = minus1199081199083119906119906 + 1199081199082119906119906
2119880119880
2 2 2 22 3 2 3 2 3( ) 2 ( ) 2h h u u w wminus = minus minus minus
2 2 2 2 2 23 2 3 2 3 2
2 2 2e
I II III
u u c c w wW minus minus minus= + minus
2 2 2 22 3 2 3
2 2 2 2 2 22 3 2 3 2 3
( ) ( )( ) ( ) ( )
minus minus minus=
minus + minus minus minusu u w wR
c c u u w w
3 2= =U u u
Expliqueacutee + tard
2 23 2
03 02 3 2 2ec cW h h h h minus
= minus = minus +
119877119877 =ℎ2 minus ℎ3ℎ1 minus ℎ3
=ℎ2 minus ℎ3ℎ02 minus ℎ03
Degreacute de reacuteaction
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
( )3u 2uw wR2U+
= minus
3 212
u uc cRU+
= minus
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
Puisque le degreacute de reacuteaction R peut aussi srsquoexprimer enfonction de variables cineacutematiques on peut chercher sarepreacutesentation dans le triangle adimensionnel de vitesses
Turbine
α2
β2 α3
β3
C2 W3
C3W2
C3uC2u
U
C2xC3x
W2u
c2u ndash c3u= w2u - w3u
2Ψ minusR
frasl119934119934120784120784119958119958 119932119932 = frasl1198621198622119906119906 119880119880 minus frasl119880119880 119880119880Ψ = frasl1198621198622119906119906 minus 1198621198623119906119906 119880119880
119877119877 = 1 minus (1198621198623119906119906 + 1198621198622119906119906 frasl) 2119880119880Φ = frasl1198621198622119909119909 119880119880 = frasl1198621198623119909119909 119880119880
= (119914119914120784120784119958119958 minus 119914119914120785120785119958119958 frasl) 2119880119880 minus 1 + (119914119914120784120784119958119958 + 119914119914120785120785119958119958 frasl) 2119880119880
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α2
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
Les repreacutesentations des triangles de vitesses adimensionnellessont tregraves riches en information et contiennent plusieurs formules
Par exemple pour un compresseur on deacuteduit aiseacutement que
Avec 120595120595 = 1198821198821198901198901198801198802
Φ = 119888119888119909119909119880119880
et que pour 119888119888119901119901 = 119888119888119888119888119888119888119888119888119888119888 119882119882119890119890 = 119888119888119901119901(11987911987902minus11987911987901)on trouve
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
(11987911987902minus11987911987901) =119880119880119888119888119909119909119888119888119901119901
(tan1205731205731 minus tan1205731205732)
Alors si lrsquoon veut obtenir une plus grande variation detempeacuterature 11987911987902 minus 11987911987901 indirectement un plus grand rapport(1199011199010211990111990101) on peut augmenter
bull la vitesse peacuteripheacuterique 119880119880 (la vitesse de rotation)bull la vitesse axiale 119888119888119909119909 (le deacutebit massique)bull lrsquoangle de deacuteflection de la pale(tan1205731205731 minus tan1205731205732)
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
U2
Stator
C2
U3
W3
Rotor
Stator
Rotor
U2
W2
Rotor
Stator
C2
U3
W3C3
Stator
Rotor
Effets 3DEffets visqueuxDistribution de pressionTraicircneacutee PortanceEacutecoulement instationnaireInteacuteraction fuide structureTurbulence
Extrados
Intrados
L
D
Profil isoleacuteGrille drsquoaubes
Pour analyser lrsquoeacutechange drsquoeacutenergie entre le fluide et le rotor drsquouneturbomachine on utilise la variation de la quantiteacute demouvement (changement de direction etou de vitesse)
Lrsquoeacutetude classique unidimensionnelle utilise le concept de trianglede vitesse Ce modegravele est appliqueacute agrave lrsquoentreacutee et agrave la sortie ducanal inter-aube (grille drsquoaubes)
Dans celui-ci on retrouve la vitesse absolue exprimeacutee dans unrepegravere lieacute aux parties fixes (le stator) accompagneacutee de lavitesse relative lieacutee aux partie tournantes ( le rotor)
Approche classique
Lrsquoapplication du triangle de vitesse permet drsquoestimer
bull lrsquoangle drsquoincidence agrave lrsquoentreacutee du rotor
bull lrsquoangle au bord de fuite du rotor
bull la deacuteflection de lrsquoeacutecoulement dans le stator
bull le travail produit ou consommeacute par la roue
Approche classique
C vitesse absolue de lrsquoeacutecoulement
W vitesse relative de lrsquoeacutecoulement
U vitesse peacuteripheacuterique du rotor
Cu Cm Cx composante tangentielle radiale et axiale de lavitesse absolue du fluide
Wu Wm Wx composante tangentielle radiale et axiale de lavitesse relative du fluide
α lrsquoangle des vitesses absolues mesureacutees par rapport agrave la directionaxiale
β lrsquoangle des vitesses relatives mesureacutees par rapport agrave la directionaxiale
La forme des pales du rotor deacutepend des angles 120631120631(des vitesses relatives dans le repegravere du rotor)
La forme des pales du stator deacutepend des angles 120630120630(des vitesses absolues dans le repegravere fixe)
Si lrsquoon veut garder la vitesse axiale 119940119940119961119961 = 119940119940119940119940119940119940119940119940119940119940 lrsquoeacutequation deconservation de la masse 119950 = 120646120646120488120488119940119940119961119961 = 119940119940119940119940119940119940119940119940119940119940 impose 120646120646120488120488 = 119940119940119940119940119940119940119940119940119940119940
Compresseur
Cu
Cm
Cu
Plan normal
Plan meacuteridional
BABF
W2
C2
U
U
W3C3
u
Les vecteurs vitesse (vitesse absolue 119940119940 pour le stator etrelative 119960119960 pour le rotor) sont consideacutereacutes tangents agrave laligne de squelette de lrsquoaube
β2
β3W3
W2
Pour les machines axiales (u= uent = usort ) le trianglede vitesse donne lieu agrave des eacutequations scalaires tellesque
cu= wu+u
cx= wxcx= = wx c
U
w
wu
cu
Pour les machines agrave eacutetages multiples on suppose que lavitesse axiale demeure constante Crsquoest agrave dire que119940119940120784120784119961119961= 119940119940120785120785119961119961= 119940119940119961119961= 119940119940119940119940119940119940119940119940119940119940
Cette condition jumeleacutee agrave 119932119932120784120784= 119932119932120785120785= 119932119932 megravene agrave lrsquohypothegravese drsquoune cineacutematique reacutepeacutetitive En pratique
Les triangles de vitesse agrave lrsquoentreacutee et agrave la sortie drsquouneacutetage peuvent ecirctre superposeacutes
Nous commenccedilons par la repreacutesentation de triangles de vitessespour le cas drsquoune turbineNous rappelons que par hypothegravese le rayon ne varie pas dans leplan inter-aube et que la vitesse axiale demeure constante dansle sens axialCes conditions permettent drsquoeffectuer la superposition destriangles de vitesse au bord drsquoattaque et bord de fuite du rotoravec la vitesse 119932119932 commune
c1w1
u
c2
u2
w2
➊
➋
➌ c3w3
u
Turbine
C2
W2 C3
W3
Stator Rotor
C2u C3u
W3u
α2β2 β3
α3
∆Cu= ∆Wu
U
W2u
119940119940119961119961119880119880 = 119888119888119888119888119888119888119888119888119888119888119888119888119909119909 = 119888119888119888119888119888119888119888119888119888119888
1198881198883 = 1198881198881
1205721205723 = 1205721205721
Cineacutematique reacutepeacutetitive
➁ ➂
➁U➀
Turbine
119914119914120784120784
119914119914120785120785
119934119934120785120785
119932119932119934119934120784120784
119932119932
119914119914120783120783120630120630120783120783
120630120630120784120784120631120631120784120784
120630120630120785120785
120631120631120785120785
Stator Rotor
➀ ➁ ➂
Le cas du compresseur est similaire agrave celui drsquoune turbineLe rayon ne varie pas (lrsquoeacutetude se deacuteroule dans le plan inter-aube) et la vitesse axiale demeure constante dans le sens axialComme pour la turbine nous pouvons superposer les triangles devitesse au bord drsquoattaque et bord de fuite du rotor
W2 C2
C3W3
U
C1W1
U
U
Compresseur
➊
➋
➌
α2 β1
C2 W1
C1W2
U
β2α1
W2U C1U
CX
( )2 1 = minuse u uW U C C
Lrsquoopeacuteration est vectorielle
C2U
119914119914120784120784119958119958 minus 119914119914120783120783119958119958 =- 119934119934120784120784119958119958 minus119934119934120783120783119958119958
➀
➁
➁
Compresseur
119914119914120784120784119914119914120785120785
119934119934120783120783
119932119932119914119914120783120783120630120630120783120783 120630120630120784120784
120631120631120784120784120630120630120785120785
Rotor Stator
119934119934120784120784120631120631120783120783
119932119932
➀ ➁ ➂
Trois quantiteacutes fondamentales sont utiliseacutees dans laconception des turbomachines
bull le coefficient de charge ψ
bull le coefficient de deacutebit Φ
bull le degreacute de reacuteaction R
Approche classique
Parfois ψ est consideacutereacute positif pour les turbines et neacutegatifpour les compresseurs 119882119882119890119890 = frasl119882 119888
Coefficient de charge
2Ψ = eWU
Coefficient de charge
2e
UW
=ψ 02 01 02 2
h h hU Uminus ∆
= = 02
s shU
η ∆=
=
eWWm
120578120578119904119904 =∆ℎ0∆ℎ0119904119904
Turbine
Coefficient de deacutebit
xcU
Φ =
xCU
Φ =ρRT
pρ
= xC RTRTU p
=RT m RT
U Ap
xCU
Φ =119888 = ρ119862119862119909119909119860119860
11988811987711987711987911987901199011199010119860119860
= 119872119872 120574120574 1 +120574120574 minus 1
21198721198722
minus 120574120574+1)2(120574120574minus1
Afin de geacuteneacuteraliser le triangle de vitesses drsquoune machineaxiale on effectue une mise agrave lrsquoeacutechelle en divisant chaquecocircteacute par la vitesse tangentielle U
Cette opeacuteration permettra de retrouver directement lescoefficients de charge ψ et de deacutebit Φ
Par la suite un nouveau paramegravetre R qursquoon appelle degreacutede reacuteaction sera introduit pour compleacuteter la caracteacuterisationde lrsquoaubage
Approche classique
α2
β2 α3
β3C2 W3
W2
c3u - c2u= w3u - w2u
W3uW2u
U
CX CX
Lrsquoopeacuteration est vectorielle
C3
120511120511 = frasl119934119934119940119940 119932119932120784120784 = frasl119914119914120785120785119958119958 minus 119914119914120784120784119958119958 119932119932
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
Turbine
3
2
3 3
2 2
12
12
12
12
u
u
u u
u u
c RUc RU
w c RU Uw c RU U
Ψ= minus +
Ψ= minus +
Ψ= minus = +
Ψ= minus = minus
2 22 23 2
2 22 23 2
1 12 2
2 2
c cR RU U
w wR RU U
Ψ Ψ = Φ + minus + = Φ + minus +
Ψ Ψ = Φ + + = Φ + minus
3 2
3 2
1 2 1 2
2 2
R Ratan atan
R Ratan atan
minus + Ψ minus + Ψ = = Φ Φ
Ψ + Ψ minus = = Φ Φ
α α
β β
α2β1
C2U
W1U
C1UW2U β2 α1
Compresseur
ΦC2U
C3UW3U
W2U
ψ
1
ψ
ΦC2UC3U
W3U
W2U
1
Il faut deacutefinir un troisiegraveme paramegravetre pour diffeacuterencier ces deux triangles de vitesses ayant le mecircme 120509120509 et le mecircme 120537120537
Les coefficients 120509120509 et 120537120537 ne suffisent pas pour deacutefinir de maniegravereunivoque le triangle de vitesses
Le degreacute de reacuteaction est un paramegravetre adimensionnel quiexprime la variation drsquoenthalpie subie par le fluide dans le rotorpar rapport agrave la chute totale drsquoenthalpie dans lrsquoeacutetage
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
Variation d enthalpie dans le rotorRVariation d enthalpie dans l eacutetage
=
❸❶
❷
Rotor
Stator
R=0 machine drsquoaction (Δhrotor=0 Δheacutetage=Δhstator)
R=1 machine agrave reacuteaction (Δhstator=0 Δheacutetage=Δhrotor)
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
R indique la reacutepartition du saut drsquoenthalpie (de pression) statique entre le stator et le rotor
Le degreacute de reacuteaction R deacutefini en fonction de variablesthermodynamiques (lrsquoenthalpie) peut aussi srsquoexprimer en fonctionde variables cineacutematiques preacutesentes dans le triangle de vitessesNotamment en fonction de la vitesse peacuteripheacuterique U et descomposantes des vitesses absolues et relatives
La vitesse absolue agrave lrsquoentreacutee du stator est eacutegale agrave celle agrave la sortie du rotor (c1= c3 eacutetage reacutepeacutetitif )
On neacuteglige les pertes et le transfert de chaleur dans le stator
1 2 3
ℎ3 minus ℎ1= ℎ3 + frasl11988811988832 2 minus ℎ1 minus frasl11988811988812 2 = ℎ03 minus ℎ01 = 120549120549ℎ0
ℎ03 minus ℎ01 = ℎ03 minus ℎ02 = 120549120549ℎ )0(3minus2
ℎ03 minus ℎ02 = ℎ3 minus ℎ1
119945119945120782120782120783120783 = 119945119945120782120782120784120784
frasl11988811988832 2 = frasl11988811988812 2
3 2
3 1
rotor
eacutetage
h h hRh h h
∆∆
minus= =
minus
2 20 3 2 2 3
0 3 2
2 2( )
( )
h c c h
∆∆
minus
minus
+ minus=
ℎ03 ℎ02
(ℎ01= ℎ02 1198881198881= 1198881198883)
120549120549ℎ )0(3minus2 = (ℎ3 + frasl11988811988832 2) minus (ℎ2 + frasl11988811988822 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
0(3 2) 3 2( )∆ minus = minusu uh U c c
= 1 minusfrasl11988811988832 2 minus frasl11988811988822 2120549120549ℎ )0(3minus2
119877119877 =120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
120549120549ℎ )0(3minus2
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
3 212+
= minus u uc cRU
119888119888119909119909 = 1198881198882119909119909 = 1198881198883119909119909 = 119888119888119888119888119888119888119888119888119888119888
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 11988811988821199061199061198881198882 = 1198881198881199091199092+1198881198881199061199062
119877119877 = 1 minus11988811988831199061199062 minus 11988811988821199061199062
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
Indirectement lrsquoeacutequation pour R deacutecrite en fonction de composantes cineacutematiques eacutetablit un lien entre les angles drsquoentreacutee et de sortie
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
= 1 minus1198881198883119906119906 + 1198881198882119906119906
2119880119880
2232
2 32 2wwh h+ = +
2 23 2
2 3 2 2w wh hminus = minus
2 2 2 23 3 2 2
2 3 2 2u x u xw w w wh h + +
minus = minus
Machine axiale U2 = U3
La composante axiale de la vitesse demeure constante w3x = w 2x
( eacutequation justifieacutee plus tard lors de la rothalpie)
2 23 2
2 3 2 2u uw wh hminus = minus
3 2 3 22 3
( )( )2
u u u uw w w wh h minus +minus =
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
Eacutetage reacutepeacutetitif c3 = c1
Si les pertes dans le stator sont neacutegligeacutees
c2u ndash c3u= w2u - w3u
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
3 2 3 23 2
( )( )2
u u u uw w w wh h minus +minus = minus
c2u ndash c3u= w2u - w3u
119877119877 = minus1199081199083119906119906 + 1199081199082119906119906
2119880119880
2 2 2 22 3 2 3 2 3( ) 2 ( ) 2h h u u w wminus = minus minus minus
2 2 2 2 2 23 2 3 2 3 2
2 2 2e
I II III
u u c c w wW minus minus minus= + minus
2 2 2 22 3 2 3
2 2 2 2 2 22 3 2 3 2 3
( ) ( )( ) ( ) ( )
minus minus minus=
minus + minus minus minusu u w wR
c c u u w w
3 2= =U u u
Expliqueacutee + tard
2 23 2
03 02 3 2 2ec cW h h h h minus
= minus = minus +
119877119877 =ℎ2 minus ℎ3ℎ1 minus ℎ3
=ℎ2 minus ℎ3ℎ02 minus ℎ03
Degreacute de reacuteaction
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
( )3u 2uw wR2U+
= minus
3 212
u uc cRU+
= minus
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
Puisque le degreacute de reacuteaction R peut aussi srsquoexprimer enfonction de variables cineacutematiques on peut chercher sarepreacutesentation dans le triangle adimensionnel de vitesses
Turbine
α2
β2 α3
β3
C2 W3
C3W2
C3uC2u
U
C2xC3x
W2u
c2u ndash c3u= w2u - w3u
2Ψ minusR
frasl119934119934120784120784119958119958 119932119932 = frasl1198621198622119906119906 119880119880 minus frasl119880119880 119880119880Ψ = frasl1198621198622119906119906 minus 1198621198623119906119906 119880119880
119877119877 = 1 minus (1198621198623119906119906 + 1198621198622119906119906 frasl) 2119880119880Φ = frasl1198621198622119909119909 119880119880 = frasl1198621198623119909119909 119880119880
= (119914119914120784120784119958119958 minus 119914119914120785120785119958119958 frasl) 2119880119880 minus 1 + (119914119914120784120784119958119958 + 119914119914120785120785119958119958 frasl) 2119880119880
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α2
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
Les repreacutesentations des triangles de vitesses adimensionnellessont tregraves riches en information et contiennent plusieurs formules
Par exemple pour un compresseur on deacuteduit aiseacutement que
Avec 120595120595 = 1198821198821198901198901198801198802
Φ = 119888119888119909119909119880119880
et que pour 119888119888119901119901 = 119888119888119888119888119888119888119888119888119888119888 119882119882119890119890 = 119888119888119901119901(11987911987902minus11987911987901)on trouve
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
(11987911987902minus11987911987901) =119880119880119888119888119909119909119888119888119901119901
(tan1205731205731 minus tan1205731205732)
Alors si lrsquoon veut obtenir une plus grande variation detempeacuterature 11987911987902 minus 11987911987901 indirectement un plus grand rapport(1199011199010211990111990101) on peut augmenter
bull la vitesse peacuteripheacuterique 119880119880 (la vitesse de rotation)bull la vitesse axiale 119888119888119909119909 (le deacutebit massique)bull lrsquoangle de deacuteflection de la pale(tan1205731205731 minus tan1205731205732)
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
U2
W2
Rotor
Stator
C2
U3
W3C3
Stator
Rotor
Effets 3DEffets visqueuxDistribution de pressionTraicircneacutee PortanceEacutecoulement instationnaireInteacuteraction fuide structureTurbulence
Extrados
Intrados
L
D
Profil isoleacuteGrille drsquoaubes
Pour analyser lrsquoeacutechange drsquoeacutenergie entre le fluide et le rotor drsquouneturbomachine on utilise la variation de la quantiteacute demouvement (changement de direction etou de vitesse)
Lrsquoeacutetude classique unidimensionnelle utilise le concept de trianglede vitesse Ce modegravele est appliqueacute agrave lrsquoentreacutee et agrave la sortie ducanal inter-aube (grille drsquoaubes)
Dans celui-ci on retrouve la vitesse absolue exprimeacutee dans unrepegravere lieacute aux parties fixes (le stator) accompagneacutee de lavitesse relative lieacutee aux partie tournantes ( le rotor)
Approche classique
Lrsquoapplication du triangle de vitesse permet drsquoestimer
bull lrsquoangle drsquoincidence agrave lrsquoentreacutee du rotor
bull lrsquoangle au bord de fuite du rotor
bull la deacuteflection de lrsquoeacutecoulement dans le stator
bull le travail produit ou consommeacute par la roue
Approche classique
C vitesse absolue de lrsquoeacutecoulement
W vitesse relative de lrsquoeacutecoulement
U vitesse peacuteripheacuterique du rotor
Cu Cm Cx composante tangentielle radiale et axiale de lavitesse absolue du fluide
Wu Wm Wx composante tangentielle radiale et axiale de lavitesse relative du fluide
α lrsquoangle des vitesses absolues mesureacutees par rapport agrave la directionaxiale
β lrsquoangle des vitesses relatives mesureacutees par rapport agrave la directionaxiale
La forme des pales du rotor deacutepend des angles 120631120631(des vitesses relatives dans le repegravere du rotor)
La forme des pales du stator deacutepend des angles 120630120630(des vitesses absolues dans le repegravere fixe)
Si lrsquoon veut garder la vitesse axiale 119940119940119961119961 = 119940119940119940119940119940119940119940119940119940119940 lrsquoeacutequation deconservation de la masse 119950 = 120646120646120488120488119940119940119961119961 = 119940119940119940119940119940119940119940119940119940119940 impose 120646120646120488120488 = 119940119940119940119940119940119940119940119940119940119940
Compresseur
Cu
Cm
Cu
Plan normal
Plan meacuteridional
BABF
W2
C2
U
U
W3C3
u
Les vecteurs vitesse (vitesse absolue 119940119940 pour le stator etrelative 119960119960 pour le rotor) sont consideacutereacutes tangents agrave laligne de squelette de lrsquoaube
β2
β3W3
W2
Pour les machines axiales (u= uent = usort ) le trianglede vitesse donne lieu agrave des eacutequations scalaires tellesque
cu= wu+u
cx= wxcx= = wx c
U
w
wu
cu
Pour les machines agrave eacutetages multiples on suppose que lavitesse axiale demeure constante Crsquoest agrave dire que119940119940120784120784119961119961= 119940119940120785120785119961119961= 119940119940119961119961= 119940119940119940119940119940119940119940119940119940119940
Cette condition jumeleacutee agrave 119932119932120784120784= 119932119932120785120785= 119932119932 megravene agrave lrsquohypothegravese drsquoune cineacutematique reacutepeacutetitive En pratique
Les triangles de vitesse agrave lrsquoentreacutee et agrave la sortie drsquouneacutetage peuvent ecirctre superposeacutes
Nous commenccedilons par la repreacutesentation de triangles de vitessespour le cas drsquoune turbineNous rappelons que par hypothegravese le rayon ne varie pas dans leplan inter-aube et que la vitesse axiale demeure constante dansle sens axialCes conditions permettent drsquoeffectuer la superposition destriangles de vitesse au bord drsquoattaque et bord de fuite du rotoravec la vitesse 119932119932 commune
c1w1
u
c2
u2
w2
➊
➋
➌ c3w3
u
Turbine
C2
W2 C3
W3
Stator Rotor
C2u C3u
W3u
α2β2 β3
α3
∆Cu= ∆Wu
U
W2u
119940119940119961119961119880119880 = 119888119888119888119888119888119888119888119888119888119888119888119888119909119909 = 119888119888119888119888119888119888119888119888119888119888
1198881198883 = 1198881198881
1205721205723 = 1205721205721
Cineacutematique reacutepeacutetitive
➁ ➂
➁U➀
Turbine
119914119914120784120784
119914119914120785120785
119934119934120785120785
119932119932119934119934120784120784
119932119932
119914119914120783120783120630120630120783120783
120630120630120784120784120631120631120784120784
120630120630120785120785
120631120631120785120785
Stator Rotor
➀ ➁ ➂
Le cas du compresseur est similaire agrave celui drsquoune turbineLe rayon ne varie pas (lrsquoeacutetude se deacuteroule dans le plan inter-aube) et la vitesse axiale demeure constante dans le sens axialComme pour la turbine nous pouvons superposer les triangles devitesse au bord drsquoattaque et bord de fuite du rotor
W2 C2
C3W3
U
C1W1
U
U
Compresseur
➊
➋
➌
α2 β1
C2 W1
C1W2
U
β2α1
W2U C1U
CX
( )2 1 = minuse u uW U C C
Lrsquoopeacuteration est vectorielle
C2U
119914119914120784120784119958119958 minus 119914119914120783120783119958119958 =- 119934119934120784120784119958119958 minus119934119934120783120783119958119958
➀
➁
➁
Compresseur
119914119914120784120784119914119914120785120785
119934119934120783120783
119932119932119914119914120783120783120630120630120783120783 120630120630120784120784
120631120631120784120784120630120630120785120785
Rotor Stator
119934119934120784120784120631120631120783120783
119932119932
➀ ➁ ➂
Trois quantiteacutes fondamentales sont utiliseacutees dans laconception des turbomachines
bull le coefficient de charge ψ
bull le coefficient de deacutebit Φ
bull le degreacute de reacuteaction R
Approche classique
Parfois ψ est consideacutereacute positif pour les turbines et neacutegatifpour les compresseurs 119882119882119890119890 = frasl119882 119888
Coefficient de charge
2Ψ = eWU
Coefficient de charge
2e
UW
=ψ 02 01 02 2
h h hU Uminus ∆
= = 02
s shU
η ∆=
=
eWWm
120578120578119904119904 =∆ℎ0∆ℎ0119904119904
Turbine
Coefficient de deacutebit
xcU
Φ =
xCU
Φ =ρRT
pρ
= xC RTRTU p
=RT m RT
U Ap
xCU
Φ =119888 = ρ119862119862119909119909119860119860
11988811987711987711987911987901199011199010119860119860
= 119872119872 120574120574 1 +120574120574 minus 1
21198721198722
minus 120574120574+1)2(120574120574minus1
Afin de geacuteneacuteraliser le triangle de vitesses drsquoune machineaxiale on effectue une mise agrave lrsquoeacutechelle en divisant chaquecocircteacute par la vitesse tangentielle U
Cette opeacuteration permettra de retrouver directement lescoefficients de charge ψ et de deacutebit Φ
Par la suite un nouveau paramegravetre R qursquoon appelle degreacutede reacuteaction sera introduit pour compleacuteter la caracteacuterisationde lrsquoaubage
Approche classique
α2
β2 α3
β3C2 W3
W2
c3u - c2u= w3u - w2u
W3uW2u
U
CX CX
Lrsquoopeacuteration est vectorielle
C3
120511120511 = frasl119934119934119940119940 119932119932120784120784 = frasl119914119914120785120785119958119958 minus 119914119914120784120784119958119958 119932119932
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
Turbine
3
2
3 3
2 2
12
12
12
12
u
u
u u
u u
c RUc RU
w c RU Uw c RU U
Ψ= minus +
Ψ= minus +
Ψ= minus = +
Ψ= minus = minus
2 22 23 2
2 22 23 2
1 12 2
2 2
c cR RU U
w wR RU U
Ψ Ψ = Φ + minus + = Φ + minus +
Ψ Ψ = Φ + + = Φ + minus
3 2
3 2
1 2 1 2
2 2
R Ratan atan
R Ratan atan
minus + Ψ minus + Ψ = = Φ Φ
Ψ + Ψ minus = = Φ Φ
α α
β β
α2β1
C2U
W1U
C1UW2U β2 α1
Compresseur
ΦC2U
C3UW3U
W2U
ψ
1
ψ
ΦC2UC3U
W3U
W2U
1
Il faut deacutefinir un troisiegraveme paramegravetre pour diffeacuterencier ces deux triangles de vitesses ayant le mecircme 120509120509 et le mecircme 120537120537
Les coefficients 120509120509 et 120537120537 ne suffisent pas pour deacutefinir de maniegravereunivoque le triangle de vitesses
Le degreacute de reacuteaction est un paramegravetre adimensionnel quiexprime la variation drsquoenthalpie subie par le fluide dans le rotorpar rapport agrave la chute totale drsquoenthalpie dans lrsquoeacutetage
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
Variation d enthalpie dans le rotorRVariation d enthalpie dans l eacutetage
=
❸❶
❷
Rotor
Stator
R=0 machine drsquoaction (Δhrotor=0 Δheacutetage=Δhstator)
R=1 machine agrave reacuteaction (Δhstator=0 Δheacutetage=Δhrotor)
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
R indique la reacutepartition du saut drsquoenthalpie (de pression) statique entre le stator et le rotor
Le degreacute de reacuteaction R deacutefini en fonction de variablesthermodynamiques (lrsquoenthalpie) peut aussi srsquoexprimer en fonctionde variables cineacutematiques preacutesentes dans le triangle de vitessesNotamment en fonction de la vitesse peacuteripheacuterique U et descomposantes des vitesses absolues et relatives
La vitesse absolue agrave lrsquoentreacutee du stator est eacutegale agrave celle agrave la sortie du rotor (c1= c3 eacutetage reacutepeacutetitif )
On neacuteglige les pertes et le transfert de chaleur dans le stator
1 2 3
ℎ3 minus ℎ1= ℎ3 + frasl11988811988832 2 minus ℎ1 minus frasl11988811988812 2 = ℎ03 minus ℎ01 = 120549120549ℎ0
ℎ03 minus ℎ01 = ℎ03 minus ℎ02 = 120549120549ℎ )0(3minus2
ℎ03 minus ℎ02 = ℎ3 minus ℎ1
119945119945120782120782120783120783 = 119945119945120782120782120784120784
frasl11988811988832 2 = frasl11988811988812 2
3 2
3 1
rotor
eacutetage
h h hRh h h
∆∆
minus= =
minus
2 20 3 2 2 3
0 3 2
2 2( )
( )
h c c h
∆∆
minus
minus
+ minus=
ℎ03 ℎ02
(ℎ01= ℎ02 1198881198881= 1198881198883)
120549120549ℎ )0(3minus2 = (ℎ3 + frasl11988811988832 2) minus (ℎ2 + frasl11988811988822 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
0(3 2) 3 2( )∆ minus = minusu uh U c c
= 1 minusfrasl11988811988832 2 minus frasl11988811988822 2120549120549ℎ )0(3minus2
119877119877 =120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
120549120549ℎ )0(3minus2
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
3 212+
= minus u uc cRU
119888119888119909119909 = 1198881198882119909119909 = 1198881198883119909119909 = 119888119888119888119888119888119888119888119888119888119888
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 11988811988821199061199061198881198882 = 1198881198881199091199092+1198881198881199061199062
119877119877 = 1 minus11988811988831199061199062 minus 11988811988821199061199062
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
Indirectement lrsquoeacutequation pour R deacutecrite en fonction de composantes cineacutematiques eacutetablit un lien entre les angles drsquoentreacutee et de sortie
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
= 1 minus1198881198883119906119906 + 1198881198882119906119906
2119880119880
2232
2 32 2wwh h+ = +
2 23 2
2 3 2 2w wh hminus = minus
2 2 2 23 3 2 2
2 3 2 2u x u xw w w wh h + +
minus = minus
Machine axiale U2 = U3
La composante axiale de la vitesse demeure constante w3x = w 2x
( eacutequation justifieacutee plus tard lors de la rothalpie)
2 23 2
2 3 2 2u uw wh hminus = minus
3 2 3 22 3
( )( )2
u u u uw w w wh h minus +minus =
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
Eacutetage reacutepeacutetitif c3 = c1
Si les pertes dans le stator sont neacutegligeacutees
c2u ndash c3u= w2u - w3u
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
3 2 3 23 2
( )( )2
u u u uw w w wh h minus +minus = minus
c2u ndash c3u= w2u - w3u
119877119877 = minus1199081199083119906119906 + 1199081199082119906119906
2119880119880
2 2 2 22 3 2 3 2 3( ) 2 ( ) 2h h u u w wminus = minus minus minus
2 2 2 2 2 23 2 3 2 3 2
2 2 2e
I II III
u u c c w wW minus minus minus= + minus
2 2 2 22 3 2 3
2 2 2 2 2 22 3 2 3 2 3
( ) ( )( ) ( ) ( )
minus minus minus=
minus + minus minus minusu u w wR
c c u u w w
3 2= =U u u
Expliqueacutee + tard
2 23 2
03 02 3 2 2ec cW h h h h minus
= minus = minus +
119877119877 =ℎ2 minus ℎ3ℎ1 minus ℎ3
=ℎ2 minus ℎ3ℎ02 minus ℎ03
Degreacute de reacuteaction
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
( )3u 2uw wR2U+
= minus
3 212
u uc cRU+
= minus
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
Puisque le degreacute de reacuteaction R peut aussi srsquoexprimer enfonction de variables cineacutematiques on peut chercher sarepreacutesentation dans le triangle adimensionnel de vitesses
Turbine
α2
β2 α3
β3
C2 W3
C3W2
C3uC2u
U
C2xC3x
W2u
c2u ndash c3u= w2u - w3u
2Ψ minusR
frasl119934119934120784120784119958119958 119932119932 = frasl1198621198622119906119906 119880119880 minus frasl119880119880 119880119880Ψ = frasl1198621198622119906119906 minus 1198621198623119906119906 119880119880
119877119877 = 1 minus (1198621198623119906119906 + 1198621198622119906119906 frasl) 2119880119880Φ = frasl1198621198622119909119909 119880119880 = frasl1198621198623119909119909 119880119880
= (119914119914120784120784119958119958 minus 119914119914120785120785119958119958 frasl) 2119880119880 minus 1 + (119914119914120784120784119958119958 + 119914119914120785120785119958119958 frasl) 2119880119880
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α2
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
Les repreacutesentations des triangles de vitesses adimensionnellessont tregraves riches en information et contiennent plusieurs formules
Par exemple pour un compresseur on deacuteduit aiseacutement que
Avec 120595120595 = 1198821198821198901198901198801198802
Φ = 119888119888119909119909119880119880
et que pour 119888119888119901119901 = 119888119888119888119888119888119888119888119888119888119888 119882119882119890119890 = 119888119888119901119901(11987911987902minus11987911987901)on trouve
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
(11987911987902minus11987911987901) =119880119880119888119888119909119909119888119888119901119901
(tan1205731205731 minus tan1205731205732)
Alors si lrsquoon veut obtenir une plus grande variation detempeacuterature 11987911987902 minus 11987911987901 indirectement un plus grand rapport(1199011199010211990111990101) on peut augmenter
bull la vitesse peacuteripheacuterique 119880119880 (la vitesse de rotation)bull la vitesse axiale 119888119888119909119909 (le deacutebit massique)bull lrsquoangle de deacuteflection de la pale(tan1205731205731 minus tan1205731205732)
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
Download Sound Effects - SoundDogs - Voice Box
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
Effets 3DEffets visqueuxDistribution de pressionTraicircneacutee PortanceEacutecoulement instationnaireInteacuteraction fuide structureTurbulence
Extrados
Intrados
L
D
Profil isoleacuteGrille drsquoaubes
Pour analyser lrsquoeacutechange drsquoeacutenergie entre le fluide et le rotor drsquouneturbomachine on utilise la variation de la quantiteacute demouvement (changement de direction etou de vitesse)
Lrsquoeacutetude classique unidimensionnelle utilise le concept de trianglede vitesse Ce modegravele est appliqueacute agrave lrsquoentreacutee et agrave la sortie ducanal inter-aube (grille drsquoaubes)
Dans celui-ci on retrouve la vitesse absolue exprimeacutee dans unrepegravere lieacute aux parties fixes (le stator) accompagneacutee de lavitesse relative lieacutee aux partie tournantes ( le rotor)
Approche classique
Lrsquoapplication du triangle de vitesse permet drsquoestimer
bull lrsquoangle drsquoincidence agrave lrsquoentreacutee du rotor
bull lrsquoangle au bord de fuite du rotor
bull la deacuteflection de lrsquoeacutecoulement dans le stator
bull le travail produit ou consommeacute par la roue
Approche classique
C vitesse absolue de lrsquoeacutecoulement
W vitesse relative de lrsquoeacutecoulement
U vitesse peacuteripheacuterique du rotor
Cu Cm Cx composante tangentielle radiale et axiale de lavitesse absolue du fluide
Wu Wm Wx composante tangentielle radiale et axiale de lavitesse relative du fluide
α lrsquoangle des vitesses absolues mesureacutees par rapport agrave la directionaxiale
β lrsquoangle des vitesses relatives mesureacutees par rapport agrave la directionaxiale
La forme des pales du rotor deacutepend des angles 120631120631(des vitesses relatives dans le repegravere du rotor)
La forme des pales du stator deacutepend des angles 120630120630(des vitesses absolues dans le repegravere fixe)
Si lrsquoon veut garder la vitesse axiale 119940119940119961119961 = 119940119940119940119940119940119940119940119940119940119940 lrsquoeacutequation deconservation de la masse 119950 = 120646120646120488120488119940119940119961119961 = 119940119940119940119940119940119940119940119940119940119940 impose 120646120646120488120488 = 119940119940119940119940119940119940119940119940119940119940
Compresseur
Cu
Cm
Cu
Plan normal
Plan meacuteridional
BABF
W2
C2
U
U
W3C3
u
Les vecteurs vitesse (vitesse absolue 119940119940 pour le stator etrelative 119960119960 pour le rotor) sont consideacutereacutes tangents agrave laligne de squelette de lrsquoaube
β2
β3W3
W2
Pour les machines axiales (u= uent = usort ) le trianglede vitesse donne lieu agrave des eacutequations scalaires tellesque
cu= wu+u
cx= wxcx= = wx c
U
w
wu
cu
Pour les machines agrave eacutetages multiples on suppose que lavitesse axiale demeure constante Crsquoest agrave dire que119940119940120784120784119961119961= 119940119940120785120785119961119961= 119940119940119961119961= 119940119940119940119940119940119940119940119940119940119940
Cette condition jumeleacutee agrave 119932119932120784120784= 119932119932120785120785= 119932119932 megravene agrave lrsquohypothegravese drsquoune cineacutematique reacutepeacutetitive En pratique
Les triangles de vitesse agrave lrsquoentreacutee et agrave la sortie drsquouneacutetage peuvent ecirctre superposeacutes
Nous commenccedilons par la repreacutesentation de triangles de vitessespour le cas drsquoune turbineNous rappelons que par hypothegravese le rayon ne varie pas dans leplan inter-aube et que la vitesse axiale demeure constante dansle sens axialCes conditions permettent drsquoeffectuer la superposition destriangles de vitesse au bord drsquoattaque et bord de fuite du rotoravec la vitesse 119932119932 commune
c1w1
u
c2
u2
w2
➊
➋
➌ c3w3
u
Turbine
C2
W2 C3
W3
Stator Rotor
C2u C3u
W3u
α2β2 β3
α3
∆Cu= ∆Wu
U
W2u
119940119940119961119961119880119880 = 119888119888119888119888119888119888119888119888119888119888119888119888119909119909 = 119888119888119888119888119888119888119888119888119888119888
1198881198883 = 1198881198881
1205721205723 = 1205721205721
Cineacutematique reacutepeacutetitive
➁ ➂
➁U➀
Turbine
119914119914120784120784
119914119914120785120785
119934119934120785120785
119932119932119934119934120784120784
119932119932
119914119914120783120783120630120630120783120783
120630120630120784120784120631120631120784120784
120630120630120785120785
120631120631120785120785
Stator Rotor
➀ ➁ ➂
Le cas du compresseur est similaire agrave celui drsquoune turbineLe rayon ne varie pas (lrsquoeacutetude se deacuteroule dans le plan inter-aube) et la vitesse axiale demeure constante dans le sens axialComme pour la turbine nous pouvons superposer les triangles devitesse au bord drsquoattaque et bord de fuite du rotor
W2 C2
C3W3
U
C1W1
U
U
Compresseur
➊
➋
➌
α2 β1
C2 W1
C1W2
U
β2α1
W2U C1U
CX
( )2 1 = minuse u uW U C C
Lrsquoopeacuteration est vectorielle
C2U
119914119914120784120784119958119958 minus 119914119914120783120783119958119958 =- 119934119934120784120784119958119958 minus119934119934120783120783119958119958
➀
➁
➁
Compresseur
119914119914120784120784119914119914120785120785
119934119934120783120783
119932119932119914119914120783120783120630120630120783120783 120630120630120784120784
120631120631120784120784120630120630120785120785
Rotor Stator
119934119934120784120784120631120631120783120783
119932119932
➀ ➁ ➂
Trois quantiteacutes fondamentales sont utiliseacutees dans laconception des turbomachines
bull le coefficient de charge ψ
bull le coefficient de deacutebit Φ
bull le degreacute de reacuteaction R
Approche classique
Parfois ψ est consideacutereacute positif pour les turbines et neacutegatifpour les compresseurs 119882119882119890119890 = frasl119882 119888
Coefficient de charge
2Ψ = eWU
Coefficient de charge
2e
UW
=ψ 02 01 02 2
h h hU Uminus ∆
= = 02
s shU
η ∆=
=
eWWm
120578120578119904119904 =∆ℎ0∆ℎ0119904119904
Turbine
Coefficient de deacutebit
xcU
Φ =
xCU
Φ =ρRT
pρ
= xC RTRTU p
=RT m RT
U Ap
xCU
Φ =119888 = ρ119862119862119909119909119860119860
11988811987711987711987911987901199011199010119860119860
= 119872119872 120574120574 1 +120574120574 minus 1
21198721198722
minus 120574120574+1)2(120574120574minus1
Afin de geacuteneacuteraliser le triangle de vitesses drsquoune machineaxiale on effectue une mise agrave lrsquoeacutechelle en divisant chaquecocircteacute par la vitesse tangentielle U
Cette opeacuteration permettra de retrouver directement lescoefficients de charge ψ et de deacutebit Φ
Par la suite un nouveau paramegravetre R qursquoon appelle degreacutede reacuteaction sera introduit pour compleacuteter la caracteacuterisationde lrsquoaubage
Approche classique
α2
β2 α3
β3C2 W3
W2
c3u - c2u= w3u - w2u
W3uW2u
U
CX CX
Lrsquoopeacuteration est vectorielle
C3
120511120511 = frasl119934119934119940119940 119932119932120784120784 = frasl119914119914120785120785119958119958 minus 119914119914120784120784119958119958 119932119932
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
Turbine
3
2
3 3
2 2
12
12
12
12
u
u
u u
u u
c RUc RU
w c RU Uw c RU U
Ψ= minus +
Ψ= minus +
Ψ= minus = +
Ψ= minus = minus
2 22 23 2
2 22 23 2
1 12 2
2 2
c cR RU U
w wR RU U
Ψ Ψ = Φ + minus + = Φ + minus +
Ψ Ψ = Φ + + = Φ + minus
3 2
3 2
1 2 1 2
2 2
R Ratan atan
R Ratan atan
minus + Ψ minus + Ψ = = Φ Φ
Ψ + Ψ minus = = Φ Φ
α α
β β
α2β1
C2U
W1U
C1UW2U β2 α1
Compresseur
ΦC2U
C3UW3U
W2U
ψ
1
ψ
ΦC2UC3U
W3U
W2U
1
Il faut deacutefinir un troisiegraveme paramegravetre pour diffeacuterencier ces deux triangles de vitesses ayant le mecircme 120509120509 et le mecircme 120537120537
Les coefficients 120509120509 et 120537120537 ne suffisent pas pour deacutefinir de maniegravereunivoque le triangle de vitesses
Le degreacute de reacuteaction est un paramegravetre adimensionnel quiexprime la variation drsquoenthalpie subie par le fluide dans le rotorpar rapport agrave la chute totale drsquoenthalpie dans lrsquoeacutetage
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
Variation d enthalpie dans le rotorRVariation d enthalpie dans l eacutetage
=
❸❶
❷
Rotor
Stator
R=0 machine drsquoaction (Δhrotor=0 Δheacutetage=Δhstator)
R=1 machine agrave reacuteaction (Δhstator=0 Δheacutetage=Δhrotor)
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
R indique la reacutepartition du saut drsquoenthalpie (de pression) statique entre le stator et le rotor
Le degreacute de reacuteaction R deacutefini en fonction de variablesthermodynamiques (lrsquoenthalpie) peut aussi srsquoexprimer en fonctionde variables cineacutematiques preacutesentes dans le triangle de vitessesNotamment en fonction de la vitesse peacuteripheacuterique U et descomposantes des vitesses absolues et relatives
La vitesse absolue agrave lrsquoentreacutee du stator est eacutegale agrave celle agrave la sortie du rotor (c1= c3 eacutetage reacutepeacutetitif )
On neacuteglige les pertes et le transfert de chaleur dans le stator
1 2 3
ℎ3 minus ℎ1= ℎ3 + frasl11988811988832 2 minus ℎ1 minus frasl11988811988812 2 = ℎ03 minus ℎ01 = 120549120549ℎ0
ℎ03 minus ℎ01 = ℎ03 minus ℎ02 = 120549120549ℎ )0(3minus2
ℎ03 minus ℎ02 = ℎ3 minus ℎ1
119945119945120782120782120783120783 = 119945119945120782120782120784120784
frasl11988811988832 2 = frasl11988811988812 2
3 2
3 1
rotor
eacutetage
h h hRh h h
∆∆
minus= =
minus
2 20 3 2 2 3
0 3 2
2 2( )
( )
h c c h
∆∆
minus
minus
+ minus=
ℎ03 ℎ02
(ℎ01= ℎ02 1198881198881= 1198881198883)
120549120549ℎ )0(3minus2 = (ℎ3 + frasl11988811988832 2) minus (ℎ2 + frasl11988811988822 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
0(3 2) 3 2( )∆ minus = minusu uh U c c
= 1 minusfrasl11988811988832 2 minus frasl11988811988822 2120549120549ℎ )0(3minus2
119877119877 =120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
120549120549ℎ )0(3minus2
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
3 212+
= minus u uc cRU
119888119888119909119909 = 1198881198882119909119909 = 1198881198883119909119909 = 119888119888119888119888119888119888119888119888119888119888
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 11988811988821199061199061198881198882 = 1198881198881199091199092+1198881198881199061199062
119877119877 = 1 minus11988811988831199061199062 minus 11988811988821199061199062
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
Indirectement lrsquoeacutequation pour R deacutecrite en fonction de composantes cineacutematiques eacutetablit un lien entre les angles drsquoentreacutee et de sortie
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
= 1 minus1198881198883119906119906 + 1198881198882119906119906
2119880119880
2232
2 32 2wwh h+ = +
2 23 2
2 3 2 2w wh hminus = minus
2 2 2 23 3 2 2
2 3 2 2u x u xw w w wh h + +
minus = minus
Machine axiale U2 = U3
La composante axiale de la vitesse demeure constante w3x = w 2x
( eacutequation justifieacutee plus tard lors de la rothalpie)
2 23 2
2 3 2 2u uw wh hminus = minus
3 2 3 22 3
( )( )2
u u u uw w w wh h minus +minus =
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
Eacutetage reacutepeacutetitif c3 = c1
Si les pertes dans le stator sont neacutegligeacutees
c2u ndash c3u= w2u - w3u
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
3 2 3 23 2
( )( )2
u u u uw w w wh h minus +minus = minus
c2u ndash c3u= w2u - w3u
119877119877 = minus1199081199083119906119906 + 1199081199082119906119906
2119880119880
2 2 2 22 3 2 3 2 3( ) 2 ( ) 2h h u u w wminus = minus minus minus
2 2 2 2 2 23 2 3 2 3 2
2 2 2e
I II III
u u c c w wW minus minus minus= + minus
2 2 2 22 3 2 3
2 2 2 2 2 22 3 2 3 2 3
( ) ( )( ) ( ) ( )
minus minus minus=
minus + minus minus minusu u w wR
c c u u w w
3 2= =U u u
Expliqueacutee + tard
2 23 2
03 02 3 2 2ec cW h h h h minus
= minus = minus +
119877119877 =ℎ2 minus ℎ3ℎ1 minus ℎ3
=ℎ2 minus ℎ3ℎ02 minus ℎ03
Degreacute de reacuteaction
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
( )3u 2uw wR2U+
= minus
3 212
u uc cRU+
= minus
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
Puisque le degreacute de reacuteaction R peut aussi srsquoexprimer enfonction de variables cineacutematiques on peut chercher sarepreacutesentation dans le triangle adimensionnel de vitesses
Turbine
α2
β2 α3
β3
C2 W3
C3W2
C3uC2u
U
C2xC3x
W2u
c2u ndash c3u= w2u - w3u
2Ψ minusR
frasl119934119934120784120784119958119958 119932119932 = frasl1198621198622119906119906 119880119880 minus frasl119880119880 119880119880Ψ = frasl1198621198622119906119906 minus 1198621198623119906119906 119880119880
119877119877 = 1 minus (1198621198623119906119906 + 1198621198622119906119906 frasl) 2119880119880Φ = frasl1198621198622119909119909 119880119880 = frasl1198621198623119909119909 119880119880
= (119914119914120784120784119958119958 minus 119914119914120785120785119958119958 frasl) 2119880119880 minus 1 + (119914119914120784120784119958119958 + 119914119914120785120785119958119958 frasl) 2119880119880
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α2
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
Les repreacutesentations des triangles de vitesses adimensionnellessont tregraves riches en information et contiennent plusieurs formules
Par exemple pour un compresseur on deacuteduit aiseacutement que
Avec 120595120595 = 1198821198821198901198901198801198802
Φ = 119888119888119909119909119880119880
et que pour 119888119888119901119901 = 119888119888119888119888119888119888119888119888119888119888 119882119882119890119890 = 119888119888119901119901(11987911987902minus11987911987901)on trouve
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
(11987911987902minus11987911987901) =119880119880119888119888119909119909119888119888119901119901
(tan1205731205731 minus tan1205731205732)
Alors si lrsquoon veut obtenir une plus grande variation detempeacuterature 11987911987902 minus 11987911987901 indirectement un plus grand rapport(1199011199010211990111990101) on peut augmenter
bull la vitesse peacuteripheacuterique 119880119880 (la vitesse de rotation)bull la vitesse axiale 119888119888119909119909 (le deacutebit massique)bull lrsquoangle de deacuteflection de la pale(tan1205731205731 minus tan1205731205732)
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
Extrados
Intrados
L
D
Profil isoleacuteGrille drsquoaubes
Pour analyser lrsquoeacutechange drsquoeacutenergie entre le fluide et le rotor drsquouneturbomachine on utilise la variation de la quantiteacute demouvement (changement de direction etou de vitesse)
Lrsquoeacutetude classique unidimensionnelle utilise le concept de trianglede vitesse Ce modegravele est appliqueacute agrave lrsquoentreacutee et agrave la sortie ducanal inter-aube (grille drsquoaubes)
Dans celui-ci on retrouve la vitesse absolue exprimeacutee dans unrepegravere lieacute aux parties fixes (le stator) accompagneacutee de lavitesse relative lieacutee aux partie tournantes ( le rotor)
Approche classique
Lrsquoapplication du triangle de vitesse permet drsquoestimer
bull lrsquoangle drsquoincidence agrave lrsquoentreacutee du rotor
bull lrsquoangle au bord de fuite du rotor
bull la deacuteflection de lrsquoeacutecoulement dans le stator
bull le travail produit ou consommeacute par la roue
Approche classique
C vitesse absolue de lrsquoeacutecoulement
W vitesse relative de lrsquoeacutecoulement
U vitesse peacuteripheacuterique du rotor
Cu Cm Cx composante tangentielle radiale et axiale de lavitesse absolue du fluide
Wu Wm Wx composante tangentielle radiale et axiale de lavitesse relative du fluide
α lrsquoangle des vitesses absolues mesureacutees par rapport agrave la directionaxiale
β lrsquoangle des vitesses relatives mesureacutees par rapport agrave la directionaxiale
La forme des pales du rotor deacutepend des angles 120631120631(des vitesses relatives dans le repegravere du rotor)
La forme des pales du stator deacutepend des angles 120630120630(des vitesses absolues dans le repegravere fixe)
Si lrsquoon veut garder la vitesse axiale 119940119940119961119961 = 119940119940119940119940119940119940119940119940119940119940 lrsquoeacutequation deconservation de la masse 119950 = 120646120646120488120488119940119940119961119961 = 119940119940119940119940119940119940119940119940119940119940 impose 120646120646120488120488 = 119940119940119940119940119940119940119940119940119940119940
Compresseur
Cu
Cm
Cu
Plan normal
Plan meacuteridional
BABF
W2
C2
U
U
W3C3
u
Les vecteurs vitesse (vitesse absolue 119940119940 pour le stator etrelative 119960119960 pour le rotor) sont consideacutereacutes tangents agrave laligne de squelette de lrsquoaube
β2
β3W3
W2
Pour les machines axiales (u= uent = usort ) le trianglede vitesse donne lieu agrave des eacutequations scalaires tellesque
cu= wu+u
cx= wxcx= = wx c
U
w
wu
cu
Pour les machines agrave eacutetages multiples on suppose que lavitesse axiale demeure constante Crsquoest agrave dire que119940119940120784120784119961119961= 119940119940120785120785119961119961= 119940119940119961119961= 119940119940119940119940119940119940119940119940119940119940
Cette condition jumeleacutee agrave 119932119932120784120784= 119932119932120785120785= 119932119932 megravene agrave lrsquohypothegravese drsquoune cineacutematique reacutepeacutetitive En pratique
Les triangles de vitesse agrave lrsquoentreacutee et agrave la sortie drsquouneacutetage peuvent ecirctre superposeacutes
Nous commenccedilons par la repreacutesentation de triangles de vitessespour le cas drsquoune turbineNous rappelons que par hypothegravese le rayon ne varie pas dans leplan inter-aube et que la vitesse axiale demeure constante dansle sens axialCes conditions permettent drsquoeffectuer la superposition destriangles de vitesse au bord drsquoattaque et bord de fuite du rotoravec la vitesse 119932119932 commune
c1w1
u
c2
u2
w2
➊
➋
➌ c3w3
u
Turbine
C2
W2 C3
W3
Stator Rotor
C2u C3u
W3u
α2β2 β3
α3
∆Cu= ∆Wu
U
W2u
119940119940119961119961119880119880 = 119888119888119888119888119888119888119888119888119888119888119888119888119909119909 = 119888119888119888119888119888119888119888119888119888119888
1198881198883 = 1198881198881
1205721205723 = 1205721205721
Cineacutematique reacutepeacutetitive
➁ ➂
➁U➀
Turbine
119914119914120784120784
119914119914120785120785
119934119934120785120785
119932119932119934119934120784120784
119932119932
119914119914120783120783120630120630120783120783
120630120630120784120784120631120631120784120784
120630120630120785120785
120631120631120785120785
Stator Rotor
➀ ➁ ➂
Le cas du compresseur est similaire agrave celui drsquoune turbineLe rayon ne varie pas (lrsquoeacutetude se deacuteroule dans le plan inter-aube) et la vitesse axiale demeure constante dans le sens axialComme pour la turbine nous pouvons superposer les triangles devitesse au bord drsquoattaque et bord de fuite du rotor
W2 C2
C3W3
U
C1W1
U
U
Compresseur
➊
➋
➌
α2 β1
C2 W1
C1W2
U
β2α1
W2U C1U
CX
( )2 1 = minuse u uW U C C
Lrsquoopeacuteration est vectorielle
C2U
119914119914120784120784119958119958 minus 119914119914120783120783119958119958 =- 119934119934120784120784119958119958 minus119934119934120783120783119958119958
➀
➁
➁
Compresseur
119914119914120784120784119914119914120785120785
119934119934120783120783
119932119932119914119914120783120783120630120630120783120783 120630120630120784120784
120631120631120784120784120630120630120785120785
Rotor Stator
119934119934120784120784120631120631120783120783
119932119932
➀ ➁ ➂
Trois quantiteacutes fondamentales sont utiliseacutees dans laconception des turbomachines
bull le coefficient de charge ψ
bull le coefficient de deacutebit Φ
bull le degreacute de reacuteaction R
Approche classique
Parfois ψ est consideacutereacute positif pour les turbines et neacutegatifpour les compresseurs 119882119882119890119890 = frasl119882 119888
Coefficient de charge
2Ψ = eWU
Coefficient de charge
2e
UW
=ψ 02 01 02 2
h h hU Uminus ∆
= = 02
s shU
η ∆=
=
eWWm
120578120578119904119904 =∆ℎ0∆ℎ0119904119904
Turbine
Coefficient de deacutebit
xcU
Φ =
xCU
Φ =ρRT
pρ
= xC RTRTU p
=RT m RT
U Ap
xCU
Φ =119888 = ρ119862119862119909119909119860119860
11988811987711987711987911987901199011199010119860119860
= 119872119872 120574120574 1 +120574120574 minus 1
21198721198722
minus 120574120574+1)2(120574120574minus1
Afin de geacuteneacuteraliser le triangle de vitesses drsquoune machineaxiale on effectue une mise agrave lrsquoeacutechelle en divisant chaquecocircteacute par la vitesse tangentielle U
Cette opeacuteration permettra de retrouver directement lescoefficients de charge ψ et de deacutebit Φ
Par la suite un nouveau paramegravetre R qursquoon appelle degreacutede reacuteaction sera introduit pour compleacuteter la caracteacuterisationde lrsquoaubage
Approche classique
α2
β2 α3
β3C2 W3
W2
c3u - c2u= w3u - w2u
W3uW2u
U
CX CX
Lrsquoopeacuteration est vectorielle
C3
120511120511 = frasl119934119934119940119940 119932119932120784120784 = frasl119914119914120785120785119958119958 minus 119914119914120784120784119958119958 119932119932
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
Turbine
3
2
3 3
2 2
12
12
12
12
u
u
u u
u u
c RUc RU
w c RU Uw c RU U
Ψ= minus +
Ψ= minus +
Ψ= minus = +
Ψ= minus = minus
2 22 23 2
2 22 23 2
1 12 2
2 2
c cR RU U
w wR RU U
Ψ Ψ = Φ + minus + = Φ + minus +
Ψ Ψ = Φ + + = Φ + minus
3 2
3 2
1 2 1 2
2 2
R Ratan atan
R Ratan atan
minus + Ψ minus + Ψ = = Φ Φ
Ψ + Ψ minus = = Φ Φ
α α
β β
α2β1
C2U
W1U
C1UW2U β2 α1
Compresseur
ΦC2U
C3UW3U
W2U
ψ
1
ψ
ΦC2UC3U
W3U
W2U
1
Il faut deacutefinir un troisiegraveme paramegravetre pour diffeacuterencier ces deux triangles de vitesses ayant le mecircme 120509120509 et le mecircme 120537120537
Les coefficients 120509120509 et 120537120537 ne suffisent pas pour deacutefinir de maniegravereunivoque le triangle de vitesses
Le degreacute de reacuteaction est un paramegravetre adimensionnel quiexprime la variation drsquoenthalpie subie par le fluide dans le rotorpar rapport agrave la chute totale drsquoenthalpie dans lrsquoeacutetage
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
Variation d enthalpie dans le rotorRVariation d enthalpie dans l eacutetage
=
❸❶
❷
Rotor
Stator
R=0 machine drsquoaction (Δhrotor=0 Δheacutetage=Δhstator)
R=1 machine agrave reacuteaction (Δhstator=0 Δheacutetage=Δhrotor)
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
R indique la reacutepartition du saut drsquoenthalpie (de pression) statique entre le stator et le rotor
Le degreacute de reacuteaction R deacutefini en fonction de variablesthermodynamiques (lrsquoenthalpie) peut aussi srsquoexprimer en fonctionde variables cineacutematiques preacutesentes dans le triangle de vitessesNotamment en fonction de la vitesse peacuteripheacuterique U et descomposantes des vitesses absolues et relatives
La vitesse absolue agrave lrsquoentreacutee du stator est eacutegale agrave celle agrave la sortie du rotor (c1= c3 eacutetage reacutepeacutetitif )
On neacuteglige les pertes et le transfert de chaleur dans le stator
1 2 3
ℎ3 minus ℎ1= ℎ3 + frasl11988811988832 2 minus ℎ1 minus frasl11988811988812 2 = ℎ03 minus ℎ01 = 120549120549ℎ0
ℎ03 minus ℎ01 = ℎ03 minus ℎ02 = 120549120549ℎ )0(3minus2
ℎ03 minus ℎ02 = ℎ3 minus ℎ1
119945119945120782120782120783120783 = 119945119945120782120782120784120784
frasl11988811988832 2 = frasl11988811988812 2
3 2
3 1
rotor
eacutetage
h h hRh h h
∆∆
minus= =
minus
2 20 3 2 2 3
0 3 2
2 2( )
( )
h c c h
∆∆
minus
minus
+ minus=
ℎ03 ℎ02
(ℎ01= ℎ02 1198881198881= 1198881198883)
120549120549ℎ )0(3minus2 = (ℎ3 + frasl11988811988832 2) minus (ℎ2 + frasl11988811988822 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
0(3 2) 3 2( )∆ minus = minusu uh U c c
= 1 minusfrasl11988811988832 2 minus frasl11988811988822 2120549120549ℎ )0(3minus2
119877119877 =120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
120549120549ℎ )0(3minus2
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
3 212+
= minus u uc cRU
119888119888119909119909 = 1198881198882119909119909 = 1198881198883119909119909 = 119888119888119888119888119888119888119888119888119888119888
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 11988811988821199061199061198881198882 = 1198881198881199091199092+1198881198881199061199062
119877119877 = 1 minus11988811988831199061199062 minus 11988811988821199061199062
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
Indirectement lrsquoeacutequation pour R deacutecrite en fonction de composantes cineacutematiques eacutetablit un lien entre les angles drsquoentreacutee et de sortie
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
= 1 minus1198881198883119906119906 + 1198881198882119906119906
2119880119880
2232
2 32 2wwh h+ = +
2 23 2
2 3 2 2w wh hminus = minus
2 2 2 23 3 2 2
2 3 2 2u x u xw w w wh h + +
minus = minus
Machine axiale U2 = U3
La composante axiale de la vitesse demeure constante w3x = w 2x
( eacutequation justifieacutee plus tard lors de la rothalpie)
2 23 2
2 3 2 2u uw wh hminus = minus
3 2 3 22 3
( )( )2
u u u uw w w wh h minus +minus =
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
Eacutetage reacutepeacutetitif c3 = c1
Si les pertes dans le stator sont neacutegligeacutees
c2u ndash c3u= w2u - w3u
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
3 2 3 23 2
( )( )2
u u u uw w w wh h minus +minus = minus
c2u ndash c3u= w2u - w3u
119877119877 = minus1199081199083119906119906 + 1199081199082119906119906
2119880119880
2 2 2 22 3 2 3 2 3( ) 2 ( ) 2h h u u w wminus = minus minus minus
2 2 2 2 2 23 2 3 2 3 2
2 2 2e
I II III
u u c c w wW minus minus minus= + minus
2 2 2 22 3 2 3
2 2 2 2 2 22 3 2 3 2 3
( ) ( )( ) ( ) ( )
minus minus minus=
minus + minus minus minusu u w wR
c c u u w w
3 2= =U u u
Expliqueacutee + tard
2 23 2
03 02 3 2 2ec cW h h h h minus
= minus = minus +
119877119877 =ℎ2 minus ℎ3ℎ1 minus ℎ3
=ℎ2 minus ℎ3ℎ02 minus ℎ03
Degreacute de reacuteaction
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
( )3u 2uw wR2U+
= minus
3 212
u uc cRU+
= minus
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
Puisque le degreacute de reacuteaction R peut aussi srsquoexprimer enfonction de variables cineacutematiques on peut chercher sarepreacutesentation dans le triangle adimensionnel de vitesses
Turbine
α2
β2 α3
β3
C2 W3
C3W2
C3uC2u
U
C2xC3x
W2u
c2u ndash c3u= w2u - w3u
2Ψ minusR
frasl119934119934120784120784119958119958 119932119932 = frasl1198621198622119906119906 119880119880 minus frasl119880119880 119880119880Ψ = frasl1198621198622119906119906 minus 1198621198623119906119906 119880119880
119877119877 = 1 minus (1198621198623119906119906 + 1198621198622119906119906 frasl) 2119880119880Φ = frasl1198621198622119909119909 119880119880 = frasl1198621198623119909119909 119880119880
= (119914119914120784120784119958119958 minus 119914119914120785120785119958119958 frasl) 2119880119880 minus 1 + (119914119914120784120784119958119958 + 119914119914120785120785119958119958 frasl) 2119880119880
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α2
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
Les repreacutesentations des triangles de vitesses adimensionnellessont tregraves riches en information et contiennent plusieurs formules
Par exemple pour un compresseur on deacuteduit aiseacutement que
Avec 120595120595 = 1198821198821198901198901198801198802
Φ = 119888119888119909119909119880119880
et que pour 119888119888119901119901 = 119888119888119888119888119888119888119888119888119888119888 119882119882119890119890 = 119888119888119901119901(11987911987902minus11987911987901)on trouve
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
(11987911987902minus11987911987901) =119880119880119888119888119909119909119888119888119901119901
(tan1205731205731 minus tan1205731205732)
Alors si lrsquoon veut obtenir une plus grande variation detempeacuterature 11987911987902 minus 11987911987901 indirectement un plus grand rapport(1199011199010211990111990101) on peut augmenter
bull la vitesse peacuteripheacuterique 119880119880 (la vitesse de rotation)bull la vitesse axiale 119888119888119909119909 (le deacutebit massique)bull lrsquoangle de deacuteflection de la pale(tan1205731205731 minus tan1205731205732)
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
Profil isoleacuteGrille drsquoaubes
Pour analyser lrsquoeacutechange drsquoeacutenergie entre le fluide et le rotor drsquouneturbomachine on utilise la variation de la quantiteacute demouvement (changement de direction etou de vitesse)
Lrsquoeacutetude classique unidimensionnelle utilise le concept de trianglede vitesse Ce modegravele est appliqueacute agrave lrsquoentreacutee et agrave la sortie ducanal inter-aube (grille drsquoaubes)
Dans celui-ci on retrouve la vitesse absolue exprimeacutee dans unrepegravere lieacute aux parties fixes (le stator) accompagneacutee de lavitesse relative lieacutee aux partie tournantes ( le rotor)
Approche classique
Lrsquoapplication du triangle de vitesse permet drsquoestimer
bull lrsquoangle drsquoincidence agrave lrsquoentreacutee du rotor
bull lrsquoangle au bord de fuite du rotor
bull la deacuteflection de lrsquoeacutecoulement dans le stator
bull le travail produit ou consommeacute par la roue
Approche classique
C vitesse absolue de lrsquoeacutecoulement
W vitesse relative de lrsquoeacutecoulement
U vitesse peacuteripheacuterique du rotor
Cu Cm Cx composante tangentielle radiale et axiale de lavitesse absolue du fluide
Wu Wm Wx composante tangentielle radiale et axiale de lavitesse relative du fluide
α lrsquoangle des vitesses absolues mesureacutees par rapport agrave la directionaxiale
β lrsquoangle des vitesses relatives mesureacutees par rapport agrave la directionaxiale
La forme des pales du rotor deacutepend des angles 120631120631(des vitesses relatives dans le repegravere du rotor)
La forme des pales du stator deacutepend des angles 120630120630(des vitesses absolues dans le repegravere fixe)
Si lrsquoon veut garder la vitesse axiale 119940119940119961119961 = 119940119940119940119940119940119940119940119940119940119940 lrsquoeacutequation deconservation de la masse 119950 = 120646120646120488120488119940119940119961119961 = 119940119940119940119940119940119940119940119940119940119940 impose 120646120646120488120488 = 119940119940119940119940119940119940119940119940119940119940
Compresseur
Cu
Cm
Cu
Plan normal
Plan meacuteridional
BABF
W2
C2
U
U
W3C3
u
Les vecteurs vitesse (vitesse absolue 119940119940 pour le stator etrelative 119960119960 pour le rotor) sont consideacutereacutes tangents agrave laligne de squelette de lrsquoaube
β2
β3W3
W2
Pour les machines axiales (u= uent = usort ) le trianglede vitesse donne lieu agrave des eacutequations scalaires tellesque
cu= wu+u
cx= wxcx= = wx c
U
w
wu
cu
Pour les machines agrave eacutetages multiples on suppose que lavitesse axiale demeure constante Crsquoest agrave dire que119940119940120784120784119961119961= 119940119940120785120785119961119961= 119940119940119961119961= 119940119940119940119940119940119940119940119940119940119940
Cette condition jumeleacutee agrave 119932119932120784120784= 119932119932120785120785= 119932119932 megravene agrave lrsquohypothegravese drsquoune cineacutematique reacutepeacutetitive En pratique
Les triangles de vitesse agrave lrsquoentreacutee et agrave la sortie drsquouneacutetage peuvent ecirctre superposeacutes
Nous commenccedilons par la repreacutesentation de triangles de vitessespour le cas drsquoune turbineNous rappelons que par hypothegravese le rayon ne varie pas dans leplan inter-aube et que la vitesse axiale demeure constante dansle sens axialCes conditions permettent drsquoeffectuer la superposition destriangles de vitesse au bord drsquoattaque et bord de fuite du rotoravec la vitesse 119932119932 commune
c1w1
u
c2
u2
w2
➊
➋
➌ c3w3
u
Turbine
C2
W2 C3
W3
Stator Rotor
C2u C3u
W3u
α2β2 β3
α3
∆Cu= ∆Wu
U
W2u
119940119940119961119961119880119880 = 119888119888119888119888119888119888119888119888119888119888119888119888119909119909 = 119888119888119888119888119888119888119888119888119888119888
1198881198883 = 1198881198881
1205721205723 = 1205721205721
Cineacutematique reacutepeacutetitive
➁ ➂
➁U➀
Turbine
119914119914120784120784
119914119914120785120785
119934119934120785120785
119932119932119934119934120784120784
119932119932
119914119914120783120783120630120630120783120783
120630120630120784120784120631120631120784120784
120630120630120785120785
120631120631120785120785
Stator Rotor
➀ ➁ ➂
Le cas du compresseur est similaire agrave celui drsquoune turbineLe rayon ne varie pas (lrsquoeacutetude se deacuteroule dans le plan inter-aube) et la vitesse axiale demeure constante dans le sens axialComme pour la turbine nous pouvons superposer les triangles devitesse au bord drsquoattaque et bord de fuite du rotor
W2 C2
C3W3
U
C1W1
U
U
Compresseur
➊
➋
➌
α2 β1
C2 W1
C1W2
U
β2α1
W2U C1U
CX
( )2 1 = minuse u uW U C C
Lrsquoopeacuteration est vectorielle
C2U
119914119914120784120784119958119958 minus 119914119914120783120783119958119958 =- 119934119934120784120784119958119958 minus119934119934120783120783119958119958
➀
➁
➁
Compresseur
119914119914120784120784119914119914120785120785
119934119934120783120783
119932119932119914119914120783120783120630120630120783120783 120630120630120784120784
120631120631120784120784120630120630120785120785
Rotor Stator
119934119934120784120784120631120631120783120783
119932119932
➀ ➁ ➂
Trois quantiteacutes fondamentales sont utiliseacutees dans laconception des turbomachines
bull le coefficient de charge ψ
bull le coefficient de deacutebit Φ
bull le degreacute de reacuteaction R
Approche classique
Parfois ψ est consideacutereacute positif pour les turbines et neacutegatifpour les compresseurs 119882119882119890119890 = frasl119882 119888
Coefficient de charge
2Ψ = eWU
Coefficient de charge
2e
UW
=ψ 02 01 02 2
h h hU Uminus ∆
= = 02
s shU
η ∆=
=
eWWm
120578120578119904119904 =∆ℎ0∆ℎ0119904119904
Turbine
Coefficient de deacutebit
xcU
Φ =
xCU
Φ =ρRT
pρ
= xC RTRTU p
=RT m RT
U Ap
xCU
Φ =119888 = ρ119862119862119909119909119860119860
11988811987711987711987911987901199011199010119860119860
= 119872119872 120574120574 1 +120574120574 minus 1
21198721198722
minus 120574120574+1)2(120574120574minus1
Afin de geacuteneacuteraliser le triangle de vitesses drsquoune machineaxiale on effectue une mise agrave lrsquoeacutechelle en divisant chaquecocircteacute par la vitesse tangentielle U
Cette opeacuteration permettra de retrouver directement lescoefficients de charge ψ et de deacutebit Φ
Par la suite un nouveau paramegravetre R qursquoon appelle degreacutede reacuteaction sera introduit pour compleacuteter la caracteacuterisationde lrsquoaubage
Approche classique
α2
β2 α3
β3C2 W3
W2
c3u - c2u= w3u - w2u
W3uW2u
U
CX CX
Lrsquoopeacuteration est vectorielle
C3
120511120511 = frasl119934119934119940119940 119932119932120784120784 = frasl119914119914120785120785119958119958 minus 119914119914120784120784119958119958 119932119932
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
Turbine
3
2
3 3
2 2
12
12
12
12
u
u
u u
u u
c RUc RU
w c RU Uw c RU U
Ψ= minus +
Ψ= minus +
Ψ= minus = +
Ψ= minus = minus
2 22 23 2
2 22 23 2
1 12 2
2 2
c cR RU U
w wR RU U
Ψ Ψ = Φ + minus + = Φ + minus +
Ψ Ψ = Φ + + = Φ + minus
3 2
3 2
1 2 1 2
2 2
R Ratan atan
R Ratan atan
minus + Ψ minus + Ψ = = Φ Φ
Ψ + Ψ minus = = Φ Φ
α α
β β
α2β1
C2U
W1U
C1UW2U β2 α1
Compresseur
ΦC2U
C3UW3U
W2U
ψ
1
ψ
ΦC2UC3U
W3U
W2U
1
Il faut deacutefinir un troisiegraveme paramegravetre pour diffeacuterencier ces deux triangles de vitesses ayant le mecircme 120509120509 et le mecircme 120537120537
Les coefficients 120509120509 et 120537120537 ne suffisent pas pour deacutefinir de maniegravereunivoque le triangle de vitesses
Le degreacute de reacuteaction est un paramegravetre adimensionnel quiexprime la variation drsquoenthalpie subie par le fluide dans le rotorpar rapport agrave la chute totale drsquoenthalpie dans lrsquoeacutetage
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
Variation d enthalpie dans le rotorRVariation d enthalpie dans l eacutetage
=
❸❶
❷
Rotor
Stator
R=0 machine drsquoaction (Δhrotor=0 Δheacutetage=Δhstator)
R=1 machine agrave reacuteaction (Δhstator=0 Δheacutetage=Δhrotor)
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
R indique la reacutepartition du saut drsquoenthalpie (de pression) statique entre le stator et le rotor
Le degreacute de reacuteaction R deacutefini en fonction de variablesthermodynamiques (lrsquoenthalpie) peut aussi srsquoexprimer en fonctionde variables cineacutematiques preacutesentes dans le triangle de vitessesNotamment en fonction de la vitesse peacuteripheacuterique U et descomposantes des vitesses absolues et relatives
La vitesse absolue agrave lrsquoentreacutee du stator est eacutegale agrave celle agrave la sortie du rotor (c1= c3 eacutetage reacutepeacutetitif )
On neacuteglige les pertes et le transfert de chaleur dans le stator
1 2 3
ℎ3 minus ℎ1= ℎ3 + frasl11988811988832 2 minus ℎ1 minus frasl11988811988812 2 = ℎ03 minus ℎ01 = 120549120549ℎ0
ℎ03 minus ℎ01 = ℎ03 minus ℎ02 = 120549120549ℎ )0(3minus2
ℎ03 minus ℎ02 = ℎ3 minus ℎ1
119945119945120782120782120783120783 = 119945119945120782120782120784120784
frasl11988811988832 2 = frasl11988811988812 2
3 2
3 1
rotor
eacutetage
h h hRh h h
∆∆
minus= =
minus
2 20 3 2 2 3
0 3 2
2 2( )
( )
h c c h
∆∆
minus
minus
+ minus=
ℎ03 ℎ02
(ℎ01= ℎ02 1198881198881= 1198881198883)
120549120549ℎ )0(3minus2 = (ℎ3 + frasl11988811988832 2) minus (ℎ2 + frasl11988811988822 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
0(3 2) 3 2( )∆ minus = minusu uh U c c
= 1 minusfrasl11988811988832 2 minus frasl11988811988822 2120549120549ℎ )0(3minus2
119877119877 =120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
120549120549ℎ )0(3minus2
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
3 212+
= minus u uc cRU
119888119888119909119909 = 1198881198882119909119909 = 1198881198883119909119909 = 119888119888119888119888119888119888119888119888119888119888
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 11988811988821199061199061198881198882 = 1198881198881199091199092+1198881198881199061199062
119877119877 = 1 minus11988811988831199061199062 minus 11988811988821199061199062
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
Indirectement lrsquoeacutequation pour R deacutecrite en fonction de composantes cineacutematiques eacutetablit un lien entre les angles drsquoentreacutee et de sortie
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
= 1 minus1198881198883119906119906 + 1198881198882119906119906
2119880119880
2232
2 32 2wwh h+ = +
2 23 2
2 3 2 2w wh hminus = minus
2 2 2 23 3 2 2
2 3 2 2u x u xw w w wh h + +
minus = minus
Machine axiale U2 = U3
La composante axiale de la vitesse demeure constante w3x = w 2x
( eacutequation justifieacutee plus tard lors de la rothalpie)
2 23 2
2 3 2 2u uw wh hminus = minus
3 2 3 22 3
( )( )2
u u u uw w w wh h minus +minus =
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
Eacutetage reacutepeacutetitif c3 = c1
Si les pertes dans le stator sont neacutegligeacutees
c2u ndash c3u= w2u - w3u
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
3 2 3 23 2
( )( )2
u u u uw w w wh h minus +minus = minus
c2u ndash c3u= w2u - w3u
119877119877 = minus1199081199083119906119906 + 1199081199082119906119906
2119880119880
2 2 2 22 3 2 3 2 3( ) 2 ( ) 2h h u u w wminus = minus minus minus
2 2 2 2 2 23 2 3 2 3 2
2 2 2e
I II III
u u c c w wW minus minus minus= + minus
2 2 2 22 3 2 3
2 2 2 2 2 22 3 2 3 2 3
( ) ( )( ) ( ) ( )
minus minus minus=
minus + minus minus minusu u w wR
c c u u w w
3 2= =U u u
Expliqueacutee + tard
2 23 2
03 02 3 2 2ec cW h h h h minus
= minus = minus +
119877119877 =ℎ2 minus ℎ3ℎ1 minus ℎ3
=ℎ2 minus ℎ3ℎ02 minus ℎ03
Degreacute de reacuteaction
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
( )3u 2uw wR2U+
= minus
3 212
u uc cRU+
= minus
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
Puisque le degreacute de reacuteaction R peut aussi srsquoexprimer enfonction de variables cineacutematiques on peut chercher sarepreacutesentation dans le triangle adimensionnel de vitesses
Turbine
α2
β2 α3
β3
C2 W3
C3W2
C3uC2u
U
C2xC3x
W2u
c2u ndash c3u= w2u - w3u
2Ψ minusR
frasl119934119934120784120784119958119958 119932119932 = frasl1198621198622119906119906 119880119880 minus frasl119880119880 119880119880Ψ = frasl1198621198622119906119906 minus 1198621198623119906119906 119880119880
119877119877 = 1 minus (1198621198623119906119906 + 1198621198622119906119906 frasl) 2119880119880Φ = frasl1198621198622119909119909 119880119880 = frasl1198621198623119909119909 119880119880
= (119914119914120784120784119958119958 minus 119914119914120785120785119958119958 frasl) 2119880119880 minus 1 + (119914119914120784120784119958119958 + 119914119914120785120785119958119958 frasl) 2119880119880
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α2
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
Les repreacutesentations des triangles de vitesses adimensionnellessont tregraves riches en information et contiennent plusieurs formules
Par exemple pour un compresseur on deacuteduit aiseacutement que
Avec 120595120595 = 1198821198821198901198901198801198802
Φ = 119888119888119909119909119880119880
et que pour 119888119888119901119901 = 119888119888119888119888119888119888119888119888119888119888 119882119882119890119890 = 119888119888119901119901(11987911987902minus11987911987901)on trouve
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
(11987911987902minus11987911987901) =119880119880119888119888119909119909119888119888119901119901
(tan1205731205731 minus tan1205731205732)
Alors si lrsquoon veut obtenir une plus grande variation detempeacuterature 11987911987902 minus 11987911987901 indirectement un plus grand rapport(1199011199010211990111990101) on peut augmenter
bull la vitesse peacuteripheacuterique 119880119880 (la vitesse de rotation)bull la vitesse axiale 119888119888119909119909 (le deacutebit massique)bull lrsquoangle de deacuteflection de la pale(tan1205731205731 minus tan1205731205732)
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
Pour analyser lrsquoeacutechange drsquoeacutenergie entre le fluide et le rotor drsquouneturbomachine on utilise la variation de la quantiteacute demouvement (changement de direction etou de vitesse)
Lrsquoeacutetude classique unidimensionnelle utilise le concept de trianglede vitesse Ce modegravele est appliqueacute agrave lrsquoentreacutee et agrave la sortie ducanal inter-aube (grille drsquoaubes)
Dans celui-ci on retrouve la vitesse absolue exprimeacutee dans unrepegravere lieacute aux parties fixes (le stator) accompagneacutee de lavitesse relative lieacutee aux partie tournantes ( le rotor)
Approche classique
Lrsquoapplication du triangle de vitesse permet drsquoestimer
bull lrsquoangle drsquoincidence agrave lrsquoentreacutee du rotor
bull lrsquoangle au bord de fuite du rotor
bull la deacuteflection de lrsquoeacutecoulement dans le stator
bull le travail produit ou consommeacute par la roue
Approche classique
C vitesse absolue de lrsquoeacutecoulement
W vitesse relative de lrsquoeacutecoulement
U vitesse peacuteripheacuterique du rotor
Cu Cm Cx composante tangentielle radiale et axiale de lavitesse absolue du fluide
Wu Wm Wx composante tangentielle radiale et axiale de lavitesse relative du fluide
α lrsquoangle des vitesses absolues mesureacutees par rapport agrave la directionaxiale
β lrsquoangle des vitesses relatives mesureacutees par rapport agrave la directionaxiale
La forme des pales du rotor deacutepend des angles 120631120631(des vitesses relatives dans le repegravere du rotor)
La forme des pales du stator deacutepend des angles 120630120630(des vitesses absolues dans le repegravere fixe)
Si lrsquoon veut garder la vitesse axiale 119940119940119961119961 = 119940119940119940119940119940119940119940119940119940119940 lrsquoeacutequation deconservation de la masse 119950 = 120646120646120488120488119940119940119961119961 = 119940119940119940119940119940119940119940119940119940119940 impose 120646120646120488120488 = 119940119940119940119940119940119940119940119940119940119940
Compresseur
Cu
Cm
Cu
Plan normal
Plan meacuteridional
BABF
W2
C2
U
U
W3C3
u
Les vecteurs vitesse (vitesse absolue 119940119940 pour le stator etrelative 119960119960 pour le rotor) sont consideacutereacutes tangents agrave laligne de squelette de lrsquoaube
β2
β3W3
W2
Pour les machines axiales (u= uent = usort ) le trianglede vitesse donne lieu agrave des eacutequations scalaires tellesque
cu= wu+u
cx= wxcx= = wx c
U
w
wu
cu
Pour les machines agrave eacutetages multiples on suppose que lavitesse axiale demeure constante Crsquoest agrave dire que119940119940120784120784119961119961= 119940119940120785120785119961119961= 119940119940119961119961= 119940119940119940119940119940119940119940119940119940119940
Cette condition jumeleacutee agrave 119932119932120784120784= 119932119932120785120785= 119932119932 megravene agrave lrsquohypothegravese drsquoune cineacutematique reacutepeacutetitive En pratique
Les triangles de vitesse agrave lrsquoentreacutee et agrave la sortie drsquouneacutetage peuvent ecirctre superposeacutes
Nous commenccedilons par la repreacutesentation de triangles de vitessespour le cas drsquoune turbineNous rappelons que par hypothegravese le rayon ne varie pas dans leplan inter-aube et que la vitesse axiale demeure constante dansle sens axialCes conditions permettent drsquoeffectuer la superposition destriangles de vitesse au bord drsquoattaque et bord de fuite du rotoravec la vitesse 119932119932 commune
c1w1
u
c2
u2
w2
➊
➋
➌ c3w3
u
Turbine
C2
W2 C3
W3
Stator Rotor
C2u C3u
W3u
α2β2 β3
α3
∆Cu= ∆Wu
U
W2u
119940119940119961119961119880119880 = 119888119888119888119888119888119888119888119888119888119888119888119888119909119909 = 119888119888119888119888119888119888119888119888119888119888
1198881198883 = 1198881198881
1205721205723 = 1205721205721
Cineacutematique reacutepeacutetitive
➁ ➂
➁U➀
Turbine
119914119914120784120784
119914119914120785120785
119934119934120785120785
119932119932119934119934120784120784
119932119932
119914119914120783120783120630120630120783120783
120630120630120784120784120631120631120784120784
120630120630120785120785
120631120631120785120785
Stator Rotor
➀ ➁ ➂
Le cas du compresseur est similaire agrave celui drsquoune turbineLe rayon ne varie pas (lrsquoeacutetude se deacuteroule dans le plan inter-aube) et la vitesse axiale demeure constante dans le sens axialComme pour la turbine nous pouvons superposer les triangles devitesse au bord drsquoattaque et bord de fuite du rotor
W2 C2
C3W3
U
C1W1
U
U
Compresseur
➊
➋
➌
α2 β1
C2 W1
C1W2
U
β2α1
W2U C1U
CX
( )2 1 = minuse u uW U C C
Lrsquoopeacuteration est vectorielle
C2U
119914119914120784120784119958119958 minus 119914119914120783120783119958119958 =- 119934119934120784120784119958119958 minus119934119934120783120783119958119958
➀
➁
➁
Compresseur
119914119914120784120784119914119914120785120785
119934119934120783120783
119932119932119914119914120783120783120630120630120783120783 120630120630120784120784
120631120631120784120784120630120630120785120785
Rotor Stator
119934119934120784120784120631120631120783120783
119932119932
➀ ➁ ➂
Trois quantiteacutes fondamentales sont utiliseacutees dans laconception des turbomachines
bull le coefficient de charge ψ
bull le coefficient de deacutebit Φ
bull le degreacute de reacuteaction R
Approche classique
Parfois ψ est consideacutereacute positif pour les turbines et neacutegatifpour les compresseurs 119882119882119890119890 = frasl119882 119888
Coefficient de charge
2Ψ = eWU
Coefficient de charge
2e
UW
=ψ 02 01 02 2
h h hU Uminus ∆
= = 02
s shU
η ∆=
=
eWWm
120578120578119904119904 =∆ℎ0∆ℎ0119904119904
Turbine
Coefficient de deacutebit
xcU
Φ =
xCU
Φ =ρRT
pρ
= xC RTRTU p
=RT m RT
U Ap
xCU
Φ =119888 = ρ119862119862119909119909119860119860
11988811987711987711987911987901199011199010119860119860
= 119872119872 120574120574 1 +120574120574 minus 1
21198721198722
minus 120574120574+1)2(120574120574minus1
Afin de geacuteneacuteraliser le triangle de vitesses drsquoune machineaxiale on effectue une mise agrave lrsquoeacutechelle en divisant chaquecocircteacute par la vitesse tangentielle U
Cette opeacuteration permettra de retrouver directement lescoefficients de charge ψ et de deacutebit Φ
Par la suite un nouveau paramegravetre R qursquoon appelle degreacutede reacuteaction sera introduit pour compleacuteter la caracteacuterisationde lrsquoaubage
Approche classique
α2
β2 α3
β3C2 W3
W2
c3u - c2u= w3u - w2u
W3uW2u
U
CX CX
Lrsquoopeacuteration est vectorielle
C3
120511120511 = frasl119934119934119940119940 119932119932120784120784 = frasl119914119914120785120785119958119958 minus 119914119914120784120784119958119958 119932119932
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
Turbine
3
2
3 3
2 2
12
12
12
12
u
u
u u
u u
c RUc RU
w c RU Uw c RU U
Ψ= minus +
Ψ= minus +
Ψ= minus = +
Ψ= minus = minus
2 22 23 2
2 22 23 2
1 12 2
2 2
c cR RU U
w wR RU U
Ψ Ψ = Φ + minus + = Φ + minus +
Ψ Ψ = Φ + + = Φ + minus
3 2
3 2
1 2 1 2
2 2
R Ratan atan
R Ratan atan
minus + Ψ minus + Ψ = = Φ Φ
Ψ + Ψ minus = = Φ Φ
α α
β β
α2β1
C2U
W1U
C1UW2U β2 α1
Compresseur
ΦC2U
C3UW3U
W2U
ψ
1
ψ
ΦC2UC3U
W3U
W2U
1
Il faut deacutefinir un troisiegraveme paramegravetre pour diffeacuterencier ces deux triangles de vitesses ayant le mecircme 120509120509 et le mecircme 120537120537
Les coefficients 120509120509 et 120537120537 ne suffisent pas pour deacutefinir de maniegravereunivoque le triangle de vitesses
Le degreacute de reacuteaction est un paramegravetre adimensionnel quiexprime la variation drsquoenthalpie subie par le fluide dans le rotorpar rapport agrave la chute totale drsquoenthalpie dans lrsquoeacutetage
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
Variation d enthalpie dans le rotorRVariation d enthalpie dans l eacutetage
=
❸❶
❷
Rotor
Stator
R=0 machine drsquoaction (Δhrotor=0 Δheacutetage=Δhstator)
R=1 machine agrave reacuteaction (Δhstator=0 Δheacutetage=Δhrotor)
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
R indique la reacutepartition du saut drsquoenthalpie (de pression) statique entre le stator et le rotor
Le degreacute de reacuteaction R deacutefini en fonction de variablesthermodynamiques (lrsquoenthalpie) peut aussi srsquoexprimer en fonctionde variables cineacutematiques preacutesentes dans le triangle de vitessesNotamment en fonction de la vitesse peacuteripheacuterique U et descomposantes des vitesses absolues et relatives
La vitesse absolue agrave lrsquoentreacutee du stator est eacutegale agrave celle agrave la sortie du rotor (c1= c3 eacutetage reacutepeacutetitif )
On neacuteglige les pertes et le transfert de chaleur dans le stator
1 2 3
ℎ3 minus ℎ1= ℎ3 + frasl11988811988832 2 minus ℎ1 minus frasl11988811988812 2 = ℎ03 minus ℎ01 = 120549120549ℎ0
ℎ03 minus ℎ01 = ℎ03 minus ℎ02 = 120549120549ℎ )0(3minus2
ℎ03 minus ℎ02 = ℎ3 minus ℎ1
119945119945120782120782120783120783 = 119945119945120782120782120784120784
frasl11988811988832 2 = frasl11988811988812 2
3 2
3 1
rotor
eacutetage
h h hRh h h
∆∆
minus= =
minus
2 20 3 2 2 3
0 3 2
2 2( )
( )
h c c h
∆∆
minus
minus
+ minus=
ℎ03 ℎ02
(ℎ01= ℎ02 1198881198881= 1198881198883)
120549120549ℎ )0(3minus2 = (ℎ3 + frasl11988811988832 2) minus (ℎ2 + frasl11988811988822 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
0(3 2) 3 2( )∆ minus = minusu uh U c c
= 1 minusfrasl11988811988832 2 minus frasl11988811988822 2120549120549ℎ )0(3minus2
119877119877 =120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
120549120549ℎ )0(3minus2
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
3 212+
= minus u uc cRU
119888119888119909119909 = 1198881198882119909119909 = 1198881198883119909119909 = 119888119888119888119888119888119888119888119888119888119888
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 11988811988821199061199061198881198882 = 1198881198881199091199092+1198881198881199061199062
119877119877 = 1 minus11988811988831199061199062 minus 11988811988821199061199062
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
Indirectement lrsquoeacutequation pour R deacutecrite en fonction de composantes cineacutematiques eacutetablit un lien entre les angles drsquoentreacutee et de sortie
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
= 1 minus1198881198883119906119906 + 1198881198882119906119906
2119880119880
2232
2 32 2wwh h+ = +
2 23 2
2 3 2 2w wh hminus = minus
2 2 2 23 3 2 2
2 3 2 2u x u xw w w wh h + +
minus = minus
Machine axiale U2 = U3
La composante axiale de la vitesse demeure constante w3x = w 2x
( eacutequation justifieacutee plus tard lors de la rothalpie)
2 23 2
2 3 2 2u uw wh hminus = minus
3 2 3 22 3
( )( )2
u u u uw w w wh h minus +minus =
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
Eacutetage reacutepeacutetitif c3 = c1
Si les pertes dans le stator sont neacutegligeacutees
c2u ndash c3u= w2u - w3u
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
3 2 3 23 2
( )( )2
u u u uw w w wh h minus +minus = minus
c2u ndash c3u= w2u - w3u
119877119877 = minus1199081199083119906119906 + 1199081199082119906119906
2119880119880
2 2 2 22 3 2 3 2 3( ) 2 ( ) 2h h u u w wminus = minus minus minus
2 2 2 2 2 23 2 3 2 3 2
2 2 2e
I II III
u u c c w wW minus minus minus= + minus
2 2 2 22 3 2 3
2 2 2 2 2 22 3 2 3 2 3
( ) ( )( ) ( ) ( )
minus minus minus=
minus + minus minus minusu u w wR
c c u u w w
3 2= =U u u
Expliqueacutee + tard
2 23 2
03 02 3 2 2ec cW h h h h minus
= minus = minus +
119877119877 =ℎ2 minus ℎ3ℎ1 minus ℎ3
=ℎ2 minus ℎ3ℎ02 minus ℎ03
Degreacute de reacuteaction
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
( )3u 2uw wR2U+
= minus
3 212
u uc cRU+
= minus
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
Puisque le degreacute de reacuteaction R peut aussi srsquoexprimer enfonction de variables cineacutematiques on peut chercher sarepreacutesentation dans le triangle adimensionnel de vitesses
Turbine
α2
β2 α3
β3
C2 W3
C3W2
C3uC2u
U
C2xC3x
W2u
c2u ndash c3u= w2u - w3u
2Ψ minusR
frasl119934119934120784120784119958119958 119932119932 = frasl1198621198622119906119906 119880119880 minus frasl119880119880 119880119880Ψ = frasl1198621198622119906119906 minus 1198621198623119906119906 119880119880
119877119877 = 1 minus (1198621198623119906119906 + 1198621198622119906119906 frasl) 2119880119880Φ = frasl1198621198622119909119909 119880119880 = frasl1198621198623119909119909 119880119880
= (119914119914120784120784119958119958 minus 119914119914120785120785119958119958 frasl) 2119880119880 minus 1 + (119914119914120784120784119958119958 + 119914119914120785120785119958119958 frasl) 2119880119880
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α2
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
Les repreacutesentations des triangles de vitesses adimensionnellessont tregraves riches en information et contiennent plusieurs formules
Par exemple pour un compresseur on deacuteduit aiseacutement que
Avec 120595120595 = 1198821198821198901198901198801198802
Φ = 119888119888119909119909119880119880
et que pour 119888119888119901119901 = 119888119888119888119888119888119888119888119888119888119888 119882119882119890119890 = 119888119888119901119901(11987911987902minus11987911987901)on trouve
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
(11987911987902minus11987911987901) =119880119880119888119888119909119909119888119888119901119901
(tan1205731205731 minus tan1205731205732)
Alors si lrsquoon veut obtenir une plus grande variation detempeacuterature 11987911987902 minus 11987911987901 indirectement un plus grand rapport(1199011199010211990111990101) on peut augmenter
bull la vitesse peacuteripheacuterique 119880119880 (la vitesse de rotation)bull la vitesse axiale 119888119888119909119909 (le deacutebit massique)bull lrsquoangle de deacuteflection de la pale(tan1205731205731 minus tan1205731205732)
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
Lrsquoapplication du triangle de vitesse permet drsquoestimer
bull lrsquoangle drsquoincidence agrave lrsquoentreacutee du rotor
bull lrsquoangle au bord de fuite du rotor
bull la deacuteflection de lrsquoeacutecoulement dans le stator
bull le travail produit ou consommeacute par la roue
Approche classique
C vitesse absolue de lrsquoeacutecoulement
W vitesse relative de lrsquoeacutecoulement
U vitesse peacuteripheacuterique du rotor
Cu Cm Cx composante tangentielle radiale et axiale de lavitesse absolue du fluide
Wu Wm Wx composante tangentielle radiale et axiale de lavitesse relative du fluide
α lrsquoangle des vitesses absolues mesureacutees par rapport agrave la directionaxiale
β lrsquoangle des vitesses relatives mesureacutees par rapport agrave la directionaxiale
La forme des pales du rotor deacutepend des angles 120631120631(des vitesses relatives dans le repegravere du rotor)
La forme des pales du stator deacutepend des angles 120630120630(des vitesses absolues dans le repegravere fixe)
Si lrsquoon veut garder la vitesse axiale 119940119940119961119961 = 119940119940119940119940119940119940119940119940119940119940 lrsquoeacutequation deconservation de la masse 119950 = 120646120646120488120488119940119940119961119961 = 119940119940119940119940119940119940119940119940119940119940 impose 120646120646120488120488 = 119940119940119940119940119940119940119940119940119940119940
Compresseur
Cu
Cm
Cu
Plan normal
Plan meacuteridional
BABF
W2
C2
U
U
W3C3
u
Les vecteurs vitesse (vitesse absolue 119940119940 pour le stator etrelative 119960119960 pour le rotor) sont consideacutereacutes tangents agrave laligne de squelette de lrsquoaube
β2
β3W3
W2
Pour les machines axiales (u= uent = usort ) le trianglede vitesse donne lieu agrave des eacutequations scalaires tellesque
cu= wu+u
cx= wxcx= = wx c
U
w
wu
cu
Pour les machines agrave eacutetages multiples on suppose que lavitesse axiale demeure constante Crsquoest agrave dire que119940119940120784120784119961119961= 119940119940120785120785119961119961= 119940119940119961119961= 119940119940119940119940119940119940119940119940119940119940
Cette condition jumeleacutee agrave 119932119932120784120784= 119932119932120785120785= 119932119932 megravene agrave lrsquohypothegravese drsquoune cineacutematique reacutepeacutetitive En pratique
Les triangles de vitesse agrave lrsquoentreacutee et agrave la sortie drsquouneacutetage peuvent ecirctre superposeacutes
Nous commenccedilons par la repreacutesentation de triangles de vitessespour le cas drsquoune turbineNous rappelons que par hypothegravese le rayon ne varie pas dans leplan inter-aube et que la vitesse axiale demeure constante dansle sens axialCes conditions permettent drsquoeffectuer la superposition destriangles de vitesse au bord drsquoattaque et bord de fuite du rotoravec la vitesse 119932119932 commune
c1w1
u
c2
u2
w2
➊
➋
➌ c3w3
u
Turbine
C2
W2 C3
W3
Stator Rotor
C2u C3u
W3u
α2β2 β3
α3
∆Cu= ∆Wu
U
W2u
119940119940119961119961119880119880 = 119888119888119888119888119888119888119888119888119888119888119888119888119909119909 = 119888119888119888119888119888119888119888119888119888119888
1198881198883 = 1198881198881
1205721205723 = 1205721205721
Cineacutematique reacutepeacutetitive
➁ ➂
➁U➀
Turbine
119914119914120784120784
119914119914120785120785
119934119934120785120785
119932119932119934119934120784120784
119932119932
119914119914120783120783120630120630120783120783
120630120630120784120784120631120631120784120784
120630120630120785120785
120631120631120785120785
Stator Rotor
➀ ➁ ➂
Le cas du compresseur est similaire agrave celui drsquoune turbineLe rayon ne varie pas (lrsquoeacutetude se deacuteroule dans le plan inter-aube) et la vitesse axiale demeure constante dans le sens axialComme pour la turbine nous pouvons superposer les triangles devitesse au bord drsquoattaque et bord de fuite du rotor
W2 C2
C3W3
U
C1W1
U
U
Compresseur
➊
➋
➌
α2 β1
C2 W1
C1W2
U
β2α1
W2U C1U
CX
( )2 1 = minuse u uW U C C
Lrsquoopeacuteration est vectorielle
C2U
119914119914120784120784119958119958 minus 119914119914120783120783119958119958 =- 119934119934120784120784119958119958 minus119934119934120783120783119958119958
➀
➁
➁
Compresseur
119914119914120784120784119914119914120785120785
119934119934120783120783
119932119932119914119914120783120783120630120630120783120783 120630120630120784120784
120631120631120784120784120630120630120785120785
Rotor Stator
119934119934120784120784120631120631120783120783
119932119932
➀ ➁ ➂
Trois quantiteacutes fondamentales sont utiliseacutees dans laconception des turbomachines
bull le coefficient de charge ψ
bull le coefficient de deacutebit Φ
bull le degreacute de reacuteaction R
Approche classique
Parfois ψ est consideacutereacute positif pour les turbines et neacutegatifpour les compresseurs 119882119882119890119890 = frasl119882 119888
Coefficient de charge
2Ψ = eWU
Coefficient de charge
2e
UW
=ψ 02 01 02 2
h h hU Uminus ∆
= = 02
s shU
η ∆=
=
eWWm
120578120578119904119904 =∆ℎ0∆ℎ0119904119904
Turbine
Coefficient de deacutebit
xcU
Φ =
xCU
Φ =ρRT
pρ
= xC RTRTU p
=RT m RT
U Ap
xCU
Φ =119888 = ρ119862119862119909119909119860119860
11988811987711987711987911987901199011199010119860119860
= 119872119872 120574120574 1 +120574120574 minus 1
21198721198722
minus 120574120574+1)2(120574120574minus1
Afin de geacuteneacuteraliser le triangle de vitesses drsquoune machineaxiale on effectue une mise agrave lrsquoeacutechelle en divisant chaquecocircteacute par la vitesse tangentielle U
Cette opeacuteration permettra de retrouver directement lescoefficients de charge ψ et de deacutebit Φ
Par la suite un nouveau paramegravetre R qursquoon appelle degreacutede reacuteaction sera introduit pour compleacuteter la caracteacuterisationde lrsquoaubage
Approche classique
α2
β2 α3
β3C2 W3
W2
c3u - c2u= w3u - w2u
W3uW2u
U
CX CX
Lrsquoopeacuteration est vectorielle
C3
120511120511 = frasl119934119934119940119940 119932119932120784120784 = frasl119914119914120785120785119958119958 minus 119914119914120784120784119958119958 119932119932
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
Turbine
3
2
3 3
2 2
12
12
12
12
u
u
u u
u u
c RUc RU
w c RU Uw c RU U
Ψ= minus +
Ψ= minus +
Ψ= minus = +
Ψ= minus = minus
2 22 23 2
2 22 23 2
1 12 2
2 2
c cR RU U
w wR RU U
Ψ Ψ = Φ + minus + = Φ + minus +
Ψ Ψ = Φ + + = Φ + minus
3 2
3 2
1 2 1 2
2 2
R Ratan atan
R Ratan atan
minus + Ψ minus + Ψ = = Φ Φ
Ψ + Ψ minus = = Φ Φ
α α
β β
α2β1
C2U
W1U
C1UW2U β2 α1
Compresseur
ΦC2U
C3UW3U
W2U
ψ
1
ψ
ΦC2UC3U
W3U
W2U
1
Il faut deacutefinir un troisiegraveme paramegravetre pour diffeacuterencier ces deux triangles de vitesses ayant le mecircme 120509120509 et le mecircme 120537120537
Les coefficients 120509120509 et 120537120537 ne suffisent pas pour deacutefinir de maniegravereunivoque le triangle de vitesses
Le degreacute de reacuteaction est un paramegravetre adimensionnel quiexprime la variation drsquoenthalpie subie par le fluide dans le rotorpar rapport agrave la chute totale drsquoenthalpie dans lrsquoeacutetage
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
Variation d enthalpie dans le rotorRVariation d enthalpie dans l eacutetage
=
❸❶
❷
Rotor
Stator
R=0 machine drsquoaction (Δhrotor=0 Δheacutetage=Δhstator)
R=1 machine agrave reacuteaction (Δhstator=0 Δheacutetage=Δhrotor)
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
R indique la reacutepartition du saut drsquoenthalpie (de pression) statique entre le stator et le rotor
Le degreacute de reacuteaction R deacutefini en fonction de variablesthermodynamiques (lrsquoenthalpie) peut aussi srsquoexprimer en fonctionde variables cineacutematiques preacutesentes dans le triangle de vitessesNotamment en fonction de la vitesse peacuteripheacuterique U et descomposantes des vitesses absolues et relatives
La vitesse absolue agrave lrsquoentreacutee du stator est eacutegale agrave celle agrave la sortie du rotor (c1= c3 eacutetage reacutepeacutetitif )
On neacuteglige les pertes et le transfert de chaleur dans le stator
1 2 3
ℎ3 minus ℎ1= ℎ3 + frasl11988811988832 2 minus ℎ1 minus frasl11988811988812 2 = ℎ03 minus ℎ01 = 120549120549ℎ0
ℎ03 minus ℎ01 = ℎ03 minus ℎ02 = 120549120549ℎ )0(3minus2
ℎ03 minus ℎ02 = ℎ3 minus ℎ1
119945119945120782120782120783120783 = 119945119945120782120782120784120784
frasl11988811988832 2 = frasl11988811988812 2
3 2
3 1
rotor
eacutetage
h h hRh h h
∆∆
minus= =
minus
2 20 3 2 2 3
0 3 2
2 2( )
( )
h c c h
∆∆
minus
minus
+ minus=
ℎ03 ℎ02
(ℎ01= ℎ02 1198881198881= 1198881198883)
120549120549ℎ )0(3minus2 = (ℎ3 + frasl11988811988832 2) minus (ℎ2 + frasl11988811988822 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
0(3 2) 3 2( )∆ minus = minusu uh U c c
= 1 minusfrasl11988811988832 2 minus frasl11988811988822 2120549120549ℎ )0(3minus2
119877119877 =120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
120549120549ℎ )0(3minus2
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
3 212+
= minus u uc cRU
119888119888119909119909 = 1198881198882119909119909 = 1198881198883119909119909 = 119888119888119888119888119888119888119888119888119888119888
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 11988811988821199061199061198881198882 = 1198881198881199091199092+1198881198881199061199062
119877119877 = 1 minus11988811988831199061199062 minus 11988811988821199061199062
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
Indirectement lrsquoeacutequation pour R deacutecrite en fonction de composantes cineacutematiques eacutetablit un lien entre les angles drsquoentreacutee et de sortie
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
= 1 minus1198881198883119906119906 + 1198881198882119906119906
2119880119880
2232
2 32 2wwh h+ = +
2 23 2
2 3 2 2w wh hminus = minus
2 2 2 23 3 2 2
2 3 2 2u x u xw w w wh h + +
minus = minus
Machine axiale U2 = U3
La composante axiale de la vitesse demeure constante w3x = w 2x
( eacutequation justifieacutee plus tard lors de la rothalpie)
2 23 2
2 3 2 2u uw wh hminus = minus
3 2 3 22 3
( )( )2
u u u uw w w wh h minus +minus =
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
Eacutetage reacutepeacutetitif c3 = c1
Si les pertes dans le stator sont neacutegligeacutees
c2u ndash c3u= w2u - w3u
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
3 2 3 23 2
( )( )2
u u u uw w w wh h minus +minus = minus
c2u ndash c3u= w2u - w3u
119877119877 = minus1199081199083119906119906 + 1199081199082119906119906
2119880119880
2 2 2 22 3 2 3 2 3( ) 2 ( ) 2h h u u w wminus = minus minus minus
2 2 2 2 2 23 2 3 2 3 2
2 2 2e
I II III
u u c c w wW minus minus minus= + minus
2 2 2 22 3 2 3
2 2 2 2 2 22 3 2 3 2 3
( ) ( )( ) ( ) ( )
minus minus minus=
minus + minus minus minusu u w wR
c c u u w w
3 2= =U u u
Expliqueacutee + tard
2 23 2
03 02 3 2 2ec cW h h h h minus
= minus = minus +
119877119877 =ℎ2 minus ℎ3ℎ1 minus ℎ3
=ℎ2 minus ℎ3ℎ02 minus ℎ03
Degreacute de reacuteaction
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
( )3u 2uw wR2U+
= minus
3 212
u uc cRU+
= minus
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
Puisque le degreacute de reacuteaction R peut aussi srsquoexprimer enfonction de variables cineacutematiques on peut chercher sarepreacutesentation dans le triangle adimensionnel de vitesses
Turbine
α2
β2 α3
β3
C2 W3
C3W2
C3uC2u
U
C2xC3x
W2u
c2u ndash c3u= w2u - w3u
2Ψ minusR
frasl119934119934120784120784119958119958 119932119932 = frasl1198621198622119906119906 119880119880 minus frasl119880119880 119880119880Ψ = frasl1198621198622119906119906 minus 1198621198623119906119906 119880119880
119877119877 = 1 minus (1198621198623119906119906 + 1198621198622119906119906 frasl) 2119880119880Φ = frasl1198621198622119909119909 119880119880 = frasl1198621198623119909119909 119880119880
= (119914119914120784120784119958119958 minus 119914119914120785120785119958119958 frasl) 2119880119880 minus 1 + (119914119914120784120784119958119958 + 119914119914120785120785119958119958 frasl) 2119880119880
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α2
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
Les repreacutesentations des triangles de vitesses adimensionnellessont tregraves riches en information et contiennent plusieurs formules
Par exemple pour un compresseur on deacuteduit aiseacutement que
Avec 120595120595 = 1198821198821198901198901198801198802
Φ = 119888119888119909119909119880119880
et que pour 119888119888119901119901 = 119888119888119888119888119888119888119888119888119888119888 119882119882119890119890 = 119888119888119901119901(11987911987902minus11987911987901)on trouve
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
(11987911987902minus11987911987901) =119880119880119888119888119909119909119888119888119901119901
(tan1205731205731 minus tan1205731205732)
Alors si lrsquoon veut obtenir une plus grande variation detempeacuterature 11987911987902 minus 11987911987901 indirectement un plus grand rapport(1199011199010211990111990101) on peut augmenter
bull la vitesse peacuteripheacuterique 119880119880 (la vitesse de rotation)bull la vitesse axiale 119888119888119909119909 (le deacutebit massique)bull lrsquoangle de deacuteflection de la pale(tan1205731205731 minus tan1205731205732)
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
C vitesse absolue de lrsquoeacutecoulement
W vitesse relative de lrsquoeacutecoulement
U vitesse peacuteripheacuterique du rotor
Cu Cm Cx composante tangentielle radiale et axiale de lavitesse absolue du fluide
Wu Wm Wx composante tangentielle radiale et axiale de lavitesse relative du fluide
α lrsquoangle des vitesses absolues mesureacutees par rapport agrave la directionaxiale
β lrsquoangle des vitesses relatives mesureacutees par rapport agrave la directionaxiale
La forme des pales du rotor deacutepend des angles 120631120631(des vitesses relatives dans le repegravere du rotor)
La forme des pales du stator deacutepend des angles 120630120630(des vitesses absolues dans le repegravere fixe)
Si lrsquoon veut garder la vitesse axiale 119940119940119961119961 = 119940119940119940119940119940119940119940119940119940119940 lrsquoeacutequation deconservation de la masse 119950 = 120646120646120488120488119940119940119961119961 = 119940119940119940119940119940119940119940119940119940119940 impose 120646120646120488120488 = 119940119940119940119940119940119940119940119940119940119940
Compresseur
Cu
Cm
Cu
Plan normal
Plan meacuteridional
BABF
W2
C2
U
U
W3C3
u
Les vecteurs vitesse (vitesse absolue 119940119940 pour le stator etrelative 119960119960 pour le rotor) sont consideacutereacutes tangents agrave laligne de squelette de lrsquoaube
β2
β3W3
W2
Pour les machines axiales (u= uent = usort ) le trianglede vitesse donne lieu agrave des eacutequations scalaires tellesque
cu= wu+u
cx= wxcx= = wx c
U
w
wu
cu
Pour les machines agrave eacutetages multiples on suppose que lavitesse axiale demeure constante Crsquoest agrave dire que119940119940120784120784119961119961= 119940119940120785120785119961119961= 119940119940119961119961= 119940119940119940119940119940119940119940119940119940119940
Cette condition jumeleacutee agrave 119932119932120784120784= 119932119932120785120785= 119932119932 megravene agrave lrsquohypothegravese drsquoune cineacutematique reacutepeacutetitive En pratique
Les triangles de vitesse agrave lrsquoentreacutee et agrave la sortie drsquouneacutetage peuvent ecirctre superposeacutes
Nous commenccedilons par la repreacutesentation de triangles de vitessespour le cas drsquoune turbineNous rappelons que par hypothegravese le rayon ne varie pas dans leplan inter-aube et que la vitesse axiale demeure constante dansle sens axialCes conditions permettent drsquoeffectuer la superposition destriangles de vitesse au bord drsquoattaque et bord de fuite du rotoravec la vitesse 119932119932 commune
c1w1
u
c2
u2
w2
➊
➋
➌ c3w3
u
Turbine
C2
W2 C3
W3
Stator Rotor
C2u C3u
W3u
α2β2 β3
α3
∆Cu= ∆Wu
U
W2u
119940119940119961119961119880119880 = 119888119888119888119888119888119888119888119888119888119888119888119888119909119909 = 119888119888119888119888119888119888119888119888119888119888
1198881198883 = 1198881198881
1205721205723 = 1205721205721
Cineacutematique reacutepeacutetitive
➁ ➂
➁U➀
Turbine
119914119914120784120784
119914119914120785120785
119934119934120785120785
119932119932119934119934120784120784
119932119932
119914119914120783120783120630120630120783120783
120630120630120784120784120631120631120784120784
120630120630120785120785
120631120631120785120785
Stator Rotor
➀ ➁ ➂
Le cas du compresseur est similaire agrave celui drsquoune turbineLe rayon ne varie pas (lrsquoeacutetude se deacuteroule dans le plan inter-aube) et la vitesse axiale demeure constante dans le sens axialComme pour la turbine nous pouvons superposer les triangles devitesse au bord drsquoattaque et bord de fuite du rotor
W2 C2
C3W3
U
C1W1
U
U
Compresseur
➊
➋
➌
α2 β1
C2 W1
C1W2
U
β2α1
W2U C1U
CX
( )2 1 = minuse u uW U C C
Lrsquoopeacuteration est vectorielle
C2U
119914119914120784120784119958119958 minus 119914119914120783120783119958119958 =- 119934119934120784120784119958119958 minus119934119934120783120783119958119958
➀
➁
➁
Compresseur
119914119914120784120784119914119914120785120785
119934119934120783120783
119932119932119914119914120783120783120630120630120783120783 120630120630120784120784
120631120631120784120784120630120630120785120785
Rotor Stator
119934119934120784120784120631120631120783120783
119932119932
➀ ➁ ➂
Trois quantiteacutes fondamentales sont utiliseacutees dans laconception des turbomachines
bull le coefficient de charge ψ
bull le coefficient de deacutebit Φ
bull le degreacute de reacuteaction R
Approche classique
Parfois ψ est consideacutereacute positif pour les turbines et neacutegatifpour les compresseurs 119882119882119890119890 = frasl119882 119888
Coefficient de charge
2Ψ = eWU
Coefficient de charge
2e
UW
=ψ 02 01 02 2
h h hU Uminus ∆
= = 02
s shU
η ∆=
=
eWWm
120578120578119904119904 =∆ℎ0∆ℎ0119904119904
Turbine
Coefficient de deacutebit
xcU
Φ =
xCU
Φ =ρRT
pρ
= xC RTRTU p
=RT m RT
U Ap
xCU
Φ =119888 = ρ119862119862119909119909119860119860
11988811987711987711987911987901199011199010119860119860
= 119872119872 120574120574 1 +120574120574 minus 1
21198721198722
minus 120574120574+1)2(120574120574minus1
Afin de geacuteneacuteraliser le triangle de vitesses drsquoune machineaxiale on effectue une mise agrave lrsquoeacutechelle en divisant chaquecocircteacute par la vitesse tangentielle U
Cette opeacuteration permettra de retrouver directement lescoefficients de charge ψ et de deacutebit Φ
Par la suite un nouveau paramegravetre R qursquoon appelle degreacutede reacuteaction sera introduit pour compleacuteter la caracteacuterisationde lrsquoaubage
Approche classique
α2
β2 α3
β3C2 W3
W2
c3u - c2u= w3u - w2u
W3uW2u
U
CX CX
Lrsquoopeacuteration est vectorielle
C3
120511120511 = frasl119934119934119940119940 119932119932120784120784 = frasl119914119914120785120785119958119958 minus 119914119914120784120784119958119958 119932119932
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
Turbine
3
2
3 3
2 2
12
12
12
12
u
u
u u
u u
c RUc RU
w c RU Uw c RU U
Ψ= minus +
Ψ= minus +
Ψ= minus = +
Ψ= minus = minus
2 22 23 2
2 22 23 2
1 12 2
2 2
c cR RU U
w wR RU U
Ψ Ψ = Φ + minus + = Φ + minus +
Ψ Ψ = Φ + + = Φ + minus
3 2
3 2
1 2 1 2
2 2
R Ratan atan
R Ratan atan
minus + Ψ minus + Ψ = = Φ Φ
Ψ + Ψ minus = = Φ Φ
α α
β β
α2β1
C2U
W1U
C1UW2U β2 α1
Compresseur
ΦC2U
C3UW3U
W2U
ψ
1
ψ
ΦC2UC3U
W3U
W2U
1
Il faut deacutefinir un troisiegraveme paramegravetre pour diffeacuterencier ces deux triangles de vitesses ayant le mecircme 120509120509 et le mecircme 120537120537
Les coefficients 120509120509 et 120537120537 ne suffisent pas pour deacutefinir de maniegravereunivoque le triangle de vitesses
Le degreacute de reacuteaction est un paramegravetre adimensionnel quiexprime la variation drsquoenthalpie subie par le fluide dans le rotorpar rapport agrave la chute totale drsquoenthalpie dans lrsquoeacutetage
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
Variation d enthalpie dans le rotorRVariation d enthalpie dans l eacutetage
=
❸❶
❷
Rotor
Stator
R=0 machine drsquoaction (Δhrotor=0 Δheacutetage=Δhstator)
R=1 machine agrave reacuteaction (Δhstator=0 Δheacutetage=Δhrotor)
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
R indique la reacutepartition du saut drsquoenthalpie (de pression) statique entre le stator et le rotor
Le degreacute de reacuteaction R deacutefini en fonction de variablesthermodynamiques (lrsquoenthalpie) peut aussi srsquoexprimer en fonctionde variables cineacutematiques preacutesentes dans le triangle de vitessesNotamment en fonction de la vitesse peacuteripheacuterique U et descomposantes des vitesses absolues et relatives
La vitesse absolue agrave lrsquoentreacutee du stator est eacutegale agrave celle agrave la sortie du rotor (c1= c3 eacutetage reacutepeacutetitif )
On neacuteglige les pertes et le transfert de chaleur dans le stator
1 2 3
ℎ3 minus ℎ1= ℎ3 + frasl11988811988832 2 minus ℎ1 minus frasl11988811988812 2 = ℎ03 minus ℎ01 = 120549120549ℎ0
ℎ03 minus ℎ01 = ℎ03 minus ℎ02 = 120549120549ℎ )0(3minus2
ℎ03 minus ℎ02 = ℎ3 minus ℎ1
119945119945120782120782120783120783 = 119945119945120782120782120784120784
frasl11988811988832 2 = frasl11988811988812 2
3 2
3 1
rotor
eacutetage
h h hRh h h
∆∆
minus= =
minus
2 20 3 2 2 3
0 3 2
2 2( )
( )
h c c h
∆∆
minus
minus
+ minus=
ℎ03 ℎ02
(ℎ01= ℎ02 1198881198881= 1198881198883)
120549120549ℎ )0(3minus2 = (ℎ3 + frasl11988811988832 2) minus (ℎ2 + frasl11988811988822 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
0(3 2) 3 2( )∆ minus = minusu uh U c c
= 1 minusfrasl11988811988832 2 minus frasl11988811988822 2120549120549ℎ )0(3minus2
119877119877 =120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
120549120549ℎ )0(3minus2
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
3 212+
= minus u uc cRU
119888119888119909119909 = 1198881198882119909119909 = 1198881198883119909119909 = 119888119888119888119888119888119888119888119888119888119888
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 11988811988821199061199061198881198882 = 1198881198881199091199092+1198881198881199061199062
119877119877 = 1 minus11988811988831199061199062 minus 11988811988821199061199062
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
Indirectement lrsquoeacutequation pour R deacutecrite en fonction de composantes cineacutematiques eacutetablit un lien entre les angles drsquoentreacutee et de sortie
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
= 1 minus1198881198883119906119906 + 1198881198882119906119906
2119880119880
2232
2 32 2wwh h+ = +
2 23 2
2 3 2 2w wh hminus = minus
2 2 2 23 3 2 2
2 3 2 2u x u xw w w wh h + +
minus = minus
Machine axiale U2 = U3
La composante axiale de la vitesse demeure constante w3x = w 2x
( eacutequation justifieacutee plus tard lors de la rothalpie)
2 23 2
2 3 2 2u uw wh hminus = minus
3 2 3 22 3
( )( )2
u u u uw w w wh h minus +minus =
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
Eacutetage reacutepeacutetitif c3 = c1
Si les pertes dans le stator sont neacutegligeacutees
c2u ndash c3u= w2u - w3u
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
3 2 3 23 2
( )( )2
u u u uw w w wh h minus +minus = minus
c2u ndash c3u= w2u - w3u
119877119877 = minus1199081199083119906119906 + 1199081199082119906119906
2119880119880
2 2 2 22 3 2 3 2 3( ) 2 ( ) 2h h u u w wminus = minus minus minus
2 2 2 2 2 23 2 3 2 3 2
2 2 2e
I II III
u u c c w wW minus minus minus= + minus
2 2 2 22 3 2 3
2 2 2 2 2 22 3 2 3 2 3
( ) ( )( ) ( ) ( )
minus minus minus=
minus + minus minus minusu u w wR
c c u u w w
3 2= =U u u
Expliqueacutee + tard
2 23 2
03 02 3 2 2ec cW h h h h minus
= minus = minus +
119877119877 =ℎ2 minus ℎ3ℎ1 minus ℎ3
=ℎ2 minus ℎ3ℎ02 minus ℎ03
Degreacute de reacuteaction
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
( )3u 2uw wR2U+
= minus
3 212
u uc cRU+
= minus
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
Puisque le degreacute de reacuteaction R peut aussi srsquoexprimer enfonction de variables cineacutematiques on peut chercher sarepreacutesentation dans le triangle adimensionnel de vitesses
Turbine
α2
β2 α3
β3
C2 W3
C3W2
C3uC2u
U
C2xC3x
W2u
c2u ndash c3u= w2u - w3u
2Ψ minusR
frasl119934119934120784120784119958119958 119932119932 = frasl1198621198622119906119906 119880119880 minus frasl119880119880 119880119880Ψ = frasl1198621198622119906119906 minus 1198621198623119906119906 119880119880
119877119877 = 1 minus (1198621198623119906119906 + 1198621198622119906119906 frasl) 2119880119880Φ = frasl1198621198622119909119909 119880119880 = frasl1198621198623119909119909 119880119880
= (119914119914120784120784119958119958 minus 119914119914120785120785119958119958 frasl) 2119880119880 minus 1 + (119914119914120784120784119958119958 + 119914119914120785120785119958119958 frasl) 2119880119880
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α2
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
Les repreacutesentations des triangles de vitesses adimensionnellessont tregraves riches en information et contiennent plusieurs formules
Par exemple pour un compresseur on deacuteduit aiseacutement que
Avec 120595120595 = 1198821198821198901198901198801198802
Φ = 119888119888119909119909119880119880
et que pour 119888119888119901119901 = 119888119888119888119888119888119888119888119888119888119888 119882119882119890119890 = 119888119888119901119901(11987911987902minus11987911987901)on trouve
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
(11987911987902minus11987911987901) =119880119880119888119888119909119909119888119888119901119901
(tan1205731205731 minus tan1205731205732)
Alors si lrsquoon veut obtenir une plus grande variation detempeacuterature 11987911987902 minus 11987911987901 indirectement un plus grand rapport(1199011199010211990111990101) on peut augmenter
bull la vitesse peacuteripheacuterique 119880119880 (la vitesse de rotation)bull la vitesse axiale 119888119888119909119909 (le deacutebit massique)bull lrsquoangle de deacuteflection de la pale(tan1205731205731 minus tan1205731205732)
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
α lrsquoangle des vitesses absolues mesureacutees par rapport agrave la directionaxiale
β lrsquoangle des vitesses relatives mesureacutees par rapport agrave la directionaxiale
La forme des pales du rotor deacutepend des angles 120631120631(des vitesses relatives dans le repegravere du rotor)
La forme des pales du stator deacutepend des angles 120630120630(des vitesses absolues dans le repegravere fixe)
Si lrsquoon veut garder la vitesse axiale 119940119940119961119961 = 119940119940119940119940119940119940119940119940119940119940 lrsquoeacutequation deconservation de la masse 119950 = 120646120646120488120488119940119940119961119961 = 119940119940119940119940119940119940119940119940119940119940 impose 120646120646120488120488 = 119940119940119940119940119940119940119940119940119940119940
Compresseur
Cu
Cm
Cu
Plan normal
Plan meacuteridional
BABF
W2
C2
U
U
W3C3
u
Les vecteurs vitesse (vitesse absolue 119940119940 pour le stator etrelative 119960119960 pour le rotor) sont consideacutereacutes tangents agrave laligne de squelette de lrsquoaube
β2
β3W3
W2
Pour les machines axiales (u= uent = usort ) le trianglede vitesse donne lieu agrave des eacutequations scalaires tellesque
cu= wu+u
cx= wxcx= = wx c
U
w
wu
cu
Pour les machines agrave eacutetages multiples on suppose que lavitesse axiale demeure constante Crsquoest agrave dire que119940119940120784120784119961119961= 119940119940120785120785119961119961= 119940119940119961119961= 119940119940119940119940119940119940119940119940119940119940
Cette condition jumeleacutee agrave 119932119932120784120784= 119932119932120785120785= 119932119932 megravene agrave lrsquohypothegravese drsquoune cineacutematique reacutepeacutetitive En pratique
Les triangles de vitesse agrave lrsquoentreacutee et agrave la sortie drsquouneacutetage peuvent ecirctre superposeacutes
Nous commenccedilons par la repreacutesentation de triangles de vitessespour le cas drsquoune turbineNous rappelons que par hypothegravese le rayon ne varie pas dans leplan inter-aube et que la vitesse axiale demeure constante dansle sens axialCes conditions permettent drsquoeffectuer la superposition destriangles de vitesse au bord drsquoattaque et bord de fuite du rotoravec la vitesse 119932119932 commune
c1w1
u
c2
u2
w2
➊
➋
➌ c3w3
u
Turbine
C2
W2 C3
W3
Stator Rotor
C2u C3u
W3u
α2β2 β3
α3
∆Cu= ∆Wu
U
W2u
119940119940119961119961119880119880 = 119888119888119888119888119888119888119888119888119888119888119888119888119909119909 = 119888119888119888119888119888119888119888119888119888119888
1198881198883 = 1198881198881
1205721205723 = 1205721205721
Cineacutematique reacutepeacutetitive
➁ ➂
➁U➀
Turbine
119914119914120784120784
119914119914120785120785
119934119934120785120785
119932119932119934119934120784120784
119932119932
119914119914120783120783120630120630120783120783
120630120630120784120784120631120631120784120784
120630120630120785120785
120631120631120785120785
Stator Rotor
➀ ➁ ➂
Le cas du compresseur est similaire agrave celui drsquoune turbineLe rayon ne varie pas (lrsquoeacutetude se deacuteroule dans le plan inter-aube) et la vitesse axiale demeure constante dans le sens axialComme pour la turbine nous pouvons superposer les triangles devitesse au bord drsquoattaque et bord de fuite du rotor
W2 C2
C3W3
U
C1W1
U
U
Compresseur
➊
➋
➌
α2 β1
C2 W1
C1W2
U
β2α1
W2U C1U
CX
( )2 1 = minuse u uW U C C
Lrsquoopeacuteration est vectorielle
C2U
119914119914120784120784119958119958 minus 119914119914120783120783119958119958 =- 119934119934120784120784119958119958 minus119934119934120783120783119958119958
➀
➁
➁
Compresseur
119914119914120784120784119914119914120785120785
119934119934120783120783
119932119932119914119914120783120783120630120630120783120783 120630120630120784120784
120631120631120784120784120630120630120785120785
Rotor Stator
119934119934120784120784120631120631120783120783
119932119932
➀ ➁ ➂
Trois quantiteacutes fondamentales sont utiliseacutees dans laconception des turbomachines
bull le coefficient de charge ψ
bull le coefficient de deacutebit Φ
bull le degreacute de reacuteaction R
Approche classique
Parfois ψ est consideacutereacute positif pour les turbines et neacutegatifpour les compresseurs 119882119882119890119890 = frasl119882 119888
Coefficient de charge
2Ψ = eWU
Coefficient de charge
2e
UW
=ψ 02 01 02 2
h h hU Uminus ∆
= = 02
s shU
η ∆=
=
eWWm
120578120578119904119904 =∆ℎ0∆ℎ0119904119904
Turbine
Coefficient de deacutebit
xcU
Φ =
xCU
Φ =ρRT
pρ
= xC RTRTU p
=RT m RT
U Ap
xCU
Φ =119888 = ρ119862119862119909119909119860119860
11988811987711987711987911987901199011199010119860119860
= 119872119872 120574120574 1 +120574120574 minus 1
21198721198722
minus 120574120574+1)2(120574120574minus1
Afin de geacuteneacuteraliser le triangle de vitesses drsquoune machineaxiale on effectue une mise agrave lrsquoeacutechelle en divisant chaquecocircteacute par la vitesse tangentielle U
Cette opeacuteration permettra de retrouver directement lescoefficients de charge ψ et de deacutebit Φ
Par la suite un nouveau paramegravetre R qursquoon appelle degreacutede reacuteaction sera introduit pour compleacuteter la caracteacuterisationde lrsquoaubage
Approche classique
α2
β2 α3
β3C2 W3
W2
c3u - c2u= w3u - w2u
W3uW2u
U
CX CX
Lrsquoopeacuteration est vectorielle
C3
120511120511 = frasl119934119934119940119940 119932119932120784120784 = frasl119914119914120785120785119958119958 minus 119914119914120784120784119958119958 119932119932
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
Turbine
3
2
3 3
2 2
12
12
12
12
u
u
u u
u u
c RUc RU
w c RU Uw c RU U
Ψ= minus +
Ψ= minus +
Ψ= minus = +
Ψ= minus = minus
2 22 23 2
2 22 23 2
1 12 2
2 2
c cR RU U
w wR RU U
Ψ Ψ = Φ + minus + = Φ + minus +
Ψ Ψ = Φ + + = Φ + minus
3 2
3 2
1 2 1 2
2 2
R Ratan atan
R Ratan atan
minus + Ψ minus + Ψ = = Φ Φ
Ψ + Ψ minus = = Φ Φ
α α
β β
α2β1
C2U
W1U
C1UW2U β2 α1
Compresseur
ΦC2U
C3UW3U
W2U
ψ
1
ψ
ΦC2UC3U
W3U
W2U
1
Il faut deacutefinir un troisiegraveme paramegravetre pour diffeacuterencier ces deux triangles de vitesses ayant le mecircme 120509120509 et le mecircme 120537120537
Les coefficients 120509120509 et 120537120537 ne suffisent pas pour deacutefinir de maniegravereunivoque le triangle de vitesses
Le degreacute de reacuteaction est un paramegravetre adimensionnel quiexprime la variation drsquoenthalpie subie par le fluide dans le rotorpar rapport agrave la chute totale drsquoenthalpie dans lrsquoeacutetage
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
Variation d enthalpie dans le rotorRVariation d enthalpie dans l eacutetage
=
❸❶
❷
Rotor
Stator
R=0 machine drsquoaction (Δhrotor=0 Δheacutetage=Δhstator)
R=1 machine agrave reacuteaction (Δhstator=0 Δheacutetage=Δhrotor)
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
R indique la reacutepartition du saut drsquoenthalpie (de pression) statique entre le stator et le rotor
Le degreacute de reacuteaction R deacutefini en fonction de variablesthermodynamiques (lrsquoenthalpie) peut aussi srsquoexprimer en fonctionde variables cineacutematiques preacutesentes dans le triangle de vitessesNotamment en fonction de la vitesse peacuteripheacuterique U et descomposantes des vitesses absolues et relatives
La vitesse absolue agrave lrsquoentreacutee du stator est eacutegale agrave celle agrave la sortie du rotor (c1= c3 eacutetage reacutepeacutetitif )
On neacuteglige les pertes et le transfert de chaleur dans le stator
1 2 3
ℎ3 minus ℎ1= ℎ3 + frasl11988811988832 2 minus ℎ1 minus frasl11988811988812 2 = ℎ03 minus ℎ01 = 120549120549ℎ0
ℎ03 minus ℎ01 = ℎ03 minus ℎ02 = 120549120549ℎ )0(3minus2
ℎ03 minus ℎ02 = ℎ3 minus ℎ1
119945119945120782120782120783120783 = 119945119945120782120782120784120784
frasl11988811988832 2 = frasl11988811988812 2
3 2
3 1
rotor
eacutetage
h h hRh h h
∆∆
minus= =
minus
2 20 3 2 2 3
0 3 2
2 2( )
( )
h c c h
∆∆
minus
minus
+ minus=
ℎ03 ℎ02
(ℎ01= ℎ02 1198881198881= 1198881198883)
120549120549ℎ )0(3minus2 = (ℎ3 + frasl11988811988832 2) minus (ℎ2 + frasl11988811988822 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
0(3 2) 3 2( )∆ minus = minusu uh U c c
= 1 minusfrasl11988811988832 2 minus frasl11988811988822 2120549120549ℎ )0(3minus2
119877119877 =120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
120549120549ℎ )0(3minus2
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
3 212+
= minus u uc cRU
119888119888119909119909 = 1198881198882119909119909 = 1198881198883119909119909 = 119888119888119888119888119888119888119888119888119888119888
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 11988811988821199061199061198881198882 = 1198881198881199091199092+1198881198881199061199062
119877119877 = 1 minus11988811988831199061199062 minus 11988811988821199061199062
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
Indirectement lrsquoeacutequation pour R deacutecrite en fonction de composantes cineacutematiques eacutetablit un lien entre les angles drsquoentreacutee et de sortie
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
= 1 minus1198881198883119906119906 + 1198881198882119906119906
2119880119880
2232
2 32 2wwh h+ = +
2 23 2
2 3 2 2w wh hminus = minus
2 2 2 23 3 2 2
2 3 2 2u x u xw w w wh h + +
minus = minus
Machine axiale U2 = U3
La composante axiale de la vitesse demeure constante w3x = w 2x
( eacutequation justifieacutee plus tard lors de la rothalpie)
2 23 2
2 3 2 2u uw wh hminus = minus
3 2 3 22 3
( )( )2
u u u uw w w wh h minus +minus =
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
Eacutetage reacutepeacutetitif c3 = c1
Si les pertes dans le stator sont neacutegligeacutees
c2u ndash c3u= w2u - w3u
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
3 2 3 23 2
( )( )2
u u u uw w w wh h minus +minus = minus
c2u ndash c3u= w2u - w3u
119877119877 = minus1199081199083119906119906 + 1199081199082119906119906
2119880119880
2 2 2 22 3 2 3 2 3( ) 2 ( ) 2h h u u w wminus = minus minus minus
2 2 2 2 2 23 2 3 2 3 2
2 2 2e
I II III
u u c c w wW minus minus minus= + minus
2 2 2 22 3 2 3
2 2 2 2 2 22 3 2 3 2 3
( ) ( )( ) ( ) ( )
minus minus minus=
minus + minus minus minusu u w wR
c c u u w w
3 2= =U u u
Expliqueacutee + tard
2 23 2
03 02 3 2 2ec cW h h h h minus
= minus = minus +
119877119877 =ℎ2 minus ℎ3ℎ1 minus ℎ3
=ℎ2 minus ℎ3ℎ02 minus ℎ03
Degreacute de reacuteaction
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
( )3u 2uw wR2U+
= minus
3 212
u uc cRU+
= minus
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
Puisque le degreacute de reacuteaction R peut aussi srsquoexprimer enfonction de variables cineacutematiques on peut chercher sarepreacutesentation dans le triangle adimensionnel de vitesses
Turbine
α2
β2 α3
β3
C2 W3
C3W2
C3uC2u
U
C2xC3x
W2u
c2u ndash c3u= w2u - w3u
2Ψ minusR
frasl119934119934120784120784119958119958 119932119932 = frasl1198621198622119906119906 119880119880 minus frasl119880119880 119880119880Ψ = frasl1198621198622119906119906 minus 1198621198623119906119906 119880119880
119877119877 = 1 minus (1198621198623119906119906 + 1198621198622119906119906 frasl) 2119880119880Φ = frasl1198621198622119909119909 119880119880 = frasl1198621198623119909119909 119880119880
= (119914119914120784120784119958119958 minus 119914119914120785120785119958119958 frasl) 2119880119880 minus 1 + (119914119914120784120784119958119958 + 119914119914120785120785119958119958 frasl) 2119880119880
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α2
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
Les repreacutesentations des triangles de vitesses adimensionnellessont tregraves riches en information et contiennent plusieurs formules
Par exemple pour un compresseur on deacuteduit aiseacutement que
Avec 120595120595 = 1198821198821198901198901198801198802
Φ = 119888119888119909119909119880119880
et que pour 119888119888119901119901 = 119888119888119888119888119888119888119888119888119888119888 119882119882119890119890 = 119888119888119901119901(11987911987902minus11987911987901)on trouve
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
(11987911987902minus11987911987901) =119880119880119888119888119909119909119888119888119901119901
(tan1205731205731 minus tan1205731205732)
Alors si lrsquoon veut obtenir une plus grande variation detempeacuterature 11987911987902 minus 11987911987901 indirectement un plus grand rapport(1199011199010211990111990101) on peut augmenter
bull la vitesse peacuteripheacuterique 119880119880 (la vitesse de rotation)bull la vitesse axiale 119888119888119909119909 (le deacutebit massique)bull lrsquoangle de deacuteflection de la pale(tan1205731205731 minus tan1205731205732)
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
Compresseur
Cu
Cm
Cu
Plan normal
Plan meacuteridional
BABF
W2
C2
U
U
W3C3
u
Les vecteurs vitesse (vitesse absolue 119940119940 pour le stator etrelative 119960119960 pour le rotor) sont consideacutereacutes tangents agrave laligne de squelette de lrsquoaube
β2
β3W3
W2
Pour les machines axiales (u= uent = usort ) le trianglede vitesse donne lieu agrave des eacutequations scalaires tellesque
cu= wu+u
cx= wxcx= = wx c
U
w
wu
cu
Pour les machines agrave eacutetages multiples on suppose que lavitesse axiale demeure constante Crsquoest agrave dire que119940119940120784120784119961119961= 119940119940120785120785119961119961= 119940119940119961119961= 119940119940119940119940119940119940119940119940119940119940
Cette condition jumeleacutee agrave 119932119932120784120784= 119932119932120785120785= 119932119932 megravene agrave lrsquohypothegravese drsquoune cineacutematique reacutepeacutetitive En pratique
Les triangles de vitesse agrave lrsquoentreacutee et agrave la sortie drsquouneacutetage peuvent ecirctre superposeacutes
Nous commenccedilons par la repreacutesentation de triangles de vitessespour le cas drsquoune turbineNous rappelons que par hypothegravese le rayon ne varie pas dans leplan inter-aube et que la vitesse axiale demeure constante dansle sens axialCes conditions permettent drsquoeffectuer la superposition destriangles de vitesse au bord drsquoattaque et bord de fuite du rotoravec la vitesse 119932119932 commune
c1w1
u
c2
u2
w2
➊
➋
➌ c3w3
u
Turbine
C2
W2 C3
W3
Stator Rotor
C2u C3u
W3u
α2β2 β3
α3
∆Cu= ∆Wu
U
W2u
119940119940119961119961119880119880 = 119888119888119888119888119888119888119888119888119888119888119888119888119909119909 = 119888119888119888119888119888119888119888119888119888119888
1198881198883 = 1198881198881
1205721205723 = 1205721205721
Cineacutematique reacutepeacutetitive
➁ ➂
➁U➀
Turbine
119914119914120784120784
119914119914120785120785
119934119934120785120785
119932119932119934119934120784120784
119932119932
119914119914120783120783120630120630120783120783
120630120630120784120784120631120631120784120784
120630120630120785120785
120631120631120785120785
Stator Rotor
➀ ➁ ➂
Le cas du compresseur est similaire agrave celui drsquoune turbineLe rayon ne varie pas (lrsquoeacutetude se deacuteroule dans le plan inter-aube) et la vitesse axiale demeure constante dans le sens axialComme pour la turbine nous pouvons superposer les triangles devitesse au bord drsquoattaque et bord de fuite du rotor
W2 C2
C3W3
U
C1W1
U
U
Compresseur
➊
➋
➌
α2 β1
C2 W1
C1W2
U
β2α1
W2U C1U
CX
( )2 1 = minuse u uW U C C
Lrsquoopeacuteration est vectorielle
C2U
119914119914120784120784119958119958 minus 119914119914120783120783119958119958 =- 119934119934120784120784119958119958 minus119934119934120783120783119958119958
➀
➁
➁
Compresseur
119914119914120784120784119914119914120785120785
119934119934120783120783
119932119932119914119914120783120783120630120630120783120783 120630120630120784120784
120631120631120784120784120630120630120785120785
Rotor Stator
119934119934120784120784120631120631120783120783
119932119932
➀ ➁ ➂
Trois quantiteacutes fondamentales sont utiliseacutees dans laconception des turbomachines
bull le coefficient de charge ψ
bull le coefficient de deacutebit Φ
bull le degreacute de reacuteaction R
Approche classique
Parfois ψ est consideacutereacute positif pour les turbines et neacutegatifpour les compresseurs 119882119882119890119890 = frasl119882 119888
Coefficient de charge
2Ψ = eWU
Coefficient de charge
2e
UW
=ψ 02 01 02 2
h h hU Uminus ∆
= = 02
s shU
η ∆=
=
eWWm
120578120578119904119904 =∆ℎ0∆ℎ0119904119904
Turbine
Coefficient de deacutebit
xcU
Φ =
xCU
Φ =ρRT
pρ
= xC RTRTU p
=RT m RT
U Ap
xCU
Φ =119888 = ρ119862119862119909119909119860119860
11988811987711987711987911987901199011199010119860119860
= 119872119872 120574120574 1 +120574120574 minus 1
21198721198722
minus 120574120574+1)2(120574120574minus1
Afin de geacuteneacuteraliser le triangle de vitesses drsquoune machineaxiale on effectue une mise agrave lrsquoeacutechelle en divisant chaquecocircteacute par la vitesse tangentielle U
Cette opeacuteration permettra de retrouver directement lescoefficients de charge ψ et de deacutebit Φ
Par la suite un nouveau paramegravetre R qursquoon appelle degreacutede reacuteaction sera introduit pour compleacuteter la caracteacuterisationde lrsquoaubage
Approche classique
α2
β2 α3
β3C2 W3
W2
c3u - c2u= w3u - w2u
W3uW2u
U
CX CX
Lrsquoopeacuteration est vectorielle
C3
120511120511 = frasl119934119934119940119940 119932119932120784120784 = frasl119914119914120785120785119958119958 minus 119914119914120784120784119958119958 119932119932
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
Turbine
3
2
3 3
2 2
12
12
12
12
u
u
u u
u u
c RUc RU
w c RU Uw c RU U
Ψ= minus +
Ψ= minus +
Ψ= minus = +
Ψ= minus = minus
2 22 23 2
2 22 23 2
1 12 2
2 2
c cR RU U
w wR RU U
Ψ Ψ = Φ + minus + = Φ + minus +
Ψ Ψ = Φ + + = Φ + minus
3 2
3 2
1 2 1 2
2 2
R Ratan atan
R Ratan atan
minus + Ψ minus + Ψ = = Φ Φ
Ψ + Ψ minus = = Φ Φ
α α
β β
α2β1
C2U
W1U
C1UW2U β2 α1
Compresseur
ΦC2U
C3UW3U
W2U
ψ
1
ψ
ΦC2UC3U
W3U
W2U
1
Il faut deacutefinir un troisiegraveme paramegravetre pour diffeacuterencier ces deux triangles de vitesses ayant le mecircme 120509120509 et le mecircme 120537120537
Les coefficients 120509120509 et 120537120537 ne suffisent pas pour deacutefinir de maniegravereunivoque le triangle de vitesses
Le degreacute de reacuteaction est un paramegravetre adimensionnel quiexprime la variation drsquoenthalpie subie par le fluide dans le rotorpar rapport agrave la chute totale drsquoenthalpie dans lrsquoeacutetage
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
Variation d enthalpie dans le rotorRVariation d enthalpie dans l eacutetage
=
❸❶
❷
Rotor
Stator
R=0 machine drsquoaction (Δhrotor=0 Δheacutetage=Δhstator)
R=1 machine agrave reacuteaction (Δhstator=0 Δheacutetage=Δhrotor)
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
R indique la reacutepartition du saut drsquoenthalpie (de pression) statique entre le stator et le rotor
Le degreacute de reacuteaction R deacutefini en fonction de variablesthermodynamiques (lrsquoenthalpie) peut aussi srsquoexprimer en fonctionde variables cineacutematiques preacutesentes dans le triangle de vitessesNotamment en fonction de la vitesse peacuteripheacuterique U et descomposantes des vitesses absolues et relatives
La vitesse absolue agrave lrsquoentreacutee du stator est eacutegale agrave celle agrave la sortie du rotor (c1= c3 eacutetage reacutepeacutetitif )
On neacuteglige les pertes et le transfert de chaleur dans le stator
1 2 3
ℎ3 minus ℎ1= ℎ3 + frasl11988811988832 2 minus ℎ1 minus frasl11988811988812 2 = ℎ03 minus ℎ01 = 120549120549ℎ0
ℎ03 minus ℎ01 = ℎ03 minus ℎ02 = 120549120549ℎ )0(3minus2
ℎ03 minus ℎ02 = ℎ3 minus ℎ1
119945119945120782120782120783120783 = 119945119945120782120782120784120784
frasl11988811988832 2 = frasl11988811988812 2
3 2
3 1
rotor
eacutetage
h h hRh h h
∆∆
minus= =
minus
2 20 3 2 2 3
0 3 2
2 2( )
( )
h c c h
∆∆
minus
minus
+ minus=
ℎ03 ℎ02
(ℎ01= ℎ02 1198881198881= 1198881198883)
120549120549ℎ )0(3minus2 = (ℎ3 + frasl11988811988832 2) minus (ℎ2 + frasl11988811988822 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
0(3 2) 3 2( )∆ minus = minusu uh U c c
= 1 minusfrasl11988811988832 2 minus frasl11988811988822 2120549120549ℎ )0(3minus2
119877119877 =120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
120549120549ℎ )0(3minus2
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
3 212+
= minus u uc cRU
119888119888119909119909 = 1198881198882119909119909 = 1198881198883119909119909 = 119888119888119888119888119888119888119888119888119888119888
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 11988811988821199061199061198881198882 = 1198881198881199091199092+1198881198881199061199062
119877119877 = 1 minus11988811988831199061199062 minus 11988811988821199061199062
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
Indirectement lrsquoeacutequation pour R deacutecrite en fonction de composantes cineacutematiques eacutetablit un lien entre les angles drsquoentreacutee et de sortie
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
= 1 minus1198881198883119906119906 + 1198881198882119906119906
2119880119880
2232
2 32 2wwh h+ = +
2 23 2
2 3 2 2w wh hminus = minus
2 2 2 23 3 2 2
2 3 2 2u x u xw w w wh h + +
minus = minus
Machine axiale U2 = U3
La composante axiale de la vitesse demeure constante w3x = w 2x
( eacutequation justifieacutee plus tard lors de la rothalpie)
2 23 2
2 3 2 2u uw wh hminus = minus
3 2 3 22 3
( )( )2
u u u uw w w wh h minus +minus =
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
Eacutetage reacutepeacutetitif c3 = c1
Si les pertes dans le stator sont neacutegligeacutees
c2u ndash c3u= w2u - w3u
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
3 2 3 23 2
( )( )2
u u u uw w w wh h minus +minus = minus
c2u ndash c3u= w2u - w3u
119877119877 = minus1199081199083119906119906 + 1199081199082119906119906
2119880119880
2 2 2 22 3 2 3 2 3( ) 2 ( ) 2h h u u w wminus = minus minus minus
2 2 2 2 2 23 2 3 2 3 2
2 2 2e
I II III
u u c c w wW minus minus minus= + minus
2 2 2 22 3 2 3
2 2 2 2 2 22 3 2 3 2 3
( ) ( )( ) ( ) ( )
minus minus minus=
minus + minus minus minusu u w wR
c c u u w w
3 2= =U u u
Expliqueacutee + tard
2 23 2
03 02 3 2 2ec cW h h h h minus
= minus = minus +
119877119877 =ℎ2 minus ℎ3ℎ1 minus ℎ3
=ℎ2 minus ℎ3ℎ02 minus ℎ03
Degreacute de reacuteaction
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
( )3u 2uw wR2U+
= minus
3 212
u uc cRU+
= minus
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
Puisque le degreacute de reacuteaction R peut aussi srsquoexprimer enfonction de variables cineacutematiques on peut chercher sarepreacutesentation dans le triangle adimensionnel de vitesses
Turbine
α2
β2 α3
β3
C2 W3
C3W2
C3uC2u
U
C2xC3x
W2u
c2u ndash c3u= w2u - w3u
2Ψ minusR
frasl119934119934120784120784119958119958 119932119932 = frasl1198621198622119906119906 119880119880 minus frasl119880119880 119880119880Ψ = frasl1198621198622119906119906 minus 1198621198623119906119906 119880119880
119877119877 = 1 minus (1198621198623119906119906 + 1198621198622119906119906 frasl) 2119880119880Φ = frasl1198621198622119909119909 119880119880 = frasl1198621198623119909119909 119880119880
= (119914119914120784120784119958119958 minus 119914119914120785120785119958119958 frasl) 2119880119880 minus 1 + (119914119914120784120784119958119958 + 119914119914120785120785119958119958 frasl) 2119880119880
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α2
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
Les repreacutesentations des triangles de vitesses adimensionnellessont tregraves riches en information et contiennent plusieurs formules
Par exemple pour un compresseur on deacuteduit aiseacutement que
Avec 120595120595 = 1198821198821198901198901198801198802
Φ = 119888119888119909119909119880119880
et que pour 119888119888119901119901 = 119888119888119888119888119888119888119888119888119888119888 119882119882119890119890 = 119888119888119901119901(11987911987902minus11987911987901)on trouve
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
(11987911987902minus11987911987901) =119880119880119888119888119909119909119888119888119901119901
(tan1205731205731 minus tan1205731205732)
Alors si lrsquoon veut obtenir une plus grande variation detempeacuterature 11987911987902 minus 11987911987901 indirectement un plus grand rapport(1199011199010211990111990101) on peut augmenter
bull la vitesse peacuteripheacuterique 119880119880 (la vitesse de rotation)bull la vitesse axiale 119888119888119909119909 (le deacutebit massique)bull lrsquoangle de deacuteflection de la pale(tan1205731205731 minus tan1205731205732)
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
W2
C2
U
U
W3C3
u
Les vecteurs vitesse (vitesse absolue 119940119940 pour le stator etrelative 119960119960 pour le rotor) sont consideacutereacutes tangents agrave laligne de squelette de lrsquoaube
β2
β3W3
W2
Pour les machines axiales (u= uent = usort ) le trianglede vitesse donne lieu agrave des eacutequations scalaires tellesque
cu= wu+u
cx= wxcx= = wx c
U
w
wu
cu
Pour les machines agrave eacutetages multiples on suppose que lavitesse axiale demeure constante Crsquoest agrave dire que119940119940120784120784119961119961= 119940119940120785120785119961119961= 119940119940119961119961= 119940119940119940119940119940119940119940119940119940119940
Cette condition jumeleacutee agrave 119932119932120784120784= 119932119932120785120785= 119932119932 megravene agrave lrsquohypothegravese drsquoune cineacutematique reacutepeacutetitive En pratique
Les triangles de vitesse agrave lrsquoentreacutee et agrave la sortie drsquouneacutetage peuvent ecirctre superposeacutes
Nous commenccedilons par la repreacutesentation de triangles de vitessespour le cas drsquoune turbineNous rappelons que par hypothegravese le rayon ne varie pas dans leplan inter-aube et que la vitesse axiale demeure constante dansle sens axialCes conditions permettent drsquoeffectuer la superposition destriangles de vitesse au bord drsquoattaque et bord de fuite du rotoravec la vitesse 119932119932 commune
c1w1
u
c2
u2
w2
➊
➋
➌ c3w3
u
Turbine
C2
W2 C3
W3
Stator Rotor
C2u C3u
W3u
α2β2 β3
α3
∆Cu= ∆Wu
U
W2u
119940119940119961119961119880119880 = 119888119888119888119888119888119888119888119888119888119888119888119888119909119909 = 119888119888119888119888119888119888119888119888119888119888
1198881198883 = 1198881198881
1205721205723 = 1205721205721
Cineacutematique reacutepeacutetitive
➁ ➂
➁U➀
Turbine
119914119914120784120784
119914119914120785120785
119934119934120785120785
119932119932119934119934120784120784
119932119932
119914119914120783120783120630120630120783120783
120630120630120784120784120631120631120784120784
120630120630120785120785
120631120631120785120785
Stator Rotor
➀ ➁ ➂
Le cas du compresseur est similaire agrave celui drsquoune turbineLe rayon ne varie pas (lrsquoeacutetude se deacuteroule dans le plan inter-aube) et la vitesse axiale demeure constante dans le sens axialComme pour la turbine nous pouvons superposer les triangles devitesse au bord drsquoattaque et bord de fuite du rotor
W2 C2
C3W3
U
C1W1
U
U
Compresseur
➊
➋
➌
α2 β1
C2 W1
C1W2
U
β2α1
W2U C1U
CX
( )2 1 = minuse u uW U C C
Lrsquoopeacuteration est vectorielle
C2U
119914119914120784120784119958119958 minus 119914119914120783120783119958119958 =- 119934119934120784120784119958119958 minus119934119934120783120783119958119958
➀
➁
➁
Compresseur
119914119914120784120784119914119914120785120785
119934119934120783120783
119932119932119914119914120783120783120630120630120783120783 120630120630120784120784
120631120631120784120784120630120630120785120785
Rotor Stator
119934119934120784120784120631120631120783120783
119932119932
➀ ➁ ➂
Trois quantiteacutes fondamentales sont utiliseacutees dans laconception des turbomachines
bull le coefficient de charge ψ
bull le coefficient de deacutebit Φ
bull le degreacute de reacuteaction R
Approche classique
Parfois ψ est consideacutereacute positif pour les turbines et neacutegatifpour les compresseurs 119882119882119890119890 = frasl119882 119888
Coefficient de charge
2Ψ = eWU
Coefficient de charge
2e
UW
=ψ 02 01 02 2
h h hU Uminus ∆
= = 02
s shU
η ∆=
=
eWWm
120578120578119904119904 =∆ℎ0∆ℎ0119904119904
Turbine
Coefficient de deacutebit
xcU
Φ =
xCU
Φ =ρRT
pρ
= xC RTRTU p
=RT m RT
U Ap
xCU
Φ =119888 = ρ119862119862119909119909119860119860
11988811987711987711987911987901199011199010119860119860
= 119872119872 120574120574 1 +120574120574 minus 1
21198721198722
minus 120574120574+1)2(120574120574minus1
Afin de geacuteneacuteraliser le triangle de vitesses drsquoune machineaxiale on effectue une mise agrave lrsquoeacutechelle en divisant chaquecocircteacute par la vitesse tangentielle U
Cette opeacuteration permettra de retrouver directement lescoefficients de charge ψ et de deacutebit Φ
Par la suite un nouveau paramegravetre R qursquoon appelle degreacutede reacuteaction sera introduit pour compleacuteter la caracteacuterisationde lrsquoaubage
Approche classique
α2
β2 α3
β3C2 W3
W2
c3u - c2u= w3u - w2u
W3uW2u
U
CX CX
Lrsquoopeacuteration est vectorielle
C3
120511120511 = frasl119934119934119940119940 119932119932120784120784 = frasl119914119914120785120785119958119958 minus 119914119914120784120784119958119958 119932119932
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
Turbine
3
2
3 3
2 2
12
12
12
12
u
u
u u
u u
c RUc RU
w c RU Uw c RU U
Ψ= minus +
Ψ= minus +
Ψ= minus = +
Ψ= minus = minus
2 22 23 2
2 22 23 2
1 12 2
2 2
c cR RU U
w wR RU U
Ψ Ψ = Φ + minus + = Φ + minus +
Ψ Ψ = Φ + + = Φ + minus
3 2
3 2
1 2 1 2
2 2
R Ratan atan
R Ratan atan
minus + Ψ minus + Ψ = = Φ Φ
Ψ + Ψ minus = = Φ Φ
α α
β β
α2β1
C2U
W1U
C1UW2U β2 α1
Compresseur
ΦC2U
C3UW3U
W2U
ψ
1
ψ
ΦC2UC3U
W3U
W2U
1
Il faut deacutefinir un troisiegraveme paramegravetre pour diffeacuterencier ces deux triangles de vitesses ayant le mecircme 120509120509 et le mecircme 120537120537
Les coefficients 120509120509 et 120537120537 ne suffisent pas pour deacutefinir de maniegravereunivoque le triangle de vitesses
Le degreacute de reacuteaction est un paramegravetre adimensionnel quiexprime la variation drsquoenthalpie subie par le fluide dans le rotorpar rapport agrave la chute totale drsquoenthalpie dans lrsquoeacutetage
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
Variation d enthalpie dans le rotorRVariation d enthalpie dans l eacutetage
=
❸❶
❷
Rotor
Stator
R=0 machine drsquoaction (Δhrotor=0 Δheacutetage=Δhstator)
R=1 machine agrave reacuteaction (Δhstator=0 Δheacutetage=Δhrotor)
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
R indique la reacutepartition du saut drsquoenthalpie (de pression) statique entre le stator et le rotor
Le degreacute de reacuteaction R deacutefini en fonction de variablesthermodynamiques (lrsquoenthalpie) peut aussi srsquoexprimer en fonctionde variables cineacutematiques preacutesentes dans le triangle de vitessesNotamment en fonction de la vitesse peacuteripheacuterique U et descomposantes des vitesses absolues et relatives
La vitesse absolue agrave lrsquoentreacutee du stator est eacutegale agrave celle agrave la sortie du rotor (c1= c3 eacutetage reacutepeacutetitif )
On neacuteglige les pertes et le transfert de chaleur dans le stator
1 2 3
ℎ3 minus ℎ1= ℎ3 + frasl11988811988832 2 minus ℎ1 minus frasl11988811988812 2 = ℎ03 minus ℎ01 = 120549120549ℎ0
ℎ03 minus ℎ01 = ℎ03 minus ℎ02 = 120549120549ℎ )0(3minus2
ℎ03 minus ℎ02 = ℎ3 minus ℎ1
119945119945120782120782120783120783 = 119945119945120782120782120784120784
frasl11988811988832 2 = frasl11988811988812 2
3 2
3 1
rotor
eacutetage
h h hRh h h
∆∆
minus= =
minus
2 20 3 2 2 3
0 3 2
2 2( )
( )
h c c h
∆∆
minus
minus
+ minus=
ℎ03 ℎ02
(ℎ01= ℎ02 1198881198881= 1198881198883)
120549120549ℎ )0(3minus2 = (ℎ3 + frasl11988811988832 2) minus (ℎ2 + frasl11988811988822 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
0(3 2) 3 2( )∆ minus = minusu uh U c c
= 1 minusfrasl11988811988832 2 minus frasl11988811988822 2120549120549ℎ )0(3minus2
119877119877 =120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
120549120549ℎ )0(3minus2
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
3 212+
= minus u uc cRU
119888119888119909119909 = 1198881198882119909119909 = 1198881198883119909119909 = 119888119888119888119888119888119888119888119888119888119888
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 11988811988821199061199061198881198882 = 1198881198881199091199092+1198881198881199061199062
119877119877 = 1 minus11988811988831199061199062 minus 11988811988821199061199062
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
Indirectement lrsquoeacutequation pour R deacutecrite en fonction de composantes cineacutematiques eacutetablit un lien entre les angles drsquoentreacutee et de sortie
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
= 1 minus1198881198883119906119906 + 1198881198882119906119906
2119880119880
2232
2 32 2wwh h+ = +
2 23 2
2 3 2 2w wh hminus = minus
2 2 2 23 3 2 2
2 3 2 2u x u xw w w wh h + +
minus = minus
Machine axiale U2 = U3
La composante axiale de la vitesse demeure constante w3x = w 2x
( eacutequation justifieacutee plus tard lors de la rothalpie)
2 23 2
2 3 2 2u uw wh hminus = minus
3 2 3 22 3
( )( )2
u u u uw w w wh h minus +minus =
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
Eacutetage reacutepeacutetitif c3 = c1
Si les pertes dans le stator sont neacutegligeacutees
c2u ndash c3u= w2u - w3u
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
3 2 3 23 2
( )( )2
u u u uw w w wh h minus +minus = minus
c2u ndash c3u= w2u - w3u
119877119877 = minus1199081199083119906119906 + 1199081199082119906119906
2119880119880
2 2 2 22 3 2 3 2 3( ) 2 ( ) 2h h u u w wminus = minus minus minus
2 2 2 2 2 23 2 3 2 3 2
2 2 2e
I II III
u u c c w wW minus minus minus= + minus
2 2 2 22 3 2 3
2 2 2 2 2 22 3 2 3 2 3
( ) ( )( ) ( ) ( )
minus minus minus=
minus + minus minus minusu u w wR
c c u u w w
3 2= =U u u
Expliqueacutee + tard
2 23 2
03 02 3 2 2ec cW h h h h minus
= minus = minus +
119877119877 =ℎ2 minus ℎ3ℎ1 minus ℎ3
=ℎ2 minus ℎ3ℎ02 minus ℎ03
Degreacute de reacuteaction
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
( )3u 2uw wR2U+
= minus
3 212
u uc cRU+
= minus
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
Puisque le degreacute de reacuteaction R peut aussi srsquoexprimer enfonction de variables cineacutematiques on peut chercher sarepreacutesentation dans le triangle adimensionnel de vitesses
Turbine
α2
β2 α3
β3
C2 W3
C3W2
C3uC2u
U
C2xC3x
W2u
c2u ndash c3u= w2u - w3u
2Ψ minusR
frasl119934119934120784120784119958119958 119932119932 = frasl1198621198622119906119906 119880119880 minus frasl119880119880 119880119880Ψ = frasl1198621198622119906119906 minus 1198621198623119906119906 119880119880
119877119877 = 1 minus (1198621198623119906119906 + 1198621198622119906119906 frasl) 2119880119880Φ = frasl1198621198622119909119909 119880119880 = frasl1198621198623119909119909 119880119880
= (119914119914120784120784119958119958 minus 119914119914120785120785119958119958 frasl) 2119880119880 minus 1 + (119914119914120784120784119958119958 + 119914119914120785120785119958119958 frasl) 2119880119880
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α2
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
Les repreacutesentations des triangles de vitesses adimensionnellessont tregraves riches en information et contiennent plusieurs formules
Par exemple pour un compresseur on deacuteduit aiseacutement que
Avec 120595120595 = 1198821198821198901198901198801198802
Φ = 119888119888119909119909119880119880
et que pour 119888119888119901119901 = 119888119888119888119888119888119888119888119888119888119888 119882119882119890119890 = 119888119888119901119901(11987911987902minus11987911987901)on trouve
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
(11987911987902minus11987911987901) =119880119880119888119888119909119909119888119888119901119901
(tan1205731205731 minus tan1205731205732)
Alors si lrsquoon veut obtenir une plus grande variation detempeacuterature 11987911987902 minus 11987911987901 indirectement un plus grand rapport(1199011199010211990111990101) on peut augmenter
bull la vitesse peacuteripheacuterique 119880119880 (la vitesse de rotation)bull la vitesse axiale 119888119888119909119909 (le deacutebit massique)bull lrsquoangle de deacuteflection de la pale(tan1205731205731 minus tan1205731205732)
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
Download Sound Effects - SoundDogs - Voice Box
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
Les vecteurs vitesse (vitesse absolue 119940119940 pour le stator etrelative 119960119960 pour le rotor) sont consideacutereacutes tangents agrave laligne de squelette de lrsquoaube
β2
β3W3
W2
Pour les machines axiales (u= uent = usort ) le trianglede vitesse donne lieu agrave des eacutequations scalaires tellesque
cu= wu+u
cx= wxcx= = wx c
U
w
wu
cu
Pour les machines agrave eacutetages multiples on suppose que lavitesse axiale demeure constante Crsquoest agrave dire que119940119940120784120784119961119961= 119940119940120785120785119961119961= 119940119940119961119961= 119940119940119940119940119940119940119940119940119940119940
Cette condition jumeleacutee agrave 119932119932120784120784= 119932119932120785120785= 119932119932 megravene agrave lrsquohypothegravese drsquoune cineacutematique reacutepeacutetitive En pratique
Les triangles de vitesse agrave lrsquoentreacutee et agrave la sortie drsquouneacutetage peuvent ecirctre superposeacutes
Nous commenccedilons par la repreacutesentation de triangles de vitessespour le cas drsquoune turbineNous rappelons que par hypothegravese le rayon ne varie pas dans leplan inter-aube et que la vitesse axiale demeure constante dansle sens axialCes conditions permettent drsquoeffectuer la superposition destriangles de vitesse au bord drsquoattaque et bord de fuite du rotoravec la vitesse 119932119932 commune
c1w1
u
c2
u2
w2
➊
➋
➌ c3w3
u
Turbine
C2
W2 C3
W3
Stator Rotor
C2u C3u
W3u
α2β2 β3
α3
∆Cu= ∆Wu
U
W2u
119940119940119961119961119880119880 = 119888119888119888119888119888119888119888119888119888119888119888119888119909119909 = 119888119888119888119888119888119888119888119888119888119888
1198881198883 = 1198881198881
1205721205723 = 1205721205721
Cineacutematique reacutepeacutetitive
➁ ➂
➁U➀
Turbine
119914119914120784120784
119914119914120785120785
119934119934120785120785
119932119932119934119934120784120784
119932119932
119914119914120783120783120630120630120783120783
120630120630120784120784120631120631120784120784
120630120630120785120785
120631120631120785120785
Stator Rotor
➀ ➁ ➂
Le cas du compresseur est similaire agrave celui drsquoune turbineLe rayon ne varie pas (lrsquoeacutetude se deacuteroule dans le plan inter-aube) et la vitesse axiale demeure constante dans le sens axialComme pour la turbine nous pouvons superposer les triangles devitesse au bord drsquoattaque et bord de fuite du rotor
W2 C2
C3W3
U
C1W1
U
U
Compresseur
➊
➋
➌
α2 β1
C2 W1
C1W2
U
β2α1
W2U C1U
CX
( )2 1 = minuse u uW U C C
Lrsquoopeacuteration est vectorielle
C2U
119914119914120784120784119958119958 minus 119914119914120783120783119958119958 =- 119934119934120784120784119958119958 minus119934119934120783120783119958119958
➀
➁
➁
Compresseur
119914119914120784120784119914119914120785120785
119934119934120783120783
119932119932119914119914120783120783120630120630120783120783 120630120630120784120784
120631120631120784120784120630120630120785120785
Rotor Stator
119934119934120784120784120631120631120783120783
119932119932
➀ ➁ ➂
Trois quantiteacutes fondamentales sont utiliseacutees dans laconception des turbomachines
bull le coefficient de charge ψ
bull le coefficient de deacutebit Φ
bull le degreacute de reacuteaction R
Approche classique
Parfois ψ est consideacutereacute positif pour les turbines et neacutegatifpour les compresseurs 119882119882119890119890 = frasl119882 119888
Coefficient de charge
2Ψ = eWU
Coefficient de charge
2e
UW
=ψ 02 01 02 2
h h hU Uminus ∆
= = 02
s shU
η ∆=
=
eWWm
120578120578119904119904 =∆ℎ0∆ℎ0119904119904
Turbine
Coefficient de deacutebit
xcU
Φ =
xCU
Φ =ρRT
pρ
= xC RTRTU p
=RT m RT
U Ap
xCU
Φ =119888 = ρ119862119862119909119909119860119860
11988811987711987711987911987901199011199010119860119860
= 119872119872 120574120574 1 +120574120574 minus 1
21198721198722
minus 120574120574+1)2(120574120574minus1
Afin de geacuteneacuteraliser le triangle de vitesses drsquoune machineaxiale on effectue une mise agrave lrsquoeacutechelle en divisant chaquecocircteacute par la vitesse tangentielle U
Cette opeacuteration permettra de retrouver directement lescoefficients de charge ψ et de deacutebit Φ
Par la suite un nouveau paramegravetre R qursquoon appelle degreacutede reacuteaction sera introduit pour compleacuteter la caracteacuterisationde lrsquoaubage
Approche classique
α2
β2 α3
β3C2 W3
W2
c3u - c2u= w3u - w2u
W3uW2u
U
CX CX
Lrsquoopeacuteration est vectorielle
C3
120511120511 = frasl119934119934119940119940 119932119932120784120784 = frasl119914119914120785120785119958119958 minus 119914119914120784120784119958119958 119932119932
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
Turbine
3
2
3 3
2 2
12
12
12
12
u
u
u u
u u
c RUc RU
w c RU Uw c RU U
Ψ= minus +
Ψ= minus +
Ψ= minus = +
Ψ= minus = minus
2 22 23 2
2 22 23 2
1 12 2
2 2
c cR RU U
w wR RU U
Ψ Ψ = Φ + minus + = Φ + minus +
Ψ Ψ = Φ + + = Φ + minus
3 2
3 2
1 2 1 2
2 2
R Ratan atan
R Ratan atan
minus + Ψ minus + Ψ = = Φ Φ
Ψ + Ψ minus = = Φ Φ
α α
β β
α2β1
C2U
W1U
C1UW2U β2 α1
Compresseur
ΦC2U
C3UW3U
W2U
ψ
1
ψ
ΦC2UC3U
W3U
W2U
1
Il faut deacutefinir un troisiegraveme paramegravetre pour diffeacuterencier ces deux triangles de vitesses ayant le mecircme 120509120509 et le mecircme 120537120537
Les coefficients 120509120509 et 120537120537 ne suffisent pas pour deacutefinir de maniegravereunivoque le triangle de vitesses
Le degreacute de reacuteaction est un paramegravetre adimensionnel quiexprime la variation drsquoenthalpie subie par le fluide dans le rotorpar rapport agrave la chute totale drsquoenthalpie dans lrsquoeacutetage
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
Variation d enthalpie dans le rotorRVariation d enthalpie dans l eacutetage
=
❸❶
❷
Rotor
Stator
R=0 machine drsquoaction (Δhrotor=0 Δheacutetage=Δhstator)
R=1 machine agrave reacuteaction (Δhstator=0 Δheacutetage=Δhrotor)
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
R indique la reacutepartition du saut drsquoenthalpie (de pression) statique entre le stator et le rotor
Le degreacute de reacuteaction R deacutefini en fonction de variablesthermodynamiques (lrsquoenthalpie) peut aussi srsquoexprimer en fonctionde variables cineacutematiques preacutesentes dans le triangle de vitessesNotamment en fonction de la vitesse peacuteripheacuterique U et descomposantes des vitesses absolues et relatives
La vitesse absolue agrave lrsquoentreacutee du stator est eacutegale agrave celle agrave la sortie du rotor (c1= c3 eacutetage reacutepeacutetitif )
On neacuteglige les pertes et le transfert de chaleur dans le stator
1 2 3
ℎ3 minus ℎ1= ℎ3 + frasl11988811988832 2 minus ℎ1 minus frasl11988811988812 2 = ℎ03 minus ℎ01 = 120549120549ℎ0
ℎ03 minus ℎ01 = ℎ03 minus ℎ02 = 120549120549ℎ )0(3minus2
ℎ03 minus ℎ02 = ℎ3 minus ℎ1
119945119945120782120782120783120783 = 119945119945120782120782120784120784
frasl11988811988832 2 = frasl11988811988812 2
3 2
3 1
rotor
eacutetage
h h hRh h h
∆∆
minus= =
minus
2 20 3 2 2 3
0 3 2
2 2( )
( )
h c c h
∆∆
minus
minus
+ minus=
ℎ03 ℎ02
(ℎ01= ℎ02 1198881198881= 1198881198883)
120549120549ℎ )0(3minus2 = (ℎ3 + frasl11988811988832 2) minus (ℎ2 + frasl11988811988822 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
0(3 2) 3 2( )∆ minus = minusu uh U c c
= 1 minusfrasl11988811988832 2 minus frasl11988811988822 2120549120549ℎ )0(3minus2
119877119877 =120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
120549120549ℎ )0(3minus2
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
3 212+
= minus u uc cRU
119888119888119909119909 = 1198881198882119909119909 = 1198881198883119909119909 = 119888119888119888119888119888119888119888119888119888119888
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 11988811988821199061199061198881198882 = 1198881198881199091199092+1198881198881199061199062
119877119877 = 1 minus11988811988831199061199062 minus 11988811988821199061199062
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
Indirectement lrsquoeacutequation pour R deacutecrite en fonction de composantes cineacutematiques eacutetablit un lien entre les angles drsquoentreacutee et de sortie
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
= 1 minus1198881198883119906119906 + 1198881198882119906119906
2119880119880
2232
2 32 2wwh h+ = +
2 23 2
2 3 2 2w wh hminus = minus
2 2 2 23 3 2 2
2 3 2 2u x u xw w w wh h + +
minus = minus
Machine axiale U2 = U3
La composante axiale de la vitesse demeure constante w3x = w 2x
( eacutequation justifieacutee plus tard lors de la rothalpie)
2 23 2
2 3 2 2u uw wh hminus = minus
3 2 3 22 3
( )( )2
u u u uw w w wh h minus +minus =
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
Eacutetage reacutepeacutetitif c3 = c1
Si les pertes dans le stator sont neacutegligeacutees
c2u ndash c3u= w2u - w3u
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
3 2 3 23 2
( )( )2
u u u uw w w wh h minus +minus = minus
c2u ndash c3u= w2u - w3u
119877119877 = minus1199081199083119906119906 + 1199081199082119906119906
2119880119880
2 2 2 22 3 2 3 2 3( ) 2 ( ) 2h h u u w wminus = minus minus minus
2 2 2 2 2 23 2 3 2 3 2
2 2 2e
I II III
u u c c w wW minus minus minus= + minus
2 2 2 22 3 2 3
2 2 2 2 2 22 3 2 3 2 3
( ) ( )( ) ( ) ( )
minus minus minus=
minus + minus minus minusu u w wR
c c u u w w
3 2= =U u u
Expliqueacutee + tard
2 23 2
03 02 3 2 2ec cW h h h h minus
= minus = minus +
119877119877 =ℎ2 minus ℎ3ℎ1 minus ℎ3
=ℎ2 minus ℎ3ℎ02 minus ℎ03
Degreacute de reacuteaction
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
( )3u 2uw wR2U+
= minus
3 212
u uc cRU+
= minus
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
Puisque le degreacute de reacuteaction R peut aussi srsquoexprimer enfonction de variables cineacutematiques on peut chercher sarepreacutesentation dans le triangle adimensionnel de vitesses
Turbine
α2
β2 α3
β3
C2 W3
C3W2
C3uC2u
U
C2xC3x
W2u
c2u ndash c3u= w2u - w3u
2Ψ minusR
frasl119934119934120784120784119958119958 119932119932 = frasl1198621198622119906119906 119880119880 minus frasl119880119880 119880119880Ψ = frasl1198621198622119906119906 minus 1198621198623119906119906 119880119880
119877119877 = 1 minus (1198621198623119906119906 + 1198621198622119906119906 frasl) 2119880119880Φ = frasl1198621198622119909119909 119880119880 = frasl1198621198623119909119909 119880119880
= (119914119914120784120784119958119958 minus 119914119914120785120785119958119958 frasl) 2119880119880 minus 1 + (119914119914120784120784119958119958 + 119914119914120785120785119958119958 frasl) 2119880119880
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α2
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
Les repreacutesentations des triangles de vitesses adimensionnellessont tregraves riches en information et contiennent plusieurs formules
Par exemple pour un compresseur on deacuteduit aiseacutement que
Avec 120595120595 = 1198821198821198901198901198801198802
Φ = 119888119888119909119909119880119880
et que pour 119888119888119901119901 = 119888119888119888119888119888119888119888119888119888119888 119882119882119890119890 = 119888119888119901119901(11987911987902minus11987911987901)on trouve
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
(11987911987902minus11987911987901) =119880119880119888119888119909119909119888119888119901119901
(tan1205731205731 minus tan1205731205732)
Alors si lrsquoon veut obtenir une plus grande variation detempeacuterature 11987911987902 minus 11987911987901 indirectement un plus grand rapport(1199011199010211990111990101) on peut augmenter
bull la vitesse peacuteripheacuterique 119880119880 (la vitesse de rotation)bull la vitesse axiale 119888119888119909119909 (le deacutebit massique)bull lrsquoangle de deacuteflection de la pale(tan1205731205731 minus tan1205731205732)
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
Pour les machines axiales (u= uent = usort ) le trianglede vitesse donne lieu agrave des eacutequations scalaires tellesque
cu= wu+u
cx= wxcx= = wx c
U
w
wu
cu
Pour les machines agrave eacutetages multiples on suppose que lavitesse axiale demeure constante Crsquoest agrave dire que119940119940120784120784119961119961= 119940119940120785120785119961119961= 119940119940119961119961= 119940119940119940119940119940119940119940119940119940119940
Cette condition jumeleacutee agrave 119932119932120784120784= 119932119932120785120785= 119932119932 megravene agrave lrsquohypothegravese drsquoune cineacutematique reacutepeacutetitive En pratique
Les triangles de vitesse agrave lrsquoentreacutee et agrave la sortie drsquouneacutetage peuvent ecirctre superposeacutes
Nous commenccedilons par la repreacutesentation de triangles de vitessespour le cas drsquoune turbineNous rappelons que par hypothegravese le rayon ne varie pas dans leplan inter-aube et que la vitesse axiale demeure constante dansle sens axialCes conditions permettent drsquoeffectuer la superposition destriangles de vitesse au bord drsquoattaque et bord de fuite du rotoravec la vitesse 119932119932 commune
c1w1
u
c2
u2
w2
➊
➋
➌ c3w3
u
Turbine
C2
W2 C3
W3
Stator Rotor
C2u C3u
W3u
α2β2 β3
α3
∆Cu= ∆Wu
U
W2u
119940119940119961119961119880119880 = 119888119888119888119888119888119888119888119888119888119888119888119888119909119909 = 119888119888119888119888119888119888119888119888119888119888
1198881198883 = 1198881198881
1205721205723 = 1205721205721
Cineacutematique reacutepeacutetitive
➁ ➂
➁U➀
Turbine
119914119914120784120784
119914119914120785120785
119934119934120785120785
119932119932119934119934120784120784
119932119932
119914119914120783120783120630120630120783120783
120630120630120784120784120631120631120784120784
120630120630120785120785
120631120631120785120785
Stator Rotor
➀ ➁ ➂
Le cas du compresseur est similaire agrave celui drsquoune turbineLe rayon ne varie pas (lrsquoeacutetude se deacuteroule dans le plan inter-aube) et la vitesse axiale demeure constante dans le sens axialComme pour la turbine nous pouvons superposer les triangles devitesse au bord drsquoattaque et bord de fuite du rotor
W2 C2
C3W3
U
C1W1
U
U
Compresseur
➊
➋
➌
α2 β1
C2 W1
C1W2
U
β2α1
W2U C1U
CX
( )2 1 = minuse u uW U C C
Lrsquoopeacuteration est vectorielle
C2U
119914119914120784120784119958119958 minus 119914119914120783120783119958119958 =- 119934119934120784120784119958119958 minus119934119934120783120783119958119958
➀
➁
➁
Compresseur
119914119914120784120784119914119914120785120785
119934119934120783120783
119932119932119914119914120783120783120630120630120783120783 120630120630120784120784
120631120631120784120784120630120630120785120785
Rotor Stator
119934119934120784120784120631120631120783120783
119932119932
➀ ➁ ➂
Trois quantiteacutes fondamentales sont utiliseacutees dans laconception des turbomachines
bull le coefficient de charge ψ
bull le coefficient de deacutebit Φ
bull le degreacute de reacuteaction R
Approche classique
Parfois ψ est consideacutereacute positif pour les turbines et neacutegatifpour les compresseurs 119882119882119890119890 = frasl119882 119888
Coefficient de charge
2Ψ = eWU
Coefficient de charge
2e
UW
=ψ 02 01 02 2
h h hU Uminus ∆
= = 02
s shU
η ∆=
=
eWWm
120578120578119904119904 =∆ℎ0∆ℎ0119904119904
Turbine
Coefficient de deacutebit
xcU
Φ =
xCU
Φ =ρRT
pρ
= xC RTRTU p
=RT m RT
U Ap
xCU
Φ =119888 = ρ119862119862119909119909119860119860
11988811987711987711987911987901199011199010119860119860
= 119872119872 120574120574 1 +120574120574 minus 1
21198721198722
minus 120574120574+1)2(120574120574minus1
Afin de geacuteneacuteraliser le triangle de vitesses drsquoune machineaxiale on effectue une mise agrave lrsquoeacutechelle en divisant chaquecocircteacute par la vitesse tangentielle U
Cette opeacuteration permettra de retrouver directement lescoefficients de charge ψ et de deacutebit Φ
Par la suite un nouveau paramegravetre R qursquoon appelle degreacutede reacuteaction sera introduit pour compleacuteter la caracteacuterisationde lrsquoaubage
Approche classique
α2
β2 α3
β3C2 W3
W2
c3u - c2u= w3u - w2u
W3uW2u
U
CX CX
Lrsquoopeacuteration est vectorielle
C3
120511120511 = frasl119934119934119940119940 119932119932120784120784 = frasl119914119914120785120785119958119958 minus 119914119914120784120784119958119958 119932119932
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
Turbine
3
2
3 3
2 2
12
12
12
12
u
u
u u
u u
c RUc RU
w c RU Uw c RU U
Ψ= minus +
Ψ= minus +
Ψ= minus = +
Ψ= minus = minus
2 22 23 2
2 22 23 2
1 12 2
2 2
c cR RU U
w wR RU U
Ψ Ψ = Φ + minus + = Φ + minus +
Ψ Ψ = Φ + + = Φ + minus
3 2
3 2
1 2 1 2
2 2
R Ratan atan
R Ratan atan
minus + Ψ minus + Ψ = = Φ Φ
Ψ + Ψ minus = = Φ Φ
α α
β β
α2β1
C2U
W1U
C1UW2U β2 α1
Compresseur
ΦC2U
C3UW3U
W2U
ψ
1
ψ
ΦC2UC3U
W3U
W2U
1
Il faut deacutefinir un troisiegraveme paramegravetre pour diffeacuterencier ces deux triangles de vitesses ayant le mecircme 120509120509 et le mecircme 120537120537
Les coefficients 120509120509 et 120537120537 ne suffisent pas pour deacutefinir de maniegravereunivoque le triangle de vitesses
Le degreacute de reacuteaction est un paramegravetre adimensionnel quiexprime la variation drsquoenthalpie subie par le fluide dans le rotorpar rapport agrave la chute totale drsquoenthalpie dans lrsquoeacutetage
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
Variation d enthalpie dans le rotorRVariation d enthalpie dans l eacutetage
=
❸❶
❷
Rotor
Stator
R=0 machine drsquoaction (Δhrotor=0 Δheacutetage=Δhstator)
R=1 machine agrave reacuteaction (Δhstator=0 Δheacutetage=Δhrotor)
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
R indique la reacutepartition du saut drsquoenthalpie (de pression) statique entre le stator et le rotor
Le degreacute de reacuteaction R deacutefini en fonction de variablesthermodynamiques (lrsquoenthalpie) peut aussi srsquoexprimer en fonctionde variables cineacutematiques preacutesentes dans le triangle de vitessesNotamment en fonction de la vitesse peacuteripheacuterique U et descomposantes des vitesses absolues et relatives
La vitesse absolue agrave lrsquoentreacutee du stator est eacutegale agrave celle agrave la sortie du rotor (c1= c3 eacutetage reacutepeacutetitif )
On neacuteglige les pertes et le transfert de chaleur dans le stator
1 2 3
ℎ3 minus ℎ1= ℎ3 + frasl11988811988832 2 minus ℎ1 minus frasl11988811988812 2 = ℎ03 minus ℎ01 = 120549120549ℎ0
ℎ03 minus ℎ01 = ℎ03 minus ℎ02 = 120549120549ℎ )0(3minus2
ℎ03 minus ℎ02 = ℎ3 minus ℎ1
119945119945120782120782120783120783 = 119945119945120782120782120784120784
frasl11988811988832 2 = frasl11988811988812 2
3 2
3 1
rotor
eacutetage
h h hRh h h
∆∆
minus= =
minus
2 20 3 2 2 3
0 3 2
2 2( )
( )
h c c h
∆∆
minus
minus
+ minus=
ℎ03 ℎ02
(ℎ01= ℎ02 1198881198881= 1198881198883)
120549120549ℎ )0(3minus2 = (ℎ3 + frasl11988811988832 2) minus (ℎ2 + frasl11988811988822 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
0(3 2) 3 2( )∆ minus = minusu uh U c c
= 1 minusfrasl11988811988832 2 minus frasl11988811988822 2120549120549ℎ )0(3minus2
119877119877 =120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
120549120549ℎ )0(3minus2
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
3 212+
= minus u uc cRU
119888119888119909119909 = 1198881198882119909119909 = 1198881198883119909119909 = 119888119888119888119888119888119888119888119888119888119888
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 11988811988821199061199061198881198882 = 1198881198881199091199092+1198881198881199061199062
119877119877 = 1 minus11988811988831199061199062 minus 11988811988821199061199062
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
Indirectement lrsquoeacutequation pour R deacutecrite en fonction de composantes cineacutematiques eacutetablit un lien entre les angles drsquoentreacutee et de sortie
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
= 1 minus1198881198883119906119906 + 1198881198882119906119906
2119880119880
2232
2 32 2wwh h+ = +
2 23 2
2 3 2 2w wh hminus = minus
2 2 2 23 3 2 2
2 3 2 2u x u xw w w wh h + +
minus = minus
Machine axiale U2 = U3
La composante axiale de la vitesse demeure constante w3x = w 2x
( eacutequation justifieacutee plus tard lors de la rothalpie)
2 23 2
2 3 2 2u uw wh hminus = minus
3 2 3 22 3
( )( )2
u u u uw w w wh h minus +minus =
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
Eacutetage reacutepeacutetitif c3 = c1
Si les pertes dans le stator sont neacutegligeacutees
c2u ndash c3u= w2u - w3u
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
3 2 3 23 2
( )( )2
u u u uw w w wh h minus +minus = minus
c2u ndash c3u= w2u - w3u
119877119877 = minus1199081199083119906119906 + 1199081199082119906119906
2119880119880
2 2 2 22 3 2 3 2 3( ) 2 ( ) 2h h u u w wminus = minus minus minus
2 2 2 2 2 23 2 3 2 3 2
2 2 2e
I II III
u u c c w wW minus minus minus= + minus
2 2 2 22 3 2 3
2 2 2 2 2 22 3 2 3 2 3
( ) ( )( ) ( ) ( )
minus minus minus=
minus + minus minus minusu u w wR
c c u u w w
3 2= =U u u
Expliqueacutee + tard
2 23 2
03 02 3 2 2ec cW h h h h minus
= minus = minus +
119877119877 =ℎ2 minus ℎ3ℎ1 minus ℎ3
=ℎ2 minus ℎ3ℎ02 minus ℎ03
Degreacute de reacuteaction
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
( )3u 2uw wR2U+
= minus
3 212
u uc cRU+
= minus
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
Puisque le degreacute de reacuteaction R peut aussi srsquoexprimer enfonction de variables cineacutematiques on peut chercher sarepreacutesentation dans le triangle adimensionnel de vitesses
Turbine
α2
β2 α3
β3
C2 W3
C3W2
C3uC2u
U
C2xC3x
W2u
c2u ndash c3u= w2u - w3u
2Ψ minusR
frasl119934119934120784120784119958119958 119932119932 = frasl1198621198622119906119906 119880119880 minus frasl119880119880 119880119880Ψ = frasl1198621198622119906119906 minus 1198621198623119906119906 119880119880
119877119877 = 1 minus (1198621198623119906119906 + 1198621198622119906119906 frasl) 2119880119880Φ = frasl1198621198622119909119909 119880119880 = frasl1198621198623119909119909 119880119880
= (119914119914120784120784119958119958 minus 119914119914120785120785119958119958 frasl) 2119880119880 minus 1 + (119914119914120784120784119958119958 + 119914119914120785120785119958119958 frasl) 2119880119880
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α2
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
Les repreacutesentations des triangles de vitesses adimensionnellessont tregraves riches en information et contiennent plusieurs formules
Par exemple pour un compresseur on deacuteduit aiseacutement que
Avec 120595120595 = 1198821198821198901198901198801198802
Φ = 119888119888119909119909119880119880
et que pour 119888119888119901119901 = 119888119888119888119888119888119888119888119888119888119888 119882119882119890119890 = 119888119888119901119901(11987911987902minus11987911987901)on trouve
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
(11987911987902minus11987911987901) =119880119880119888119888119909119909119888119888119901119901
(tan1205731205731 minus tan1205731205732)
Alors si lrsquoon veut obtenir une plus grande variation detempeacuterature 11987911987902 minus 11987911987901 indirectement un plus grand rapport(1199011199010211990111990101) on peut augmenter
bull la vitesse peacuteripheacuterique 119880119880 (la vitesse de rotation)bull la vitesse axiale 119888119888119909119909 (le deacutebit massique)bull lrsquoangle de deacuteflection de la pale(tan1205731205731 minus tan1205731205732)
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
Pour les machines agrave eacutetages multiples on suppose que lavitesse axiale demeure constante Crsquoest agrave dire que119940119940120784120784119961119961= 119940119940120785120785119961119961= 119940119940119961119961= 119940119940119940119940119940119940119940119940119940119940
Cette condition jumeleacutee agrave 119932119932120784120784= 119932119932120785120785= 119932119932 megravene agrave lrsquohypothegravese drsquoune cineacutematique reacutepeacutetitive En pratique
Les triangles de vitesse agrave lrsquoentreacutee et agrave la sortie drsquouneacutetage peuvent ecirctre superposeacutes
Nous commenccedilons par la repreacutesentation de triangles de vitessespour le cas drsquoune turbineNous rappelons que par hypothegravese le rayon ne varie pas dans leplan inter-aube et que la vitesse axiale demeure constante dansle sens axialCes conditions permettent drsquoeffectuer la superposition destriangles de vitesse au bord drsquoattaque et bord de fuite du rotoravec la vitesse 119932119932 commune
c1w1
u
c2
u2
w2
➊
➋
➌ c3w3
u
Turbine
C2
W2 C3
W3
Stator Rotor
C2u C3u
W3u
α2β2 β3
α3
∆Cu= ∆Wu
U
W2u
119940119940119961119961119880119880 = 119888119888119888119888119888119888119888119888119888119888119888119888119909119909 = 119888119888119888119888119888119888119888119888119888119888
1198881198883 = 1198881198881
1205721205723 = 1205721205721
Cineacutematique reacutepeacutetitive
➁ ➂
➁U➀
Turbine
119914119914120784120784
119914119914120785120785
119934119934120785120785
119932119932119934119934120784120784
119932119932
119914119914120783120783120630120630120783120783
120630120630120784120784120631120631120784120784
120630120630120785120785
120631120631120785120785
Stator Rotor
➀ ➁ ➂
Le cas du compresseur est similaire agrave celui drsquoune turbineLe rayon ne varie pas (lrsquoeacutetude se deacuteroule dans le plan inter-aube) et la vitesse axiale demeure constante dans le sens axialComme pour la turbine nous pouvons superposer les triangles devitesse au bord drsquoattaque et bord de fuite du rotor
W2 C2
C3W3
U
C1W1
U
U
Compresseur
➊
➋
➌
α2 β1
C2 W1
C1W2
U
β2α1
W2U C1U
CX
( )2 1 = minuse u uW U C C
Lrsquoopeacuteration est vectorielle
C2U
119914119914120784120784119958119958 minus 119914119914120783120783119958119958 =- 119934119934120784120784119958119958 minus119934119934120783120783119958119958
➀
➁
➁
Compresseur
119914119914120784120784119914119914120785120785
119934119934120783120783
119932119932119914119914120783120783120630120630120783120783 120630120630120784120784
120631120631120784120784120630120630120785120785
Rotor Stator
119934119934120784120784120631120631120783120783
119932119932
➀ ➁ ➂
Trois quantiteacutes fondamentales sont utiliseacutees dans laconception des turbomachines
bull le coefficient de charge ψ
bull le coefficient de deacutebit Φ
bull le degreacute de reacuteaction R
Approche classique
Parfois ψ est consideacutereacute positif pour les turbines et neacutegatifpour les compresseurs 119882119882119890119890 = frasl119882 119888
Coefficient de charge
2Ψ = eWU
Coefficient de charge
2e
UW
=ψ 02 01 02 2
h h hU Uminus ∆
= = 02
s shU
η ∆=
=
eWWm
120578120578119904119904 =∆ℎ0∆ℎ0119904119904
Turbine
Coefficient de deacutebit
xcU
Φ =
xCU
Φ =ρRT
pρ
= xC RTRTU p
=RT m RT
U Ap
xCU
Φ =119888 = ρ119862119862119909119909119860119860
11988811987711987711987911987901199011199010119860119860
= 119872119872 120574120574 1 +120574120574 minus 1
21198721198722
minus 120574120574+1)2(120574120574minus1
Afin de geacuteneacuteraliser le triangle de vitesses drsquoune machineaxiale on effectue une mise agrave lrsquoeacutechelle en divisant chaquecocircteacute par la vitesse tangentielle U
Cette opeacuteration permettra de retrouver directement lescoefficients de charge ψ et de deacutebit Φ
Par la suite un nouveau paramegravetre R qursquoon appelle degreacutede reacuteaction sera introduit pour compleacuteter la caracteacuterisationde lrsquoaubage
Approche classique
α2
β2 α3
β3C2 W3
W2
c3u - c2u= w3u - w2u
W3uW2u
U
CX CX
Lrsquoopeacuteration est vectorielle
C3
120511120511 = frasl119934119934119940119940 119932119932120784120784 = frasl119914119914120785120785119958119958 minus 119914119914120784120784119958119958 119932119932
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
Turbine
3
2
3 3
2 2
12
12
12
12
u
u
u u
u u
c RUc RU
w c RU Uw c RU U
Ψ= minus +
Ψ= minus +
Ψ= minus = +
Ψ= minus = minus
2 22 23 2
2 22 23 2
1 12 2
2 2
c cR RU U
w wR RU U
Ψ Ψ = Φ + minus + = Φ + minus +
Ψ Ψ = Φ + + = Φ + minus
3 2
3 2
1 2 1 2
2 2
R Ratan atan
R Ratan atan
minus + Ψ minus + Ψ = = Φ Φ
Ψ + Ψ minus = = Φ Φ
α α
β β
α2β1
C2U
W1U
C1UW2U β2 α1
Compresseur
ΦC2U
C3UW3U
W2U
ψ
1
ψ
ΦC2UC3U
W3U
W2U
1
Il faut deacutefinir un troisiegraveme paramegravetre pour diffeacuterencier ces deux triangles de vitesses ayant le mecircme 120509120509 et le mecircme 120537120537
Les coefficients 120509120509 et 120537120537 ne suffisent pas pour deacutefinir de maniegravereunivoque le triangle de vitesses
Le degreacute de reacuteaction est un paramegravetre adimensionnel quiexprime la variation drsquoenthalpie subie par le fluide dans le rotorpar rapport agrave la chute totale drsquoenthalpie dans lrsquoeacutetage
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
Variation d enthalpie dans le rotorRVariation d enthalpie dans l eacutetage
=
❸❶
❷
Rotor
Stator
R=0 machine drsquoaction (Δhrotor=0 Δheacutetage=Δhstator)
R=1 machine agrave reacuteaction (Δhstator=0 Δheacutetage=Δhrotor)
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
R indique la reacutepartition du saut drsquoenthalpie (de pression) statique entre le stator et le rotor
Le degreacute de reacuteaction R deacutefini en fonction de variablesthermodynamiques (lrsquoenthalpie) peut aussi srsquoexprimer en fonctionde variables cineacutematiques preacutesentes dans le triangle de vitessesNotamment en fonction de la vitesse peacuteripheacuterique U et descomposantes des vitesses absolues et relatives
La vitesse absolue agrave lrsquoentreacutee du stator est eacutegale agrave celle agrave la sortie du rotor (c1= c3 eacutetage reacutepeacutetitif )
On neacuteglige les pertes et le transfert de chaleur dans le stator
1 2 3
ℎ3 minus ℎ1= ℎ3 + frasl11988811988832 2 minus ℎ1 minus frasl11988811988812 2 = ℎ03 minus ℎ01 = 120549120549ℎ0
ℎ03 minus ℎ01 = ℎ03 minus ℎ02 = 120549120549ℎ )0(3minus2
ℎ03 minus ℎ02 = ℎ3 minus ℎ1
119945119945120782120782120783120783 = 119945119945120782120782120784120784
frasl11988811988832 2 = frasl11988811988812 2
3 2
3 1
rotor
eacutetage
h h hRh h h
∆∆
minus= =
minus
2 20 3 2 2 3
0 3 2
2 2( )
( )
h c c h
∆∆
minus
minus
+ minus=
ℎ03 ℎ02
(ℎ01= ℎ02 1198881198881= 1198881198883)
120549120549ℎ )0(3minus2 = (ℎ3 + frasl11988811988832 2) minus (ℎ2 + frasl11988811988822 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
0(3 2) 3 2( )∆ minus = minusu uh U c c
= 1 minusfrasl11988811988832 2 minus frasl11988811988822 2120549120549ℎ )0(3minus2
119877119877 =120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
120549120549ℎ )0(3minus2
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
3 212+
= minus u uc cRU
119888119888119909119909 = 1198881198882119909119909 = 1198881198883119909119909 = 119888119888119888119888119888119888119888119888119888119888
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 11988811988821199061199061198881198882 = 1198881198881199091199092+1198881198881199061199062
119877119877 = 1 minus11988811988831199061199062 minus 11988811988821199061199062
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
Indirectement lrsquoeacutequation pour R deacutecrite en fonction de composantes cineacutematiques eacutetablit un lien entre les angles drsquoentreacutee et de sortie
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
= 1 minus1198881198883119906119906 + 1198881198882119906119906
2119880119880
2232
2 32 2wwh h+ = +
2 23 2
2 3 2 2w wh hminus = minus
2 2 2 23 3 2 2
2 3 2 2u x u xw w w wh h + +
minus = minus
Machine axiale U2 = U3
La composante axiale de la vitesse demeure constante w3x = w 2x
( eacutequation justifieacutee plus tard lors de la rothalpie)
2 23 2
2 3 2 2u uw wh hminus = minus
3 2 3 22 3
( )( )2
u u u uw w w wh h minus +minus =
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
Eacutetage reacutepeacutetitif c3 = c1
Si les pertes dans le stator sont neacutegligeacutees
c2u ndash c3u= w2u - w3u
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
3 2 3 23 2
( )( )2
u u u uw w w wh h minus +minus = minus
c2u ndash c3u= w2u - w3u
119877119877 = minus1199081199083119906119906 + 1199081199082119906119906
2119880119880
2 2 2 22 3 2 3 2 3( ) 2 ( ) 2h h u u w wminus = minus minus minus
2 2 2 2 2 23 2 3 2 3 2
2 2 2e
I II III
u u c c w wW minus minus minus= + minus
2 2 2 22 3 2 3
2 2 2 2 2 22 3 2 3 2 3
( ) ( )( ) ( ) ( )
minus minus minus=
minus + minus minus minusu u w wR
c c u u w w
3 2= =U u u
Expliqueacutee + tard
2 23 2
03 02 3 2 2ec cW h h h h minus
= minus = minus +
119877119877 =ℎ2 minus ℎ3ℎ1 minus ℎ3
=ℎ2 minus ℎ3ℎ02 minus ℎ03
Degreacute de reacuteaction
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
( )3u 2uw wR2U+
= minus
3 212
u uc cRU+
= minus
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
Puisque le degreacute de reacuteaction R peut aussi srsquoexprimer enfonction de variables cineacutematiques on peut chercher sarepreacutesentation dans le triangle adimensionnel de vitesses
Turbine
α2
β2 α3
β3
C2 W3
C3W2
C3uC2u
U
C2xC3x
W2u
c2u ndash c3u= w2u - w3u
2Ψ minusR
frasl119934119934120784120784119958119958 119932119932 = frasl1198621198622119906119906 119880119880 minus frasl119880119880 119880119880Ψ = frasl1198621198622119906119906 minus 1198621198623119906119906 119880119880
119877119877 = 1 minus (1198621198623119906119906 + 1198621198622119906119906 frasl) 2119880119880Φ = frasl1198621198622119909119909 119880119880 = frasl1198621198623119909119909 119880119880
= (119914119914120784120784119958119958 minus 119914119914120785120785119958119958 frasl) 2119880119880 minus 1 + (119914119914120784120784119958119958 + 119914119914120785120785119958119958 frasl) 2119880119880
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α2
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
Les repreacutesentations des triangles de vitesses adimensionnellessont tregraves riches en information et contiennent plusieurs formules
Par exemple pour un compresseur on deacuteduit aiseacutement que
Avec 120595120595 = 1198821198821198901198901198801198802
Φ = 119888119888119909119909119880119880
et que pour 119888119888119901119901 = 119888119888119888119888119888119888119888119888119888119888 119882119882119890119890 = 119888119888119901119901(11987911987902minus11987911987901)on trouve
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
(11987911987902minus11987911987901) =119880119880119888119888119909119909119888119888119901119901
(tan1205731205731 minus tan1205731205732)
Alors si lrsquoon veut obtenir une plus grande variation detempeacuterature 11987911987902 minus 11987911987901 indirectement un plus grand rapport(1199011199010211990111990101) on peut augmenter
bull la vitesse peacuteripheacuterique 119880119880 (la vitesse de rotation)bull la vitesse axiale 119888119888119909119909 (le deacutebit massique)bull lrsquoangle de deacuteflection de la pale(tan1205731205731 minus tan1205731205732)
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
Nous commenccedilons par la repreacutesentation de triangles de vitessespour le cas drsquoune turbineNous rappelons que par hypothegravese le rayon ne varie pas dans leplan inter-aube et que la vitesse axiale demeure constante dansle sens axialCes conditions permettent drsquoeffectuer la superposition destriangles de vitesse au bord drsquoattaque et bord de fuite du rotoravec la vitesse 119932119932 commune
c1w1
u
c2
u2
w2
➊
➋
➌ c3w3
u
Turbine
C2
W2 C3
W3
Stator Rotor
C2u C3u
W3u
α2β2 β3
α3
∆Cu= ∆Wu
U
W2u
119940119940119961119961119880119880 = 119888119888119888119888119888119888119888119888119888119888119888119888119909119909 = 119888119888119888119888119888119888119888119888119888119888
1198881198883 = 1198881198881
1205721205723 = 1205721205721
Cineacutematique reacutepeacutetitive
➁ ➂
➁U➀
Turbine
119914119914120784120784
119914119914120785120785
119934119934120785120785
119932119932119934119934120784120784
119932119932
119914119914120783120783120630120630120783120783
120630120630120784120784120631120631120784120784
120630120630120785120785
120631120631120785120785
Stator Rotor
➀ ➁ ➂
Le cas du compresseur est similaire agrave celui drsquoune turbineLe rayon ne varie pas (lrsquoeacutetude se deacuteroule dans le plan inter-aube) et la vitesse axiale demeure constante dans le sens axialComme pour la turbine nous pouvons superposer les triangles devitesse au bord drsquoattaque et bord de fuite du rotor
W2 C2
C3W3
U
C1W1
U
U
Compresseur
➊
➋
➌
α2 β1
C2 W1
C1W2
U
β2α1
W2U C1U
CX
( )2 1 = minuse u uW U C C
Lrsquoopeacuteration est vectorielle
C2U
119914119914120784120784119958119958 minus 119914119914120783120783119958119958 =- 119934119934120784120784119958119958 minus119934119934120783120783119958119958
➀
➁
➁
Compresseur
119914119914120784120784119914119914120785120785
119934119934120783120783
119932119932119914119914120783120783120630120630120783120783 120630120630120784120784
120631120631120784120784120630120630120785120785
Rotor Stator
119934119934120784120784120631120631120783120783
119932119932
➀ ➁ ➂
Trois quantiteacutes fondamentales sont utiliseacutees dans laconception des turbomachines
bull le coefficient de charge ψ
bull le coefficient de deacutebit Φ
bull le degreacute de reacuteaction R
Approche classique
Parfois ψ est consideacutereacute positif pour les turbines et neacutegatifpour les compresseurs 119882119882119890119890 = frasl119882 119888
Coefficient de charge
2Ψ = eWU
Coefficient de charge
2e
UW
=ψ 02 01 02 2
h h hU Uminus ∆
= = 02
s shU
η ∆=
=
eWWm
120578120578119904119904 =∆ℎ0∆ℎ0119904119904
Turbine
Coefficient de deacutebit
xcU
Φ =
xCU
Φ =ρRT
pρ
= xC RTRTU p
=RT m RT
U Ap
xCU
Φ =119888 = ρ119862119862119909119909119860119860
11988811987711987711987911987901199011199010119860119860
= 119872119872 120574120574 1 +120574120574 minus 1
21198721198722
minus 120574120574+1)2(120574120574minus1
Afin de geacuteneacuteraliser le triangle de vitesses drsquoune machineaxiale on effectue une mise agrave lrsquoeacutechelle en divisant chaquecocircteacute par la vitesse tangentielle U
Cette opeacuteration permettra de retrouver directement lescoefficients de charge ψ et de deacutebit Φ
Par la suite un nouveau paramegravetre R qursquoon appelle degreacutede reacuteaction sera introduit pour compleacuteter la caracteacuterisationde lrsquoaubage
Approche classique
α2
β2 α3
β3C2 W3
W2
c3u - c2u= w3u - w2u
W3uW2u
U
CX CX
Lrsquoopeacuteration est vectorielle
C3
120511120511 = frasl119934119934119940119940 119932119932120784120784 = frasl119914119914120785120785119958119958 minus 119914119914120784120784119958119958 119932119932
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
Turbine
3
2
3 3
2 2
12
12
12
12
u
u
u u
u u
c RUc RU
w c RU Uw c RU U
Ψ= minus +
Ψ= minus +
Ψ= minus = +
Ψ= minus = minus
2 22 23 2
2 22 23 2
1 12 2
2 2
c cR RU U
w wR RU U
Ψ Ψ = Φ + minus + = Φ + minus +
Ψ Ψ = Φ + + = Φ + minus
3 2
3 2
1 2 1 2
2 2
R Ratan atan
R Ratan atan
minus + Ψ minus + Ψ = = Φ Φ
Ψ + Ψ minus = = Φ Φ
α α
β β
α2β1
C2U
W1U
C1UW2U β2 α1
Compresseur
ΦC2U
C3UW3U
W2U
ψ
1
ψ
ΦC2UC3U
W3U
W2U
1
Il faut deacutefinir un troisiegraveme paramegravetre pour diffeacuterencier ces deux triangles de vitesses ayant le mecircme 120509120509 et le mecircme 120537120537
Les coefficients 120509120509 et 120537120537 ne suffisent pas pour deacutefinir de maniegravereunivoque le triangle de vitesses
Le degreacute de reacuteaction est un paramegravetre adimensionnel quiexprime la variation drsquoenthalpie subie par le fluide dans le rotorpar rapport agrave la chute totale drsquoenthalpie dans lrsquoeacutetage
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
Variation d enthalpie dans le rotorRVariation d enthalpie dans l eacutetage
=
❸❶
❷
Rotor
Stator
R=0 machine drsquoaction (Δhrotor=0 Δheacutetage=Δhstator)
R=1 machine agrave reacuteaction (Δhstator=0 Δheacutetage=Δhrotor)
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
R indique la reacutepartition du saut drsquoenthalpie (de pression) statique entre le stator et le rotor
Le degreacute de reacuteaction R deacutefini en fonction de variablesthermodynamiques (lrsquoenthalpie) peut aussi srsquoexprimer en fonctionde variables cineacutematiques preacutesentes dans le triangle de vitessesNotamment en fonction de la vitesse peacuteripheacuterique U et descomposantes des vitesses absolues et relatives
La vitesse absolue agrave lrsquoentreacutee du stator est eacutegale agrave celle agrave la sortie du rotor (c1= c3 eacutetage reacutepeacutetitif )
On neacuteglige les pertes et le transfert de chaleur dans le stator
1 2 3
ℎ3 minus ℎ1= ℎ3 + frasl11988811988832 2 minus ℎ1 minus frasl11988811988812 2 = ℎ03 minus ℎ01 = 120549120549ℎ0
ℎ03 minus ℎ01 = ℎ03 minus ℎ02 = 120549120549ℎ )0(3minus2
ℎ03 minus ℎ02 = ℎ3 minus ℎ1
119945119945120782120782120783120783 = 119945119945120782120782120784120784
frasl11988811988832 2 = frasl11988811988812 2
3 2
3 1
rotor
eacutetage
h h hRh h h
∆∆
minus= =
minus
2 20 3 2 2 3
0 3 2
2 2( )
( )
h c c h
∆∆
minus
minus
+ minus=
ℎ03 ℎ02
(ℎ01= ℎ02 1198881198881= 1198881198883)
120549120549ℎ )0(3minus2 = (ℎ3 + frasl11988811988832 2) minus (ℎ2 + frasl11988811988822 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
0(3 2) 3 2( )∆ minus = minusu uh U c c
= 1 minusfrasl11988811988832 2 minus frasl11988811988822 2120549120549ℎ )0(3minus2
119877119877 =120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
120549120549ℎ )0(3minus2
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
3 212+
= minus u uc cRU
119888119888119909119909 = 1198881198882119909119909 = 1198881198883119909119909 = 119888119888119888119888119888119888119888119888119888119888
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 11988811988821199061199061198881198882 = 1198881198881199091199092+1198881198881199061199062
119877119877 = 1 minus11988811988831199061199062 minus 11988811988821199061199062
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
Indirectement lrsquoeacutequation pour R deacutecrite en fonction de composantes cineacutematiques eacutetablit un lien entre les angles drsquoentreacutee et de sortie
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
= 1 minus1198881198883119906119906 + 1198881198882119906119906
2119880119880
2232
2 32 2wwh h+ = +
2 23 2
2 3 2 2w wh hminus = minus
2 2 2 23 3 2 2
2 3 2 2u x u xw w w wh h + +
minus = minus
Machine axiale U2 = U3
La composante axiale de la vitesse demeure constante w3x = w 2x
( eacutequation justifieacutee plus tard lors de la rothalpie)
2 23 2
2 3 2 2u uw wh hminus = minus
3 2 3 22 3
( )( )2
u u u uw w w wh h minus +minus =
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
Eacutetage reacutepeacutetitif c3 = c1
Si les pertes dans le stator sont neacutegligeacutees
c2u ndash c3u= w2u - w3u
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
3 2 3 23 2
( )( )2
u u u uw w w wh h minus +minus = minus
c2u ndash c3u= w2u - w3u
119877119877 = minus1199081199083119906119906 + 1199081199082119906119906
2119880119880
2 2 2 22 3 2 3 2 3( ) 2 ( ) 2h h u u w wminus = minus minus minus
2 2 2 2 2 23 2 3 2 3 2
2 2 2e
I II III
u u c c w wW minus minus minus= + minus
2 2 2 22 3 2 3
2 2 2 2 2 22 3 2 3 2 3
( ) ( )( ) ( ) ( )
minus minus minus=
minus + minus minus minusu u w wR
c c u u w w
3 2= =U u u
Expliqueacutee + tard
2 23 2
03 02 3 2 2ec cW h h h h minus
= minus = minus +
119877119877 =ℎ2 minus ℎ3ℎ1 minus ℎ3
=ℎ2 minus ℎ3ℎ02 minus ℎ03
Degreacute de reacuteaction
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
( )3u 2uw wR2U+
= minus
3 212
u uc cRU+
= minus
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
Puisque le degreacute de reacuteaction R peut aussi srsquoexprimer enfonction de variables cineacutematiques on peut chercher sarepreacutesentation dans le triangle adimensionnel de vitesses
Turbine
α2
β2 α3
β3
C2 W3
C3W2
C3uC2u
U
C2xC3x
W2u
c2u ndash c3u= w2u - w3u
2Ψ minusR
frasl119934119934120784120784119958119958 119932119932 = frasl1198621198622119906119906 119880119880 minus frasl119880119880 119880119880Ψ = frasl1198621198622119906119906 minus 1198621198623119906119906 119880119880
119877119877 = 1 minus (1198621198623119906119906 + 1198621198622119906119906 frasl) 2119880119880Φ = frasl1198621198622119909119909 119880119880 = frasl1198621198623119909119909 119880119880
= (119914119914120784120784119958119958 minus 119914119914120785120785119958119958 frasl) 2119880119880 minus 1 + (119914119914120784120784119958119958 + 119914119914120785120785119958119958 frasl) 2119880119880
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α2
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
Les repreacutesentations des triangles de vitesses adimensionnellessont tregraves riches en information et contiennent plusieurs formules
Par exemple pour un compresseur on deacuteduit aiseacutement que
Avec 120595120595 = 1198821198821198901198901198801198802
Φ = 119888119888119909119909119880119880
et que pour 119888119888119901119901 = 119888119888119888119888119888119888119888119888119888119888 119882119882119890119890 = 119888119888119901119901(11987911987902minus11987911987901)on trouve
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
(11987911987902minus11987911987901) =119880119880119888119888119909119909119888119888119901119901
(tan1205731205731 minus tan1205731205732)
Alors si lrsquoon veut obtenir une plus grande variation detempeacuterature 11987911987902 minus 11987911987901 indirectement un plus grand rapport(1199011199010211990111990101) on peut augmenter
bull la vitesse peacuteripheacuterique 119880119880 (la vitesse de rotation)bull la vitesse axiale 119888119888119909119909 (le deacutebit massique)bull lrsquoangle de deacuteflection de la pale(tan1205731205731 minus tan1205731205732)
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
c1w1
u
c2
u2
w2
➊
➋
➌ c3w3
u
Turbine
C2
W2 C3
W3
Stator Rotor
C2u C3u
W3u
α2β2 β3
α3
∆Cu= ∆Wu
U
W2u
119940119940119961119961119880119880 = 119888119888119888119888119888119888119888119888119888119888119888119888119909119909 = 119888119888119888119888119888119888119888119888119888119888
1198881198883 = 1198881198881
1205721205723 = 1205721205721
Cineacutematique reacutepeacutetitive
➁ ➂
➁U➀
Turbine
119914119914120784120784
119914119914120785120785
119934119934120785120785
119932119932119934119934120784120784
119932119932
119914119914120783120783120630120630120783120783
120630120630120784120784120631120631120784120784
120630120630120785120785
120631120631120785120785
Stator Rotor
➀ ➁ ➂
Le cas du compresseur est similaire agrave celui drsquoune turbineLe rayon ne varie pas (lrsquoeacutetude se deacuteroule dans le plan inter-aube) et la vitesse axiale demeure constante dans le sens axialComme pour la turbine nous pouvons superposer les triangles devitesse au bord drsquoattaque et bord de fuite du rotor
W2 C2
C3W3
U
C1W1
U
U
Compresseur
➊
➋
➌
α2 β1
C2 W1
C1W2
U
β2α1
W2U C1U
CX
( )2 1 = minuse u uW U C C
Lrsquoopeacuteration est vectorielle
C2U
119914119914120784120784119958119958 minus 119914119914120783120783119958119958 =- 119934119934120784120784119958119958 minus119934119934120783120783119958119958
➀
➁
➁
Compresseur
119914119914120784120784119914119914120785120785
119934119934120783120783
119932119932119914119914120783120783120630120630120783120783 120630120630120784120784
120631120631120784120784120630120630120785120785
Rotor Stator
119934119934120784120784120631120631120783120783
119932119932
➀ ➁ ➂
Trois quantiteacutes fondamentales sont utiliseacutees dans laconception des turbomachines
bull le coefficient de charge ψ
bull le coefficient de deacutebit Φ
bull le degreacute de reacuteaction R
Approche classique
Parfois ψ est consideacutereacute positif pour les turbines et neacutegatifpour les compresseurs 119882119882119890119890 = frasl119882 119888
Coefficient de charge
2Ψ = eWU
Coefficient de charge
2e
UW
=ψ 02 01 02 2
h h hU Uminus ∆
= = 02
s shU
η ∆=
=
eWWm
120578120578119904119904 =∆ℎ0∆ℎ0119904119904
Turbine
Coefficient de deacutebit
xcU
Φ =
xCU
Φ =ρRT
pρ
= xC RTRTU p
=RT m RT
U Ap
xCU
Φ =119888 = ρ119862119862119909119909119860119860
11988811987711987711987911987901199011199010119860119860
= 119872119872 120574120574 1 +120574120574 minus 1
21198721198722
minus 120574120574+1)2(120574120574minus1
Afin de geacuteneacuteraliser le triangle de vitesses drsquoune machineaxiale on effectue une mise agrave lrsquoeacutechelle en divisant chaquecocircteacute par la vitesse tangentielle U
Cette opeacuteration permettra de retrouver directement lescoefficients de charge ψ et de deacutebit Φ
Par la suite un nouveau paramegravetre R qursquoon appelle degreacutede reacuteaction sera introduit pour compleacuteter la caracteacuterisationde lrsquoaubage
Approche classique
α2
β2 α3
β3C2 W3
W2
c3u - c2u= w3u - w2u
W3uW2u
U
CX CX
Lrsquoopeacuteration est vectorielle
C3
120511120511 = frasl119934119934119940119940 119932119932120784120784 = frasl119914119914120785120785119958119958 minus 119914119914120784120784119958119958 119932119932
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
Turbine
3
2
3 3
2 2
12
12
12
12
u
u
u u
u u
c RUc RU
w c RU Uw c RU U
Ψ= minus +
Ψ= minus +
Ψ= minus = +
Ψ= minus = minus
2 22 23 2
2 22 23 2
1 12 2
2 2
c cR RU U
w wR RU U
Ψ Ψ = Φ + minus + = Φ + minus +
Ψ Ψ = Φ + + = Φ + minus
3 2
3 2
1 2 1 2
2 2
R Ratan atan
R Ratan atan
minus + Ψ minus + Ψ = = Φ Φ
Ψ + Ψ minus = = Φ Φ
α α
β β
α2β1
C2U
W1U
C1UW2U β2 α1
Compresseur
ΦC2U
C3UW3U
W2U
ψ
1
ψ
ΦC2UC3U
W3U
W2U
1
Il faut deacutefinir un troisiegraveme paramegravetre pour diffeacuterencier ces deux triangles de vitesses ayant le mecircme 120509120509 et le mecircme 120537120537
Les coefficients 120509120509 et 120537120537 ne suffisent pas pour deacutefinir de maniegravereunivoque le triangle de vitesses
Le degreacute de reacuteaction est un paramegravetre adimensionnel quiexprime la variation drsquoenthalpie subie par le fluide dans le rotorpar rapport agrave la chute totale drsquoenthalpie dans lrsquoeacutetage
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
Variation d enthalpie dans le rotorRVariation d enthalpie dans l eacutetage
=
❸❶
❷
Rotor
Stator
R=0 machine drsquoaction (Δhrotor=0 Δheacutetage=Δhstator)
R=1 machine agrave reacuteaction (Δhstator=0 Δheacutetage=Δhrotor)
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
R indique la reacutepartition du saut drsquoenthalpie (de pression) statique entre le stator et le rotor
Le degreacute de reacuteaction R deacutefini en fonction de variablesthermodynamiques (lrsquoenthalpie) peut aussi srsquoexprimer en fonctionde variables cineacutematiques preacutesentes dans le triangle de vitessesNotamment en fonction de la vitesse peacuteripheacuterique U et descomposantes des vitesses absolues et relatives
La vitesse absolue agrave lrsquoentreacutee du stator est eacutegale agrave celle agrave la sortie du rotor (c1= c3 eacutetage reacutepeacutetitif )
On neacuteglige les pertes et le transfert de chaleur dans le stator
1 2 3
ℎ3 minus ℎ1= ℎ3 + frasl11988811988832 2 minus ℎ1 minus frasl11988811988812 2 = ℎ03 minus ℎ01 = 120549120549ℎ0
ℎ03 minus ℎ01 = ℎ03 minus ℎ02 = 120549120549ℎ )0(3minus2
ℎ03 minus ℎ02 = ℎ3 minus ℎ1
119945119945120782120782120783120783 = 119945119945120782120782120784120784
frasl11988811988832 2 = frasl11988811988812 2
3 2
3 1
rotor
eacutetage
h h hRh h h
∆∆
minus= =
minus
2 20 3 2 2 3
0 3 2
2 2( )
( )
h c c h
∆∆
minus
minus
+ minus=
ℎ03 ℎ02
(ℎ01= ℎ02 1198881198881= 1198881198883)
120549120549ℎ )0(3minus2 = (ℎ3 + frasl11988811988832 2) minus (ℎ2 + frasl11988811988822 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
0(3 2) 3 2( )∆ minus = minusu uh U c c
= 1 minusfrasl11988811988832 2 minus frasl11988811988822 2120549120549ℎ )0(3minus2
119877119877 =120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
120549120549ℎ )0(3minus2
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
3 212+
= minus u uc cRU
119888119888119909119909 = 1198881198882119909119909 = 1198881198883119909119909 = 119888119888119888119888119888119888119888119888119888119888
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 11988811988821199061199061198881198882 = 1198881198881199091199092+1198881198881199061199062
119877119877 = 1 minus11988811988831199061199062 minus 11988811988821199061199062
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
Indirectement lrsquoeacutequation pour R deacutecrite en fonction de composantes cineacutematiques eacutetablit un lien entre les angles drsquoentreacutee et de sortie
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
= 1 minus1198881198883119906119906 + 1198881198882119906119906
2119880119880
2232
2 32 2wwh h+ = +
2 23 2
2 3 2 2w wh hminus = minus
2 2 2 23 3 2 2
2 3 2 2u x u xw w w wh h + +
minus = minus
Machine axiale U2 = U3
La composante axiale de la vitesse demeure constante w3x = w 2x
( eacutequation justifieacutee plus tard lors de la rothalpie)
2 23 2
2 3 2 2u uw wh hminus = minus
3 2 3 22 3
( )( )2
u u u uw w w wh h minus +minus =
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
Eacutetage reacutepeacutetitif c3 = c1
Si les pertes dans le stator sont neacutegligeacutees
c2u ndash c3u= w2u - w3u
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
3 2 3 23 2
( )( )2
u u u uw w w wh h minus +minus = minus
c2u ndash c3u= w2u - w3u
119877119877 = minus1199081199083119906119906 + 1199081199082119906119906
2119880119880
2 2 2 22 3 2 3 2 3( ) 2 ( ) 2h h u u w wminus = minus minus minus
2 2 2 2 2 23 2 3 2 3 2
2 2 2e
I II III
u u c c w wW minus minus minus= + minus
2 2 2 22 3 2 3
2 2 2 2 2 22 3 2 3 2 3
( ) ( )( ) ( ) ( )
minus minus minus=
minus + minus minus minusu u w wR
c c u u w w
3 2= =U u u
Expliqueacutee + tard
2 23 2
03 02 3 2 2ec cW h h h h minus
= minus = minus +
119877119877 =ℎ2 minus ℎ3ℎ1 minus ℎ3
=ℎ2 minus ℎ3ℎ02 minus ℎ03
Degreacute de reacuteaction
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
( )3u 2uw wR2U+
= minus
3 212
u uc cRU+
= minus
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
Puisque le degreacute de reacuteaction R peut aussi srsquoexprimer enfonction de variables cineacutematiques on peut chercher sarepreacutesentation dans le triangle adimensionnel de vitesses
Turbine
α2
β2 α3
β3
C2 W3
C3W2
C3uC2u
U
C2xC3x
W2u
c2u ndash c3u= w2u - w3u
2Ψ minusR
frasl119934119934120784120784119958119958 119932119932 = frasl1198621198622119906119906 119880119880 minus frasl119880119880 119880119880Ψ = frasl1198621198622119906119906 minus 1198621198623119906119906 119880119880
119877119877 = 1 minus (1198621198623119906119906 + 1198621198622119906119906 frasl) 2119880119880Φ = frasl1198621198622119909119909 119880119880 = frasl1198621198623119909119909 119880119880
= (119914119914120784120784119958119958 minus 119914119914120785120785119958119958 frasl) 2119880119880 minus 1 + (119914119914120784120784119958119958 + 119914119914120785120785119958119958 frasl) 2119880119880
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α2
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
Les repreacutesentations des triangles de vitesses adimensionnellessont tregraves riches en information et contiennent plusieurs formules
Par exemple pour un compresseur on deacuteduit aiseacutement que
Avec 120595120595 = 1198821198821198901198901198801198802
Φ = 119888119888119909119909119880119880
et que pour 119888119888119901119901 = 119888119888119888119888119888119888119888119888119888119888 119882119882119890119890 = 119888119888119901119901(11987911987902minus11987911987901)on trouve
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
(11987911987902minus11987911987901) =119880119880119888119888119909119909119888119888119901119901
(tan1205731205731 minus tan1205731205732)
Alors si lrsquoon veut obtenir une plus grande variation detempeacuterature 11987911987902 minus 11987911987901 indirectement un plus grand rapport(1199011199010211990111990101) on peut augmenter
bull la vitesse peacuteripheacuterique 119880119880 (la vitesse de rotation)bull la vitesse axiale 119888119888119909119909 (le deacutebit massique)bull lrsquoangle de deacuteflection de la pale(tan1205731205731 minus tan1205731205732)
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
C2
W2 C3
W3
Stator Rotor
C2u C3u
W3u
α2β2 β3
α3
∆Cu= ∆Wu
U
W2u
119940119940119961119961119880119880 = 119888119888119888119888119888119888119888119888119888119888119888119888119909119909 = 119888119888119888119888119888119888119888119888119888119888
1198881198883 = 1198881198881
1205721205723 = 1205721205721
Cineacutematique reacutepeacutetitive
➁ ➂
➁U➀
Turbine
119914119914120784120784
119914119914120785120785
119934119934120785120785
119932119932119934119934120784120784
119932119932
119914119914120783120783120630120630120783120783
120630120630120784120784120631120631120784120784
120630120630120785120785
120631120631120785120785
Stator Rotor
➀ ➁ ➂
Le cas du compresseur est similaire agrave celui drsquoune turbineLe rayon ne varie pas (lrsquoeacutetude se deacuteroule dans le plan inter-aube) et la vitesse axiale demeure constante dans le sens axialComme pour la turbine nous pouvons superposer les triangles devitesse au bord drsquoattaque et bord de fuite du rotor
W2 C2
C3W3
U
C1W1
U
U
Compresseur
➊
➋
➌
α2 β1
C2 W1
C1W2
U
β2α1
W2U C1U
CX
( )2 1 = minuse u uW U C C
Lrsquoopeacuteration est vectorielle
C2U
119914119914120784120784119958119958 minus 119914119914120783120783119958119958 =- 119934119934120784120784119958119958 minus119934119934120783120783119958119958
➀
➁
➁
Compresseur
119914119914120784120784119914119914120785120785
119934119934120783120783
119932119932119914119914120783120783120630120630120783120783 120630120630120784120784
120631120631120784120784120630120630120785120785
Rotor Stator
119934119934120784120784120631120631120783120783
119932119932
➀ ➁ ➂
Trois quantiteacutes fondamentales sont utiliseacutees dans laconception des turbomachines
bull le coefficient de charge ψ
bull le coefficient de deacutebit Φ
bull le degreacute de reacuteaction R
Approche classique
Parfois ψ est consideacutereacute positif pour les turbines et neacutegatifpour les compresseurs 119882119882119890119890 = frasl119882 119888
Coefficient de charge
2Ψ = eWU
Coefficient de charge
2e
UW
=ψ 02 01 02 2
h h hU Uminus ∆
= = 02
s shU
η ∆=
=
eWWm
120578120578119904119904 =∆ℎ0∆ℎ0119904119904
Turbine
Coefficient de deacutebit
xcU
Φ =
xCU
Φ =ρRT
pρ
= xC RTRTU p
=RT m RT
U Ap
xCU
Φ =119888 = ρ119862119862119909119909119860119860
11988811987711987711987911987901199011199010119860119860
= 119872119872 120574120574 1 +120574120574 minus 1
21198721198722
minus 120574120574+1)2(120574120574minus1
Afin de geacuteneacuteraliser le triangle de vitesses drsquoune machineaxiale on effectue une mise agrave lrsquoeacutechelle en divisant chaquecocircteacute par la vitesse tangentielle U
Cette opeacuteration permettra de retrouver directement lescoefficients de charge ψ et de deacutebit Φ
Par la suite un nouveau paramegravetre R qursquoon appelle degreacutede reacuteaction sera introduit pour compleacuteter la caracteacuterisationde lrsquoaubage
Approche classique
α2
β2 α3
β3C2 W3
W2
c3u - c2u= w3u - w2u
W3uW2u
U
CX CX
Lrsquoopeacuteration est vectorielle
C3
120511120511 = frasl119934119934119940119940 119932119932120784120784 = frasl119914119914120785120785119958119958 minus 119914119914120784120784119958119958 119932119932
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
Turbine
3
2
3 3
2 2
12
12
12
12
u
u
u u
u u
c RUc RU
w c RU Uw c RU U
Ψ= minus +
Ψ= minus +
Ψ= minus = +
Ψ= minus = minus
2 22 23 2
2 22 23 2
1 12 2
2 2
c cR RU U
w wR RU U
Ψ Ψ = Φ + minus + = Φ + minus +
Ψ Ψ = Φ + + = Φ + minus
3 2
3 2
1 2 1 2
2 2
R Ratan atan
R Ratan atan
minus + Ψ minus + Ψ = = Φ Φ
Ψ + Ψ minus = = Φ Φ
α α
β β
α2β1
C2U
W1U
C1UW2U β2 α1
Compresseur
ΦC2U
C3UW3U
W2U
ψ
1
ψ
ΦC2UC3U
W3U
W2U
1
Il faut deacutefinir un troisiegraveme paramegravetre pour diffeacuterencier ces deux triangles de vitesses ayant le mecircme 120509120509 et le mecircme 120537120537
Les coefficients 120509120509 et 120537120537 ne suffisent pas pour deacutefinir de maniegravereunivoque le triangle de vitesses
Le degreacute de reacuteaction est un paramegravetre adimensionnel quiexprime la variation drsquoenthalpie subie par le fluide dans le rotorpar rapport agrave la chute totale drsquoenthalpie dans lrsquoeacutetage
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
Variation d enthalpie dans le rotorRVariation d enthalpie dans l eacutetage
=
❸❶
❷
Rotor
Stator
R=0 machine drsquoaction (Δhrotor=0 Δheacutetage=Δhstator)
R=1 machine agrave reacuteaction (Δhstator=0 Δheacutetage=Δhrotor)
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
R indique la reacutepartition du saut drsquoenthalpie (de pression) statique entre le stator et le rotor
Le degreacute de reacuteaction R deacutefini en fonction de variablesthermodynamiques (lrsquoenthalpie) peut aussi srsquoexprimer en fonctionde variables cineacutematiques preacutesentes dans le triangle de vitessesNotamment en fonction de la vitesse peacuteripheacuterique U et descomposantes des vitesses absolues et relatives
La vitesse absolue agrave lrsquoentreacutee du stator est eacutegale agrave celle agrave la sortie du rotor (c1= c3 eacutetage reacutepeacutetitif )
On neacuteglige les pertes et le transfert de chaleur dans le stator
1 2 3
ℎ3 minus ℎ1= ℎ3 + frasl11988811988832 2 minus ℎ1 minus frasl11988811988812 2 = ℎ03 minus ℎ01 = 120549120549ℎ0
ℎ03 minus ℎ01 = ℎ03 minus ℎ02 = 120549120549ℎ )0(3minus2
ℎ03 minus ℎ02 = ℎ3 minus ℎ1
119945119945120782120782120783120783 = 119945119945120782120782120784120784
frasl11988811988832 2 = frasl11988811988812 2
3 2
3 1
rotor
eacutetage
h h hRh h h
∆∆
minus= =
minus
2 20 3 2 2 3
0 3 2
2 2( )
( )
h c c h
∆∆
minus
minus
+ minus=
ℎ03 ℎ02
(ℎ01= ℎ02 1198881198881= 1198881198883)
120549120549ℎ )0(3minus2 = (ℎ3 + frasl11988811988832 2) minus (ℎ2 + frasl11988811988822 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
0(3 2) 3 2( )∆ minus = minusu uh U c c
= 1 minusfrasl11988811988832 2 minus frasl11988811988822 2120549120549ℎ )0(3minus2
119877119877 =120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
120549120549ℎ )0(3minus2
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
3 212+
= minus u uc cRU
119888119888119909119909 = 1198881198882119909119909 = 1198881198883119909119909 = 119888119888119888119888119888119888119888119888119888119888
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 11988811988821199061199061198881198882 = 1198881198881199091199092+1198881198881199061199062
119877119877 = 1 minus11988811988831199061199062 minus 11988811988821199061199062
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
Indirectement lrsquoeacutequation pour R deacutecrite en fonction de composantes cineacutematiques eacutetablit un lien entre les angles drsquoentreacutee et de sortie
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
= 1 minus1198881198883119906119906 + 1198881198882119906119906
2119880119880
2232
2 32 2wwh h+ = +
2 23 2
2 3 2 2w wh hminus = minus
2 2 2 23 3 2 2
2 3 2 2u x u xw w w wh h + +
minus = minus
Machine axiale U2 = U3
La composante axiale de la vitesse demeure constante w3x = w 2x
( eacutequation justifieacutee plus tard lors de la rothalpie)
2 23 2
2 3 2 2u uw wh hminus = minus
3 2 3 22 3
( )( )2
u u u uw w w wh h minus +minus =
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
Eacutetage reacutepeacutetitif c3 = c1
Si les pertes dans le stator sont neacutegligeacutees
c2u ndash c3u= w2u - w3u
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
3 2 3 23 2
( )( )2
u u u uw w w wh h minus +minus = minus
c2u ndash c3u= w2u - w3u
119877119877 = minus1199081199083119906119906 + 1199081199082119906119906
2119880119880
2 2 2 22 3 2 3 2 3( ) 2 ( ) 2h h u u w wminus = minus minus minus
2 2 2 2 2 23 2 3 2 3 2
2 2 2e
I II III
u u c c w wW minus minus minus= + minus
2 2 2 22 3 2 3
2 2 2 2 2 22 3 2 3 2 3
( ) ( )( ) ( ) ( )
minus minus minus=
minus + minus minus minusu u w wR
c c u u w w
3 2= =U u u
Expliqueacutee + tard
2 23 2
03 02 3 2 2ec cW h h h h minus
= minus = minus +
119877119877 =ℎ2 minus ℎ3ℎ1 minus ℎ3
=ℎ2 minus ℎ3ℎ02 minus ℎ03
Degreacute de reacuteaction
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
( )3u 2uw wR2U+
= minus
3 212
u uc cRU+
= minus
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
Puisque le degreacute de reacuteaction R peut aussi srsquoexprimer enfonction de variables cineacutematiques on peut chercher sarepreacutesentation dans le triangle adimensionnel de vitesses
Turbine
α2
β2 α3
β3
C2 W3
C3W2
C3uC2u
U
C2xC3x
W2u
c2u ndash c3u= w2u - w3u
2Ψ minusR
frasl119934119934120784120784119958119958 119932119932 = frasl1198621198622119906119906 119880119880 minus frasl119880119880 119880119880Ψ = frasl1198621198622119906119906 minus 1198621198623119906119906 119880119880
119877119877 = 1 minus (1198621198623119906119906 + 1198621198622119906119906 frasl) 2119880119880Φ = frasl1198621198622119909119909 119880119880 = frasl1198621198623119909119909 119880119880
= (119914119914120784120784119958119958 minus 119914119914120785120785119958119958 frasl) 2119880119880 minus 1 + (119914119914120784120784119958119958 + 119914119914120785120785119958119958 frasl) 2119880119880
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α2
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
Les repreacutesentations des triangles de vitesses adimensionnellessont tregraves riches en information et contiennent plusieurs formules
Par exemple pour un compresseur on deacuteduit aiseacutement que
Avec 120595120595 = 1198821198821198901198901198801198802
Φ = 119888119888119909119909119880119880
et que pour 119888119888119901119901 = 119888119888119888119888119888119888119888119888119888119888 119882119882119890119890 = 119888119888119901119901(11987911987902minus11987911987901)on trouve
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
(11987911987902minus11987911987901) =119880119880119888119888119909119909119888119888119901119901
(tan1205731205731 minus tan1205731205732)
Alors si lrsquoon veut obtenir une plus grande variation detempeacuterature 11987911987902 minus 11987911987901 indirectement un plus grand rapport(1199011199010211990111990101) on peut augmenter
bull la vitesse peacuteripheacuterique 119880119880 (la vitesse de rotation)bull la vitesse axiale 119888119888119909119909 (le deacutebit massique)bull lrsquoangle de deacuteflection de la pale(tan1205731205731 minus tan1205731205732)
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
Turbine
119914119914120784120784
119914119914120785120785
119934119934120785120785
119932119932119934119934120784120784
119932119932
119914119914120783120783120630120630120783120783
120630120630120784120784120631120631120784120784
120630120630120785120785
120631120631120785120785
Stator Rotor
➀ ➁ ➂
Le cas du compresseur est similaire agrave celui drsquoune turbineLe rayon ne varie pas (lrsquoeacutetude se deacuteroule dans le plan inter-aube) et la vitesse axiale demeure constante dans le sens axialComme pour la turbine nous pouvons superposer les triangles devitesse au bord drsquoattaque et bord de fuite du rotor
W2 C2
C3W3
U
C1W1
U
U
Compresseur
➊
➋
➌
α2 β1
C2 W1
C1W2
U
β2α1
W2U C1U
CX
( )2 1 = minuse u uW U C C
Lrsquoopeacuteration est vectorielle
C2U
119914119914120784120784119958119958 minus 119914119914120783120783119958119958 =- 119934119934120784120784119958119958 minus119934119934120783120783119958119958
➀
➁
➁
Compresseur
119914119914120784120784119914119914120785120785
119934119934120783120783
119932119932119914119914120783120783120630120630120783120783 120630120630120784120784
120631120631120784120784120630120630120785120785
Rotor Stator
119934119934120784120784120631120631120783120783
119932119932
➀ ➁ ➂
Trois quantiteacutes fondamentales sont utiliseacutees dans laconception des turbomachines
bull le coefficient de charge ψ
bull le coefficient de deacutebit Φ
bull le degreacute de reacuteaction R
Approche classique
Parfois ψ est consideacutereacute positif pour les turbines et neacutegatifpour les compresseurs 119882119882119890119890 = frasl119882 119888
Coefficient de charge
2Ψ = eWU
Coefficient de charge
2e
UW
=ψ 02 01 02 2
h h hU Uminus ∆
= = 02
s shU
η ∆=
=
eWWm
120578120578119904119904 =∆ℎ0∆ℎ0119904119904
Turbine
Coefficient de deacutebit
xcU
Φ =
xCU
Φ =ρRT
pρ
= xC RTRTU p
=RT m RT
U Ap
xCU
Φ =119888 = ρ119862119862119909119909119860119860
11988811987711987711987911987901199011199010119860119860
= 119872119872 120574120574 1 +120574120574 minus 1
21198721198722
minus 120574120574+1)2(120574120574minus1
Afin de geacuteneacuteraliser le triangle de vitesses drsquoune machineaxiale on effectue une mise agrave lrsquoeacutechelle en divisant chaquecocircteacute par la vitesse tangentielle U
Cette opeacuteration permettra de retrouver directement lescoefficients de charge ψ et de deacutebit Φ
Par la suite un nouveau paramegravetre R qursquoon appelle degreacutede reacuteaction sera introduit pour compleacuteter la caracteacuterisationde lrsquoaubage
Approche classique
α2
β2 α3
β3C2 W3
W2
c3u - c2u= w3u - w2u
W3uW2u
U
CX CX
Lrsquoopeacuteration est vectorielle
C3
120511120511 = frasl119934119934119940119940 119932119932120784120784 = frasl119914119914120785120785119958119958 minus 119914119914120784120784119958119958 119932119932
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
Turbine
3
2
3 3
2 2
12
12
12
12
u
u
u u
u u
c RUc RU
w c RU Uw c RU U
Ψ= minus +
Ψ= minus +
Ψ= minus = +
Ψ= minus = minus
2 22 23 2
2 22 23 2
1 12 2
2 2
c cR RU U
w wR RU U
Ψ Ψ = Φ + minus + = Φ + minus +
Ψ Ψ = Φ + + = Φ + minus
3 2
3 2
1 2 1 2
2 2
R Ratan atan
R Ratan atan
minus + Ψ minus + Ψ = = Φ Φ
Ψ + Ψ minus = = Φ Φ
α α
β β
α2β1
C2U
W1U
C1UW2U β2 α1
Compresseur
ΦC2U
C3UW3U
W2U
ψ
1
ψ
ΦC2UC3U
W3U
W2U
1
Il faut deacutefinir un troisiegraveme paramegravetre pour diffeacuterencier ces deux triangles de vitesses ayant le mecircme 120509120509 et le mecircme 120537120537
Les coefficients 120509120509 et 120537120537 ne suffisent pas pour deacutefinir de maniegravereunivoque le triangle de vitesses
Le degreacute de reacuteaction est un paramegravetre adimensionnel quiexprime la variation drsquoenthalpie subie par le fluide dans le rotorpar rapport agrave la chute totale drsquoenthalpie dans lrsquoeacutetage
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
Variation d enthalpie dans le rotorRVariation d enthalpie dans l eacutetage
=
❸❶
❷
Rotor
Stator
R=0 machine drsquoaction (Δhrotor=0 Δheacutetage=Δhstator)
R=1 machine agrave reacuteaction (Δhstator=0 Δheacutetage=Δhrotor)
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
R indique la reacutepartition du saut drsquoenthalpie (de pression) statique entre le stator et le rotor
Le degreacute de reacuteaction R deacutefini en fonction de variablesthermodynamiques (lrsquoenthalpie) peut aussi srsquoexprimer en fonctionde variables cineacutematiques preacutesentes dans le triangle de vitessesNotamment en fonction de la vitesse peacuteripheacuterique U et descomposantes des vitesses absolues et relatives
La vitesse absolue agrave lrsquoentreacutee du stator est eacutegale agrave celle agrave la sortie du rotor (c1= c3 eacutetage reacutepeacutetitif )
On neacuteglige les pertes et le transfert de chaleur dans le stator
1 2 3
ℎ3 minus ℎ1= ℎ3 + frasl11988811988832 2 minus ℎ1 minus frasl11988811988812 2 = ℎ03 minus ℎ01 = 120549120549ℎ0
ℎ03 minus ℎ01 = ℎ03 minus ℎ02 = 120549120549ℎ )0(3minus2
ℎ03 minus ℎ02 = ℎ3 minus ℎ1
119945119945120782120782120783120783 = 119945119945120782120782120784120784
frasl11988811988832 2 = frasl11988811988812 2
3 2
3 1
rotor
eacutetage
h h hRh h h
∆∆
minus= =
minus
2 20 3 2 2 3
0 3 2
2 2( )
( )
h c c h
∆∆
minus
minus
+ minus=
ℎ03 ℎ02
(ℎ01= ℎ02 1198881198881= 1198881198883)
120549120549ℎ )0(3minus2 = (ℎ3 + frasl11988811988832 2) minus (ℎ2 + frasl11988811988822 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
0(3 2) 3 2( )∆ minus = minusu uh U c c
= 1 minusfrasl11988811988832 2 minus frasl11988811988822 2120549120549ℎ )0(3minus2
119877119877 =120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
120549120549ℎ )0(3minus2
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
3 212+
= minus u uc cRU
119888119888119909119909 = 1198881198882119909119909 = 1198881198883119909119909 = 119888119888119888119888119888119888119888119888119888119888
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 11988811988821199061199061198881198882 = 1198881198881199091199092+1198881198881199061199062
119877119877 = 1 minus11988811988831199061199062 minus 11988811988821199061199062
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
Indirectement lrsquoeacutequation pour R deacutecrite en fonction de composantes cineacutematiques eacutetablit un lien entre les angles drsquoentreacutee et de sortie
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
= 1 minus1198881198883119906119906 + 1198881198882119906119906
2119880119880
2232
2 32 2wwh h+ = +
2 23 2
2 3 2 2w wh hminus = minus
2 2 2 23 3 2 2
2 3 2 2u x u xw w w wh h + +
minus = minus
Machine axiale U2 = U3
La composante axiale de la vitesse demeure constante w3x = w 2x
( eacutequation justifieacutee plus tard lors de la rothalpie)
2 23 2
2 3 2 2u uw wh hminus = minus
3 2 3 22 3
( )( )2
u u u uw w w wh h minus +minus =
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
Eacutetage reacutepeacutetitif c3 = c1
Si les pertes dans le stator sont neacutegligeacutees
c2u ndash c3u= w2u - w3u
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
3 2 3 23 2
( )( )2
u u u uw w w wh h minus +minus = minus
c2u ndash c3u= w2u - w3u
119877119877 = minus1199081199083119906119906 + 1199081199082119906119906
2119880119880
2 2 2 22 3 2 3 2 3( ) 2 ( ) 2h h u u w wminus = minus minus minus
2 2 2 2 2 23 2 3 2 3 2
2 2 2e
I II III
u u c c w wW minus minus minus= + minus
2 2 2 22 3 2 3
2 2 2 2 2 22 3 2 3 2 3
( ) ( )( ) ( ) ( )
minus minus minus=
minus + minus minus minusu u w wR
c c u u w w
3 2= =U u u
Expliqueacutee + tard
2 23 2
03 02 3 2 2ec cW h h h h minus
= minus = minus +
119877119877 =ℎ2 minus ℎ3ℎ1 minus ℎ3
=ℎ2 minus ℎ3ℎ02 minus ℎ03
Degreacute de reacuteaction
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
( )3u 2uw wR2U+
= minus
3 212
u uc cRU+
= minus
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
Puisque le degreacute de reacuteaction R peut aussi srsquoexprimer enfonction de variables cineacutematiques on peut chercher sarepreacutesentation dans le triangle adimensionnel de vitesses
Turbine
α2
β2 α3
β3
C2 W3
C3W2
C3uC2u
U
C2xC3x
W2u
c2u ndash c3u= w2u - w3u
2Ψ minusR
frasl119934119934120784120784119958119958 119932119932 = frasl1198621198622119906119906 119880119880 minus frasl119880119880 119880119880Ψ = frasl1198621198622119906119906 minus 1198621198623119906119906 119880119880
119877119877 = 1 minus (1198621198623119906119906 + 1198621198622119906119906 frasl) 2119880119880Φ = frasl1198621198622119909119909 119880119880 = frasl1198621198623119909119909 119880119880
= (119914119914120784120784119958119958 minus 119914119914120785120785119958119958 frasl) 2119880119880 minus 1 + (119914119914120784120784119958119958 + 119914119914120785120785119958119958 frasl) 2119880119880
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α2
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
Les repreacutesentations des triangles de vitesses adimensionnellessont tregraves riches en information et contiennent plusieurs formules
Par exemple pour un compresseur on deacuteduit aiseacutement que
Avec 120595120595 = 1198821198821198901198901198801198802
Φ = 119888119888119909119909119880119880
et que pour 119888119888119901119901 = 119888119888119888119888119888119888119888119888119888119888 119882119882119890119890 = 119888119888119901119901(11987911987902minus11987911987901)on trouve
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
(11987911987902minus11987911987901) =119880119880119888119888119909119909119888119888119901119901
(tan1205731205731 minus tan1205731205732)
Alors si lrsquoon veut obtenir une plus grande variation detempeacuterature 11987911987902 minus 11987911987901 indirectement un plus grand rapport(1199011199010211990111990101) on peut augmenter
bull la vitesse peacuteripheacuterique 119880119880 (la vitesse de rotation)bull la vitesse axiale 119888119888119909119909 (le deacutebit massique)bull lrsquoangle de deacuteflection de la pale(tan1205731205731 minus tan1205731205732)
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
Le cas du compresseur est similaire agrave celui drsquoune turbineLe rayon ne varie pas (lrsquoeacutetude se deacuteroule dans le plan inter-aube) et la vitesse axiale demeure constante dans le sens axialComme pour la turbine nous pouvons superposer les triangles devitesse au bord drsquoattaque et bord de fuite du rotor
W2 C2
C3W3
U
C1W1
U
U
Compresseur
➊
➋
➌
α2 β1
C2 W1
C1W2
U
β2α1
W2U C1U
CX
( )2 1 = minuse u uW U C C
Lrsquoopeacuteration est vectorielle
C2U
119914119914120784120784119958119958 minus 119914119914120783120783119958119958 =- 119934119934120784120784119958119958 minus119934119934120783120783119958119958
➀
➁
➁
Compresseur
119914119914120784120784119914119914120785120785
119934119934120783120783
119932119932119914119914120783120783120630120630120783120783 120630120630120784120784
120631120631120784120784120630120630120785120785
Rotor Stator
119934119934120784120784120631120631120783120783
119932119932
➀ ➁ ➂
Trois quantiteacutes fondamentales sont utiliseacutees dans laconception des turbomachines
bull le coefficient de charge ψ
bull le coefficient de deacutebit Φ
bull le degreacute de reacuteaction R
Approche classique
Parfois ψ est consideacutereacute positif pour les turbines et neacutegatifpour les compresseurs 119882119882119890119890 = frasl119882 119888
Coefficient de charge
2Ψ = eWU
Coefficient de charge
2e
UW
=ψ 02 01 02 2
h h hU Uminus ∆
= = 02
s shU
η ∆=
=
eWWm
120578120578119904119904 =∆ℎ0∆ℎ0119904119904
Turbine
Coefficient de deacutebit
xcU
Φ =
xCU
Φ =ρRT
pρ
= xC RTRTU p
=RT m RT
U Ap
xCU
Φ =119888 = ρ119862119862119909119909119860119860
11988811987711987711987911987901199011199010119860119860
= 119872119872 120574120574 1 +120574120574 minus 1
21198721198722
minus 120574120574+1)2(120574120574minus1
Afin de geacuteneacuteraliser le triangle de vitesses drsquoune machineaxiale on effectue une mise agrave lrsquoeacutechelle en divisant chaquecocircteacute par la vitesse tangentielle U
Cette opeacuteration permettra de retrouver directement lescoefficients de charge ψ et de deacutebit Φ
Par la suite un nouveau paramegravetre R qursquoon appelle degreacutede reacuteaction sera introduit pour compleacuteter la caracteacuterisationde lrsquoaubage
Approche classique
α2
β2 α3
β3C2 W3
W2
c3u - c2u= w3u - w2u
W3uW2u
U
CX CX
Lrsquoopeacuteration est vectorielle
C3
120511120511 = frasl119934119934119940119940 119932119932120784120784 = frasl119914119914120785120785119958119958 minus 119914119914120784120784119958119958 119932119932
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
Turbine
3
2
3 3
2 2
12
12
12
12
u
u
u u
u u
c RUc RU
w c RU Uw c RU U
Ψ= minus +
Ψ= minus +
Ψ= minus = +
Ψ= minus = minus
2 22 23 2
2 22 23 2
1 12 2
2 2
c cR RU U
w wR RU U
Ψ Ψ = Φ + minus + = Φ + minus +
Ψ Ψ = Φ + + = Φ + minus
3 2
3 2
1 2 1 2
2 2
R Ratan atan
R Ratan atan
minus + Ψ minus + Ψ = = Φ Φ
Ψ + Ψ minus = = Φ Φ
α α
β β
α2β1
C2U
W1U
C1UW2U β2 α1
Compresseur
ΦC2U
C3UW3U
W2U
ψ
1
ψ
ΦC2UC3U
W3U
W2U
1
Il faut deacutefinir un troisiegraveme paramegravetre pour diffeacuterencier ces deux triangles de vitesses ayant le mecircme 120509120509 et le mecircme 120537120537
Les coefficients 120509120509 et 120537120537 ne suffisent pas pour deacutefinir de maniegravereunivoque le triangle de vitesses
Le degreacute de reacuteaction est un paramegravetre adimensionnel quiexprime la variation drsquoenthalpie subie par le fluide dans le rotorpar rapport agrave la chute totale drsquoenthalpie dans lrsquoeacutetage
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
Variation d enthalpie dans le rotorRVariation d enthalpie dans l eacutetage
=
❸❶
❷
Rotor
Stator
R=0 machine drsquoaction (Δhrotor=0 Δheacutetage=Δhstator)
R=1 machine agrave reacuteaction (Δhstator=0 Δheacutetage=Δhrotor)
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
R indique la reacutepartition du saut drsquoenthalpie (de pression) statique entre le stator et le rotor
Le degreacute de reacuteaction R deacutefini en fonction de variablesthermodynamiques (lrsquoenthalpie) peut aussi srsquoexprimer en fonctionde variables cineacutematiques preacutesentes dans le triangle de vitessesNotamment en fonction de la vitesse peacuteripheacuterique U et descomposantes des vitesses absolues et relatives
La vitesse absolue agrave lrsquoentreacutee du stator est eacutegale agrave celle agrave la sortie du rotor (c1= c3 eacutetage reacutepeacutetitif )
On neacuteglige les pertes et le transfert de chaleur dans le stator
1 2 3
ℎ3 minus ℎ1= ℎ3 + frasl11988811988832 2 minus ℎ1 minus frasl11988811988812 2 = ℎ03 minus ℎ01 = 120549120549ℎ0
ℎ03 minus ℎ01 = ℎ03 minus ℎ02 = 120549120549ℎ )0(3minus2
ℎ03 minus ℎ02 = ℎ3 minus ℎ1
119945119945120782120782120783120783 = 119945119945120782120782120784120784
frasl11988811988832 2 = frasl11988811988812 2
3 2
3 1
rotor
eacutetage
h h hRh h h
∆∆
minus= =
minus
2 20 3 2 2 3
0 3 2
2 2( )
( )
h c c h
∆∆
minus
minus
+ minus=
ℎ03 ℎ02
(ℎ01= ℎ02 1198881198881= 1198881198883)
120549120549ℎ )0(3minus2 = (ℎ3 + frasl11988811988832 2) minus (ℎ2 + frasl11988811988822 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
0(3 2) 3 2( )∆ minus = minusu uh U c c
= 1 minusfrasl11988811988832 2 minus frasl11988811988822 2120549120549ℎ )0(3minus2
119877119877 =120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
120549120549ℎ )0(3minus2
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
3 212+
= minus u uc cRU
119888119888119909119909 = 1198881198882119909119909 = 1198881198883119909119909 = 119888119888119888119888119888119888119888119888119888119888
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 11988811988821199061199061198881198882 = 1198881198881199091199092+1198881198881199061199062
119877119877 = 1 minus11988811988831199061199062 minus 11988811988821199061199062
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
Indirectement lrsquoeacutequation pour R deacutecrite en fonction de composantes cineacutematiques eacutetablit un lien entre les angles drsquoentreacutee et de sortie
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
= 1 minus1198881198883119906119906 + 1198881198882119906119906
2119880119880
2232
2 32 2wwh h+ = +
2 23 2
2 3 2 2w wh hminus = minus
2 2 2 23 3 2 2
2 3 2 2u x u xw w w wh h + +
minus = minus
Machine axiale U2 = U3
La composante axiale de la vitesse demeure constante w3x = w 2x
( eacutequation justifieacutee plus tard lors de la rothalpie)
2 23 2
2 3 2 2u uw wh hminus = minus
3 2 3 22 3
( )( )2
u u u uw w w wh h minus +minus =
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
Eacutetage reacutepeacutetitif c3 = c1
Si les pertes dans le stator sont neacutegligeacutees
c2u ndash c3u= w2u - w3u
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
3 2 3 23 2
( )( )2
u u u uw w w wh h minus +minus = minus
c2u ndash c3u= w2u - w3u
119877119877 = minus1199081199083119906119906 + 1199081199082119906119906
2119880119880
2 2 2 22 3 2 3 2 3( ) 2 ( ) 2h h u u w wminus = minus minus minus
2 2 2 2 2 23 2 3 2 3 2
2 2 2e
I II III
u u c c w wW minus minus minus= + minus
2 2 2 22 3 2 3
2 2 2 2 2 22 3 2 3 2 3
( ) ( )( ) ( ) ( )
minus minus minus=
minus + minus minus minusu u w wR
c c u u w w
3 2= =U u u
Expliqueacutee + tard
2 23 2
03 02 3 2 2ec cW h h h h minus
= minus = minus +
119877119877 =ℎ2 minus ℎ3ℎ1 minus ℎ3
=ℎ2 minus ℎ3ℎ02 minus ℎ03
Degreacute de reacuteaction
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
( )3u 2uw wR2U+
= minus
3 212
u uc cRU+
= minus
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
Puisque le degreacute de reacuteaction R peut aussi srsquoexprimer enfonction de variables cineacutematiques on peut chercher sarepreacutesentation dans le triangle adimensionnel de vitesses
Turbine
α2
β2 α3
β3
C2 W3
C3W2
C3uC2u
U
C2xC3x
W2u
c2u ndash c3u= w2u - w3u
2Ψ minusR
frasl119934119934120784120784119958119958 119932119932 = frasl1198621198622119906119906 119880119880 minus frasl119880119880 119880119880Ψ = frasl1198621198622119906119906 minus 1198621198623119906119906 119880119880
119877119877 = 1 minus (1198621198623119906119906 + 1198621198622119906119906 frasl) 2119880119880Φ = frasl1198621198622119909119909 119880119880 = frasl1198621198623119909119909 119880119880
= (119914119914120784120784119958119958 minus 119914119914120785120785119958119958 frasl) 2119880119880 minus 1 + (119914119914120784120784119958119958 + 119914119914120785120785119958119958 frasl) 2119880119880
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α2
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
Les repreacutesentations des triangles de vitesses adimensionnellessont tregraves riches en information et contiennent plusieurs formules
Par exemple pour un compresseur on deacuteduit aiseacutement que
Avec 120595120595 = 1198821198821198901198901198801198802
Φ = 119888119888119909119909119880119880
et que pour 119888119888119901119901 = 119888119888119888119888119888119888119888119888119888119888 119882119882119890119890 = 119888119888119901119901(11987911987902minus11987911987901)on trouve
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
(11987911987902minus11987911987901) =119880119880119888119888119909119909119888119888119901119901
(tan1205731205731 minus tan1205731205732)
Alors si lrsquoon veut obtenir une plus grande variation detempeacuterature 11987911987902 minus 11987911987901 indirectement un plus grand rapport(1199011199010211990111990101) on peut augmenter
bull la vitesse peacuteripheacuterique 119880119880 (la vitesse de rotation)bull la vitesse axiale 119888119888119909119909 (le deacutebit massique)bull lrsquoangle de deacuteflection de la pale(tan1205731205731 minus tan1205731205732)
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
Download Sound Effects - SoundDogs - Voice Box
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
W2 C2
C3W3
U
C1W1
U
U
Compresseur
➊
➋
➌
α2 β1
C2 W1
C1W2
U
β2α1
W2U C1U
CX
( )2 1 = minuse u uW U C C
Lrsquoopeacuteration est vectorielle
C2U
119914119914120784120784119958119958 minus 119914119914120783120783119958119958 =- 119934119934120784120784119958119958 minus119934119934120783120783119958119958
➀
➁
➁
Compresseur
119914119914120784120784119914119914120785120785
119934119934120783120783
119932119932119914119914120783120783120630120630120783120783 120630120630120784120784
120631120631120784120784120630120630120785120785
Rotor Stator
119934119934120784120784120631120631120783120783
119932119932
➀ ➁ ➂
Trois quantiteacutes fondamentales sont utiliseacutees dans laconception des turbomachines
bull le coefficient de charge ψ
bull le coefficient de deacutebit Φ
bull le degreacute de reacuteaction R
Approche classique
Parfois ψ est consideacutereacute positif pour les turbines et neacutegatifpour les compresseurs 119882119882119890119890 = frasl119882 119888
Coefficient de charge
2Ψ = eWU
Coefficient de charge
2e
UW
=ψ 02 01 02 2
h h hU Uminus ∆
= = 02
s shU
η ∆=
=
eWWm
120578120578119904119904 =∆ℎ0∆ℎ0119904119904
Turbine
Coefficient de deacutebit
xcU
Φ =
xCU
Φ =ρRT
pρ
= xC RTRTU p
=RT m RT
U Ap
xCU
Φ =119888 = ρ119862119862119909119909119860119860
11988811987711987711987911987901199011199010119860119860
= 119872119872 120574120574 1 +120574120574 minus 1
21198721198722
minus 120574120574+1)2(120574120574minus1
Afin de geacuteneacuteraliser le triangle de vitesses drsquoune machineaxiale on effectue une mise agrave lrsquoeacutechelle en divisant chaquecocircteacute par la vitesse tangentielle U
Cette opeacuteration permettra de retrouver directement lescoefficients de charge ψ et de deacutebit Φ
Par la suite un nouveau paramegravetre R qursquoon appelle degreacutede reacuteaction sera introduit pour compleacuteter la caracteacuterisationde lrsquoaubage
Approche classique
α2
β2 α3
β3C2 W3
W2
c3u - c2u= w3u - w2u
W3uW2u
U
CX CX
Lrsquoopeacuteration est vectorielle
C3
120511120511 = frasl119934119934119940119940 119932119932120784120784 = frasl119914119914120785120785119958119958 minus 119914119914120784120784119958119958 119932119932
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
Turbine
3
2
3 3
2 2
12
12
12
12
u
u
u u
u u
c RUc RU
w c RU Uw c RU U
Ψ= minus +
Ψ= minus +
Ψ= minus = +
Ψ= minus = minus
2 22 23 2
2 22 23 2
1 12 2
2 2
c cR RU U
w wR RU U
Ψ Ψ = Φ + minus + = Φ + minus +
Ψ Ψ = Φ + + = Φ + minus
3 2
3 2
1 2 1 2
2 2
R Ratan atan
R Ratan atan
minus + Ψ minus + Ψ = = Φ Φ
Ψ + Ψ minus = = Φ Φ
α α
β β
α2β1
C2U
W1U
C1UW2U β2 α1
Compresseur
ΦC2U
C3UW3U
W2U
ψ
1
ψ
ΦC2UC3U
W3U
W2U
1
Il faut deacutefinir un troisiegraveme paramegravetre pour diffeacuterencier ces deux triangles de vitesses ayant le mecircme 120509120509 et le mecircme 120537120537
Les coefficients 120509120509 et 120537120537 ne suffisent pas pour deacutefinir de maniegravereunivoque le triangle de vitesses
Le degreacute de reacuteaction est un paramegravetre adimensionnel quiexprime la variation drsquoenthalpie subie par le fluide dans le rotorpar rapport agrave la chute totale drsquoenthalpie dans lrsquoeacutetage
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
Variation d enthalpie dans le rotorRVariation d enthalpie dans l eacutetage
=
❸❶
❷
Rotor
Stator
R=0 machine drsquoaction (Δhrotor=0 Δheacutetage=Δhstator)
R=1 machine agrave reacuteaction (Δhstator=0 Δheacutetage=Δhrotor)
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
R indique la reacutepartition du saut drsquoenthalpie (de pression) statique entre le stator et le rotor
Le degreacute de reacuteaction R deacutefini en fonction de variablesthermodynamiques (lrsquoenthalpie) peut aussi srsquoexprimer en fonctionde variables cineacutematiques preacutesentes dans le triangle de vitessesNotamment en fonction de la vitesse peacuteripheacuterique U et descomposantes des vitesses absolues et relatives
La vitesse absolue agrave lrsquoentreacutee du stator est eacutegale agrave celle agrave la sortie du rotor (c1= c3 eacutetage reacutepeacutetitif )
On neacuteglige les pertes et le transfert de chaleur dans le stator
1 2 3
ℎ3 minus ℎ1= ℎ3 + frasl11988811988832 2 minus ℎ1 minus frasl11988811988812 2 = ℎ03 minus ℎ01 = 120549120549ℎ0
ℎ03 minus ℎ01 = ℎ03 minus ℎ02 = 120549120549ℎ )0(3minus2
ℎ03 minus ℎ02 = ℎ3 minus ℎ1
119945119945120782120782120783120783 = 119945119945120782120782120784120784
frasl11988811988832 2 = frasl11988811988812 2
3 2
3 1
rotor
eacutetage
h h hRh h h
∆∆
minus= =
minus
2 20 3 2 2 3
0 3 2
2 2( )
( )
h c c h
∆∆
minus
minus
+ minus=
ℎ03 ℎ02
(ℎ01= ℎ02 1198881198881= 1198881198883)
120549120549ℎ )0(3minus2 = (ℎ3 + frasl11988811988832 2) minus (ℎ2 + frasl11988811988822 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
0(3 2) 3 2( )∆ minus = minusu uh U c c
= 1 minusfrasl11988811988832 2 minus frasl11988811988822 2120549120549ℎ )0(3minus2
119877119877 =120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
120549120549ℎ )0(3minus2
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
3 212+
= minus u uc cRU
119888119888119909119909 = 1198881198882119909119909 = 1198881198883119909119909 = 119888119888119888119888119888119888119888119888119888119888
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 11988811988821199061199061198881198882 = 1198881198881199091199092+1198881198881199061199062
119877119877 = 1 minus11988811988831199061199062 minus 11988811988821199061199062
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
Indirectement lrsquoeacutequation pour R deacutecrite en fonction de composantes cineacutematiques eacutetablit un lien entre les angles drsquoentreacutee et de sortie
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
= 1 minus1198881198883119906119906 + 1198881198882119906119906
2119880119880
2232
2 32 2wwh h+ = +
2 23 2
2 3 2 2w wh hminus = minus
2 2 2 23 3 2 2
2 3 2 2u x u xw w w wh h + +
minus = minus
Machine axiale U2 = U3
La composante axiale de la vitesse demeure constante w3x = w 2x
( eacutequation justifieacutee plus tard lors de la rothalpie)
2 23 2
2 3 2 2u uw wh hminus = minus
3 2 3 22 3
( )( )2
u u u uw w w wh h minus +minus =
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
Eacutetage reacutepeacutetitif c3 = c1
Si les pertes dans le stator sont neacutegligeacutees
c2u ndash c3u= w2u - w3u
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
3 2 3 23 2
( )( )2
u u u uw w w wh h minus +minus = minus
c2u ndash c3u= w2u - w3u
119877119877 = minus1199081199083119906119906 + 1199081199082119906119906
2119880119880
2 2 2 22 3 2 3 2 3( ) 2 ( ) 2h h u u w wminus = minus minus minus
2 2 2 2 2 23 2 3 2 3 2
2 2 2e
I II III
u u c c w wW minus minus minus= + minus
2 2 2 22 3 2 3
2 2 2 2 2 22 3 2 3 2 3
( ) ( )( ) ( ) ( )
minus minus minus=
minus + minus minus minusu u w wR
c c u u w w
3 2= =U u u
Expliqueacutee + tard
2 23 2
03 02 3 2 2ec cW h h h h minus
= minus = minus +
119877119877 =ℎ2 minus ℎ3ℎ1 minus ℎ3
=ℎ2 minus ℎ3ℎ02 minus ℎ03
Degreacute de reacuteaction
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
( )3u 2uw wR2U+
= minus
3 212
u uc cRU+
= minus
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
Puisque le degreacute de reacuteaction R peut aussi srsquoexprimer enfonction de variables cineacutematiques on peut chercher sarepreacutesentation dans le triangle adimensionnel de vitesses
Turbine
α2
β2 α3
β3
C2 W3
C3W2
C3uC2u
U
C2xC3x
W2u
c2u ndash c3u= w2u - w3u
2Ψ minusR
frasl119934119934120784120784119958119958 119932119932 = frasl1198621198622119906119906 119880119880 minus frasl119880119880 119880119880Ψ = frasl1198621198622119906119906 minus 1198621198623119906119906 119880119880
119877119877 = 1 minus (1198621198623119906119906 + 1198621198622119906119906 frasl) 2119880119880Φ = frasl1198621198622119909119909 119880119880 = frasl1198621198623119909119909 119880119880
= (119914119914120784120784119958119958 minus 119914119914120785120785119958119958 frasl) 2119880119880 minus 1 + (119914119914120784120784119958119958 + 119914119914120785120785119958119958 frasl) 2119880119880
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α2
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
Les repreacutesentations des triangles de vitesses adimensionnellessont tregraves riches en information et contiennent plusieurs formules
Par exemple pour un compresseur on deacuteduit aiseacutement que
Avec 120595120595 = 1198821198821198901198901198801198802
Φ = 119888119888119909119909119880119880
et que pour 119888119888119901119901 = 119888119888119888119888119888119888119888119888119888119888 119882119882119890119890 = 119888119888119901119901(11987911987902minus11987911987901)on trouve
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
(11987911987902minus11987911987901) =119880119880119888119888119909119909119888119888119901119901
(tan1205731205731 minus tan1205731205732)
Alors si lrsquoon veut obtenir une plus grande variation detempeacuterature 11987911987902 minus 11987911987901 indirectement un plus grand rapport(1199011199010211990111990101) on peut augmenter
bull la vitesse peacuteripheacuterique 119880119880 (la vitesse de rotation)bull la vitesse axiale 119888119888119909119909 (le deacutebit massique)bull lrsquoangle de deacuteflection de la pale(tan1205731205731 minus tan1205731205732)
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
α2 β1
C2 W1
C1W2
U
β2α1
W2U C1U
CX
( )2 1 = minuse u uW U C C
Lrsquoopeacuteration est vectorielle
C2U
119914119914120784120784119958119958 minus 119914119914120783120783119958119958 =- 119934119934120784120784119958119958 minus119934119934120783120783119958119958
➀
➁
➁
Compresseur
119914119914120784120784119914119914120785120785
119934119934120783120783
119932119932119914119914120783120783120630120630120783120783 120630120630120784120784
120631120631120784120784120630120630120785120785
Rotor Stator
119934119934120784120784120631120631120783120783
119932119932
➀ ➁ ➂
Trois quantiteacutes fondamentales sont utiliseacutees dans laconception des turbomachines
bull le coefficient de charge ψ
bull le coefficient de deacutebit Φ
bull le degreacute de reacuteaction R
Approche classique
Parfois ψ est consideacutereacute positif pour les turbines et neacutegatifpour les compresseurs 119882119882119890119890 = frasl119882 119888
Coefficient de charge
2Ψ = eWU
Coefficient de charge
2e
UW
=ψ 02 01 02 2
h h hU Uminus ∆
= = 02
s shU
η ∆=
=
eWWm
120578120578119904119904 =∆ℎ0∆ℎ0119904119904
Turbine
Coefficient de deacutebit
xcU
Φ =
xCU
Φ =ρRT
pρ
= xC RTRTU p
=RT m RT
U Ap
xCU
Φ =119888 = ρ119862119862119909119909119860119860
11988811987711987711987911987901199011199010119860119860
= 119872119872 120574120574 1 +120574120574 minus 1
21198721198722
minus 120574120574+1)2(120574120574minus1
Afin de geacuteneacuteraliser le triangle de vitesses drsquoune machineaxiale on effectue une mise agrave lrsquoeacutechelle en divisant chaquecocircteacute par la vitesse tangentielle U
Cette opeacuteration permettra de retrouver directement lescoefficients de charge ψ et de deacutebit Φ
Par la suite un nouveau paramegravetre R qursquoon appelle degreacutede reacuteaction sera introduit pour compleacuteter la caracteacuterisationde lrsquoaubage
Approche classique
α2
β2 α3
β3C2 W3
W2
c3u - c2u= w3u - w2u
W3uW2u
U
CX CX
Lrsquoopeacuteration est vectorielle
C3
120511120511 = frasl119934119934119940119940 119932119932120784120784 = frasl119914119914120785120785119958119958 minus 119914119914120784120784119958119958 119932119932
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
Turbine
3
2
3 3
2 2
12
12
12
12
u
u
u u
u u
c RUc RU
w c RU Uw c RU U
Ψ= minus +
Ψ= minus +
Ψ= minus = +
Ψ= minus = minus
2 22 23 2
2 22 23 2
1 12 2
2 2
c cR RU U
w wR RU U
Ψ Ψ = Φ + minus + = Φ + minus +
Ψ Ψ = Φ + + = Φ + minus
3 2
3 2
1 2 1 2
2 2
R Ratan atan
R Ratan atan
minus + Ψ minus + Ψ = = Φ Φ
Ψ + Ψ minus = = Φ Φ
α α
β β
α2β1
C2U
W1U
C1UW2U β2 α1
Compresseur
ΦC2U
C3UW3U
W2U
ψ
1
ψ
ΦC2UC3U
W3U
W2U
1
Il faut deacutefinir un troisiegraveme paramegravetre pour diffeacuterencier ces deux triangles de vitesses ayant le mecircme 120509120509 et le mecircme 120537120537
Les coefficients 120509120509 et 120537120537 ne suffisent pas pour deacutefinir de maniegravereunivoque le triangle de vitesses
Le degreacute de reacuteaction est un paramegravetre adimensionnel quiexprime la variation drsquoenthalpie subie par le fluide dans le rotorpar rapport agrave la chute totale drsquoenthalpie dans lrsquoeacutetage
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
Variation d enthalpie dans le rotorRVariation d enthalpie dans l eacutetage
=
❸❶
❷
Rotor
Stator
R=0 machine drsquoaction (Δhrotor=0 Δheacutetage=Δhstator)
R=1 machine agrave reacuteaction (Δhstator=0 Δheacutetage=Δhrotor)
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
R indique la reacutepartition du saut drsquoenthalpie (de pression) statique entre le stator et le rotor
Le degreacute de reacuteaction R deacutefini en fonction de variablesthermodynamiques (lrsquoenthalpie) peut aussi srsquoexprimer en fonctionde variables cineacutematiques preacutesentes dans le triangle de vitessesNotamment en fonction de la vitesse peacuteripheacuterique U et descomposantes des vitesses absolues et relatives
La vitesse absolue agrave lrsquoentreacutee du stator est eacutegale agrave celle agrave la sortie du rotor (c1= c3 eacutetage reacutepeacutetitif )
On neacuteglige les pertes et le transfert de chaleur dans le stator
1 2 3
ℎ3 minus ℎ1= ℎ3 + frasl11988811988832 2 minus ℎ1 minus frasl11988811988812 2 = ℎ03 minus ℎ01 = 120549120549ℎ0
ℎ03 minus ℎ01 = ℎ03 minus ℎ02 = 120549120549ℎ )0(3minus2
ℎ03 minus ℎ02 = ℎ3 minus ℎ1
119945119945120782120782120783120783 = 119945119945120782120782120784120784
frasl11988811988832 2 = frasl11988811988812 2
3 2
3 1
rotor
eacutetage
h h hRh h h
∆∆
minus= =
minus
2 20 3 2 2 3
0 3 2
2 2( )
( )
h c c h
∆∆
minus
minus
+ minus=
ℎ03 ℎ02
(ℎ01= ℎ02 1198881198881= 1198881198883)
120549120549ℎ )0(3minus2 = (ℎ3 + frasl11988811988832 2) minus (ℎ2 + frasl11988811988822 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
0(3 2) 3 2( )∆ minus = minusu uh U c c
= 1 minusfrasl11988811988832 2 minus frasl11988811988822 2120549120549ℎ )0(3minus2
119877119877 =120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
120549120549ℎ )0(3minus2
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
3 212+
= minus u uc cRU
119888119888119909119909 = 1198881198882119909119909 = 1198881198883119909119909 = 119888119888119888119888119888119888119888119888119888119888
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 11988811988821199061199061198881198882 = 1198881198881199091199092+1198881198881199061199062
119877119877 = 1 minus11988811988831199061199062 minus 11988811988821199061199062
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
Indirectement lrsquoeacutequation pour R deacutecrite en fonction de composantes cineacutematiques eacutetablit un lien entre les angles drsquoentreacutee et de sortie
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
= 1 minus1198881198883119906119906 + 1198881198882119906119906
2119880119880
2232
2 32 2wwh h+ = +
2 23 2
2 3 2 2w wh hminus = minus
2 2 2 23 3 2 2
2 3 2 2u x u xw w w wh h + +
minus = minus
Machine axiale U2 = U3
La composante axiale de la vitesse demeure constante w3x = w 2x
( eacutequation justifieacutee plus tard lors de la rothalpie)
2 23 2
2 3 2 2u uw wh hminus = minus
3 2 3 22 3
( )( )2
u u u uw w w wh h minus +minus =
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
Eacutetage reacutepeacutetitif c3 = c1
Si les pertes dans le stator sont neacutegligeacutees
c2u ndash c3u= w2u - w3u
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
3 2 3 23 2
( )( )2
u u u uw w w wh h minus +minus = minus
c2u ndash c3u= w2u - w3u
119877119877 = minus1199081199083119906119906 + 1199081199082119906119906
2119880119880
2 2 2 22 3 2 3 2 3( ) 2 ( ) 2h h u u w wminus = minus minus minus
2 2 2 2 2 23 2 3 2 3 2
2 2 2e
I II III
u u c c w wW minus minus minus= + minus
2 2 2 22 3 2 3
2 2 2 2 2 22 3 2 3 2 3
( ) ( )( ) ( ) ( )
minus minus minus=
minus + minus minus minusu u w wR
c c u u w w
3 2= =U u u
Expliqueacutee + tard
2 23 2
03 02 3 2 2ec cW h h h h minus
= minus = minus +
119877119877 =ℎ2 minus ℎ3ℎ1 minus ℎ3
=ℎ2 minus ℎ3ℎ02 minus ℎ03
Degreacute de reacuteaction
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
( )3u 2uw wR2U+
= minus
3 212
u uc cRU+
= minus
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
Puisque le degreacute de reacuteaction R peut aussi srsquoexprimer enfonction de variables cineacutematiques on peut chercher sarepreacutesentation dans le triangle adimensionnel de vitesses
Turbine
α2
β2 α3
β3
C2 W3
C3W2
C3uC2u
U
C2xC3x
W2u
c2u ndash c3u= w2u - w3u
2Ψ minusR
frasl119934119934120784120784119958119958 119932119932 = frasl1198621198622119906119906 119880119880 minus frasl119880119880 119880119880Ψ = frasl1198621198622119906119906 minus 1198621198623119906119906 119880119880
119877119877 = 1 minus (1198621198623119906119906 + 1198621198622119906119906 frasl) 2119880119880Φ = frasl1198621198622119909119909 119880119880 = frasl1198621198623119909119909 119880119880
= (119914119914120784120784119958119958 minus 119914119914120785120785119958119958 frasl) 2119880119880 minus 1 + (119914119914120784120784119958119958 + 119914119914120785120785119958119958 frasl) 2119880119880
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α2
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
Les repreacutesentations des triangles de vitesses adimensionnellessont tregraves riches en information et contiennent plusieurs formules
Par exemple pour un compresseur on deacuteduit aiseacutement que
Avec 120595120595 = 1198821198821198901198901198801198802
Φ = 119888119888119909119909119880119880
et que pour 119888119888119901119901 = 119888119888119888119888119888119888119888119888119888119888 119882119882119890119890 = 119888119888119901119901(11987911987902minus11987911987901)on trouve
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
(11987911987902minus11987911987901) =119880119880119888119888119909119909119888119888119901119901
(tan1205731205731 minus tan1205731205732)
Alors si lrsquoon veut obtenir une plus grande variation detempeacuterature 11987911987902 minus 11987911987901 indirectement un plus grand rapport(1199011199010211990111990101) on peut augmenter
bull la vitesse peacuteripheacuterique 119880119880 (la vitesse de rotation)bull la vitesse axiale 119888119888119909119909 (le deacutebit massique)bull lrsquoangle de deacuteflection de la pale(tan1205731205731 minus tan1205731205732)
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
Compresseur
119914119914120784120784119914119914120785120785
119934119934120783120783
119932119932119914119914120783120783120630120630120783120783 120630120630120784120784
120631120631120784120784120630120630120785120785
Rotor Stator
119934119934120784120784120631120631120783120783
119932119932
➀ ➁ ➂
Trois quantiteacutes fondamentales sont utiliseacutees dans laconception des turbomachines
bull le coefficient de charge ψ
bull le coefficient de deacutebit Φ
bull le degreacute de reacuteaction R
Approche classique
Parfois ψ est consideacutereacute positif pour les turbines et neacutegatifpour les compresseurs 119882119882119890119890 = frasl119882 119888
Coefficient de charge
2Ψ = eWU
Coefficient de charge
2e
UW
=ψ 02 01 02 2
h h hU Uminus ∆
= = 02
s shU
η ∆=
=
eWWm
120578120578119904119904 =∆ℎ0∆ℎ0119904119904
Turbine
Coefficient de deacutebit
xcU
Φ =
xCU
Φ =ρRT
pρ
= xC RTRTU p
=RT m RT
U Ap
xCU
Φ =119888 = ρ119862119862119909119909119860119860
11988811987711987711987911987901199011199010119860119860
= 119872119872 120574120574 1 +120574120574 minus 1
21198721198722
minus 120574120574+1)2(120574120574minus1
Afin de geacuteneacuteraliser le triangle de vitesses drsquoune machineaxiale on effectue une mise agrave lrsquoeacutechelle en divisant chaquecocircteacute par la vitesse tangentielle U
Cette opeacuteration permettra de retrouver directement lescoefficients de charge ψ et de deacutebit Φ
Par la suite un nouveau paramegravetre R qursquoon appelle degreacutede reacuteaction sera introduit pour compleacuteter la caracteacuterisationde lrsquoaubage
Approche classique
α2
β2 α3
β3C2 W3
W2
c3u - c2u= w3u - w2u
W3uW2u
U
CX CX
Lrsquoopeacuteration est vectorielle
C3
120511120511 = frasl119934119934119940119940 119932119932120784120784 = frasl119914119914120785120785119958119958 minus 119914119914120784120784119958119958 119932119932
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
Turbine
3
2
3 3
2 2
12
12
12
12
u
u
u u
u u
c RUc RU
w c RU Uw c RU U
Ψ= minus +
Ψ= minus +
Ψ= minus = +
Ψ= minus = minus
2 22 23 2
2 22 23 2
1 12 2
2 2
c cR RU U
w wR RU U
Ψ Ψ = Φ + minus + = Φ + minus +
Ψ Ψ = Φ + + = Φ + minus
3 2
3 2
1 2 1 2
2 2
R Ratan atan
R Ratan atan
minus + Ψ minus + Ψ = = Φ Φ
Ψ + Ψ minus = = Φ Φ
α α
β β
α2β1
C2U
W1U
C1UW2U β2 α1
Compresseur
ΦC2U
C3UW3U
W2U
ψ
1
ψ
ΦC2UC3U
W3U
W2U
1
Il faut deacutefinir un troisiegraveme paramegravetre pour diffeacuterencier ces deux triangles de vitesses ayant le mecircme 120509120509 et le mecircme 120537120537
Les coefficients 120509120509 et 120537120537 ne suffisent pas pour deacutefinir de maniegravereunivoque le triangle de vitesses
Le degreacute de reacuteaction est un paramegravetre adimensionnel quiexprime la variation drsquoenthalpie subie par le fluide dans le rotorpar rapport agrave la chute totale drsquoenthalpie dans lrsquoeacutetage
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
Variation d enthalpie dans le rotorRVariation d enthalpie dans l eacutetage
=
❸❶
❷
Rotor
Stator
R=0 machine drsquoaction (Δhrotor=0 Δheacutetage=Δhstator)
R=1 machine agrave reacuteaction (Δhstator=0 Δheacutetage=Δhrotor)
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
R indique la reacutepartition du saut drsquoenthalpie (de pression) statique entre le stator et le rotor
Le degreacute de reacuteaction R deacutefini en fonction de variablesthermodynamiques (lrsquoenthalpie) peut aussi srsquoexprimer en fonctionde variables cineacutematiques preacutesentes dans le triangle de vitessesNotamment en fonction de la vitesse peacuteripheacuterique U et descomposantes des vitesses absolues et relatives
La vitesse absolue agrave lrsquoentreacutee du stator est eacutegale agrave celle agrave la sortie du rotor (c1= c3 eacutetage reacutepeacutetitif )
On neacuteglige les pertes et le transfert de chaleur dans le stator
1 2 3
ℎ3 minus ℎ1= ℎ3 + frasl11988811988832 2 minus ℎ1 minus frasl11988811988812 2 = ℎ03 minus ℎ01 = 120549120549ℎ0
ℎ03 minus ℎ01 = ℎ03 minus ℎ02 = 120549120549ℎ )0(3minus2
ℎ03 minus ℎ02 = ℎ3 minus ℎ1
119945119945120782120782120783120783 = 119945119945120782120782120784120784
frasl11988811988832 2 = frasl11988811988812 2
3 2
3 1
rotor
eacutetage
h h hRh h h
∆∆
minus= =
minus
2 20 3 2 2 3
0 3 2
2 2( )
( )
h c c h
∆∆
minus
minus
+ minus=
ℎ03 ℎ02
(ℎ01= ℎ02 1198881198881= 1198881198883)
120549120549ℎ )0(3minus2 = (ℎ3 + frasl11988811988832 2) minus (ℎ2 + frasl11988811988822 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
0(3 2) 3 2( )∆ minus = minusu uh U c c
= 1 minusfrasl11988811988832 2 minus frasl11988811988822 2120549120549ℎ )0(3minus2
119877119877 =120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
120549120549ℎ )0(3minus2
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
3 212+
= minus u uc cRU
119888119888119909119909 = 1198881198882119909119909 = 1198881198883119909119909 = 119888119888119888119888119888119888119888119888119888119888
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 11988811988821199061199061198881198882 = 1198881198881199091199092+1198881198881199061199062
119877119877 = 1 minus11988811988831199061199062 minus 11988811988821199061199062
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
Indirectement lrsquoeacutequation pour R deacutecrite en fonction de composantes cineacutematiques eacutetablit un lien entre les angles drsquoentreacutee et de sortie
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
= 1 minus1198881198883119906119906 + 1198881198882119906119906
2119880119880
2232
2 32 2wwh h+ = +
2 23 2
2 3 2 2w wh hminus = minus
2 2 2 23 3 2 2
2 3 2 2u x u xw w w wh h + +
minus = minus
Machine axiale U2 = U3
La composante axiale de la vitesse demeure constante w3x = w 2x
( eacutequation justifieacutee plus tard lors de la rothalpie)
2 23 2
2 3 2 2u uw wh hminus = minus
3 2 3 22 3
( )( )2
u u u uw w w wh h minus +minus =
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
Eacutetage reacutepeacutetitif c3 = c1
Si les pertes dans le stator sont neacutegligeacutees
c2u ndash c3u= w2u - w3u
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
3 2 3 23 2
( )( )2
u u u uw w w wh h minus +minus = minus
c2u ndash c3u= w2u - w3u
119877119877 = minus1199081199083119906119906 + 1199081199082119906119906
2119880119880
2 2 2 22 3 2 3 2 3( ) 2 ( ) 2h h u u w wminus = minus minus minus
2 2 2 2 2 23 2 3 2 3 2
2 2 2e
I II III
u u c c w wW minus minus minus= + minus
2 2 2 22 3 2 3
2 2 2 2 2 22 3 2 3 2 3
( ) ( )( ) ( ) ( )
minus minus minus=
minus + minus minus minusu u w wR
c c u u w w
3 2= =U u u
Expliqueacutee + tard
2 23 2
03 02 3 2 2ec cW h h h h minus
= minus = minus +
119877119877 =ℎ2 minus ℎ3ℎ1 minus ℎ3
=ℎ2 minus ℎ3ℎ02 minus ℎ03
Degreacute de reacuteaction
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
( )3u 2uw wR2U+
= minus
3 212
u uc cRU+
= minus
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
Puisque le degreacute de reacuteaction R peut aussi srsquoexprimer enfonction de variables cineacutematiques on peut chercher sarepreacutesentation dans le triangle adimensionnel de vitesses
Turbine
α2
β2 α3
β3
C2 W3
C3W2
C3uC2u
U
C2xC3x
W2u
c2u ndash c3u= w2u - w3u
2Ψ minusR
frasl119934119934120784120784119958119958 119932119932 = frasl1198621198622119906119906 119880119880 minus frasl119880119880 119880119880Ψ = frasl1198621198622119906119906 minus 1198621198623119906119906 119880119880
119877119877 = 1 minus (1198621198623119906119906 + 1198621198622119906119906 frasl) 2119880119880Φ = frasl1198621198622119909119909 119880119880 = frasl1198621198623119909119909 119880119880
= (119914119914120784120784119958119958 minus 119914119914120785120785119958119958 frasl) 2119880119880 minus 1 + (119914119914120784120784119958119958 + 119914119914120785120785119958119958 frasl) 2119880119880
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α2
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
Les repreacutesentations des triangles de vitesses adimensionnellessont tregraves riches en information et contiennent plusieurs formules
Par exemple pour un compresseur on deacuteduit aiseacutement que
Avec 120595120595 = 1198821198821198901198901198801198802
Φ = 119888119888119909119909119880119880
et que pour 119888119888119901119901 = 119888119888119888119888119888119888119888119888119888119888 119882119882119890119890 = 119888119888119901119901(11987911987902minus11987911987901)on trouve
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
(11987911987902minus11987911987901) =119880119880119888119888119909119909119888119888119901119901
(tan1205731205731 minus tan1205731205732)
Alors si lrsquoon veut obtenir une plus grande variation detempeacuterature 11987911987902 minus 11987911987901 indirectement un plus grand rapport(1199011199010211990111990101) on peut augmenter
bull la vitesse peacuteripheacuterique 119880119880 (la vitesse de rotation)bull la vitesse axiale 119888119888119909119909 (le deacutebit massique)bull lrsquoangle de deacuteflection de la pale(tan1205731205731 minus tan1205731205732)
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
Trois quantiteacutes fondamentales sont utiliseacutees dans laconception des turbomachines
bull le coefficient de charge ψ
bull le coefficient de deacutebit Φ
bull le degreacute de reacuteaction R
Approche classique
Parfois ψ est consideacutereacute positif pour les turbines et neacutegatifpour les compresseurs 119882119882119890119890 = frasl119882 119888
Coefficient de charge
2Ψ = eWU
Coefficient de charge
2e
UW
=ψ 02 01 02 2
h h hU Uminus ∆
= = 02
s shU
η ∆=
=
eWWm
120578120578119904119904 =∆ℎ0∆ℎ0119904119904
Turbine
Coefficient de deacutebit
xcU
Φ =
xCU
Φ =ρRT
pρ
= xC RTRTU p
=RT m RT
U Ap
xCU
Φ =119888 = ρ119862119862119909119909119860119860
11988811987711987711987911987901199011199010119860119860
= 119872119872 120574120574 1 +120574120574 minus 1
21198721198722
minus 120574120574+1)2(120574120574minus1
Afin de geacuteneacuteraliser le triangle de vitesses drsquoune machineaxiale on effectue une mise agrave lrsquoeacutechelle en divisant chaquecocircteacute par la vitesse tangentielle U
Cette opeacuteration permettra de retrouver directement lescoefficients de charge ψ et de deacutebit Φ
Par la suite un nouveau paramegravetre R qursquoon appelle degreacutede reacuteaction sera introduit pour compleacuteter la caracteacuterisationde lrsquoaubage
Approche classique
α2
β2 α3
β3C2 W3
W2
c3u - c2u= w3u - w2u
W3uW2u
U
CX CX
Lrsquoopeacuteration est vectorielle
C3
120511120511 = frasl119934119934119940119940 119932119932120784120784 = frasl119914119914120785120785119958119958 minus 119914119914120784120784119958119958 119932119932
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
Turbine
3
2
3 3
2 2
12
12
12
12
u
u
u u
u u
c RUc RU
w c RU Uw c RU U
Ψ= minus +
Ψ= minus +
Ψ= minus = +
Ψ= minus = minus
2 22 23 2
2 22 23 2
1 12 2
2 2
c cR RU U
w wR RU U
Ψ Ψ = Φ + minus + = Φ + minus +
Ψ Ψ = Φ + + = Φ + minus
3 2
3 2
1 2 1 2
2 2
R Ratan atan
R Ratan atan
minus + Ψ minus + Ψ = = Φ Φ
Ψ + Ψ minus = = Φ Φ
α α
β β
α2β1
C2U
W1U
C1UW2U β2 α1
Compresseur
ΦC2U
C3UW3U
W2U
ψ
1
ψ
ΦC2UC3U
W3U
W2U
1
Il faut deacutefinir un troisiegraveme paramegravetre pour diffeacuterencier ces deux triangles de vitesses ayant le mecircme 120509120509 et le mecircme 120537120537
Les coefficients 120509120509 et 120537120537 ne suffisent pas pour deacutefinir de maniegravereunivoque le triangle de vitesses
Le degreacute de reacuteaction est un paramegravetre adimensionnel quiexprime la variation drsquoenthalpie subie par le fluide dans le rotorpar rapport agrave la chute totale drsquoenthalpie dans lrsquoeacutetage
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
Variation d enthalpie dans le rotorRVariation d enthalpie dans l eacutetage
=
❸❶
❷
Rotor
Stator
R=0 machine drsquoaction (Δhrotor=0 Δheacutetage=Δhstator)
R=1 machine agrave reacuteaction (Δhstator=0 Δheacutetage=Δhrotor)
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
R indique la reacutepartition du saut drsquoenthalpie (de pression) statique entre le stator et le rotor
Le degreacute de reacuteaction R deacutefini en fonction de variablesthermodynamiques (lrsquoenthalpie) peut aussi srsquoexprimer en fonctionde variables cineacutematiques preacutesentes dans le triangle de vitessesNotamment en fonction de la vitesse peacuteripheacuterique U et descomposantes des vitesses absolues et relatives
La vitesse absolue agrave lrsquoentreacutee du stator est eacutegale agrave celle agrave la sortie du rotor (c1= c3 eacutetage reacutepeacutetitif )
On neacuteglige les pertes et le transfert de chaleur dans le stator
1 2 3
ℎ3 minus ℎ1= ℎ3 + frasl11988811988832 2 minus ℎ1 minus frasl11988811988812 2 = ℎ03 minus ℎ01 = 120549120549ℎ0
ℎ03 minus ℎ01 = ℎ03 minus ℎ02 = 120549120549ℎ )0(3minus2
ℎ03 minus ℎ02 = ℎ3 minus ℎ1
119945119945120782120782120783120783 = 119945119945120782120782120784120784
frasl11988811988832 2 = frasl11988811988812 2
3 2
3 1
rotor
eacutetage
h h hRh h h
∆∆
minus= =
minus
2 20 3 2 2 3
0 3 2
2 2( )
( )
h c c h
∆∆
minus
minus
+ minus=
ℎ03 ℎ02
(ℎ01= ℎ02 1198881198881= 1198881198883)
120549120549ℎ )0(3minus2 = (ℎ3 + frasl11988811988832 2) minus (ℎ2 + frasl11988811988822 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
0(3 2) 3 2( )∆ minus = minusu uh U c c
= 1 minusfrasl11988811988832 2 minus frasl11988811988822 2120549120549ℎ )0(3minus2
119877119877 =120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
120549120549ℎ )0(3minus2
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
3 212+
= minus u uc cRU
119888119888119909119909 = 1198881198882119909119909 = 1198881198883119909119909 = 119888119888119888119888119888119888119888119888119888119888
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 11988811988821199061199061198881198882 = 1198881198881199091199092+1198881198881199061199062
119877119877 = 1 minus11988811988831199061199062 minus 11988811988821199061199062
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
Indirectement lrsquoeacutequation pour R deacutecrite en fonction de composantes cineacutematiques eacutetablit un lien entre les angles drsquoentreacutee et de sortie
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
= 1 minus1198881198883119906119906 + 1198881198882119906119906
2119880119880
2232
2 32 2wwh h+ = +
2 23 2
2 3 2 2w wh hminus = minus
2 2 2 23 3 2 2
2 3 2 2u x u xw w w wh h + +
minus = minus
Machine axiale U2 = U3
La composante axiale de la vitesse demeure constante w3x = w 2x
( eacutequation justifieacutee plus tard lors de la rothalpie)
2 23 2
2 3 2 2u uw wh hminus = minus
3 2 3 22 3
( )( )2
u u u uw w w wh h minus +minus =
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
Eacutetage reacutepeacutetitif c3 = c1
Si les pertes dans le stator sont neacutegligeacutees
c2u ndash c3u= w2u - w3u
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
3 2 3 23 2
( )( )2
u u u uw w w wh h minus +minus = minus
c2u ndash c3u= w2u - w3u
119877119877 = minus1199081199083119906119906 + 1199081199082119906119906
2119880119880
2 2 2 22 3 2 3 2 3( ) 2 ( ) 2h h u u w wminus = minus minus minus
2 2 2 2 2 23 2 3 2 3 2
2 2 2e
I II III
u u c c w wW minus minus minus= + minus
2 2 2 22 3 2 3
2 2 2 2 2 22 3 2 3 2 3
( ) ( )( ) ( ) ( )
minus minus minus=
minus + minus minus minusu u w wR
c c u u w w
3 2= =U u u
Expliqueacutee + tard
2 23 2
03 02 3 2 2ec cW h h h h minus
= minus = minus +
119877119877 =ℎ2 minus ℎ3ℎ1 minus ℎ3
=ℎ2 minus ℎ3ℎ02 minus ℎ03
Degreacute de reacuteaction
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
( )3u 2uw wR2U+
= minus
3 212
u uc cRU+
= minus
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
Puisque le degreacute de reacuteaction R peut aussi srsquoexprimer enfonction de variables cineacutematiques on peut chercher sarepreacutesentation dans le triangle adimensionnel de vitesses
Turbine
α2
β2 α3
β3
C2 W3
C3W2
C3uC2u
U
C2xC3x
W2u
c2u ndash c3u= w2u - w3u
2Ψ minusR
frasl119934119934120784120784119958119958 119932119932 = frasl1198621198622119906119906 119880119880 minus frasl119880119880 119880119880Ψ = frasl1198621198622119906119906 minus 1198621198623119906119906 119880119880
119877119877 = 1 minus (1198621198623119906119906 + 1198621198622119906119906 frasl) 2119880119880Φ = frasl1198621198622119909119909 119880119880 = frasl1198621198623119909119909 119880119880
= (119914119914120784120784119958119958 minus 119914119914120785120785119958119958 frasl) 2119880119880 minus 1 + (119914119914120784120784119958119958 + 119914119914120785120785119958119958 frasl) 2119880119880
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α2
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
Les repreacutesentations des triangles de vitesses adimensionnellessont tregraves riches en information et contiennent plusieurs formules
Par exemple pour un compresseur on deacuteduit aiseacutement que
Avec 120595120595 = 1198821198821198901198901198801198802
Φ = 119888119888119909119909119880119880
et que pour 119888119888119901119901 = 119888119888119888119888119888119888119888119888119888119888 119882119882119890119890 = 119888119888119901119901(11987911987902minus11987911987901)on trouve
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
(11987911987902minus11987911987901) =119880119880119888119888119909119909119888119888119901119901
(tan1205731205731 minus tan1205731205732)
Alors si lrsquoon veut obtenir une plus grande variation detempeacuterature 11987911987902 minus 11987911987901 indirectement un plus grand rapport(1199011199010211990111990101) on peut augmenter
bull la vitesse peacuteripheacuterique 119880119880 (la vitesse de rotation)bull la vitesse axiale 119888119888119909119909 (le deacutebit massique)bull lrsquoangle de deacuteflection de la pale(tan1205731205731 minus tan1205731205732)
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
Parfois ψ est consideacutereacute positif pour les turbines et neacutegatifpour les compresseurs 119882119882119890119890 = frasl119882 119888
Coefficient de charge
2Ψ = eWU
Coefficient de charge
2e
UW
=ψ 02 01 02 2
h h hU Uminus ∆
= = 02
s shU
η ∆=
=
eWWm
120578120578119904119904 =∆ℎ0∆ℎ0119904119904
Turbine
Coefficient de deacutebit
xcU
Φ =
xCU
Φ =ρRT
pρ
= xC RTRTU p
=RT m RT
U Ap
xCU
Φ =119888 = ρ119862119862119909119909119860119860
11988811987711987711987911987901199011199010119860119860
= 119872119872 120574120574 1 +120574120574 minus 1
21198721198722
minus 120574120574+1)2(120574120574minus1
Afin de geacuteneacuteraliser le triangle de vitesses drsquoune machineaxiale on effectue une mise agrave lrsquoeacutechelle en divisant chaquecocircteacute par la vitesse tangentielle U
Cette opeacuteration permettra de retrouver directement lescoefficients de charge ψ et de deacutebit Φ
Par la suite un nouveau paramegravetre R qursquoon appelle degreacutede reacuteaction sera introduit pour compleacuteter la caracteacuterisationde lrsquoaubage
Approche classique
α2
β2 α3
β3C2 W3
W2
c3u - c2u= w3u - w2u
W3uW2u
U
CX CX
Lrsquoopeacuteration est vectorielle
C3
120511120511 = frasl119934119934119940119940 119932119932120784120784 = frasl119914119914120785120785119958119958 minus 119914119914120784120784119958119958 119932119932
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
Turbine
3
2
3 3
2 2
12
12
12
12
u
u
u u
u u
c RUc RU
w c RU Uw c RU U
Ψ= minus +
Ψ= minus +
Ψ= minus = +
Ψ= minus = minus
2 22 23 2
2 22 23 2
1 12 2
2 2
c cR RU U
w wR RU U
Ψ Ψ = Φ + minus + = Φ + minus +
Ψ Ψ = Φ + + = Φ + minus
3 2
3 2
1 2 1 2
2 2
R Ratan atan
R Ratan atan
minus + Ψ minus + Ψ = = Φ Φ
Ψ + Ψ minus = = Φ Φ
α α
β β
α2β1
C2U
W1U
C1UW2U β2 α1
Compresseur
ΦC2U
C3UW3U
W2U
ψ
1
ψ
ΦC2UC3U
W3U
W2U
1
Il faut deacutefinir un troisiegraveme paramegravetre pour diffeacuterencier ces deux triangles de vitesses ayant le mecircme 120509120509 et le mecircme 120537120537
Les coefficients 120509120509 et 120537120537 ne suffisent pas pour deacutefinir de maniegravereunivoque le triangle de vitesses
Le degreacute de reacuteaction est un paramegravetre adimensionnel quiexprime la variation drsquoenthalpie subie par le fluide dans le rotorpar rapport agrave la chute totale drsquoenthalpie dans lrsquoeacutetage
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
Variation d enthalpie dans le rotorRVariation d enthalpie dans l eacutetage
=
❸❶
❷
Rotor
Stator
R=0 machine drsquoaction (Δhrotor=0 Δheacutetage=Δhstator)
R=1 machine agrave reacuteaction (Δhstator=0 Δheacutetage=Δhrotor)
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
R indique la reacutepartition du saut drsquoenthalpie (de pression) statique entre le stator et le rotor
Le degreacute de reacuteaction R deacutefini en fonction de variablesthermodynamiques (lrsquoenthalpie) peut aussi srsquoexprimer en fonctionde variables cineacutematiques preacutesentes dans le triangle de vitessesNotamment en fonction de la vitesse peacuteripheacuterique U et descomposantes des vitesses absolues et relatives
La vitesse absolue agrave lrsquoentreacutee du stator est eacutegale agrave celle agrave la sortie du rotor (c1= c3 eacutetage reacutepeacutetitif )
On neacuteglige les pertes et le transfert de chaleur dans le stator
1 2 3
ℎ3 minus ℎ1= ℎ3 + frasl11988811988832 2 minus ℎ1 minus frasl11988811988812 2 = ℎ03 minus ℎ01 = 120549120549ℎ0
ℎ03 minus ℎ01 = ℎ03 minus ℎ02 = 120549120549ℎ )0(3minus2
ℎ03 minus ℎ02 = ℎ3 minus ℎ1
119945119945120782120782120783120783 = 119945119945120782120782120784120784
frasl11988811988832 2 = frasl11988811988812 2
3 2
3 1
rotor
eacutetage
h h hRh h h
∆∆
minus= =
minus
2 20 3 2 2 3
0 3 2
2 2( )
( )
h c c h
∆∆
minus
minus
+ minus=
ℎ03 ℎ02
(ℎ01= ℎ02 1198881198881= 1198881198883)
120549120549ℎ )0(3minus2 = (ℎ3 + frasl11988811988832 2) minus (ℎ2 + frasl11988811988822 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
0(3 2) 3 2( )∆ minus = minusu uh U c c
= 1 minusfrasl11988811988832 2 minus frasl11988811988822 2120549120549ℎ )0(3minus2
119877119877 =120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
120549120549ℎ )0(3minus2
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
3 212+
= minus u uc cRU
119888119888119909119909 = 1198881198882119909119909 = 1198881198883119909119909 = 119888119888119888119888119888119888119888119888119888119888
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 11988811988821199061199061198881198882 = 1198881198881199091199092+1198881198881199061199062
119877119877 = 1 minus11988811988831199061199062 minus 11988811988821199061199062
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
Indirectement lrsquoeacutequation pour R deacutecrite en fonction de composantes cineacutematiques eacutetablit un lien entre les angles drsquoentreacutee et de sortie
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
= 1 minus1198881198883119906119906 + 1198881198882119906119906
2119880119880
2232
2 32 2wwh h+ = +
2 23 2
2 3 2 2w wh hminus = minus
2 2 2 23 3 2 2
2 3 2 2u x u xw w w wh h + +
minus = minus
Machine axiale U2 = U3
La composante axiale de la vitesse demeure constante w3x = w 2x
( eacutequation justifieacutee plus tard lors de la rothalpie)
2 23 2
2 3 2 2u uw wh hminus = minus
3 2 3 22 3
( )( )2
u u u uw w w wh h minus +minus =
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
Eacutetage reacutepeacutetitif c3 = c1
Si les pertes dans le stator sont neacutegligeacutees
c2u ndash c3u= w2u - w3u
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
3 2 3 23 2
( )( )2
u u u uw w w wh h minus +minus = minus
c2u ndash c3u= w2u - w3u
119877119877 = minus1199081199083119906119906 + 1199081199082119906119906
2119880119880
2 2 2 22 3 2 3 2 3( ) 2 ( ) 2h h u u w wminus = minus minus minus
2 2 2 2 2 23 2 3 2 3 2
2 2 2e
I II III
u u c c w wW minus minus minus= + minus
2 2 2 22 3 2 3
2 2 2 2 2 22 3 2 3 2 3
( ) ( )( ) ( ) ( )
minus minus minus=
minus + minus minus minusu u w wR
c c u u w w
3 2= =U u u
Expliqueacutee + tard
2 23 2
03 02 3 2 2ec cW h h h h minus
= minus = minus +
119877119877 =ℎ2 minus ℎ3ℎ1 minus ℎ3
=ℎ2 minus ℎ3ℎ02 minus ℎ03
Degreacute de reacuteaction
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
( )3u 2uw wR2U+
= minus
3 212
u uc cRU+
= minus
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
Puisque le degreacute de reacuteaction R peut aussi srsquoexprimer enfonction de variables cineacutematiques on peut chercher sarepreacutesentation dans le triangle adimensionnel de vitesses
Turbine
α2
β2 α3
β3
C2 W3
C3W2
C3uC2u
U
C2xC3x
W2u
c2u ndash c3u= w2u - w3u
2Ψ minusR
frasl119934119934120784120784119958119958 119932119932 = frasl1198621198622119906119906 119880119880 minus frasl119880119880 119880119880Ψ = frasl1198621198622119906119906 minus 1198621198623119906119906 119880119880
119877119877 = 1 minus (1198621198623119906119906 + 1198621198622119906119906 frasl) 2119880119880Φ = frasl1198621198622119909119909 119880119880 = frasl1198621198623119909119909 119880119880
= (119914119914120784120784119958119958 minus 119914119914120785120785119958119958 frasl) 2119880119880 minus 1 + (119914119914120784120784119958119958 + 119914119914120785120785119958119958 frasl) 2119880119880
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α2
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
Les repreacutesentations des triangles de vitesses adimensionnellessont tregraves riches en information et contiennent plusieurs formules
Par exemple pour un compresseur on deacuteduit aiseacutement que
Avec 120595120595 = 1198821198821198901198901198801198802
Φ = 119888119888119909119909119880119880
et que pour 119888119888119901119901 = 119888119888119888119888119888119888119888119888119888119888 119882119882119890119890 = 119888119888119901119901(11987911987902minus11987911987901)on trouve
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
(11987911987902minus11987911987901) =119880119880119888119888119909119909119888119888119901119901
(tan1205731205731 minus tan1205731205732)
Alors si lrsquoon veut obtenir une plus grande variation detempeacuterature 11987911987902 minus 11987911987901 indirectement un plus grand rapport(1199011199010211990111990101) on peut augmenter
bull la vitesse peacuteripheacuterique 119880119880 (la vitesse de rotation)bull la vitesse axiale 119888119888119909119909 (le deacutebit massique)bull lrsquoangle de deacuteflection de la pale(tan1205731205731 minus tan1205731205732)
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
Coefficient de charge
2e
UW
=ψ 02 01 02 2
h h hU Uminus ∆
= = 02
s shU
η ∆=
=
eWWm
120578120578119904119904 =∆ℎ0∆ℎ0119904119904
Turbine
Coefficient de deacutebit
xcU
Φ =
xCU
Φ =ρRT
pρ
= xC RTRTU p
=RT m RT
U Ap
xCU
Φ =119888 = ρ119862119862119909119909119860119860
11988811987711987711987911987901199011199010119860119860
= 119872119872 120574120574 1 +120574120574 minus 1
21198721198722
minus 120574120574+1)2(120574120574minus1
Afin de geacuteneacuteraliser le triangle de vitesses drsquoune machineaxiale on effectue une mise agrave lrsquoeacutechelle en divisant chaquecocircteacute par la vitesse tangentielle U
Cette opeacuteration permettra de retrouver directement lescoefficients de charge ψ et de deacutebit Φ
Par la suite un nouveau paramegravetre R qursquoon appelle degreacutede reacuteaction sera introduit pour compleacuteter la caracteacuterisationde lrsquoaubage
Approche classique
α2
β2 α3
β3C2 W3
W2
c3u - c2u= w3u - w2u
W3uW2u
U
CX CX
Lrsquoopeacuteration est vectorielle
C3
120511120511 = frasl119934119934119940119940 119932119932120784120784 = frasl119914119914120785120785119958119958 minus 119914119914120784120784119958119958 119932119932
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
Turbine
3
2
3 3
2 2
12
12
12
12
u
u
u u
u u
c RUc RU
w c RU Uw c RU U
Ψ= minus +
Ψ= minus +
Ψ= minus = +
Ψ= minus = minus
2 22 23 2
2 22 23 2
1 12 2
2 2
c cR RU U
w wR RU U
Ψ Ψ = Φ + minus + = Φ + minus +
Ψ Ψ = Φ + + = Φ + minus
3 2
3 2
1 2 1 2
2 2
R Ratan atan
R Ratan atan
minus + Ψ minus + Ψ = = Φ Φ
Ψ + Ψ minus = = Φ Φ
α α
β β
α2β1
C2U
W1U
C1UW2U β2 α1
Compresseur
ΦC2U
C3UW3U
W2U
ψ
1
ψ
ΦC2UC3U
W3U
W2U
1
Il faut deacutefinir un troisiegraveme paramegravetre pour diffeacuterencier ces deux triangles de vitesses ayant le mecircme 120509120509 et le mecircme 120537120537
Les coefficients 120509120509 et 120537120537 ne suffisent pas pour deacutefinir de maniegravereunivoque le triangle de vitesses
Le degreacute de reacuteaction est un paramegravetre adimensionnel quiexprime la variation drsquoenthalpie subie par le fluide dans le rotorpar rapport agrave la chute totale drsquoenthalpie dans lrsquoeacutetage
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
Variation d enthalpie dans le rotorRVariation d enthalpie dans l eacutetage
=
❸❶
❷
Rotor
Stator
R=0 machine drsquoaction (Δhrotor=0 Δheacutetage=Δhstator)
R=1 machine agrave reacuteaction (Δhstator=0 Δheacutetage=Δhrotor)
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
R indique la reacutepartition du saut drsquoenthalpie (de pression) statique entre le stator et le rotor
Le degreacute de reacuteaction R deacutefini en fonction de variablesthermodynamiques (lrsquoenthalpie) peut aussi srsquoexprimer en fonctionde variables cineacutematiques preacutesentes dans le triangle de vitessesNotamment en fonction de la vitesse peacuteripheacuterique U et descomposantes des vitesses absolues et relatives
La vitesse absolue agrave lrsquoentreacutee du stator est eacutegale agrave celle agrave la sortie du rotor (c1= c3 eacutetage reacutepeacutetitif )
On neacuteglige les pertes et le transfert de chaleur dans le stator
1 2 3
ℎ3 minus ℎ1= ℎ3 + frasl11988811988832 2 minus ℎ1 minus frasl11988811988812 2 = ℎ03 minus ℎ01 = 120549120549ℎ0
ℎ03 minus ℎ01 = ℎ03 minus ℎ02 = 120549120549ℎ )0(3minus2
ℎ03 minus ℎ02 = ℎ3 minus ℎ1
119945119945120782120782120783120783 = 119945119945120782120782120784120784
frasl11988811988832 2 = frasl11988811988812 2
3 2
3 1
rotor
eacutetage
h h hRh h h
∆∆
minus= =
minus
2 20 3 2 2 3
0 3 2
2 2( )
( )
h c c h
∆∆
minus
minus
+ minus=
ℎ03 ℎ02
(ℎ01= ℎ02 1198881198881= 1198881198883)
120549120549ℎ )0(3minus2 = (ℎ3 + frasl11988811988832 2) minus (ℎ2 + frasl11988811988822 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
0(3 2) 3 2( )∆ minus = minusu uh U c c
= 1 minusfrasl11988811988832 2 minus frasl11988811988822 2120549120549ℎ )0(3minus2
119877119877 =120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
120549120549ℎ )0(3minus2
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
3 212+
= minus u uc cRU
119888119888119909119909 = 1198881198882119909119909 = 1198881198883119909119909 = 119888119888119888119888119888119888119888119888119888119888
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 11988811988821199061199061198881198882 = 1198881198881199091199092+1198881198881199061199062
119877119877 = 1 minus11988811988831199061199062 minus 11988811988821199061199062
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
Indirectement lrsquoeacutequation pour R deacutecrite en fonction de composantes cineacutematiques eacutetablit un lien entre les angles drsquoentreacutee et de sortie
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
= 1 minus1198881198883119906119906 + 1198881198882119906119906
2119880119880
2232
2 32 2wwh h+ = +
2 23 2
2 3 2 2w wh hminus = minus
2 2 2 23 3 2 2
2 3 2 2u x u xw w w wh h + +
minus = minus
Machine axiale U2 = U3
La composante axiale de la vitesse demeure constante w3x = w 2x
( eacutequation justifieacutee plus tard lors de la rothalpie)
2 23 2
2 3 2 2u uw wh hminus = minus
3 2 3 22 3
( )( )2
u u u uw w w wh h minus +minus =
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
Eacutetage reacutepeacutetitif c3 = c1
Si les pertes dans le stator sont neacutegligeacutees
c2u ndash c3u= w2u - w3u
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
3 2 3 23 2
( )( )2
u u u uw w w wh h minus +minus = minus
c2u ndash c3u= w2u - w3u
119877119877 = minus1199081199083119906119906 + 1199081199082119906119906
2119880119880
2 2 2 22 3 2 3 2 3( ) 2 ( ) 2h h u u w wminus = minus minus minus
2 2 2 2 2 23 2 3 2 3 2
2 2 2e
I II III
u u c c w wW minus minus minus= + minus
2 2 2 22 3 2 3
2 2 2 2 2 22 3 2 3 2 3
( ) ( )( ) ( ) ( )
minus minus minus=
minus + minus minus minusu u w wR
c c u u w w
3 2= =U u u
Expliqueacutee + tard
2 23 2
03 02 3 2 2ec cW h h h h minus
= minus = minus +
119877119877 =ℎ2 minus ℎ3ℎ1 minus ℎ3
=ℎ2 minus ℎ3ℎ02 minus ℎ03
Degreacute de reacuteaction
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
( )3u 2uw wR2U+
= minus
3 212
u uc cRU+
= minus
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
Puisque le degreacute de reacuteaction R peut aussi srsquoexprimer enfonction de variables cineacutematiques on peut chercher sarepreacutesentation dans le triangle adimensionnel de vitesses
Turbine
α2
β2 α3
β3
C2 W3
C3W2
C3uC2u
U
C2xC3x
W2u
c2u ndash c3u= w2u - w3u
2Ψ minusR
frasl119934119934120784120784119958119958 119932119932 = frasl1198621198622119906119906 119880119880 minus frasl119880119880 119880119880Ψ = frasl1198621198622119906119906 minus 1198621198623119906119906 119880119880
119877119877 = 1 minus (1198621198623119906119906 + 1198621198622119906119906 frasl) 2119880119880Φ = frasl1198621198622119909119909 119880119880 = frasl1198621198623119909119909 119880119880
= (119914119914120784120784119958119958 minus 119914119914120785120785119958119958 frasl) 2119880119880 minus 1 + (119914119914120784120784119958119958 + 119914119914120785120785119958119958 frasl) 2119880119880
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α2
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
Les repreacutesentations des triangles de vitesses adimensionnellessont tregraves riches en information et contiennent plusieurs formules
Par exemple pour un compresseur on deacuteduit aiseacutement que
Avec 120595120595 = 1198821198821198901198901198801198802
Φ = 119888119888119909119909119880119880
et que pour 119888119888119901119901 = 119888119888119888119888119888119888119888119888119888119888 119882119882119890119890 = 119888119888119901119901(11987911987902minus11987911987901)on trouve
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
(11987911987902minus11987911987901) =119880119880119888119888119909119909119888119888119901119901
(tan1205731205731 minus tan1205731205732)
Alors si lrsquoon veut obtenir une plus grande variation detempeacuterature 11987911987902 minus 11987911987901 indirectement un plus grand rapport(1199011199010211990111990101) on peut augmenter
bull la vitesse peacuteripheacuterique 119880119880 (la vitesse de rotation)bull la vitesse axiale 119888119888119909119909 (le deacutebit massique)bull lrsquoangle de deacuteflection de la pale(tan1205731205731 minus tan1205731205732)
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
Coefficient de deacutebit
xcU
Φ =
xCU
Φ =ρRT
pρ
= xC RTRTU p
=RT m RT
U Ap
xCU
Φ =119888 = ρ119862119862119909119909119860119860
11988811987711987711987911987901199011199010119860119860
= 119872119872 120574120574 1 +120574120574 minus 1
21198721198722
minus 120574120574+1)2(120574120574minus1
Afin de geacuteneacuteraliser le triangle de vitesses drsquoune machineaxiale on effectue une mise agrave lrsquoeacutechelle en divisant chaquecocircteacute par la vitesse tangentielle U
Cette opeacuteration permettra de retrouver directement lescoefficients de charge ψ et de deacutebit Φ
Par la suite un nouveau paramegravetre R qursquoon appelle degreacutede reacuteaction sera introduit pour compleacuteter la caracteacuterisationde lrsquoaubage
Approche classique
α2
β2 α3
β3C2 W3
W2
c3u - c2u= w3u - w2u
W3uW2u
U
CX CX
Lrsquoopeacuteration est vectorielle
C3
120511120511 = frasl119934119934119940119940 119932119932120784120784 = frasl119914119914120785120785119958119958 minus 119914119914120784120784119958119958 119932119932
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
Turbine
3
2
3 3
2 2
12
12
12
12
u
u
u u
u u
c RUc RU
w c RU Uw c RU U
Ψ= minus +
Ψ= minus +
Ψ= minus = +
Ψ= minus = minus
2 22 23 2
2 22 23 2
1 12 2
2 2
c cR RU U
w wR RU U
Ψ Ψ = Φ + minus + = Φ + minus +
Ψ Ψ = Φ + + = Φ + minus
3 2
3 2
1 2 1 2
2 2
R Ratan atan
R Ratan atan
minus + Ψ minus + Ψ = = Φ Φ
Ψ + Ψ minus = = Φ Φ
α α
β β
α2β1
C2U
W1U
C1UW2U β2 α1
Compresseur
ΦC2U
C3UW3U
W2U
ψ
1
ψ
ΦC2UC3U
W3U
W2U
1
Il faut deacutefinir un troisiegraveme paramegravetre pour diffeacuterencier ces deux triangles de vitesses ayant le mecircme 120509120509 et le mecircme 120537120537
Les coefficients 120509120509 et 120537120537 ne suffisent pas pour deacutefinir de maniegravereunivoque le triangle de vitesses
Le degreacute de reacuteaction est un paramegravetre adimensionnel quiexprime la variation drsquoenthalpie subie par le fluide dans le rotorpar rapport agrave la chute totale drsquoenthalpie dans lrsquoeacutetage
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
Variation d enthalpie dans le rotorRVariation d enthalpie dans l eacutetage
=
❸❶
❷
Rotor
Stator
R=0 machine drsquoaction (Δhrotor=0 Δheacutetage=Δhstator)
R=1 machine agrave reacuteaction (Δhstator=0 Δheacutetage=Δhrotor)
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
R indique la reacutepartition du saut drsquoenthalpie (de pression) statique entre le stator et le rotor
Le degreacute de reacuteaction R deacutefini en fonction de variablesthermodynamiques (lrsquoenthalpie) peut aussi srsquoexprimer en fonctionde variables cineacutematiques preacutesentes dans le triangle de vitessesNotamment en fonction de la vitesse peacuteripheacuterique U et descomposantes des vitesses absolues et relatives
La vitesse absolue agrave lrsquoentreacutee du stator est eacutegale agrave celle agrave la sortie du rotor (c1= c3 eacutetage reacutepeacutetitif )
On neacuteglige les pertes et le transfert de chaleur dans le stator
1 2 3
ℎ3 minus ℎ1= ℎ3 + frasl11988811988832 2 minus ℎ1 minus frasl11988811988812 2 = ℎ03 minus ℎ01 = 120549120549ℎ0
ℎ03 minus ℎ01 = ℎ03 minus ℎ02 = 120549120549ℎ )0(3minus2
ℎ03 minus ℎ02 = ℎ3 minus ℎ1
119945119945120782120782120783120783 = 119945119945120782120782120784120784
frasl11988811988832 2 = frasl11988811988812 2
3 2
3 1
rotor
eacutetage
h h hRh h h
∆∆
minus= =
minus
2 20 3 2 2 3
0 3 2
2 2( )
( )
h c c h
∆∆
minus
minus
+ minus=
ℎ03 ℎ02
(ℎ01= ℎ02 1198881198881= 1198881198883)
120549120549ℎ )0(3minus2 = (ℎ3 + frasl11988811988832 2) minus (ℎ2 + frasl11988811988822 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
0(3 2) 3 2( )∆ minus = minusu uh U c c
= 1 minusfrasl11988811988832 2 minus frasl11988811988822 2120549120549ℎ )0(3minus2
119877119877 =120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
120549120549ℎ )0(3minus2
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
3 212+
= minus u uc cRU
119888119888119909119909 = 1198881198882119909119909 = 1198881198883119909119909 = 119888119888119888119888119888119888119888119888119888119888
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 11988811988821199061199061198881198882 = 1198881198881199091199092+1198881198881199061199062
119877119877 = 1 minus11988811988831199061199062 minus 11988811988821199061199062
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
Indirectement lrsquoeacutequation pour R deacutecrite en fonction de composantes cineacutematiques eacutetablit un lien entre les angles drsquoentreacutee et de sortie
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
= 1 minus1198881198883119906119906 + 1198881198882119906119906
2119880119880
2232
2 32 2wwh h+ = +
2 23 2
2 3 2 2w wh hminus = minus
2 2 2 23 3 2 2
2 3 2 2u x u xw w w wh h + +
minus = minus
Machine axiale U2 = U3
La composante axiale de la vitesse demeure constante w3x = w 2x
( eacutequation justifieacutee plus tard lors de la rothalpie)
2 23 2
2 3 2 2u uw wh hminus = minus
3 2 3 22 3
( )( )2
u u u uw w w wh h minus +minus =
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
Eacutetage reacutepeacutetitif c3 = c1
Si les pertes dans le stator sont neacutegligeacutees
c2u ndash c3u= w2u - w3u
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
3 2 3 23 2
( )( )2
u u u uw w w wh h minus +minus = minus
c2u ndash c3u= w2u - w3u
119877119877 = minus1199081199083119906119906 + 1199081199082119906119906
2119880119880
2 2 2 22 3 2 3 2 3( ) 2 ( ) 2h h u u w wminus = minus minus minus
2 2 2 2 2 23 2 3 2 3 2
2 2 2e
I II III
u u c c w wW minus minus minus= + minus
2 2 2 22 3 2 3
2 2 2 2 2 22 3 2 3 2 3
( ) ( )( ) ( ) ( )
minus minus minus=
minus + minus minus minusu u w wR
c c u u w w
3 2= =U u u
Expliqueacutee + tard
2 23 2
03 02 3 2 2ec cW h h h h minus
= minus = minus +
119877119877 =ℎ2 minus ℎ3ℎ1 minus ℎ3
=ℎ2 minus ℎ3ℎ02 minus ℎ03
Degreacute de reacuteaction
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
( )3u 2uw wR2U+
= minus
3 212
u uc cRU+
= minus
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
Puisque le degreacute de reacuteaction R peut aussi srsquoexprimer enfonction de variables cineacutematiques on peut chercher sarepreacutesentation dans le triangle adimensionnel de vitesses
Turbine
α2
β2 α3
β3
C2 W3
C3W2
C3uC2u
U
C2xC3x
W2u
c2u ndash c3u= w2u - w3u
2Ψ minusR
frasl119934119934120784120784119958119958 119932119932 = frasl1198621198622119906119906 119880119880 minus frasl119880119880 119880119880Ψ = frasl1198621198622119906119906 minus 1198621198623119906119906 119880119880
119877119877 = 1 minus (1198621198623119906119906 + 1198621198622119906119906 frasl) 2119880119880Φ = frasl1198621198622119909119909 119880119880 = frasl1198621198623119909119909 119880119880
= (119914119914120784120784119958119958 minus 119914119914120785120785119958119958 frasl) 2119880119880 minus 1 + (119914119914120784120784119958119958 + 119914119914120785120785119958119958 frasl) 2119880119880
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α2
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
Les repreacutesentations des triangles de vitesses adimensionnellessont tregraves riches en information et contiennent plusieurs formules
Par exemple pour un compresseur on deacuteduit aiseacutement que
Avec 120595120595 = 1198821198821198901198901198801198802
Φ = 119888119888119909119909119880119880
et que pour 119888119888119901119901 = 119888119888119888119888119888119888119888119888119888119888 119882119882119890119890 = 119888119888119901119901(11987911987902minus11987911987901)on trouve
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
(11987911987902minus11987911987901) =119880119880119888119888119909119909119888119888119901119901
(tan1205731205731 minus tan1205731205732)
Alors si lrsquoon veut obtenir une plus grande variation detempeacuterature 11987911987902 minus 11987911987901 indirectement un plus grand rapport(1199011199010211990111990101) on peut augmenter
bull la vitesse peacuteripheacuterique 119880119880 (la vitesse de rotation)bull la vitesse axiale 119888119888119909119909 (le deacutebit massique)bull lrsquoangle de deacuteflection de la pale(tan1205731205731 minus tan1205731205732)
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
xCU
Φ =ρRT
pρ
= xC RTRTU p
=RT m RT
U Ap
xCU
Φ =119888 = ρ119862119862119909119909119860119860
11988811987711987711987911987901199011199010119860119860
= 119872119872 120574120574 1 +120574120574 minus 1
21198721198722
minus 120574120574+1)2(120574120574minus1
Afin de geacuteneacuteraliser le triangle de vitesses drsquoune machineaxiale on effectue une mise agrave lrsquoeacutechelle en divisant chaquecocircteacute par la vitesse tangentielle U
Cette opeacuteration permettra de retrouver directement lescoefficients de charge ψ et de deacutebit Φ
Par la suite un nouveau paramegravetre R qursquoon appelle degreacutede reacuteaction sera introduit pour compleacuteter la caracteacuterisationde lrsquoaubage
Approche classique
α2
β2 α3
β3C2 W3
W2
c3u - c2u= w3u - w2u
W3uW2u
U
CX CX
Lrsquoopeacuteration est vectorielle
C3
120511120511 = frasl119934119934119940119940 119932119932120784120784 = frasl119914119914120785120785119958119958 minus 119914119914120784120784119958119958 119932119932
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
Turbine
3
2
3 3
2 2
12
12
12
12
u
u
u u
u u
c RUc RU
w c RU Uw c RU U
Ψ= minus +
Ψ= minus +
Ψ= minus = +
Ψ= minus = minus
2 22 23 2
2 22 23 2
1 12 2
2 2
c cR RU U
w wR RU U
Ψ Ψ = Φ + minus + = Φ + minus +
Ψ Ψ = Φ + + = Φ + minus
3 2
3 2
1 2 1 2
2 2
R Ratan atan
R Ratan atan
minus + Ψ minus + Ψ = = Φ Φ
Ψ + Ψ minus = = Φ Φ
α α
β β
α2β1
C2U
W1U
C1UW2U β2 α1
Compresseur
ΦC2U
C3UW3U
W2U
ψ
1
ψ
ΦC2UC3U
W3U
W2U
1
Il faut deacutefinir un troisiegraveme paramegravetre pour diffeacuterencier ces deux triangles de vitesses ayant le mecircme 120509120509 et le mecircme 120537120537
Les coefficients 120509120509 et 120537120537 ne suffisent pas pour deacutefinir de maniegravereunivoque le triangle de vitesses
Le degreacute de reacuteaction est un paramegravetre adimensionnel quiexprime la variation drsquoenthalpie subie par le fluide dans le rotorpar rapport agrave la chute totale drsquoenthalpie dans lrsquoeacutetage
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
Variation d enthalpie dans le rotorRVariation d enthalpie dans l eacutetage
=
❸❶
❷
Rotor
Stator
R=0 machine drsquoaction (Δhrotor=0 Δheacutetage=Δhstator)
R=1 machine agrave reacuteaction (Δhstator=0 Δheacutetage=Δhrotor)
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
R indique la reacutepartition du saut drsquoenthalpie (de pression) statique entre le stator et le rotor
Le degreacute de reacuteaction R deacutefini en fonction de variablesthermodynamiques (lrsquoenthalpie) peut aussi srsquoexprimer en fonctionde variables cineacutematiques preacutesentes dans le triangle de vitessesNotamment en fonction de la vitesse peacuteripheacuterique U et descomposantes des vitesses absolues et relatives
La vitesse absolue agrave lrsquoentreacutee du stator est eacutegale agrave celle agrave la sortie du rotor (c1= c3 eacutetage reacutepeacutetitif )
On neacuteglige les pertes et le transfert de chaleur dans le stator
1 2 3
ℎ3 minus ℎ1= ℎ3 + frasl11988811988832 2 minus ℎ1 minus frasl11988811988812 2 = ℎ03 minus ℎ01 = 120549120549ℎ0
ℎ03 minus ℎ01 = ℎ03 minus ℎ02 = 120549120549ℎ )0(3minus2
ℎ03 minus ℎ02 = ℎ3 minus ℎ1
119945119945120782120782120783120783 = 119945119945120782120782120784120784
frasl11988811988832 2 = frasl11988811988812 2
3 2
3 1
rotor
eacutetage
h h hRh h h
∆∆
minus= =
minus
2 20 3 2 2 3
0 3 2
2 2( )
( )
h c c h
∆∆
minus
minus
+ minus=
ℎ03 ℎ02
(ℎ01= ℎ02 1198881198881= 1198881198883)
120549120549ℎ )0(3minus2 = (ℎ3 + frasl11988811988832 2) minus (ℎ2 + frasl11988811988822 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
0(3 2) 3 2( )∆ minus = minusu uh U c c
= 1 minusfrasl11988811988832 2 minus frasl11988811988822 2120549120549ℎ )0(3minus2
119877119877 =120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
120549120549ℎ )0(3minus2
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
3 212+
= minus u uc cRU
119888119888119909119909 = 1198881198882119909119909 = 1198881198883119909119909 = 119888119888119888119888119888119888119888119888119888119888
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 11988811988821199061199061198881198882 = 1198881198881199091199092+1198881198881199061199062
119877119877 = 1 minus11988811988831199061199062 minus 11988811988821199061199062
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
Indirectement lrsquoeacutequation pour R deacutecrite en fonction de composantes cineacutematiques eacutetablit un lien entre les angles drsquoentreacutee et de sortie
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
= 1 minus1198881198883119906119906 + 1198881198882119906119906
2119880119880
2232
2 32 2wwh h+ = +
2 23 2
2 3 2 2w wh hminus = minus
2 2 2 23 3 2 2
2 3 2 2u x u xw w w wh h + +
minus = minus
Machine axiale U2 = U3
La composante axiale de la vitesse demeure constante w3x = w 2x
( eacutequation justifieacutee plus tard lors de la rothalpie)
2 23 2
2 3 2 2u uw wh hminus = minus
3 2 3 22 3
( )( )2
u u u uw w w wh h minus +minus =
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
Eacutetage reacutepeacutetitif c3 = c1
Si les pertes dans le stator sont neacutegligeacutees
c2u ndash c3u= w2u - w3u
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
3 2 3 23 2
( )( )2
u u u uw w w wh h minus +minus = minus
c2u ndash c3u= w2u - w3u
119877119877 = minus1199081199083119906119906 + 1199081199082119906119906
2119880119880
2 2 2 22 3 2 3 2 3( ) 2 ( ) 2h h u u w wminus = minus minus minus
2 2 2 2 2 23 2 3 2 3 2
2 2 2e
I II III
u u c c w wW minus minus minus= + minus
2 2 2 22 3 2 3
2 2 2 2 2 22 3 2 3 2 3
( ) ( )( ) ( ) ( )
minus minus minus=
minus + minus minus minusu u w wR
c c u u w w
3 2= =U u u
Expliqueacutee + tard
2 23 2
03 02 3 2 2ec cW h h h h minus
= minus = minus +
119877119877 =ℎ2 minus ℎ3ℎ1 minus ℎ3
=ℎ2 minus ℎ3ℎ02 minus ℎ03
Degreacute de reacuteaction
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
( )3u 2uw wR2U+
= minus
3 212
u uc cRU+
= minus
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
Puisque le degreacute de reacuteaction R peut aussi srsquoexprimer enfonction de variables cineacutematiques on peut chercher sarepreacutesentation dans le triangle adimensionnel de vitesses
Turbine
α2
β2 α3
β3
C2 W3
C3W2
C3uC2u
U
C2xC3x
W2u
c2u ndash c3u= w2u - w3u
2Ψ minusR
frasl119934119934120784120784119958119958 119932119932 = frasl1198621198622119906119906 119880119880 minus frasl119880119880 119880119880Ψ = frasl1198621198622119906119906 minus 1198621198623119906119906 119880119880
119877119877 = 1 minus (1198621198623119906119906 + 1198621198622119906119906 frasl) 2119880119880Φ = frasl1198621198622119909119909 119880119880 = frasl1198621198623119909119909 119880119880
= (119914119914120784120784119958119958 minus 119914119914120785120785119958119958 frasl) 2119880119880 minus 1 + (119914119914120784120784119958119958 + 119914119914120785120785119958119958 frasl) 2119880119880
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α2
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
Les repreacutesentations des triangles de vitesses adimensionnellessont tregraves riches en information et contiennent plusieurs formules
Par exemple pour un compresseur on deacuteduit aiseacutement que
Avec 120595120595 = 1198821198821198901198901198801198802
Φ = 119888119888119909119909119880119880
et que pour 119888119888119901119901 = 119888119888119888119888119888119888119888119888119888119888 119882119882119890119890 = 119888119888119901119901(11987911987902minus11987911987901)on trouve
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
(11987911987902minus11987911987901) =119880119880119888119888119909119909119888119888119901119901
(tan1205731205731 minus tan1205731205732)
Alors si lrsquoon veut obtenir une plus grande variation detempeacuterature 11987911987902 minus 11987911987901 indirectement un plus grand rapport(1199011199010211990111990101) on peut augmenter
bull la vitesse peacuteripheacuterique 119880119880 (la vitesse de rotation)bull la vitesse axiale 119888119888119909119909 (le deacutebit massique)bull lrsquoangle de deacuteflection de la pale(tan1205731205731 minus tan1205731205732)
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
Afin de geacuteneacuteraliser le triangle de vitesses drsquoune machineaxiale on effectue une mise agrave lrsquoeacutechelle en divisant chaquecocircteacute par la vitesse tangentielle U
Cette opeacuteration permettra de retrouver directement lescoefficients de charge ψ et de deacutebit Φ
Par la suite un nouveau paramegravetre R qursquoon appelle degreacutede reacuteaction sera introduit pour compleacuteter la caracteacuterisationde lrsquoaubage
Approche classique
α2
β2 α3
β3C2 W3
W2
c3u - c2u= w3u - w2u
W3uW2u
U
CX CX
Lrsquoopeacuteration est vectorielle
C3
120511120511 = frasl119934119934119940119940 119932119932120784120784 = frasl119914119914120785120785119958119958 minus 119914119914120784120784119958119958 119932119932
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
Turbine
3
2
3 3
2 2
12
12
12
12
u
u
u u
u u
c RUc RU
w c RU Uw c RU U
Ψ= minus +
Ψ= minus +
Ψ= minus = +
Ψ= minus = minus
2 22 23 2
2 22 23 2
1 12 2
2 2
c cR RU U
w wR RU U
Ψ Ψ = Φ + minus + = Φ + minus +
Ψ Ψ = Φ + + = Φ + minus
3 2
3 2
1 2 1 2
2 2
R Ratan atan
R Ratan atan
minus + Ψ minus + Ψ = = Φ Φ
Ψ + Ψ minus = = Φ Φ
α α
β β
α2β1
C2U
W1U
C1UW2U β2 α1
Compresseur
ΦC2U
C3UW3U
W2U
ψ
1
ψ
ΦC2UC3U
W3U
W2U
1
Il faut deacutefinir un troisiegraveme paramegravetre pour diffeacuterencier ces deux triangles de vitesses ayant le mecircme 120509120509 et le mecircme 120537120537
Les coefficients 120509120509 et 120537120537 ne suffisent pas pour deacutefinir de maniegravereunivoque le triangle de vitesses
Le degreacute de reacuteaction est un paramegravetre adimensionnel quiexprime la variation drsquoenthalpie subie par le fluide dans le rotorpar rapport agrave la chute totale drsquoenthalpie dans lrsquoeacutetage
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
Variation d enthalpie dans le rotorRVariation d enthalpie dans l eacutetage
=
❸❶
❷
Rotor
Stator
R=0 machine drsquoaction (Δhrotor=0 Δheacutetage=Δhstator)
R=1 machine agrave reacuteaction (Δhstator=0 Δheacutetage=Δhrotor)
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
R indique la reacutepartition du saut drsquoenthalpie (de pression) statique entre le stator et le rotor
Le degreacute de reacuteaction R deacutefini en fonction de variablesthermodynamiques (lrsquoenthalpie) peut aussi srsquoexprimer en fonctionde variables cineacutematiques preacutesentes dans le triangle de vitessesNotamment en fonction de la vitesse peacuteripheacuterique U et descomposantes des vitesses absolues et relatives
La vitesse absolue agrave lrsquoentreacutee du stator est eacutegale agrave celle agrave la sortie du rotor (c1= c3 eacutetage reacutepeacutetitif )
On neacuteglige les pertes et le transfert de chaleur dans le stator
1 2 3
ℎ3 minus ℎ1= ℎ3 + frasl11988811988832 2 minus ℎ1 minus frasl11988811988812 2 = ℎ03 minus ℎ01 = 120549120549ℎ0
ℎ03 minus ℎ01 = ℎ03 minus ℎ02 = 120549120549ℎ )0(3minus2
ℎ03 minus ℎ02 = ℎ3 minus ℎ1
119945119945120782120782120783120783 = 119945119945120782120782120784120784
frasl11988811988832 2 = frasl11988811988812 2
3 2
3 1
rotor
eacutetage
h h hRh h h
∆∆
minus= =
minus
2 20 3 2 2 3
0 3 2
2 2( )
( )
h c c h
∆∆
minus
minus
+ minus=
ℎ03 ℎ02
(ℎ01= ℎ02 1198881198881= 1198881198883)
120549120549ℎ )0(3minus2 = (ℎ3 + frasl11988811988832 2) minus (ℎ2 + frasl11988811988822 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
0(3 2) 3 2( )∆ minus = minusu uh U c c
= 1 minusfrasl11988811988832 2 minus frasl11988811988822 2120549120549ℎ )0(3minus2
119877119877 =120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
120549120549ℎ )0(3minus2
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
3 212+
= minus u uc cRU
119888119888119909119909 = 1198881198882119909119909 = 1198881198883119909119909 = 119888119888119888119888119888119888119888119888119888119888
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 11988811988821199061199061198881198882 = 1198881198881199091199092+1198881198881199061199062
119877119877 = 1 minus11988811988831199061199062 minus 11988811988821199061199062
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
Indirectement lrsquoeacutequation pour R deacutecrite en fonction de composantes cineacutematiques eacutetablit un lien entre les angles drsquoentreacutee et de sortie
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
= 1 minus1198881198883119906119906 + 1198881198882119906119906
2119880119880
2232
2 32 2wwh h+ = +
2 23 2
2 3 2 2w wh hminus = minus
2 2 2 23 3 2 2
2 3 2 2u x u xw w w wh h + +
minus = minus
Machine axiale U2 = U3
La composante axiale de la vitesse demeure constante w3x = w 2x
( eacutequation justifieacutee plus tard lors de la rothalpie)
2 23 2
2 3 2 2u uw wh hminus = minus
3 2 3 22 3
( )( )2
u u u uw w w wh h minus +minus =
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
Eacutetage reacutepeacutetitif c3 = c1
Si les pertes dans le stator sont neacutegligeacutees
c2u ndash c3u= w2u - w3u
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
3 2 3 23 2
( )( )2
u u u uw w w wh h minus +minus = minus
c2u ndash c3u= w2u - w3u
119877119877 = minus1199081199083119906119906 + 1199081199082119906119906
2119880119880
2 2 2 22 3 2 3 2 3( ) 2 ( ) 2h h u u w wminus = minus minus minus
2 2 2 2 2 23 2 3 2 3 2
2 2 2e
I II III
u u c c w wW minus minus minus= + minus
2 2 2 22 3 2 3
2 2 2 2 2 22 3 2 3 2 3
( ) ( )( ) ( ) ( )
minus minus minus=
minus + minus minus minusu u w wR
c c u u w w
3 2= =U u u
Expliqueacutee + tard
2 23 2
03 02 3 2 2ec cW h h h h minus
= minus = minus +
119877119877 =ℎ2 minus ℎ3ℎ1 minus ℎ3
=ℎ2 minus ℎ3ℎ02 minus ℎ03
Degreacute de reacuteaction
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
( )3u 2uw wR2U+
= minus
3 212
u uc cRU+
= minus
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
Puisque le degreacute de reacuteaction R peut aussi srsquoexprimer enfonction de variables cineacutematiques on peut chercher sarepreacutesentation dans le triangle adimensionnel de vitesses
Turbine
α2
β2 α3
β3
C2 W3
C3W2
C3uC2u
U
C2xC3x
W2u
c2u ndash c3u= w2u - w3u
2Ψ minusR
frasl119934119934120784120784119958119958 119932119932 = frasl1198621198622119906119906 119880119880 minus frasl119880119880 119880119880Ψ = frasl1198621198622119906119906 minus 1198621198623119906119906 119880119880
119877119877 = 1 minus (1198621198623119906119906 + 1198621198622119906119906 frasl) 2119880119880Φ = frasl1198621198622119909119909 119880119880 = frasl1198621198623119909119909 119880119880
= (119914119914120784120784119958119958 minus 119914119914120785120785119958119958 frasl) 2119880119880 minus 1 + (119914119914120784120784119958119958 + 119914119914120785120785119958119958 frasl) 2119880119880
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α2
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
Les repreacutesentations des triangles de vitesses adimensionnellessont tregraves riches en information et contiennent plusieurs formules
Par exemple pour un compresseur on deacuteduit aiseacutement que
Avec 120595120595 = 1198821198821198901198901198801198802
Φ = 119888119888119909119909119880119880
et que pour 119888119888119901119901 = 119888119888119888119888119888119888119888119888119888119888 119882119882119890119890 = 119888119888119901119901(11987911987902minus11987911987901)on trouve
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
(11987911987902minus11987911987901) =119880119880119888119888119909119909119888119888119901119901
(tan1205731205731 minus tan1205731205732)
Alors si lrsquoon veut obtenir une plus grande variation detempeacuterature 11987911987902 minus 11987911987901 indirectement un plus grand rapport(1199011199010211990111990101) on peut augmenter
bull la vitesse peacuteripheacuterique 119880119880 (la vitesse de rotation)bull la vitesse axiale 119888119888119909119909 (le deacutebit massique)bull lrsquoangle de deacuteflection de la pale(tan1205731205731 minus tan1205731205732)
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
α2
β2 α3
β3C2 W3
W2
c3u - c2u= w3u - w2u
W3uW2u
U
CX CX
Lrsquoopeacuteration est vectorielle
C3
120511120511 = frasl119934119934119940119940 119932119932120784120784 = frasl119914119914120785120785119958119958 minus 119914119914120784120784119958119958 119932119932
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
Turbine
3
2
3 3
2 2
12
12
12
12
u
u
u u
u u
c RUc RU
w c RU Uw c RU U
Ψ= minus +
Ψ= minus +
Ψ= minus = +
Ψ= minus = minus
2 22 23 2
2 22 23 2
1 12 2
2 2
c cR RU U
w wR RU U
Ψ Ψ = Φ + minus + = Φ + minus +
Ψ Ψ = Φ + + = Φ + minus
3 2
3 2
1 2 1 2
2 2
R Ratan atan
R Ratan atan
minus + Ψ minus + Ψ = = Φ Φ
Ψ + Ψ minus = = Φ Φ
α α
β β
α2β1
C2U
W1U
C1UW2U β2 α1
Compresseur
ΦC2U
C3UW3U
W2U
ψ
1
ψ
ΦC2UC3U
W3U
W2U
1
Il faut deacutefinir un troisiegraveme paramegravetre pour diffeacuterencier ces deux triangles de vitesses ayant le mecircme 120509120509 et le mecircme 120537120537
Les coefficients 120509120509 et 120537120537 ne suffisent pas pour deacutefinir de maniegravereunivoque le triangle de vitesses
Le degreacute de reacuteaction est un paramegravetre adimensionnel quiexprime la variation drsquoenthalpie subie par le fluide dans le rotorpar rapport agrave la chute totale drsquoenthalpie dans lrsquoeacutetage
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
Variation d enthalpie dans le rotorRVariation d enthalpie dans l eacutetage
=
❸❶
❷
Rotor
Stator
R=0 machine drsquoaction (Δhrotor=0 Δheacutetage=Δhstator)
R=1 machine agrave reacuteaction (Δhstator=0 Δheacutetage=Δhrotor)
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
R indique la reacutepartition du saut drsquoenthalpie (de pression) statique entre le stator et le rotor
Le degreacute de reacuteaction R deacutefini en fonction de variablesthermodynamiques (lrsquoenthalpie) peut aussi srsquoexprimer en fonctionde variables cineacutematiques preacutesentes dans le triangle de vitessesNotamment en fonction de la vitesse peacuteripheacuterique U et descomposantes des vitesses absolues et relatives
La vitesse absolue agrave lrsquoentreacutee du stator est eacutegale agrave celle agrave la sortie du rotor (c1= c3 eacutetage reacutepeacutetitif )
On neacuteglige les pertes et le transfert de chaleur dans le stator
1 2 3
ℎ3 minus ℎ1= ℎ3 + frasl11988811988832 2 minus ℎ1 minus frasl11988811988812 2 = ℎ03 minus ℎ01 = 120549120549ℎ0
ℎ03 minus ℎ01 = ℎ03 minus ℎ02 = 120549120549ℎ )0(3minus2
ℎ03 minus ℎ02 = ℎ3 minus ℎ1
119945119945120782120782120783120783 = 119945119945120782120782120784120784
frasl11988811988832 2 = frasl11988811988812 2
3 2
3 1
rotor
eacutetage
h h hRh h h
∆∆
minus= =
minus
2 20 3 2 2 3
0 3 2
2 2( )
( )
h c c h
∆∆
minus
minus
+ minus=
ℎ03 ℎ02
(ℎ01= ℎ02 1198881198881= 1198881198883)
120549120549ℎ )0(3minus2 = (ℎ3 + frasl11988811988832 2) minus (ℎ2 + frasl11988811988822 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
0(3 2) 3 2( )∆ minus = minusu uh U c c
= 1 minusfrasl11988811988832 2 minus frasl11988811988822 2120549120549ℎ )0(3minus2
119877119877 =120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
120549120549ℎ )0(3minus2
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
3 212+
= minus u uc cRU
119888119888119909119909 = 1198881198882119909119909 = 1198881198883119909119909 = 119888119888119888119888119888119888119888119888119888119888
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 11988811988821199061199061198881198882 = 1198881198881199091199092+1198881198881199061199062
119877119877 = 1 minus11988811988831199061199062 minus 11988811988821199061199062
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
Indirectement lrsquoeacutequation pour R deacutecrite en fonction de composantes cineacutematiques eacutetablit un lien entre les angles drsquoentreacutee et de sortie
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
= 1 minus1198881198883119906119906 + 1198881198882119906119906
2119880119880
2232
2 32 2wwh h+ = +
2 23 2
2 3 2 2w wh hminus = minus
2 2 2 23 3 2 2
2 3 2 2u x u xw w w wh h + +
minus = minus
Machine axiale U2 = U3
La composante axiale de la vitesse demeure constante w3x = w 2x
( eacutequation justifieacutee plus tard lors de la rothalpie)
2 23 2
2 3 2 2u uw wh hminus = minus
3 2 3 22 3
( )( )2
u u u uw w w wh h minus +minus =
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
Eacutetage reacutepeacutetitif c3 = c1
Si les pertes dans le stator sont neacutegligeacutees
c2u ndash c3u= w2u - w3u
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
3 2 3 23 2
( )( )2
u u u uw w w wh h minus +minus = minus
c2u ndash c3u= w2u - w3u
119877119877 = minus1199081199083119906119906 + 1199081199082119906119906
2119880119880
2 2 2 22 3 2 3 2 3( ) 2 ( ) 2h h u u w wminus = minus minus minus
2 2 2 2 2 23 2 3 2 3 2
2 2 2e
I II III
u u c c w wW minus minus minus= + minus
2 2 2 22 3 2 3
2 2 2 2 2 22 3 2 3 2 3
( ) ( )( ) ( ) ( )
minus minus minus=
minus + minus minus minusu u w wR
c c u u w w
3 2= =U u u
Expliqueacutee + tard
2 23 2
03 02 3 2 2ec cW h h h h minus
= minus = minus +
119877119877 =ℎ2 minus ℎ3ℎ1 minus ℎ3
=ℎ2 minus ℎ3ℎ02 minus ℎ03
Degreacute de reacuteaction
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
( )3u 2uw wR2U+
= minus
3 212
u uc cRU+
= minus
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
Puisque le degreacute de reacuteaction R peut aussi srsquoexprimer enfonction de variables cineacutematiques on peut chercher sarepreacutesentation dans le triangle adimensionnel de vitesses
Turbine
α2
β2 α3
β3
C2 W3
C3W2
C3uC2u
U
C2xC3x
W2u
c2u ndash c3u= w2u - w3u
2Ψ minusR
frasl119934119934120784120784119958119958 119932119932 = frasl1198621198622119906119906 119880119880 minus frasl119880119880 119880119880Ψ = frasl1198621198622119906119906 minus 1198621198623119906119906 119880119880
119877119877 = 1 minus (1198621198623119906119906 + 1198621198622119906119906 frasl) 2119880119880Φ = frasl1198621198622119909119909 119880119880 = frasl1198621198623119909119909 119880119880
= (119914119914120784120784119958119958 minus 119914119914120785120785119958119958 frasl) 2119880119880 minus 1 + (119914119914120784120784119958119958 + 119914119914120785120785119958119958 frasl) 2119880119880
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α2
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
Les repreacutesentations des triangles de vitesses adimensionnellessont tregraves riches en information et contiennent plusieurs formules
Par exemple pour un compresseur on deacuteduit aiseacutement que
Avec 120595120595 = 1198821198821198901198901198801198802
Φ = 119888119888119909119909119880119880
et que pour 119888119888119901119901 = 119888119888119888119888119888119888119888119888119888119888 119882119882119890119890 = 119888119888119901119901(11987911987902minus11987911987901)on trouve
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
(11987911987902minus11987911987901) =119880119880119888119888119909119909119888119888119901119901
(tan1205731205731 minus tan1205731205732)
Alors si lrsquoon veut obtenir une plus grande variation detempeacuterature 11987911987902 minus 11987911987901 indirectement un plus grand rapport(1199011199010211990111990101) on peut augmenter
bull la vitesse peacuteripheacuterique 119880119880 (la vitesse de rotation)bull la vitesse axiale 119888119888119909119909 (le deacutebit massique)bull lrsquoangle de deacuteflection de la pale(tan1205731205731 minus tan1205731205732)
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
Turbine
3
2
3 3
2 2
12
12
12
12
u
u
u u
u u
c RUc RU
w c RU Uw c RU U
Ψ= minus +
Ψ= minus +
Ψ= minus = +
Ψ= minus = minus
2 22 23 2
2 22 23 2
1 12 2
2 2
c cR RU U
w wR RU U
Ψ Ψ = Φ + minus + = Φ + minus +
Ψ Ψ = Φ + + = Φ + minus
3 2
3 2
1 2 1 2
2 2
R Ratan atan
R Ratan atan
minus + Ψ minus + Ψ = = Φ Φ
Ψ + Ψ minus = = Φ Φ
α α
β β
α2β1
C2U
W1U
C1UW2U β2 α1
Compresseur
ΦC2U
C3UW3U
W2U
ψ
1
ψ
ΦC2UC3U
W3U
W2U
1
Il faut deacutefinir un troisiegraveme paramegravetre pour diffeacuterencier ces deux triangles de vitesses ayant le mecircme 120509120509 et le mecircme 120537120537
Les coefficients 120509120509 et 120537120537 ne suffisent pas pour deacutefinir de maniegravereunivoque le triangle de vitesses
Le degreacute de reacuteaction est un paramegravetre adimensionnel quiexprime la variation drsquoenthalpie subie par le fluide dans le rotorpar rapport agrave la chute totale drsquoenthalpie dans lrsquoeacutetage
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
Variation d enthalpie dans le rotorRVariation d enthalpie dans l eacutetage
=
❸❶
❷
Rotor
Stator
R=0 machine drsquoaction (Δhrotor=0 Δheacutetage=Δhstator)
R=1 machine agrave reacuteaction (Δhstator=0 Δheacutetage=Δhrotor)
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
R indique la reacutepartition du saut drsquoenthalpie (de pression) statique entre le stator et le rotor
Le degreacute de reacuteaction R deacutefini en fonction de variablesthermodynamiques (lrsquoenthalpie) peut aussi srsquoexprimer en fonctionde variables cineacutematiques preacutesentes dans le triangle de vitessesNotamment en fonction de la vitesse peacuteripheacuterique U et descomposantes des vitesses absolues et relatives
La vitesse absolue agrave lrsquoentreacutee du stator est eacutegale agrave celle agrave la sortie du rotor (c1= c3 eacutetage reacutepeacutetitif )
On neacuteglige les pertes et le transfert de chaleur dans le stator
1 2 3
ℎ3 minus ℎ1= ℎ3 + frasl11988811988832 2 minus ℎ1 minus frasl11988811988812 2 = ℎ03 minus ℎ01 = 120549120549ℎ0
ℎ03 minus ℎ01 = ℎ03 minus ℎ02 = 120549120549ℎ )0(3minus2
ℎ03 minus ℎ02 = ℎ3 minus ℎ1
119945119945120782120782120783120783 = 119945119945120782120782120784120784
frasl11988811988832 2 = frasl11988811988812 2
3 2
3 1
rotor
eacutetage
h h hRh h h
∆∆
minus= =
minus
2 20 3 2 2 3
0 3 2
2 2( )
( )
h c c h
∆∆
minus
minus
+ minus=
ℎ03 ℎ02
(ℎ01= ℎ02 1198881198881= 1198881198883)
120549120549ℎ )0(3minus2 = (ℎ3 + frasl11988811988832 2) minus (ℎ2 + frasl11988811988822 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
0(3 2) 3 2( )∆ minus = minusu uh U c c
= 1 minusfrasl11988811988832 2 minus frasl11988811988822 2120549120549ℎ )0(3minus2
119877119877 =120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
120549120549ℎ )0(3minus2
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
3 212+
= minus u uc cRU
119888119888119909119909 = 1198881198882119909119909 = 1198881198883119909119909 = 119888119888119888119888119888119888119888119888119888119888
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 11988811988821199061199061198881198882 = 1198881198881199091199092+1198881198881199061199062
119877119877 = 1 minus11988811988831199061199062 minus 11988811988821199061199062
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
Indirectement lrsquoeacutequation pour R deacutecrite en fonction de composantes cineacutematiques eacutetablit un lien entre les angles drsquoentreacutee et de sortie
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
= 1 minus1198881198883119906119906 + 1198881198882119906119906
2119880119880
2232
2 32 2wwh h+ = +
2 23 2
2 3 2 2w wh hminus = minus
2 2 2 23 3 2 2
2 3 2 2u x u xw w w wh h + +
minus = minus
Machine axiale U2 = U3
La composante axiale de la vitesse demeure constante w3x = w 2x
( eacutequation justifieacutee plus tard lors de la rothalpie)
2 23 2
2 3 2 2u uw wh hminus = minus
3 2 3 22 3
( )( )2
u u u uw w w wh h minus +minus =
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
Eacutetage reacutepeacutetitif c3 = c1
Si les pertes dans le stator sont neacutegligeacutees
c2u ndash c3u= w2u - w3u
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
3 2 3 23 2
( )( )2
u u u uw w w wh h minus +minus = minus
c2u ndash c3u= w2u - w3u
119877119877 = minus1199081199083119906119906 + 1199081199082119906119906
2119880119880
2 2 2 22 3 2 3 2 3( ) 2 ( ) 2h h u u w wminus = minus minus minus
2 2 2 2 2 23 2 3 2 3 2
2 2 2e
I II III
u u c c w wW minus minus minus= + minus
2 2 2 22 3 2 3
2 2 2 2 2 22 3 2 3 2 3
( ) ( )( ) ( ) ( )
minus minus minus=
minus + minus minus minusu u w wR
c c u u w w
3 2= =U u u
Expliqueacutee + tard
2 23 2
03 02 3 2 2ec cW h h h h minus
= minus = minus +
119877119877 =ℎ2 minus ℎ3ℎ1 minus ℎ3
=ℎ2 minus ℎ3ℎ02 minus ℎ03
Degreacute de reacuteaction
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
( )3u 2uw wR2U+
= minus
3 212
u uc cRU+
= minus
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
Puisque le degreacute de reacuteaction R peut aussi srsquoexprimer enfonction de variables cineacutematiques on peut chercher sarepreacutesentation dans le triangle adimensionnel de vitesses
Turbine
α2
β2 α3
β3
C2 W3
C3W2
C3uC2u
U
C2xC3x
W2u
c2u ndash c3u= w2u - w3u
2Ψ minusR
frasl119934119934120784120784119958119958 119932119932 = frasl1198621198622119906119906 119880119880 minus frasl119880119880 119880119880Ψ = frasl1198621198622119906119906 minus 1198621198623119906119906 119880119880
119877119877 = 1 minus (1198621198623119906119906 + 1198621198622119906119906 frasl) 2119880119880Φ = frasl1198621198622119909119909 119880119880 = frasl1198621198623119909119909 119880119880
= (119914119914120784120784119958119958 minus 119914119914120785120785119958119958 frasl) 2119880119880 minus 1 + (119914119914120784120784119958119958 + 119914119914120785120785119958119958 frasl) 2119880119880
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α2
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
Les repreacutesentations des triangles de vitesses adimensionnellessont tregraves riches en information et contiennent plusieurs formules
Par exemple pour un compresseur on deacuteduit aiseacutement que
Avec 120595120595 = 1198821198821198901198901198801198802
Φ = 119888119888119909119909119880119880
et que pour 119888119888119901119901 = 119888119888119888119888119888119888119888119888119888119888 119882119882119890119890 = 119888119888119901119901(11987911987902minus11987911987901)on trouve
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
(11987911987902minus11987911987901) =119880119880119888119888119909119909119888119888119901119901
(tan1205731205731 minus tan1205731205732)
Alors si lrsquoon veut obtenir une plus grande variation detempeacuterature 11987911987902 minus 11987911987901 indirectement un plus grand rapport(1199011199010211990111990101) on peut augmenter
bull la vitesse peacuteripheacuterique 119880119880 (la vitesse de rotation)bull la vitesse axiale 119888119888119909119909 (le deacutebit massique)bull lrsquoangle de deacuteflection de la pale(tan1205731205731 minus tan1205731205732)
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
3
2
3 3
2 2
12
12
12
12
u
u
u u
u u
c RUc RU
w c RU Uw c RU U
Ψ= minus +
Ψ= minus +
Ψ= minus = +
Ψ= minus = minus
2 22 23 2
2 22 23 2
1 12 2
2 2
c cR RU U
w wR RU U
Ψ Ψ = Φ + minus + = Φ + minus +
Ψ Ψ = Φ + + = Φ + minus
3 2
3 2
1 2 1 2
2 2
R Ratan atan
R Ratan atan
minus + Ψ minus + Ψ = = Φ Φ
Ψ + Ψ minus = = Φ Φ
α α
β β
α2β1
C2U
W1U
C1UW2U β2 α1
Compresseur
ΦC2U
C3UW3U
W2U
ψ
1
ψ
ΦC2UC3U
W3U
W2U
1
Il faut deacutefinir un troisiegraveme paramegravetre pour diffeacuterencier ces deux triangles de vitesses ayant le mecircme 120509120509 et le mecircme 120537120537
Les coefficients 120509120509 et 120537120537 ne suffisent pas pour deacutefinir de maniegravereunivoque le triangle de vitesses
Le degreacute de reacuteaction est un paramegravetre adimensionnel quiexprime la variation drsquoenthalpie subie par le fluide dans le rotorpar rapport agrave la chute totale drsquoenthalpie dans lrsquoeacutetage
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
Variation d enthalpie dans le rotorRVariation d enthalpie dans l eacutetage
=
❸❶
❷
Rotor
Stator
R=0 machine drsquoaction (Δhrotor=0 Δheacutetage=Δhstator)
R=1 machine agrave reacuteaction (Δhstator=0 Δheacutetage=Δhrotor)
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
R indique la reacutepartition du saut drsquoenthalpie (de pression) statique entre le stator et le rotor
Le degreacute de reacuteaction R deacutefini en fonction de variablesthermodynamiques (lrsquoenthalpie) peut aussi srsquoexprimer en fonctionde variables cineacutematiques preacutesentes dans le triangle de vitessesNotamment en fonction de la vitesse peacuteripheacuterique U et descomposantes des vitesses absolues et relatives
La vitesse absolue agrave lrsquoentreacutee du stator est eacutegale agrave celle agrave la sortie du rotor (c1= c3 eacutetage reacutepeacutetitif )
On neacuteglige les pertes et le transfert de chaleur dans le stator
1 2 3
ℎ3 minus ℎ1= ℎ3 + frasl11988811988832 2 minus ℎ1 minus frasl11988811988812 2 = ℎ03 minus ℎ01 = 120549120549ℎ0
ℎ03 minus ℎ01 = ℎ03 minus ℎ02 = 120549120549ℎ )0(3minus2
ℎ03 minus ℎ02 = ℎ3 minus ℎ1
119945119945120782120782120783120783 = 119945119945120782120782120784120784
frasl11988811988832 2 = frasl11988811988812 2
3 2
3 1
rotor
eacutetage
h h hRh h h
∆∆
minus= =
minus
2 20 3 2 2 3
0 3 2
2 2( )
( )
h c c h
∆∆
minus
minus
+ minus=
ℎ03 ℎ02
(ℎ01= ℎ02 1198881198881= 1198881198883)
120549120549ℎ )0(3minus2 = (ℎ3 + frasl11988811988832 2) minus (ℎ2 + frasl11988811988822 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
0(3 2) 3 2( )∆ minus = minusu uh U c c
= 1 minusfrasl11988811988832 2 minus frasl11988811988822 2120549120549ℎ )0(3minus2
119877119877 =120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
120549120549ℎ )0(3minus2
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
3 212+
= minus u uc cRU
119888119888119909119909 = 1198881198882119909119909 = 1198881198883119909119909 = 119888119888119888119888119888119888119888119888119888119888
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 11988811988821199061199061198881198882 = 1198881198881199091199092+1198881198881199061199062
119877119877 = 1 minus11988811988831199061199062 minus 11988811988821199061199062
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
Indirectement lrsquoeacutequation pour R deacutecrite en fonction de composantes cineacutematiques eacutetablit un lien entre les angles drsquoentreacutee et de sortie
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
= 1 minus1198881198883119906119906 + 1198881198882119906119906
2119880119880
2232
2 32 2wwh h+ = +
2 23 2
2 3 2 2w wh hminus = minus
2 2 2 23 3 2 2
2 3 2 2u x u xw w w wh h + +
minus = minus
Machine axiale U2 = U3
La composante axiale de la vitesse demeure constante w3x = w 2x
( eacutequation justifieacutee plus tard lors de la rothalpie)
2 23 2
2 3 2 2u uw wh hminus = minus
3 2 3 22 3
( )( )2
u u u uw w w wh h minus +minus =
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
Eacutetage reacutepeacutetitif c3 = c1
Si les pertes dans le stator sont neacutegligeacutees
c2u ndash c3u= w2u - w3u
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
3 2 3 23 2
( )( )2
u u u uw w w wh h minus +minus = minus
c2u ndash c3u= w2u - w3u
119877119877 = minus1199081199083119906119906 + 1199081199082119906119906
2119880119880
2 2 2 22 3 2 3 2 3( ) 2 ( ) 2h h u u w wminus = minus minus minus
2 2 2 2 2 23 2 3 2 3 2
2 2 2e
I II III
u u c c w wW minus minus minus= + minus
2 2 2 22 3 2 3
2 2 2 2 2 22 3 2 3 2 3
( ) ( )( ) ( ) ( )
minus minus minus=
minus + minus minus minusu u w wR
c c u u w w
3 2= =U u u
Expliqueacutee + tard
2 23 2
03 02 3 2 2ec cW h h h h minus
= minus = minus +
119877119877 =ℎ2 minus ℎ3ℎ1 minus ℎ3
=ℎ2 minus ℎ3ℎ02 minus ℎ03
Degreacute de reacuteaction
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
( )3u 2uw wR2U+
= minus
3 212
u uc cRU+
= minus
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
Puisque le degreacute de reacuteaction R peut aussi srsquoexprimer enfonction de variables cineacutematiques on peut chercher sarepreacutesentation dans le triangle adimensionnel de vitesses
Turbine
α2
β2 α3
β3
C2 W3
C3W2
C3uC2u
U
C2xC3x
W2u
c2u ndash c3u= w2u - w3u
2Ψ minusR
frasl119934119934120784120784119958119958 119932119932 = frasl1198621198622119906119906 119880119880 minus frasl119880119880 119880119880Ψ = frasl1198621198622119906119906 minus 1198621198623119906119906 119880119880
119877119877 = 1 minus (1198621198623119906119906 + 1198621198622119906119906 frasl) 2119880119880Φ = frasl1198621198622119909119909 119880119880 = frasl1198621198623119909119909 119880119880
= (119914119914120784120784119958119958 minus 119914119914120785120785119958119958 frasl) 2119880119880 minus 1 + (119914119914120784120784119958119958 + 119914119914120785120785119958119958 frasl) 2119880119880
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α2
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
Les repreacutesentations des triangles de vitesses adimensionnellessont tregraves riches en information et contiennent plusieurs formules
Par exemple pour un compresseur on deacuteduit aiseacutement que
Avec 120595120595 = 1198821198821198901198901198801198802
Φ = 119888119888119909119909119880119880
et que pour 119888119888119901119901 = 119888119888119888119888119888119888119888119888119888119888 119882119882119890119890 = 119888119888119901119901(11987911987902minus11987911987901)on trouve
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
(11987911987902minus11987911987901) =119880119880119888119888119909119909119888119888119901119901
(tan1205731205731 minus tan1205731205732)
Alors si lrsquoon veut obtenir une plus grande variation detempeacuterature 11987911987902 minus 11987911987901 indirectement un plus grand rapport(1199011199010211990111990101) on peut augmenter
bull la vitesse peacuteripheacuterique 119880119880 (la vitesse de rotation)bull la vitesse axiale 119888119888119909119909 (le deacutebit massique)bull lrsquoangle de deacuteflection de la pale(tan1205731205731 minus tan1205731205732)
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
α2β1
C2U
W1U
C1UW2U β2 α1
Compresseur
ΦC2U
C3UW3U
W2U
ψ
1
ψ
ΦC2UC3U
W3U
W2U
1
Il faut deacutefinir un troisiegraveme paramegravetre pour diffeacuterencier ces deux triangles de vitesses ayant le mecircme 120509120509 et le mecircme 120537120537
Les coefficients 120509120509 et 120537120537 ne suffisent pas pour deacutefinir de maniegravereunivoque le triangle de vitesses
Le degreacute de reacuteaction est un paramegravetre adimensionnel quiexprime la variation drsquoenthalpie subie par le fluide dans le rotorpar rapport agrave la chute totale drsquoenthalpie dans lrsquoeacutetage
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
Variation d enthalpie dans le rotorRVariation d enthalpie dans l eacutetage
=
❸❶
❷
Rotor
Stator
R=0 machine drsquoaction (Δhrotor=0 Δheacutetage=Δhstator)
R=1 machine agrave reacuteaction (Δhstator=0 Δheacutetage=Δhrotor)
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
R indique la reacutepartition du saut drsquoenthalpie (de pression) statique entre le stator et le rotor
Le degreacute de reacuteaction R deacutefini en fonction de variablesthermodynamiques (lrsquoenthalpie) peut aussi srsquoexprimer en fonctionde variables cineacutematiques preacutesentes dans le triangle de vitessesNotamment en fonction de la vitesse peacuteripheacuterique U et descomposantes des vitesses absolues et relatives
La vitesse absolue agrave lrsquoentreacutee du stator est eacutegale agrave celle agrave la sortie du rotor (c1= c3 eacutetage reacutepeacutetitif )
On neacuteglige les pertes et le transfert de chaleur dans le stator
1 2 3
ℎ3 minus ℎ1= ℎ3 + frasl11988811988832 2 minus ℎ1 minus frasl11988811988812 2 = ℎ03 minus ℎ01 = 120549120549ℎ0
ℎ03 minus ℎ01 = ℎ03 minus ℎ02 = 120549120549ℎ )0(3minus2
ℎ03 minus ℎ02 = ℎ3 minus ℎ1
119945119945120782120782120783120783 = 119945119945120782120782120784120784
frasl11988811988832 2 = frasl11988811988812 2
3 2
3 1
rotor
eacutetage
h h hRh h h
∆∆
minus= =
minus
2 20 3 2 2 3
0 3 2
2 2( )
( )
h c c h
∆∆
minus
minus
+ minus=
ℎ03 ℎ02
(ℎ01= ℎ02 1198881198881= 1198881198883)
120549120549ℎ )0(3minus2 = (ℎ3 + frasl11988811988832 2) minus (ℎ2 + frasl11988811988822 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
0(3 2) 3 2( )∆ minus = minusu uh U c c
= 1 minusfrasl11988811988832 2 minus frasl11988811988822 2120549120549ℎ )0(3minus2
119877119877 =120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
120549120549ℎ )0(3minus2
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
3 212+
= minus u uc cRU
119888119888119909119909 = 1198881198882119909119909 = 1198881198883119909119909 = 119888119888119888119888119888119888119888119888119888119888
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 11988811988821199061199061198881198882 = 1198881198881199091199092+1198881198881199061199062
119877119877 = 1 minus11988811988831199061199062 minus 11988811988821199061199062
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
Indirectement lrsquoeacutequation pour R deacutecrite en fonction de composantes cineacutematiques eacutetablit un lien entre les angles drsquoentreacutee et de sortie
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
= 1 minus1198881198883119906119906 + 1198881198882119906119906
2119880119880
2232
2 32 2wwh h+ = +
2 23 2
2 3 2 2w wh hminus = minus
2 2 2 23 3 2 2
2 3 2 2u x u xw w w wh h + +
minus = minus
Machine axiale U2 = U3
La composante axiale de la vitesse demeure constante w3x = w 2x
( eacutequation justifieacutee plus tard lors de la rothalpie)
2 23 2
2 3 2 2u uw wh hminus = minus
3 2 3 22 3
( )( )2
u u u uw w w wh h minus +minus =
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
Eacutetage reacutepeacutetitif c3 = c1
Si les pertes dans le stator sont neacutegligeacutees
c2u ndash c3u= w2u - w3u
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
3 2 3 23 2
( )( )2
u u u uw w w wh h minus +minus = minus
c2u ndash c3u= w2u - w3u
119877119877 = minus1199081199083119906119906 + 1199081199082119906119906
2119880119880
2 2 2 22 3 2 3 2 3( ) 2 ( ) 2h h u u w wminus = minus minus minus
2 2 2 2 2 23 2 3 2 3 2
2 2 2e
I II III
u u c c w wW minus minus minus= + minus
2 2 2 22 3 2 3
2 2 2 2 2 22 3 2 3 2 3
( ) ( )( ) ( ) ( )
minus minus minus=
minus + minus minus minusu u w wR
c c u u w w
3 2= =U u u
Expliqueacutee + tard
2 23 2
03 02 3 2 2ec cW h h h h minus
= minus = minus +
119877119877 =ℎ2 minus ℎ3ℎ1 minus ℎ3
=ℎ2 minus ℎ3ℎ02 minus ℎ03
Degreacute de reacuteaction
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
( )3u 2uw wR2U+
= minus
3 212
u uc cRU+
= minus
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
Puisque le degreacute de reacuteaction R peut aussi srsquoexprimer enfonction de variables cineacutematiques on peut chercher sarepreacutesentation dans le triangle adimensionnel de vitesses
Turbine
α2
β2 α3
β3
C2 W3
C3W2
C3uC2u
U
C2xC3x
W2u
c2u ndash c3u= w2u - w3u
2Ψ minusR
frasl119934119934120784120784119958119958 119932119932 = frasl1198621198622119906119906 119880119880 minus frasl119880119880 119880119880Ψ = frasl1198621198622119906119906 minus 1198621198623119906119906 119880119880
119877119877 = 1 minus (1198621198623119906119906 + 1198621198622119906119906 frasl) 2119880119880Φ = frasl1198621198622119909119909 119880119880 = frasl1198621198623119909119909 119880119880
= (119914119914120784120784119958119958 minus 119914119914120785120785119958119958 frasl) 2119880119880 minus 1 + (119914119914120784120784119958119958 + 119914119914120785120785119958119958 frasl) 2119880119880
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α2
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
Les repreacutesentations des triangles de vitesses adimensionnellessont tregraves riches en information et contiennent plusieurs formules
Par exemple pour un compresseur on deacuteduit aiseacutement que
Avec 120595120595 = 1198821198821198901198901198801198802
Φ = 119888119888119909119909119880119880
et que pour 119888119888119901119901 = 119888119888119888119888119888119888119888119888119888119888 119882119882119890119890 = 119888119888119901119901(11987911987902minus11987911987901)on trouve
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
(11987911987902minus11987911987901) =119880119880119888119888119909119909119888119888119901119901
(tan1205731205731 minus tan1205731205732)
Alors si lrsquoon veut obtenir une plus grande variation detempeacuterature 11987911987902 minus 11987911987901 indirectement un plus grand rapport(1199011199010211990111990101) on peut augmenter
bull la vitesse peacuteripheacuterique 119880119880 (la vitesse de rotation)bull la vitesse axiale 119888119888119909119909 (le deacutebit massique)bull lrsquoangle de deacuteflection de la pale(tan1205731205731 minus tan1205731205732)
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
ΦC2U
C3UW3U
W2U
ψ
1
ψ
ΦC2UC3U
W3U
W2U
1
Il faut deacutefinir un troisiegraveme paramegravetre pour diffeacuterencier ces deux triangles de vitesses ayant le mecircme 120509120509 et le mecircme 120537120537
Les coefficients 120509120509 et 120537120537 ne suffisent pas pour deacutefinir de maniegravereunivoque le triangle de vitesses
Le degreacute de reacuteaction est un paramegravetre adimensionnel quiexprime la variation drsquoenthalpie subie par le fluide dans le rotorpar rapport agrave la chute totale drsquoenthalpie dans lrsquoeacutetage
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
Variation d enthalpie dans le rotorRVariation d enthalpie dans l eacutetage
=
❸❶
❷
Rotor
Stator
R=0 machine drsquoaction (Δhrotor=0 Δheacutetage=Δhstator)
R=1 machine agrave reacuteaction (Δhstator=0 Δheacutetage=Δhrotor)
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
R indique la reacutepartition du saut drsquoenthalpie (de pression) statique entre le stator et le rotor
Le degreacute de reacuteaction R deacutefini en fonction de variablesthermodynamiques (lrsquoenthalpie) peut aussi srsquoexprimer en fonctionde variables cineacutematiques preacutesentes dans le triangle de vitessesNotamment en fonction de la vitesse peacuteripheacuterique U et descomposantes des vitesses absolues et relatives
La vitesse absolue agrave lrsquoentreacutee du stator est eacutegale agrave celle agrave la sortie du rotor (c1= c3 eacutetage reacutepeacutetitif )
On neacuteglige les pertes et le transfert de chaleur dans le stator
1 2 3
ℎ3 minus ℎ1= ℎ3 + frasl11988811988832 2 minus ℎ1 minus frasl11988811988812 2 = ℎ03 minus ℎ01 = 120549120549ℎ0
ℎ03 minus ℎ01 = ℎ03 minus ℎ02 = 120549120549ℎ )0(3minus2
ℎ03 minus ℎ02 = ℎ3 minus ℎ1
119945119945120782120782120783120783 = 119945119945120782120782120784120784
frasl11988811988832 2 = frasl11988811988812 2
3 2
3 1
rotor
eacutetage
h h hRh h h
∆∆
minus= =
minus
2 20 3 2 2 3
0 3 2
2 2( )
( )
h c c h
∆∆
minus
minus
+ minus=
ℎ03 ℎ02
(ℎ01= ℎ02 1198881198881= 1198881198883)
120549120549ℎ )0(3minus2 = (ℎ3 + frasl11988811988832 2) minus (ℎ2 + frasl11988811988822 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
0(3 2) 3 2( )∆ minus = minusu uh U c c
= 1 minusfrasl11988811988832 2 minus frasl11988811988822 2120549120549ℎ )0(3minus2
119877119877 =120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
120549120549ℎ )0(3minus2
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
3 212+
= minus u uc cRU
119888119888119909119909 = 1198881198882119909119909 = 1198881198883119909119909 = 119888119888119888119888119888119888119888119888119888119888
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 11988811988821199061199061198881198882 = 1198881198881199091199092+1198881198881199061199062
119877119877 = 1 minus11988811988831199061199062 minus 11988811988821199061199062
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
Indirectement lrsquoeacutequation pour R deacutecrite en fonction de composantes cineacutematiques eacutetablit un lien entre les angles drsquoentreacutee et de sortie
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
= 1 minus1198881198883119906119906 + 1198881198882119906119906
2119880119880
2232
2 32 2wwh h+ = +
2 23 2
2 3 2 2w wh hminus = minus
2 2 2 23 3 2 2
2 3 2 2u x u xw w w wh h + +
minus = minus
Machine axiale U2 = U3
La composante axiale de la vitesse demeure constante w3x = w 2x
( eacutequation justifieacutee plus tard lors de la rothalpie)
2 23 2
2 3 2 2u uw wh hminus = minus
3 2 3 22 3
( )( )2
u u u uw w w wh h minus +minus =
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
Eacutetage reacutepeacutetitif c3 = c1
Si les pertes dans le stator sont neacutegligeacutees
c2u ndash c3u= w2u - w3u
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
3 2 3 23 2
( )( )2
u u u uw w w wh h minus +minus = minus
c2u ndash c3u= w2u - w3u
119877119877 = minus1199081199083119906119906 + 1199081199082119906119906
2119880119880
2 2 2 22 3 2 3 2 3( ) 2 ( ) 2h h u u w wminus = minus minus minus
2 2 2 2 2 23 2 3 2 3 2
2 2 2e
I II III
u u c c w wW minus minus minus= + minus
2 2 2 22 3 2 3
2 2 2 2 2 22 3 2 3 2 3
( ) ( )( ) ( ) ( )
minus minus minus=
minus + minus minus minusu u w wR
c c u u w w
3 2= =U u u
Expliqueacutee + tard
2 23 2
03 02 3 2 2ec cW h h h h minus
= minus = minus +
119877119877 =ℎ2 minus ℎ3ℎ1 minus ℎ3
=ℎ2 minus ℎ3ℎ02 minus ℎ03
Degreacute de reacuteaction
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
( )3u 2uw wR2U+
= minus
3 212
u uc cRU+
= minus
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
Puisque le degreacute de reacuteaction R peut aussi srsquoexprimer enfonction de variables cineacutematiques on peut chercher sarepreacutesentation dans le triangle adimensionnel de vitesses
Turbine
α2
β2 α3
β3
C2 W3
C3W2
C3uC2u
U
C2xC3x
W2u
c2u ndash c3u= w2u - w3u
2Ψ minusR
frasl119934119934120784120784119958119958 119932119932 = frasl1198621198622119906119906 119880119880 minus frasl119880119880 119880119880Ψ = frasl1198621198622119906119906 minus 1198621198623119906119906 119880119880
119877119877 = 1 minus (1198621198623119906119906 + 1198621198622119906119906 frasl) 2119880119880Φ = frasl1198621198622119909119909 119880119880 = frasl1198621198623119909119909 119880119880
= (119914119914120784120784119958119958 minus 119914119914120785120785119958119958 frasl) 2119880119880 minus 1 + (119914119914120784120784119958119958 + 119914119914120785120785119958119958 frasl) 2119880119880
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α2
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
Les repreacutesentations des triangles de vitesses adimensionnellessont tregraves riches en information et contiennent plusieurs formules
Par exemple pour un compresseur on deacuteduit aiseacutement que
Avec 120595120595 = 1198821198821198901198901198801198802
Φ = 119888119888119909119909119880119880
et que pour 119888119888119901119901 = 119888119888119888119888119888119888119888119888119888119888 119882119882119890119890 = 119888119888119901119901(11987911987902minus11987911987901)on trouve
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
(11987911987902minus11987911987901) =119880119880119888119888119909119909119888119888119901119901
(tan1205731205731 minus tan1205731205732)
Alors si lrsquoon veut obtenir une plus grande variation detempeacuterature 11987911987902 minus 11987911987901 indirectement un plus grand rapport(1199011199010211990111990101) on peut augmenter
bull la vitesse peacuteripheacuterique 119880119880 (la vitesse de rotation)bull la vitesse axiale 119888119888119909119909 (le deacutebit massique)bull lrsquoangle de deacuteflection de la pale(tan1205731205731 minus tan1205731205732)
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
Le degreacute de reacuteaction est un paramegravetre adimensionnel quiexprime la variation drsquoenthalpie subie par le fluide dans le rotorpar rapport agrave la chute totale drsquoenthalpie dans lrsquoeacutetage
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
Variation d enthalpie dans le rotorRVariation d enthalpie dans l eacutetage
=
❸❶
❷
Rotor
Stator
R=0 machine drsquoaction (Δhrotor=0 Δheacutetage=Δhstator)
R=1 machine agrave reacuteaction (Δhstator=0 Δheacutetage=Δhrotor)
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
R indique la reacutepartition du saut drsquoenthalpie (de pression) statique entre le stator et le rotor
Le degreacute de reacuteaction R deacutefini en fonction de variablesthermodynamiques (lrsquoenthalpie) peut aussi srsquoexprimer en fonctionde variables cineacutematiques preacutesentes dans le triangle de vitessesNotamment en fonction de la vitesse peacuteripheacuterique U et descomposantes des vitesses absolues et relatives
La vitesse absolue agrave lrsquoentreacutee du stator est eacutegale agrave celle agrave la sortie du rotor (c1= c3 eacutetage reacutepeacutetitif )
On neacuteglige les pertes et le transfert de chaleur dans le stator
1 2 3
ℎ3 minus ℎ1= ℎ3 + frasl11988811988832 2 minus ℎ1 minus frasl11988811988812 2 = ℎ03 minus ℎ01 = 120549120549ℎ0
ℎ03 minus ℎ01 = ℎ03 minus ℎ02 = 120549120549ℎ )0(3minus2
ℎ03 minus ℎ02 = ℎ3 minus ℎ1
119945119945120782120782120783120783 = 119945119945120782120782120784120784
frasl11988811988832 2 = frasl11988811988812 2
3 2
3 1
rotor
eacutetage
h h hRh h h
∆∆
minus= =
minus
2 20 3 2 2 3
0 3 2
2 2( )
( )
h c c h
∆∆
minus
minus
+ minus=
ℎ03 ℎ02
(ℎ01= ℎ02 1198881198881= 1198881198883)
120549120549ℎ )0(3minus2 = (ℎ3 + frasl11988811988832 2) minus (ℎ2 + frasl11988811988822 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
0(3 2) 3 2( )∆ minus = minusu uh U c c
= 1 minusfrasl11988811988832 2 minus frasl11988811988822 2120549120549ℎ )0(3minus2
119877119877 =120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
120549120549ℎ )0(3minus2
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
3 212+
= minus u uc cRU
119888119888119909119909 = 1198881198882119909119909 = 1198881198883119909119909 = 119888119888119888119888119888119888119888119888119888119888
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 11988811988821199061199061198881198882 = 1198881198881199091199092+1198881198881199061199062
119877119877 = 1 minus11988811988831199061199062 minus 11988811988821199061199062
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
Indirectement lrsquoeacutequation pour R deacutecrite en fonction de composantes cineacutematiques eacutetablit un lien entre les angles drsquoentreacutee et de sortie
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
= 1 minus1198881198883119906119906 + 1198881198882119906119906
2119880119880
2232
2 32 2wwh h+ = +
2 23 2
2 3 2 2w wh hminus = minus
2 2 2 23 3 2 2
2 3 2 2u x u xw w w wh h + +
minus = minus
Machine axiale U2 = U3
La composante axiale de la vitesse demeure constante w3x = w 2x
( eacutequation justifieacutee plus tard lors de la rothalpie)
2 23 2
2 3 2 2u uw wh hminus = minus
3 2 3 22 3
( )( )2
u u u uw w w wh h minus +minus =
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
Eacutetage reacutepeacutetitif c3 = c1
Si les pertes dans le stator sont neacutegligeacutees
c2u ndash c3u= w2u - w3u
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
3 2 3 23 2
( )( )2
u u u uw w w wh h minus +minus = minus
c2u ndash c3u= w2u - w3u
119877119877 = minus1199081199083119906119906 + 1199081199082119906119906
2119880119880
2 2 2 22 3 2 3 2 3( ) 2 ( ) 2h h u u w wminus = minus minus minus
2 2 2 2 2 23 2 3 2 3 2
2 2 2e
I II III
u u c c w wW minus minus minus= + minus
2 2 2 22 3 2 3
2 2 2 2 2 22 3 2 3 2 3
( ) ( )( ) ( ) ( )
minus minus minus=
minus + minus minus minusu u w wR
c c u u w w
3 2= =U u u
Expliqueacutee + tard
2 23 2
03 02 3 2 2ec cW h h h h minus
= minus = minus +
119877119877 =ℎ2 minus ℎ3ℎ1 minus ℎ3
=ℎ2 minus ℎ3ℎ02 minus ℎ03
Degreacute de reacuteaction
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
( )3u 2uw wR2U+
= minus
3 212
u uc cRU+
= minus
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
Puisque le degreacute de reacuteaction R peut aussi srsquoexprimer enfonction de variables cineacutematiques on peut chercher sarepreacutesentation dans le triangle adimensionnel de vitesses
Turbine
α2
β2 α3
β3
C2 W3
C3W2
C3uC2u
U
C2xC3x
W2u
c2u ndash c3u= w2u - w3u
2Ψ minusR
frasl119934119934120784120784119958119958 119932119932 = frasl1198621198622119906119906 119880119880 minus frasl119880119880 119880119880Ψ = frasl1198621198622119906119906 minus 1198621198623119906119906 119880119880
119877119877 = 1 minus (1198621198623119906119906 + 1198621198622119906119906 frasl) 2119880119880Φ = frasl1198621198622119909119909 119880119880 = frasl1198621198623119909119909 119880119880
= (119914119914120784120784119958119958 minus 119914119914120785120785119958119958 frasl) 2119880119880 minus 1 + (119914119914120784120784119958119958 + 119914119914120785120785119958119958 frasl) 2119880119880
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α2
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
Les repreacutesentations des triangles de vitesses adimensionnellessont tregraves riches en information et contiennent plusieurs formules
Par exemple pour un compresseur on deacuteduit aiseacutement que
Avec 120595120595 = 1198821198821198901198901198801198802
Φ = 119888119888119909119909119880119880
et que pour 119888119888119901119901 = 119888119888119888119888119888119888119888119888119888119888 119882119882119890119890 = 119888119888119901119901(11987911987902minus11987911987901)on trouve
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
(11987911987902minus11987911987901) =119880119880119888119888119909119909119888119888119901119901
(tan1205731205731 minus tan1205731205732)
Alors si lrsquoon veut obtenir une plus grande variation detempeacuterature 11987911987902 minus 11987911987901 indirectement un plus grand rapport(1199011199010211990111990101) on peut augmenter
bull la vitesse peacuteripheacuterique 119880119880 (la vitesse de rotation)bull la vitesse axiale 119888119888119909119909 (le deacutebit massique)bull lrsquoangle de deacuteflection de la pale(tan1205731205731 minus tan1205731205732)
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
Download Sound Effects - SoundDogs - Voice Box
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
Variation d enthalpie dans le rotorRVariation d enthalpie dans l eacutetage
=
❸❶
❷
Rotor
Stator
R=0 machine drsquoaction (Δhrotor=0 Δheacutetage=Δhstator)
R=1 machine agrave reacuteaction (Δhstator=0 Δheacutetage=Δhrotor)
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
R indique la reacutepartition du saut drsquoenthalpie (de pression) statique entre le stator et le rotor
Le degreacute de reacuteaction R deacutefini en fonction de variablesthermodynamiques (lrsquoenthalpie) peut aussi srsquoexprimer en fonctionde variables cineacutematiques preacutesentes dans le triangle de vitessesNotamment en fonction de la vitesse peacuteripheacuterique U et descomposantes des vitesses absolues et relatives
La vitesse absolue agrave lrsquoentreacutee du stator est eacutegale agrave celle agrave la sortie du rotor (c1= c3 eacutetage reacutepeacutetitif )
On neacuteglige les pertes et le transfert de chaleur dans le stator
1 2 3
ℎ3 minus ℎ1= ℎ3 + frasl11988811988832 2 minus ℎ1 minus frasl11988811988812 2 = ℎ03 minus ℎ01 = 120549120549ℎ0
ℎ03 minus ℎ01 = ℎ03 minus ℎ02 = 120549120549ℎ )0(3minus2
ℎ03 minus ℎ02 = ℎ3 minus ℎ1
119945119945120782120782120783120783 = 119945119945120782120782120784120784
frasl11988811988832 2 = frasl11988811988812 2
3 2
3 1
rotor
eacutetage
h h hRh h h
∆∆
minus= =
minus
2 20 3 2 2 3
0 3 2
2 2( )
( )
h c c h
∆∆
minus
minus
+ minus=
ℎ03 ℎ02
(ℎ01= ℎ02 1198881198881= 1198881198883)
120549120549ℎ )0(3minus2 = (ℎ3 + frasl11988811988832 2) minus (ℎ2 + frasl11988811988822 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
0(3 2) 3 2( )∆ minus = minusu uh U c c
= 1 minusfrasl11988811988832 2 minus frasl11988811988822 2120549120549ℎ )0(3minus2
119877119877 =120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
120549120549ℎ )0(3minus2
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
3 212+
= minus u uc cRU
119888119888119909119909 = 1198881198882119909119909 = 1198881198883119909119909 = 119888119888119888119888119888119888119888119888119888119888
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 11988811988821199061199061198881198882 = 1198881198881199091199092+1198881198881199061199062
119877119877 = 1 minus11988811988831199061199062 minus 11988811988821199061199062
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
Indirectement lrsquoeacutequation pour R deacutecrite en fonction de composantes cineacutematiques eacutetablit un lien entre les angles drsquoentreacutee et de sortie
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
= 1 minus1198881198883119906119906 + 1198881198882119906119906
2119880119880
2232
2 32 2wwh h+ = +
2 23 2
2 3 2 2w wh hminus = minus
2 2 2 23 3 2 2
2 3 2 2u x u xw w w wh h + +
minus = minus
Machine axiale U2 = U3
La composante axiale de la vitesse demeure constante w3x = w 2x
( eacutequation justifieacutee plus tard lors de la rothalpie)
2 23 2
2 3 2 2u uw wh hminus = minus
3 2 3 22 3
( )( )2
u u u uw w w wh h minus +minus =
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
Eacutetage reacutepeacutetitif c3 = c1
Si les pertes dans le stator sont neacutegligeacutees
c2u ndash c3u= w2u - w3u
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
3 2 3 23 2
( )( )2
u u u uw w w wh h minus +minus = minus
c2u ndash c3u= w2u - w3u
119877119877 = minus1199081199083119906119906 + 1199081199082119906119906
2119880119880
2 2 2 22 3 2 3 2 3( ) 2 ( ) 2h h u u w wminus = minus minus minus
2 2 2 2 2 23 2 3 2 3 2
2 2 2e
I II III
u u c c w wW minus minus minus= + minus
2 2 2 22 3 2 3
2 2 2 2 2 22 3 2 3 2 3
( ) ( )( ) ( ) ( )
minus minus minus=
minus + minus minus minusu u w wR
c c u u w w
3 2= =U u u
Expliqueacutee + tard
2 23 2
03 02 3 2 2ec cW h h h h minus
= minus = minus +
119877119877 =ℎ2 minus ℎ3ℎ1 minus ℎ3
=ℎ2 minus ℎ3ℎ02 minus ℎ03
Degreacute de reacuteaction
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
( )3u 2uw wR2U+
= minus
3 212
u uc cRU+
= minus
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
Puisque le degreacute de reacuteaction R peut aussi srsquoexprimer enfonction de variables cineacutematiques on peut chercher sarepreacutesentation dans le triangle adimensionnel de vitesses
Turbine
α2
β2 α3
β3
C2 W3
C3W2
C3uC2u
U
C2xC3x
W2u
c2u ndash c3u= w2u - w3u
2Ψ minusR
frasl119934119934120784120784119958119958 119932119932 = frasl1198621198622119906119906 119880119880 minus frasl119880119880 119880119880Ψ = frasl1198621198622119906119906 minus 1198621198623119906119906 119880119880
119877119877 = 1 minus (1198621198623119906119906 + 1198621198622119906119906 frasl) 2119880119880Φ = frasl1198621198622119909119909 119880119880 = frasl1198621198623119909119909 119880119880
= (119914119914120784120784119958119958 minus 119914119914120785120785119958119958 frasl) 2119880119880 minus 1 + (119914119914120784120784119958119958 + 119914119914120785120785119958119958 frasl) 2119880119880
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α2
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
Les repreacutesentations des triangles de vitesses adimensionnellessont tregraves riches en information et contiennent plusieurs formules
Par exemple pour un compresseur on deacuteduit aiseacutement que
Avec 120595120595 = 1198821198821198901198901198801198802
Φ = 119888119888119909119909119880119880
et que pour 119888119888119901119901 = 119888119888119888119888119888119888119888119888119888119888 119882119882119890119890 = 119888119888119901119901(11987911987902minus11987911987901)on trouve
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
(11987911987902minus11987911987901) =119880119880119888119888119909119909119888119888119901119901
(tan1205731205731 minus tan1205731205732)
Alors si lrsquoon veut obtenir une plus grande variation detempeacuterature 11987911987902 minus 11987911987901 indirectement un plus grand rapport(1199011199010211990111990101) on peut augmenter
bull la vitesse peacuteripheacuterique 119880119880 (la vitesse de rotation)bull la vitesse axiale 119888119888119909119909 (le deacutebit massique)bull lrsquoangle de deacuteflection de la pale(tan1205731205731 minus tan1205731205732)
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
R=0 machine drsquoaction (Δhrotor=0 Δheacutetage=Δhstator)
R=1 machine agrave reacuteaction (Δhstator=0 Δheacutetage=Δhrotor)
rotor 3 2
eacutetage 3 1
h h hRh h h
∆ minus= =
∆ minus
R indique la reacutepartition du saut drsquoenthalpie (de pression) statique entre le stator et le rotor
Le degreacute de reacuteaction R deacutefini en fonction de variablesthermodynamiques (lrsquoenthalpie) peut aussi srsquoexprimer en fonctionde variables cineacutematiques preacutesentes dans le triangle de vitessesNotamment en fonction de la vitesse peacuteripheacuterique U et descomposantes des vitesses absolues et relatives
La vitesse absolue agrave lrsquoentreacutee du stator est eacutegale agrave celle agrave la sortie du rotor (c1= c3 eacutetage reacutepeacutetitif )
On neacuteglige les pertes et le transfert de chaleur dans le stator
1 2 3
ℎ3 minus ℎ1= ℎ3 + frasl11988811988832 2 minus ℎ1 minus frasl11988811988812 2 = ℎ03 minus ℎ01 = 120549120549ℎ0
ℎ03 minus ℎ01 = ℎ03 minus ℎ02 = 120549120549ℎ )0(3minus2
ℎ03 minus ℎ02 = ℎ3 minus ℎ1
119945119945120782120782120783120783 = 119945119945120782120782120784120784
frasl11988811988832 2 = frasl11988811988812 2
3 2
3 1
rotor
eacutetage
h h hRh h h
∆∆
minus= =
minus
2 20 3 2 2 3
0 3 2
2 2( )
( )
h c c h
∆∆
minus
minus
+ minus=
ℎ03 ℎ02
(ℎ01= ℎ02 1198881198881= 1198881198883)
120549120549ℎ )0(3minus2 = (ℎ3 + frasl11988811988832 2) minus (ℎ2 + frasl11988811988822 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
0(3 2) 3 2( )∆ minus = minusu uh U c c
= 1 minusfrasl11988811988832 2 minus frasl11988811988822 2120549120549ℎ )0(3minus2
119877119877 =120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
120549120549ℎ )0(3minus2
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
3 212+
= minus u uc cRU
119888119888119909119909 = 1198881198882119909119909 = 1198881198883119909119909 = 119888119888119888119888119888119888119888119888119888119888
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 11988811988821199061199061198881198882 = 1198881198881199091199092+1198881198881199061199062
119877119877 = 1 minus11988811988831199061199062 minus 11988811988821199061199062
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
Indirectement lrsquoeacutequation pour R deacutecrite en fonction de composantes cineacutematiques eacutetablit un lien entre les angles drsquoentreacutee et de sortie
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
= 1 minus1198881198883119906119906 + 1198881198882119906119906
2119880119880
2232
2 32 2wwh h+ = +
2 23 2
2 3 2 2w wh hminus = minus
2 2 2 23 3 2 2
2 3 2 2u x u xw w w wh h + +
minus = minus
Machine axiale U2 = U3
La composante axiale de la vitesse demeure constante w3x = w 2x
( eacutequation justifieacutee plus tard lors de la rothalpie)
2 23 2
2 3 2 2u uw wh hminus = minus
3 2 3 22 3
( )( )2
u u u uw w w wh h minus +minus =
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
Eacutetage reacutepeacutetitif c3 = c1
Si les pertes dans le stator sont neacutegligeacutees
c2u ndash c3u= w2u - w3u
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
3 2 3 23 2
( )( )2
u u u uw w w wh h minus +minus = minus
c2u ndash c3u= w2u - w3u
119877119877 = minus1199081199083119906119906 + 1199081199082119906119906
2119880119880
2 2 2 22 3 2 3 2 3( ) 2 ( ) 2h h u u w wminus = minus minus minus
2 2 2 2 2 23 2 3 2 3 2
2 2 2e
I II III
u u c c w wW minus minus minus= + minus
2 2 2 22 3 2 3
2 2 2 2 2 22 3 2 3 2 3
( ) ( )( ) ( ) ( )
minus minus minus=
minus + minus minus minusu u w wR
c c u u w w
3 2= =U u u
Expliqueacutee + tard
2 23 2
03 02 3 2 2ec cW h h h h minus
= minus = minus +
119877119877 =ℎ2 minus ℎ3ℎ1 minus ℎ3
=ℎ2 minus ℎ3ℎ02 minus ℎ03
Degreacute de reacuteaction
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
( )3u 2uw wR2U+
= minus
3 212
u uc cRU+
= minus
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
Puisque le degreacute de reacuteaction R peut aussi srsquoexprimer enfonction de variables cineacutematiques on peut chercher sarepreacutesentation dans le triangle adimensionnel de vitesses
Turbine
α2
β2 α3
β3
C2 W3
C3W2
C3uC2u
U
C2xC3x
W2u
c2u ndash c3u= w2u - w3u
2Ψ minusR
frasl119934119934120784120784119958119958 119932119932 = frasl1198621198622119906119906 119880119880 minus frasl119880119880 119880119880Ψ = frasl1198621198622119906119906 minus 1198621198623119906119906 119880119880
119877119877 = 1 minus (1198621198623119906119906 + 1198621198622119906119906 frasl) 2119880119880Φ = frasl1198621198622119909119909 119880119880 = frasl1198621198623119909119909 119880119880
= (119914119914120784120784119958119958 minus 119914119914120785120785119958119958 frasl) 2119880119880 minus 1 + (119914119914120784120784119958119958 + 119914119914120785120785119958119958 frasl) 2119880119880
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α2
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
Les repreacutesentations des triangles de vitesses adimensionnellessont tregraves riches en information et contiennent plusieurs formules
Par exemple pour un compresseur on deacuteduit aiseacutement que
Avec 120595120595 = 1198821198821198901198901198801198802
Φ = 119888119888119909119909119880119880
et que pour 119888119888119901119901 = 119888119888119888119888119888119888119888119888119888119888 119882119882119890119890 = 119888119888119901119901(11987911987902minus11987911987901)on trouve
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
(11987911987902minus11987911987901) =119880119880119888119888119909119909119888119888119901119901
(tan1205731205731 minus tan1205731205732)
Alors si lrsquoon veut obtenir une plus grande variation detempeacuterature 11987911987902 minus 11987911987901 indirectement un plus grand rapport(1199011199010211990111990101) on peut augmenter
bull la vitesse peacuteripheacuterique 119880119880 (la vitesse de rotation)bull la vitesse axiale 119888119888119909119909 (le deacutebit massique)bull lrsquoangle de deacuteflection de la pale(tan1205731205731 minus tan1205731205732)
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
Download Sound Effects - SoundDogs - Voice Box
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
Le degreacute de reacuteaction R deacutefini en fonction de variablesthermodynamiques (lrsquoenthalpie) peut aussi srsquoexprimer en fonctionde variables cineacutematiques preacutesentes dans le triangle de vitessesNotamment en fonction de la vitesse peacuteripheacuterique U et descomposantes des vitesses absolues et relatives
La vitesse absolue agrave lrsquoentreacutee du stator est eacutegale agrave celle agrave la sortie du rotor (c1= c3 eacutetage reacutepeacutetitif )
On neacuteglige les pertes et le transfert de chaleur dans le stator
1 2 3
ℎ3 minus ℎ1= ℎ3 + frasl11988811988832 2 minus ℎ1 minus frasl11988811988812 2 = ℎ03 minus ℎ01 = 120549120549ℎ0
ℎ03 minus ℎ01 = ℎ03 minus ℎ02 = 120549120549ℎ )0(3minus2
ℎ03 minus ℎ02 = ℎ3 minus ℎ1
119945119945120782120782120783120783 = 119945119945120782120782120784120784
frasl11988811988832 2 = frasl11988811988812 2
3 2
3 1
rotor
eacutetage
h h hRh h h
∆∆
minus= =
minus
2 20 3 2 2 3
0 3 2
2 2( )
( )
h c c h
∆∆
minus
minus
+ minus=
ℎ03 ℎ02
(ℎ01= ℎ02 1198881198881= 1198881198883)
120549120549ℎ )0(3minus2 = (ℎ3 + frasl11988811988832 2) minus (ℎ2 + frasl11988811988822 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
0(3 2) 3 2( )∆ minus = minusu uh U c c
= 1 minusfrasl11988811988832 2 minus frasl11988811988822 2120549120549ℎ )0(3minus2
119877119877 =120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
120549120549ℎ )0(3minus2
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
3 212+
= minus u uc cRU
119888119888119909119909 = 1198881198882119909119909 = 1198881198883119909119909 = 119888119888119888119888119888119888119888119888119888119888
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 11988811988821199061199061198881198882 = 1198881198881199091199092+1198881198881199061199062
119877119877 = 1 minus11988811988831199061199062 minus 11988811988821199061199062
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
Indirectement lrsquoeacutequation pour R deacutecrite en fonction de composantes cineacutematiques eacutetablit un lien entre les angles drsquoentreacutee et de sortie
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
= 1 minus1198881198883119906119906 + 1198881198882119906119906
2119880119880
2232
2 32 2wwh h+ = +
2 23 2
2 3 2 2w wh hminus = minus
2 2 2 23 3 2 2
2 3 2 2u x u xw w w wh h + +
minus = minus
Machine axiale U2 = U3
La composante axiale de la vitesse demeure constante w3x = w 2x
( eacutequation justifieacutee plus tard lors de la rothalpie)
2 23 2
2 3 2 2u uw wh hminus = minus
3 2 3 22 3
( )( )2
u u u uw w w wh h minus +minus =
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
Eacutetage reacutepeacutetitif c3 = c1
Si les pertes dans le stator sont neacutegligeacutees
c2u ndash c3u= w2u - w3u
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
3 2 3 23 2
( )( )2
u u u uw w w wh h minus +minus = minus
c2u ndash c3u= w2u - w3u
119877119877 = minus1199081199083119906119906 + 1199081199082119906119906
2119880119880
2 2 2 22 3 2 3 2 3( ) 2 ( ) 2h h u u w wminus = minus minus minus
2 2 2 2 2 23 2 3 2 3 2
2 2 2e
I II III
u u c c w wW minus minus minus= + minus
2 2 2 22 3 2 3
2 2 2 2 2 22 3 2 3 2 3
( ) ( )( ) ( ) ( )
minus minus minus=
minus + minus minus minusu u w wR
c c u u w w
3 2= =U u u
Expliqueacutee + tard
2 23 2
03 02 3 2 2ec cW h h h h minus
= minus = minus +
119877119877 =ℎ2 minus ℎ3ℎ1 minus ℎ3
=ℎ2 minus ℎ3ℎ02 minus ℎ03
Degreacute de reacuteaction
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
( )3u 2uw wR2U+
= minus
3 212
u uc cRU+
= minus
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
Puisque le degreacute de reacuteaction R peut aussi srsquoexprimer enfonction de variables cineacutematiques on peut chercher sarepreacutesentation dans le triangle adimensionnel de vitesses
Turbine
α2
β2 α3
β3
C2 W3
C3W2
C3uC2u
U
C2xC3x
W2u
c2u ndash c3u= w2u - w3u
2Ψ minusR
frasl119934119934120784120784119958119958 119932119932 = frasl1198621198622119906119906 119880119880 minus frasl119880119880 119880119880Ψ = frasl1198621198622119906119906 minus 1198621198623119906119906 119880119880
119877119877 = 1 minus (1198621198623119906119906 + 1198621198622119906119906 frasl) 2119880119880Φ = frasl1198621198622119909119909 119880119880 = frasl1198621198623119909119909 119880119880
= (119914119914120784120784119958119958 minus 119914119914120785120785119958119958 frasl) 2119880119880 minus 1 + (119914119914120784120784119958119958 + 119914119914120785120785119958119958 frasl) 2119880119880
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α2
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
Les repreacutesentations des triangles de vitesses adimensionnellessont tregraves riches en information et contiennent plusieurs formules
Par exemple pour un compresseur on deacuteduit aiseacutement que
Avec 120595120595 = 1198821198821198901198901198801198802
Φ = 119888119888119909119909119880119880
et que pour 119888119888119901119901 = 119888119888119888119888119888119888119888119888119888119888 119882119882119890119890 = 119888119888119901119901(11987911987902minus11987911987901)on trouve
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
(11987911987902minus11987911987901) =119880119880119888119888119909119909119888119888119901119901
(tan1205731205731 minus tan1205731205732)
Alors si lrsquoon veut obtenir une plus grande variation detempeacuterature 11987911987902 minus 11987911987901 indirectement un plus grand rapport(1199011199010211990111990101) on peut augmenter
bull la vitesse peacuteripheacuterique 119880119880 (la vitesse de rotation)bull la vitesse axiale 119888119888119909119909 (le deacutebit massique)bull lrsquoangle de deacuteflection de la pale(tan1205731205731 minus tan1205731205732)
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
La vitesse absolue agrave lrsquoentreacutee du stator est eacutegale agrave celle agrave la sortie du rotor (c1= c3 eacutetage reacutepeacutetitif )
On neacuteglige les pertes et le transfert de chaleur dans le stator
1 2 3
ℎ3 minus ℎ1= ℎ3 + frasl11988811988832 2 minus ℎ1 minus frasl11988811988812 2 = ℎ03 minus ℎ01 = 120549120549ℎ0
ℎ03 minus ℎ01 = ℎ03 minus ℎ02 = 120549120549ℎ )0(3minus2
ℎ03 minus ℎ02 = ℎ3 minus ℎ1
119945119945120782120782120783120783 = 119945119945120782120782120784120784
frasl11988811988832 2 = frasl11988811988812 2
3 2
3 1
rotor
eacutetage
h h hRh h h
∆∆
minus= =
minus
2 20 3 2 2 3
0 3 2
2 2( )
( )
h c c h
∆∆
minus
minus
+ minus=
ℎ03 ℎ02
(ℎ01= ℎ02 1198881198881= 1198881198883)
120549120549ℎ )0(3minus2 = (ℎ3 + frasl11988811988832 2) minus (ℎ2 + frasl11988811988822 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
0(3 2) 3 2( )∆ minus = minusu uh U c c
= 1 minusfrasl11988811988832 2 minus frasl11988811988822 2120549120549ℎ )0(3minus2
119877119877 =120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
120549120549ℎ )0(3minus2
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
3 212+
= minus u uc cRU
119888119888119909119909 = 1198881198882119909119909 = 1198881198883119909119909 = 119888119888119888119888119888119888119888119888119888119888
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 11988811988821199061199061198881198882 = 1198881198881199091199092+1198881198881199061199062
119877119877 = 1 minus11988811988831199061199062 minus 11988811988821199061199062
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
Indirectement lrsquoeacutequation pour R deacutecrite en fonction de composantes cineacutematiques eacutetablit un lien entre les angles drsquoentreacutee et de sortie
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
= 1 minus1198881198883119906119906 + 1198881198882119906119906
2119880119880
2232
2 32 2wwh h+ = +
2 23 2
2 3 2 2w wh hminus = minus
2 2 2 23 3 2 2
2 3 2 2u x u xw w w wh h + +
minus = minus
Machine axiale U2 = U3
La composante axiale de la vitesse demeure constante w3x = w 2x
( eacutequation justifieacutee plus tard lors de la rothalpie)
2 23 2
2 3 2 2u uw wh hminus = minus
3 2 3 22 3
( )( )2
u u u uw w w wh h minus +minus =
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
Eacutetage reacutepeacutetitif c3 = c1
Si les pertes dans le stator sont neacutegligeacutees
c2u ndash c3u= w2u - w3u
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
3 2 3 23 2
( )( )2
u u u uw w w wh h minus +minus = minus
c2u ndash c3u= w2u - w3u
119877119877 = minus1199081199083119906119906 + 1199081199082119906119906
2119880119880
2 2 2 22 3 2 3 2 3( ) 2 ( ) 2h h u u w wminus = minus minus minus
2 2 2 2 2 23 2 3 2 3 2
2 2 2e
I II III
u u c c w wW minus minus minus= + minus
2 2 2 22 3 2 3
2 2 2 2 2 22 3 2 3 2 3
( ) ( )( ) ( ) ( )
minus minus minus=
minus + minus minus minusu u w wR
c c u u w w
3 2= =U u u
Expliqueacutee + tard
2 23 2
03 02 3 2 2ec cW h h h h minus
= minus = minus +
119877119877 =ℎ2 minus ℎ3ℎ1 minus ℎ3
=ℎ2 minus ℎ3ℎ02 minus ℎ03
Degreacute de reacuteaction
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
( )3u 2uw wR2U+
= minus
3 212
u uc cRU+
= minus
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
Puisque le degreacute de reacuteaction R peut aussi srsquoexprimer enfonction de variables cineacutematiques on peut chercher sarepreacutesentation dans le triangle adimensionnel de vitesses
Turbine
α2
β2 α3
β3
C2 W3
C3W2
C3uC2u
U
C2xC3x
W2u
c2u ndash c3u= w2u - w3u
2Ψ minusR
frasl119934119934120784120784119958119958 119932119932 = frasl1198621198622119906119906 119880119880 minus frasl119880119880 119880119880Ψ = frasl1198621198622119906119906 minus 1198621198623119906119906 119880119880
119877119877 = 1 minus (1198621198623119906119906 + 1198621198622119906119906 frasl) 2119880119880Φ = frasl1198621198622119909119909 119880119880 = frasl1198621198623119909119909 119880119880
= (119914119914120784120784119958119958 minus 119914119914120785120785119958119958 frasl) 2119880119880 minus 1 + (119914119914120784120784119958119958 + 119914119914120785120785119958119958 frasl) 2119880119880
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α2
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
Les repreacutesentations des triangles de vitesses adimensionnellessont tregraves riches en information et contiennent plusieurs formules
Par exemple pour un compresseur on deacuteduit aiseacutement que
Avec 120595120595 = 1198821198821198901198901198801198802
Φ = 119888119888119909119909119880119880
et que pour 119888119888119901119901 = 119888119888119888119888119888119888119888119888119888119888 119882119882119890119890 = 119888119888119901119901(11987911987902minus11987911987901)on trouve
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
(11987911987902minus11987911987901) =119880119880119888119888119909119909119888119888119901119901
(tan1205731205731 minus tan1205731205732)
Alors si lrsquoon veut obtenir une plus grande variation detempeacuterature 11987911987902 minus 11987911987901 indirectement un plus grand rapport(1199011199010211990111990101) on peut augmenter
bull la vitesse peacuteripheacuterique 119880119880 (la vitesse de rotation)bull la vitesse axiale 119888119888119909119909 (le deacutebit massique)bull lrsquoangle de deacuteflection de la pale(tan1205731205731 minus tan1205731205732)
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
3 2
3 1
rotor
eacutetage
h h hRh h h
∆∆
minus= =
minus
2 20 3 2 2 3
0 3 2
2 2( )
( )
h c c h
∆∆
minus
minus
+ minus=
ℎ03 ℎ02
(ℎ01= ℎ02 1198881198881= 1198881198883)
120549120549ℎ )0(3minus2 = (ℎ3 + frasl11988811988832 2) minus (ℎ2 + frasl11988811988822 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
ℎ3 minus ℎ2 = 120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
0(3 2) 3 2( )∆ minus = minusu uh U c c
= 1 minusfrasl11988811988832 2 minus frasl11988811988822 2120549120549ℎ )0(3minus2
119877119877 =120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
120549120549ℎ )0(3minus2
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
3 212+
= minus u uc cRU
119888119888119909119909 = 1198881198882119909119909 = 1198881198883119909119909 = 119888119888119888119888119888119888119888119888119888119888
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 11988811988821199061199061198881198882 = 1198881198881199091199092+1198881198881199061199062
119877119877 = 1 minus11988811988831199061199062 minus 11988811988821199061199062
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
Indirectement lrsquoeacutequation pour R deacutecrite en fonction de composantes cineacutematiques eacutetablit un lien entre les angles drsquoentreacutee et de sortie
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
= 1 minus1198881198883119906119906 + 1198881198882119906119906
2119880119880
2232
2 32 2wwh h+ = +
2 23 2
2 3 2 2w wh hminus = minus
2 2 2 23 3 2 2
2 3 2 2u x u xw w w wh h + +
minus = minus
Machine axiale U2 = U3
La composante axiale de la vitesse demeure constante w3x = w 2x
( eacutequation justifieacutee plus tard lors de la rothalpie)
2 23 2
2 3 2 2u uw wh hminus = minus
3 2 3 22 3
( )( )2
u u u uw w w wh h minus +minus =
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
Eacutetage reacutepeacutetitif c3 = c1
Si les pertes dans le stator sont neacutegligeacutees
c2u ndash c3u= w2u - w3u
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
3 2 3 23 2
( )( )2
u u u uw w w wh h minus +minus = minus
c2u ndash c3u= w2u - w3u
119877119877 = minus1199081199083119906119906 + 1199081199082119906119906
2119880119880
2 2 2 22 3 2 3 2 3( ) 2 ( ) 2h h u u w wminus = minus minus minus
2 2 2 2 2 23 2 3 2 3 2
2 2 2e
I II III
u u c c w wW minus minus minus= + minus
2 2 2 22 3 2 3
2 2 2 2 2 22 3 2 3 2 3
( ) ( )( ) ( ) ( )
minus minus minus=
minus + minus minus minusu u w wR
c c u u w w
3 2= =U u u
Expliqueacutee + tard
2 23 2
03 02 3 2 2ec cW h h h h minus
= minus = minus +
119877119877 =ℎ2 minus ℎ3ℎ1 minus ℎ3
=ℎ2 minus ℎ3ℎ02 minus ℎ03
Degreacute de reacuteaction
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
( )3u 2uw wR2U+
= minus
3 212
u uc cRU+
= minus
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
Puisque le degreacute de reacuteaction R peut aussi srsquoexprimer enfonction de variables cineacutematiques on peut chercher sarepreacutesentation dans le triangle adimensionnel de vitesses
Turbine
α2
β2 α3
β3
C2 W3
C3W2
C3uC2u
U
C2xC3x
W2u
c2u ndash c3u= w2u - w3u
2Ψ minusR
frasl119934119934120784120784119958119958 119932119932 = frasl1198621198622119906119906 119880119880 minus frasl119880119880 119880119880Ψ = frasl1198621198622119906119906 minus 1198621198623119906119906 119880119880
119877119877 = 1 minus (1198621198623119906119906 + 1198621198622119906119906 frasl) 2119880119880Φ = frasl1198621198622119909119909 119880119880 = frasl1198621198623119909119909 119880119880
= (119914119914120784120784119958119958 minus 119914119914120785120785119958119958 frasl) 2119880119880 minus 1 + (119914119914120784120784119958119958 + 119914119914120785120785119958119958 frasl) 2119880119880
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α2
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
Les repreacutesentations des triangles de vitesses adimensionnellessont tregraves riches en information et contiennent plusieurs formules
Par exemple pour un compresseur on deacuteduit aiseacutement que
Avec 120595120595 = 1198821198821198901198901198801198802
Φ = 119888119888119909119909119880119880
et que pour 119888119888119901119901 = 119888119888119888119888119888119888119888119888119888119888 119882119882119890119890 = 119888119888119901119901(11987911987902minus11987911987901)on trouve
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
(11987911987902minus11987911987901) =119880119880119888119888119909119909119888119888119901119901
(tan1205731205731 minus tan1205731205732)
Alors si lrsquoon veut obtenir une plus grande variation detempeacuterature 11987911987902 minus 11987911987901 indirectement un plus grand rapport(1199011199010211990111990101) on peut augmenter
bull la vitesse peacuteripheacuterique 119880119880 (la vitesse de rotation)bull la vitesse axiale 119888119888119909119909 (le deacutebit massique)bull lrsquoangle de deacuteflection de la pale(tan1205731205731 minus tan1205731205732)
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
0(3 2) 3 2( )∆ minus = minusu uh U c c
= 1 minusfrasl11988811988832 2 minus frasl11988811988822 2120549120549ℎ )0(3minus2
119877119877 =120549120549ℎ )0(3minus2 + frasl11988811988822 2 minus frasl11988811988832 2
120549120549ℎ )0(3minus2
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
3 212+
= minus u uc cRU
119888119888119909119909 = 1198881198882119909119909 = 1198881198883119909119909 = 119888119888119888119888119888119888119888119888119888119888
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 11988811988821199061199061198881198882 = 1198881198881199091199092+1198881198881199061199062
119877119877 = 1 minus11988811988831199061199062 minus 11988811988821199061199062
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
Indirectement lrsquoeacutequation pour R deacutecrite en fonction de composantes cineacutematiques eacutetablit un lien entre les angles drsquoentreacutee et de sortie
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
= 1 minus1198881198883119906119906 + 1198881198882119906119906
2119880119880
2232
2 32 2wwh h+ = +
2 23 2
2 3 2 2w wh hminus = minus
2 2 2 23 3 2 2
2 3 2 2u x u xw w w wh h + +
minus = minus
Machine axiale U2 = U3
La composante axiale de la vitesse demeure constante w3x = w 2x
( eacutequation justifieacutee plus tard lors de la rothalpie)
2 23 2
2 3 2 2u uw wh hminus = minus
3 2 3 22 3
( )( )2
u u u uw w w wh h minus +minus =
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
Eacutetage reacutepeacutetitif c3 = c1
Si les pertes dans le stator sont neacutegligeacutees
c2u ndash c3u= w2u - w3u
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
3 2 3 23 2
( )( )2
u u u uw w w wh h minus +minus = minus
c2u ndash c3u= w2u - w3u
119877119877 = minus1199081199083119906119906 + 1199081199082119906119906
2119880119880
2 2 2 22 3 2 3 2 3( ) 2 ( ) 2h h u u w wminus = minus minus minus
2 2 2 2 2 23 2 3 2 3 2
2 2 2e
I II III
u u c c w wW minus minus minus= + minus
2 2 2 22 3 2 3
2 2 2 2 2 22 3 2 3 2 3
( ) ( )( ) ( ) ( )
minus minus minus=
minus + minus minus minusu u w wR
c c u u w w
3 2= =U u u
Expliqueacutee + tard
2 23 2
03 02 3 2 2ec cW h h h h minus
= minus = minus +
119877119877 =ℎ2 minus ℎ3ℎ1 minus ℎ3
=ℎ2 minus ℎ3ℎ02 minus ℎ03
Degreacute de reacuteaction
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
( )3u 2uw wR2U+
= minus
3 212
u uc cRU+
= minus
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
Puisque le degreacute de reacuteaction R peut aussi srsquoexprimer enfonction de variables cineacutematiques on peut chercher sarepreacutesentation dans le triangle adimensionnel de vitesses
Turbine
α2
β2 α3
β3
C2 W3
C3W2
C3uC2u
U
C2xC3x
W2u
c2u ndash c3u= w2u - w3u
2Ψ minusR
frasl119934119934120784120784119958119958 119932119932 = frasl1198621198622119906119906 119880119880 minus frasl119880119880 119880119880Ψ = frasl1198621198622119906119906 minus 1198621198623119906119906 119880119880
119877119877 = 1 minus (1198621198623119906119906 + 1198621198622119906119906 frasl) 2119880119880Φ = frasl1198621198622119909119909 119880119880 = frasl1198621198623119909119909 119880119880
= (119914119914120784120784119958119958 minus 119914119914120785120785119958119958 frasl) 2119880119880 minus 1 + (119914119914120784120784119958119958 + 119914119914120785120785119958119958 frasl) 2119880119880
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α2
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
Les repreacutesentations des triangles de vitesses adimensionnellessont tregraves riches en information et contiennent plusieurs formules
Par exemple pour un compresseur on deacuteduit aiseacutement que
Avec 120595120595 = 1198821198821198901198901198801198802
Φ = 119888119888119909119909119880119880
et que pour 119888119888119901119901 = 119888119888119888119888119888119888119888119888119888119888 119882119882119890119890 = 119888119888119901119901(11987911987902minus11987911987901)on trouve
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
(11987911987902minus11987911987901) =119880119880119888119888119909119909119888119888119901119901
(tan1205731205731 minus tan1205731205732)
Alors si lrsquoon veut obtenir une plus grande variation detempeacuterature 11987911987902 minus 11987911987901 indirectement un plus grand rapport(1199011199010211990111990101) on peut augmenter
bull la vitesse peacuteripheacuterique 119880119880 (la vitesse de rotation)bull la vitesse axiale 119888119888119909119909 (le deacutebit massique)bull lrsquoangle de deacuteflection de la pale(tan1205731205731 minus tan1205731205732)
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
3 212+
= minus u uc cRU
119888119888119909119909 = 1198881198882119909119909 = 1198881198883119909119909 = 119888119888119888119888119888119888119888119888119888119888
119877119877 = 1 minus11988811988832 minus 11988811988822
)2119880119880(1198881198883119906119906 minus 11988811988821199061199061198881198882 = 1198881198881199091199092+1198881198881199061199062
119877119877 = 1 minus11988811988831199061199062 minus 11988811988821199061199062
)2119880119880(1198881198883119906119906 minus 1198881198882119906119906
Indirectement lrsquoeacutequation pour R deacutecrite en fonction de composantes cineacutematiques eacutetablit un lien entre les angles drsquoentreacutee et de sortie
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
= 1 minus1198881198883119906119906 + 1198881198882119906119906
2119880119880
2232
2 32 2wwh h+ = +
2 23 2
2 3 2 2w wh hminus = minus
2 2 2 23 3 2 2
2 3 2 2u x u xw w w wh h + +
minus = minus
Machine axiale U2 = U3
La composante axiale de la vitesse demeure constante w3x = w 2x
( eacutequation justifieacutee plus tard lors de la rothalpie)
2 23 2
2 3 2 2u uw wh hminus = minus
3 2 3 22 3
( )( )2
u u u uw w w wh h minus +minus =
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
Eacutetage reacutepeacutetitif c3 = c1
Si les pertes dans le stator sont neacutegligeacutees
c2u ndash c3u= w2u - w3u
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
3 2 3 23 2
( )( )2
u u u uw w w wh h minus +minus = minus
c2u ndash c3u= w2u - w3u
119877119877 = minus1199081199083119906119906 + 1199081199082119906119906
2119880119880
2 2 2 22 3 2 3 2 3( ) 2 ( ) 2h h u u w wminus = minus minus minus
2 2 2 2 2 23 2 3 2 3 2
2 2 2e
I II III
u u c c w wW minus minus minus= + minus
2 2 2 22 3 2 3
2 2 2 2 2 22 3 2 3 2 3
( ) ( )( ) ( ) ( )
minus minus minus=
minus + minus minus minusu u w wR
c c u u w w
3 2= =U u u
Expliqueacutee + tard
2 23 2
03 02 3 2 2ec cW h h h h minus
= minus = minus +
119877119877 =ℎ2 minus ℎ3ℎ1 minus ℎ3
=ℎ2 minus ℎ3ℎ02 minus ℎ03
Degreacute de reacuteaction
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
( )3u 2uw wR2U+
= minus
3 212
u uc cRU+
= minus
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
Puisque le degreacute de reacuteaction R peut aussi srsquoexprimer enfonction de variables cineacutematiques on peut chercher sarepreacutesentation dans le triangle adimensionnel de vitesses
Turbine
α2
β2 α3
β3
C2 W3
C3W2
C3uC2u
U
C2xC3x
W2u
c2u ndash c3u= w2u - w3u
2Ψ minusR
frasl119934119934120784120784119958119958 119932119932 = frasl1198621198622119906119906 119880119880 minus frasl119880119880 119880119880Ψ = frasl1198621198622119906119906 minus 1198621198623119906119906 119880119880
119877119877 = 1 minus (1198621198623119906119906 + 1198621198622119906119906 frasl) 2119880119880Φ = frasl1198621198622119909119909 119880119880 = frasl1198621198623119909119909 119880119880
= (119914119914120784120784119958119958 minus 119914119914120785120785119958119958 frasl) 2119880119880 minus 1 + (119914119914120784120784119958119958 + 119914119914120785120785119958119958 frasl) 2119880119880
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α2
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
Les repreacutesentations des triangles de vitesses adimensionnellessont tregraves riches en information et contiennent plusieurs formules
Par exemple pour un compresseur on deacuteduit aiseacutement que
Avec 120595120595 = 1198821198821198901198901198801198802
Φ = 119888119888119909119909119880119880
et que pour 119888119888119901119901 = 119888119888119888119888119888119888119888119888119888119888 119882119882119890119890 = 119888119888119901119901(11987911987902minus11987911987901)on trouve
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
(11987911987902minus11987911987901) =119880119880119888119888119909119909119888119888119901119901
(tan1205731205731 minus tan1205731205732)
Alors si lrsquoon veut obtenir une plus grande variation detempeacuterature 11987911987902 minus 11987911987901 indirectement un plus grand rapport(1199011199010211990111990101) on peut augmenter
bull la vitesse peacuteripheacuterique 119880119880 (la vitesse de rotation)bull la vitesse axiale 119888119888119909119909 (le deacutebit massique)bull lrsquoangle de deacuteflection de la pale(tan1205731205731 minus tan1205731205732)
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
Indirectement lrsquoeacutequation pour R deacutecrite en fonction de composantes cineacutematiques eacutetablit un lien entre les angles drsquoentreacutee et de sortie
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
= 1 minus1198881198883119906119906 + 1198881198882119906119906
2119880119880
2232
2 32 2wwh h+ = +
2 23 2
2 3 2 2w wh hminus = minus
2 2 2 23 3 2 2
2 3 2 2u x u xw w w wh h + +
minus = minus
Machine axiale U2 = U3
La composante axiale de la vitesse demeure constante w3x = w 2x
( eacutequation justifieacutee plus tard lors de la rothalpie)
2 23 2
2 3 2 2u uw wh hminus = minus
3 2 3 22 3
( )( )2
u u u uw w w wh h minus +minus =
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
Eacutetage reacutepeacutetitif c3 = c1
Si les pertes dans le stator sont neacutegligeacutees
c2u ndash c3u= w2u - w3u
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
3 2 3 23 2
( )( )2
u u u uw w w wh h minus +minus = minus
c2u ndash c3u= w2u - w3u
119877119877 = minus1199081199083119906119906 + 1199081199082119906119906
2119880119880
2 2 2 22 3 2 3 2 3( ) 2 ( ) 2h h u u w wminus = minus minus minus
2 2 2 2 2 23 2 3 2 3 2
2 2 2e
I II III
u u c c w wW minus minus minus= + minus
2 2 2 22 3 2 3
2 2 2 2 2 22 3 2 3 2 3
( ) ( )( ) ( ) ( )
minus minus minus=
minus + minus minus minusu u w wR
c c u u w w
3 2= =U u u
Expliqueacutee + tard
2 23 2
03 02 3 2 2ec cW h h h h minus
= minus = minus +
119877119877 =ℎ2 minus ℎ3ℎ1 minus ℎ3
=ℎ2 minus ℎ3ℎ02 minus ℎ03
Degreacute de reacuteaction
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
( )3u 2uw wR2U+
= minus
3 212
u uc cRU+
= minus
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
Puisque le degreacute de reacuteaction R peut aussi srsquoexprimer enfonction de variables cineacutematiques on peut chercher sarepreacutesentation dans le triangle adimensionnel de vitesses
Turbine
α2
β2 α3
β3
C2 W3
C3W2
C3uC2u
U
C2xC3x
W2u
c2u ndash c3u= w2u - w3u
2Ψ minusR
frasl119934119934120784120784119958119958 119932119932 = frasl1198621198622119906119906 119880119880 minus frasl119880119880 119880119880Ψ = frasl1198621198622119906119906 minus 1198621198623119906119906 119880119880
119877119877 = 1 minus (1198621198623119906119906 + 1198621198622119906119906 frasl) 2119880119880Φ = frasl1198621198622119909119909 119880119880 = frasl1198621198623119909119909 119880119880
= (119914119914120784120784119958119958 minus 119914119914120785120785119958119958 frasl) 2119880119880 minus 1 + (119914119914120784120784119958119958 + 119914119914120785120785119958119958 frasl) 2119880119880
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α2
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
Les repreacutesentations des triangles de vitesses adimensionnellessont tregraves riches en information et contiennent plusieurs formules
Par exemple pour un compresseur on deacuteduit aiseacutement que
Avec 120595120595 = 1198821198821198901198901198801198802
Φ = 119888119888119909119909119880119880
et que pour 119888119888119901119901 = 119888119888119888119888119888119888119888119888119888119888 119882119882119890119890 = 119888119888119901119901(11987911987902minus11987911987901)on trouve
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
(11987911987902minus11987911987901) =119880119880119888119888119909119909119888119888119901119901
(tan1205731205731 minus tan1205731205732)
Alors si lrsquoon veut obtenir une plus grande variation detempeacuterature 11987911987902 minus 11987911987901 indirectement un plus grand rapport(1199011199010211990111990101) on peut augmenter
bull la vitesse peacuteripheacuterique 119880119880 (la vitesse de rotation)bull la vitesse axiale 119888119888119909119909 (le deacutebit massique)bull lrsquoangle de deacuteflection de la pale(tan1205731205731 minus tan1205731205732)
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
2232
2 32 2wwh h+ = +
2 23 2
2 3 2 2w wh hminus = minus
2 2 2 23 3 2 2
2 3 2 2u x u xw w w wh h + +
minus = minus
Machine axiale U2 = U3
La composante axiale de la vitesse demeure constante w3x = w 2x
( eacutequation justifieacutee plus tard lors de la rothalpie)
2 23 2
2 3 2 2u uw wh hminus = minus
3 2 3 22 3
( )( )2
u u u uw w w wh h minus +minus =
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
Eacutetage reacutepeacutetitif c3 = c1
Si les pertes dans le stator sont neacutegligeacutees
c2u ndash c3u= w2u - w3u
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
3 2 3 23 2
( )( )2
u u u uw w w wh h minus +minus = minus
c2u ndash c3u= w2u - w3u
119877119877 = minus1199081199083119906119906 + 1199081199082119906119906
2119880119880
2 2 2 22 3 2 3 2 3( ) 2 ( ) 2h h u u w wminus = minus minus minus
2 2 2 2 2 23 2 3 2 3 2
2 2 2e
I II III
u u c c w wW minus minus minus= + minus
2 2 2 22 3 2 3
2 2 2 2 2 22 3 2 3 2 3
( ) ( )( ) ( ) ( )
minus minus minus=
minus + minus minus minusu u w wR
c c u u w w
3 2= =U u u
Expliqueacutee + tard
2 23 2
03 02 3 2 2ec cW h h h h minus
= minus = minus +
119877119877 =ℎ2 minus ℎ3ℎ1 minus ℎ3
=ℎ2 minus ℎ3ℎ02 minus ℎ03
Degreacute de reacuteaction
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
( )3u 2uw wR2U+
= minus
3 212
u uc cRU+
= minus
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
Puisque le degreacute de reacuteaction R peut aussi srsquoexprimer enfonction de variables cineacutematiques on peut chercher sarepreacutesentation dans le triangle adimensionnel de vitesses
Turbine
α2
β2 α3
β3
C2 W3
C3W2
C3uC2u
U
C2xC3x
W2u
c2u ndash c3u= w2u - w3u
2Ψ minusR
frasl119934119934120784120784119958119958 119932119932 = frasl1198621198622119906119906 119880119880 minus frasl119880119880 119880119880Ψ = frasl1198621198622119906119906 minus 1198621198623119906119906 119880119880
119877119877 = 1 minus (1198621198623119906119906 + 1198621198622119906119906 frasl) 2119880119880Φ = frasl1198621198622119909119909 119880119880 = frasl1198621198623119909119909 119880119880
= (119914119914120784120784119958119958 minus 119914119914120785120785119958119958 frasl) 2119880119880 minus 1 + (119914119914120784120784119958119958 + 119914119914120785120785119958119958 frasl) 2119880119880
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α2
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
Les repreacutesentations des triangles de vitesses adimensionnellessont tregraves riches en information et contiennent plusieurs formules
Par exemple pour un compresseur on deacuteduit aiseacutement que
Avec 120595120595 = 1198821198821198901198901198801198802
Φ = 119888119888119909119909119880119880
et que pour 119888119888119901119901 = 119888119888119888119888119888119888119888119888119888119888 119882119882119890119890 = 119888119888119901119901(11987911987902minus11987911987901)on trouve
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
(11987911987902minus11987911987901) =119880119880119888119888119909119909119888119888119901119901
(tan1205731205731 minus tan1205731205732)
Alors si lrsquoon veut obtenir une plus grande variation detempeacuterature 11987911987902 minus 11987911987901 indirectement un plus grand rapport(1199011199010211990111990101) on peut augmenter
bull la vitesse peacuteripheacuterique 119880119880 (la vitesse de rotation)bull la vitesse axiale 119888119888119909119909 (le deacutebit massique)bull lrsquoangle de deacuteflection de la pale(tan1205731205731 minus tan1205731205732)
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
2 23 2
2 3 2 2u uw wh hminus = minus
3 2 3 22 3
( )( )2
u u u uw w w wh h minus +minus =
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
Eacutetage reacutepeacutetitif c3 = c1
Si les pertes dans le stator sont neacutegligeacutees
c2u ndash c3u= w2u - w3u
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
3 2 3 23 2
( )( )2
u u u uw w w wh h minus +minus = minus
c2u ndash c3u= w2u - w3u
119877119877 = minus1199081199083119906119906 + 1199081199082119906119906
2119880119880
2 2 2 22 3 2 3 2 3( ) 2 ( ) 2h h u u w wminus = minus minus minus
2 2 2 2 2 23 2 3 2 3 2
2 2 2e
I II III
u u c c w wW minus minus minus= + minus
2 2 2 22 3 2 3
2 2 2 2 2 22 3 2 3 2 3
( ) ( )( ) ( ) ( )
minus minus minus=
minus + minus minus minusu u w wR
c c u u w w
3 2= =U u u
Expliqueacutee + tard
2 23 2
03 02 3 2 2ec cW h h h h minus
= minus = minus +
119877119877 =ℎ2 minus ℎ3ℎ1 minus ℎ3
=ℎ2 minus ℎ3ℎ02 minus ℎ03
Degreacute de reacuteaction
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
( )3u 2uw wR2U+
= minus
3 212
u uc cRU+
= minus
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
Puisque le degreacute de reacuteaction R peut aussi srsquoexprimer enfonction de variables cineacutematiques on peut chercher sarepreacutesentation dans le triangle adimensionnel de vitesses
Turbine
α2
β2 α3
β3
C2 W3
C3W2
C3uC2u
U
C2xC3x
W2u
c2u ndash c3u= w2u - w3u
2Ψ minusR
frasl119934119934120784120784119958119958 119932119932 = frasl1198621198622119906119906 119880119880 minus frasl119880119880 119880119880Ψ = frasl1198621198622119906119906 minus 1198621198623119906119906 119880119880
119877119877 = 1 minus (1198621198623119906119906 + 1198621198622119906119906 frasl) 2119880119880Φ = frasl1198621198622119909119909 119880119880 = frasl1198621198623119909119909 119880119880
= (119914119914120784120784119958119958 minus 119914119914120785120785119958119958 frasl) 2119880119880 minus 1 + (119914119914120784120784119958119958 + 119914119914120785120785119958119958 frasl) 2119880119880
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α2
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
Les repreacutesentations des triangles de vitesses adimensionnellessont tregraves riches en information et contiennent plusieurs formules
Par exemple pour un compresseur on deacuteduit aiseacutement que
Avec 120595120595 = 1198821198821198901198901198801198802
Φ = 119888119888119909119909119880119880
et que pour 119888119888119901119901 = 119888119888119888119888119888119888119888119888119888119888 119882119882119890119890 = 119888119888119901119901(11987911987902minus11987911987901)on trouve
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
(11987911987902minus11987911987901) =119880119880119888119888119909119909119888119888119901119901
(tan1205731205731 minus tan1205731205732)
Alors si lrsquoon veut obtenir une plus grande variation detempeacuterature 11987911987902 minus 11987911987901 indirectement un plus grand rapport(1199011199010211990111990101) on peut augmenter
bull la vitesse peacuteripheacuterique 119880119880 (la vitesse de rotation)bull la vitesse axiale 119888119888119909119909 (le deacutebit massique)bull lrsquoangle de deacuteflection de la pale(tan1205731205731 minus tan1205731205732)
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
( ) ( )03 01 3u 2u 3u 2uh h U c c U w wminus = minus = minus
rotor 3 2 3 2
eacutetage 3 1 03 01
h h h h hRh h h h h
∆ minus minus= = =
∆ minus minus
3 2 3 23 2
( )( )2
u u u uw w w wh h minus +minus = minus
c2u ndash c3u= w2u - w3u
119877119877 = minus1199081199083119906119906 + 1199081199082119906119906
2119880119880
2 2 2 22 3 2 3 2 3( ) 2 ( ) 2h h u u w wminus = minus minus minus
2 2 2 2 2 23 2 3 2 3 2
2 2 2e
I II III
u u c c w wW minus minus minus= + minus
2 2 2 22 3 2 3
2 2 2 2 2 22 3 2 3 2 3
( ) ( )( ) ( ) ( )
minus minus minus=
minus + minus minus minusu u w wR
c c u u w w
3 2= =U u u
Expliqueacutee + tard
2 23 2
03 02 3 2 2ec cW h h h h minus
= minus = minus +
119877119877 =ℎ2 minus ℎ3ℎ1 minus ℎ3
=ℎ2 minus ℎ3ℎ02 minus ℎ03
Degreacute de reacuteaction
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
( )3u 2uw wR2U+
= minus
3 212
u uc cRU+
= minus
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
Puisque le degreacute de reacuteaction R peut aussi srsquoexprimer enfonction de variables cineacutematiques on peut chercher sarepreacutesentation dans le triangle adimensionnel de vitesses
Turbine
α2
β2 α3
β3
C2 W3
C3W2
C3uC2u
U
C2xC3x
W2u
c2u ndash c3u= w2u - w3u
2Ψ minusR
frasl119934119934120784120784119958119958 119932119932 = frasl1198621198622119906119906 119880119880 minus frasl119880119880 119880119880Ψ = frasl1198621198622119906119906 minus 1198621198623119906119906 119880119880
119877119877 = 1 minus (1198621198623119906119906 + 1198621198622119906119906 frasl) 2119880119880Φ = frasl1198621198622119909119909 119880119880 = frasl1198621198623119909119909 119880119880
= (119914119914120784120784119958119958 minus 119914119914120785120785119958119958 frasl) 2119880119880 minus 1 + (119914119914120784120784119958119958 + 119914119914120785120785119958119958 frasl) 2119880119880
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α2
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
Les repreacutesentations des triangles de vitesses adimensionnellessont tregraves riches en information et contiennent plusieurs formules
Par exemple pour un compresseur on deacuteduit aiseacutement que
Avec 120595120595 = 1198821198821198901198901198801198802
Φ = 119888119888119909119909119880119880
et que pour 119888119888119901119901 = 119888119888119888119888119888119888119888119888119888119888 119882119882119890119890 = 119888119888119901119901(11987911987902minus11987911987901)on trouve
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
(11987911987902minus11987911987901) =119880119880119888119888119909119909119888119888119901119901
(tan1205731205731 minus tan1205731205732)
Alors si lrsquoon veut obtenir une plus grande variation detempeacuterature 11987911987902 minus 11987911987901 indirectement un plus grand rapport(1199011199010211990111990101) on peut augmenter
bull la vitesse peacuteripheacuterique 119880119880 (la vitesse de rotation)bull la vitesse axiale 119888119888119909119909 (le deacutebit massique)bull lrsquoangle de deacuteflection de la pale(tan1205731205731 minus tan1205731205732)
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
2 2 2 22 3 2 3 2 3( ) 2 ( ) 2h h u u w wminus = minus minus minus
2 2 2 2 2 23 2 3 2 3 2
2 2 2e
I II III
u u c c w wW minus minus minus= + minus
2 2 2 22 3 2 3
2 2 2 2 2 22 3 2 3 2 3
( ) ( )( ) ( ) ( )
minus minus minus=
minus + minus minus minusu u w wR
c c u u w w
3 2= =U u u
Expliqueacutee + tard
2 23 2
03 02 3 2 2ec cW h h h h minus
= minus = minus +
119877119877 =ℎ2 minus ℎ3ℎ1 minus ℎ3
=ℎ2 minus ℎ3ℎ02 minus ℎ03
Degreacute de reacuteaction
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
( )3u 2uw wR2U+
= minus
3 212
u uc cRU+
= minus
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
Puisque le degreacute de reacuteaction R peut aussi srsquoexprimer enfonction de variables cineacutematiques on peut chercher sarepreacutesentation dans le triangle adimensionnel de vitesses
Turbine
α2
β2 α3
β3
C2 W3
C3W2
C3uC2u
U
C2xC3x
W2u
c2u ndash c3u= w2u - w3u
2Ψ minusR
frasl119934119934120784120784119958119958 119932119932 = frasl1198621198622119906119906 119880119880 minus frasl119880119880 119880119880Ψ = frasl1198621198622119906119906 minus 1198621198623119906119906 119880119880
119877119877 = 1 minus (1198621198623119906119906 + 1198621198622119906119906 frasl) 2119880119880Φ = frasl1198621198622119909119909 119880119880 = frasl1198621198623119909119909 119880119880
= (119914119914120784120784119958119958 minus 119914119914120785120785119958119958 frasl) 2119880119880 minus 1 + (119914119914120784120784119958119958 + 119914119914120785120785119958119958 frasl) 2119880119880
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α2
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
Les repreacutesentations des triangles de vitesses adimensionnellessont tregraves riches en information et contiennent plusieurs formules
Par exemple pour un compresseur on deacuteduit aiseacutement que
Avec 120595120595 = 1198821198821198901198901198801198802
Φ = 119888119888119909119909119880119880
et que pour 119888119888119901119901 = 119888119888119888119888119888119888119888119888119888119888 119882119882119890119890 = 119888119888119901119901(11987911987902minus11987911987901)on trouve
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
(11987911987902minus11987911987901) =119880119880119888119888119909119909119888119888119901119901
(tan1205731205731 minus tan1205731205732)
Alors si lrsquoon veut obtenir une plus grande variation detempeacuterature 11987911987902 minus 11987911987901 indirectement un plus grand rapport(1199011199010211990111990101) on peut augmenter
bull la vitesse peacuteripheacuterique 119880119880 (la vitesse de rotation)bull la vitesse axiale 119888119888119909119909 (le deacutebit massique)bull lrsquoangle de deacuteflection de la pale(tan1205731205731 minus tan1205731205732)
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
( )3u 2uw wR2U+
= minus
3 212
u uc cRU+
= minus
119877119877 =minus(11990811990822 minus 11990811990832)
11988811988822 minus 11988811988832) minus (11990811990822 minus 11990811990832
119877119877 =ℎ3 minus ℎ2ℎ3 minus ℎ1
Puisque le degreacute de reacuteaction R peut aussi srsquoexprimer enfonction de variables cineacutematiques on peut chercher sarepreacutesentation dans le triangle adimensionnel de vitesses
Turbine
α2
β2 α3
β3
C2 W3
C3W2
C3uC2u
U
C2xC3x
W2u
c2u ndash c3u= w2u - w3u
2Ψ minusR
frasl119934119934120784120784119958119958 119932119932 = frasl1198621198622119906119906 119880119880 minus frasl119880119880 119880119880Ψ = frasl1198621198622119906119906 minus 1198621198623119906119906 119880119880
119877119877 = 1 minus (1198621198623119906119906 + 1198621198622119906119906 frasl) 2119880119880Φ = frasl1198621198622119909119909 119880119880 = frasl1198621198623119909119909 119880119880
= (119914119914120784120784119958119958 minus 119914119914120785120785119958119958 frasl) 2119880119880 minus 1 + (119914119914120784120784119958119958 + 119914119914120785120785119958119958 frasl) 2119880119880
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α2
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
Les repreacutesentations des triangles de vitesses adimensionnellessont tregraves riches en information et contiennent plusieurs formules
Par exemple pour un compresseur on deacuteduit aiseacutement que
Avec 120595120595 = 1198821198821198901198901198801198802
Φ = 119888119888119909119909119880119880
et que pour 119888119888119901119901 = 119888119888119888119888119888119888119888119888119888119888 119882119882119890119890 = 119888119888119901119901(11987911987902minus11987911987901)on trouve
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
(11987911987902minus11987911987901) =119880119880119888119888119909119909119888119888119901119901
(tan1205731205731 minus tan1205731205732)
Alors si lrsquoon veut obtenir une plus grande variation detempeacuterature 11987911987902 minus 11987911987901 indirectement un plus grand rapport(1199011199010211990111990101) on peut augmenter
bull la vitesse peacuteripheacuterique 119880119880 (la vitesse de rotation)bull la vitesse axiale 119888119888119909119909 (le deacutebit massique)bull lrsquoangle de deacuteflection de la pale(tan1205731205731 minus tan1205731205732)
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
Puisque le degreacute de reacuteaction R peut aussi srsquoexprimer enfonction de variables cineacutematiques on peut chercher sarepreacutesentation dans le triangle adimensionnel de vitesses
Turbine
α2
β2 α3
β3
C2 W3
C3W2
C3uC2u
U
C2xC3x
W2u
c2u ndash c3u= w2u - w3u
2Ψ minusR
frasl119934119934120784120784119958119958 119932119932 = frasl1198621198622119906119906 119880119880 minus frasl119880119880 119880119880Ψ = frasl1198621198622119906119906 minus 1198621198623119906119906 119880119880
119877119877 = 1 minus (1198621198623119906119906 + 1198621198622119906119906 frasl) 2119880119880Φ = frasl1198621198622119909119909 119880119880 = frasl1198621198623119909119909 119880119880
= (119914119914120784120784119958119958 minus 119914119914120785120785119958119958 frasl) 2119880119880 minus 1 + (119914119914120784120784119958119958 + 119914119914120785120785119958119958 frasl) 2119880119880
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α2
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
Les repreacutesentations des triangles de vitesses adimensionnellessont tregraves riches en information et contiennent plusieurs formules
Par exemple pour un compresseur on deacuteduit aiseacutement que
Avec 120595120595 = 1198821198821198901198901198801198802
Φ = 119888119888119909119909119880119880
et que pour 119888119888119901119901 = 119888119888119888119888119888119888119888119888119888119888 119882119882119890119890 = 119888119888119901119901(11987911987902minus11987911987901)on trouve
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
(11987911987902minus11987911987901) =119880119880119888119888119909119909119888119888119901119901
(tan1205731205731 minus tan1205731205732)
Alors si lrsquoon veut obtenir une plus grande variation detempeacuterature 11987911987902 minus 11987911987901 indirectement un plus grand rapport(1199011199010211990111990101) on peut augmenter
bull la vitesse peacuteripheacuterique 119880119880 (la vitesse de rotation)bull la vitesse axiale 119888119888119909119909 (le deacutebit massique)bull lrsquoangle de deacuteflection de la pale(tan1205731205731 minus tan1205731205732)
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
Turbine
α2
β2 α3
β3
C2 W3
C3W2
C3uC2u
U
C2xC3x
W2u
c2u ndash c3u= w2u - w3u
2Ψ minusR
frasl119934119934120784120784119958119958 119932119932 = frasl1198621198622119906119906 119880119880 minus frasl119880119880 119880119880Ψ = frasl1198621198622119906119906 minus 1198621198623119906119906 119880119880
119877119877 = 1 minus (1198621198623119906119906 + 1198621198622119906119906 frasl) 2119880119880Φ = frasl1198621198622119909119909 119880119880 = frasl1198621198623119909119909 119880119880
= (119914119914120784120784119958119958 minus 119914119914120785120785119958119958 frasl) 2119880119880 minus 1 + (119914119914120784120784119958119958 + 119914119914120785120785119958119958 frasl) 2119880119880
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α2
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
Les repreacutesentations des triangles de vitesses adimensionnellessont tregraves riches en information et contiennent plusieurs formules
Par exemple pour un compresseur on deacuteduit aiseacutement que
Avec 120595120595 = 1198821198821198901198901198801198802
Φ = 119888119888119909119909119880119880
et que pour 119888119888119901119901 = 119888119888119888119888119888119888119888119888119888119888 119882119882119890119890 = 119888119888119901119901(11987911987902minus11987911987901)on trouve
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
(11987911987902minus11987911987901) =119880119880119888119888119909119909119888119888119901119901
(tan1205731205731 minus tan1205731205732)
Alors si lrsquoon veut obtenir une plus grande variation detempeacuterature 11987911987902 minus 11987911987901 indirectement un plus grand rapport(1199011199010211990111990101) on peut augmenter
bull la vitesse peacuteripheacuterique 119880119880 (la vitesse de rotation)bull la vitesse axiale 119888119888119909119909 (le deacutebit massique)bull lrsquoangle de deacuteflection de la pale(tan1205731205731 minus tan1205731205732)
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α2
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
Les repreacutesentations des triangles de vitesses adimensionnellessont tregraves riches en information et contiennent plusieurs formules
Par exemple pour un compresseur on deacuteduit aiseacutement que
Avec 120595120595 = 1198821198821198901198901198801198802
Φ = 119888119888119909119909119880119880
et que pour 119888119888119901119901 = 119888119888119888119888119888119888119888119888119888119888 119882119882119890119890 = 119888119888119901119901(11987911987902minus11987911987901)on trouve
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
(11987911987902minus11987911987901) =119880119880119888119888119909119909119888119888119901119901
(tan1205731205731 minus tan1205731205732)
Alors si lrsquoon veut obtenir une plus grande variation detempeacuterature 11987911987902 minus 11987911987901 indirectement un plus grand rapport(1199011199010211990111990101) on peut augmenter
bull la vitesse peacuteripheacuterique 119880119880 (la vitesse de rotation)bull la vitesse axiale 119888119888119909119909 (le deacutebit massique)bull lrsquoangle de deacuteflection de la pale(tan1205731205731 minus tan1205731205732)
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α2
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932 120509120509 =119940119940119961119961119932119932
120511120511 =119934119934119940119940
119932119932120784120784
Les repreacutesentations des triangles de vitesses adimensionnellessont tregraves riches en information et contiennent plusieurs formules
Par exemple pour un compresseur on deacuteduit aiseacutement que
Avec 120595120595 = 1198821198821198901198901198801198802
Φ = 119888119888119909119909119880119880
et que pour 119888119888119901119901 = 119888119888119888119888119888119888119888119888119888119888 119882119882119890119890 = 119888119888119901119901(11987911987902minus11987911987901)on trouve
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
(11987911987902minus11987911987901) =119880119880119888119888119909119909119888119888119901119901
(tan1205731205731 minus tan1205731205732)
Alors si lrsquoon veut obtenir une plus grande variation detempeacuterature 11987911987902 minus 11987911987901 indirectement un plus grand rapport(1199011199010211990111990101) on peut augmenter
bull la vitesse peacuteripheacuterique 119880119880 (la vitesse de rotation)bull la vitesse axiale 119888119888119909119909 (le deacutebit massique)bull lrsquoangle de deacuteflection de la pale(tan1205731205731 minus tan1205731205732)
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
Les repreacutesentations des triangles de vitesses adimensionnellessont tregraves riches en information et contiennent plusieurs formules
Par exemple pour un compresseur on deacuteduit aiseacutement que
Avec 120595120595 = 1198821198821198901198901198801198802
Φ = 119888119888119909119909119880119880
et que pour 119888119888119901119901 = 119888119888119888119888119888119888119888119888119888119888 119882119882119890119890 = 119888119888119901119901(11987911987902minus11987911987901)on trouve
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
(11987911987902minus11987911987901) =119880119880119888119888119909119909119888119888119901119901
(tan1205731205731 minus tan1205731205732)
Alors si lrsquoon veut obtenir une plus grande variation detempeacuterature 11987911987902 minus 11987911987901 indirectement un plus grand rapport(1199011199010211990111990101) on peut augmenter
bull la vitesse peacuteripheacuterique 119880119880 (la vitesse de rotation)bull la vitesse axiale 119888119888119909119909 (le deacutebit massique)bull lrsquoangle de deacuteflection de la pale(tan1205731205731 minus tan1205731205732)
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
Alors si lrsquoon veut obtenir une plus grande variation detempeacuterature 11987911987902 minus 11987911987901 indirectement un plus grand rapport(1199011199010211990111990101) on peut augmenter
bull la vitesse peacuteripheacuterique 119880119880 (la vitesse de rotation)bull la vitesse axiale 119888119888119909119909 (le deacutebit massique)bull lrsquoangle de deacuteflection de la pale(tan1205731205731 minus tan1205731205732)
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
De maniegravere similaire agrave partir du triangle de vitesses on peuttrouver le degreacute de reacuteaction deacutecrit par
On note que la variation adimensionnelle de tempeacuterature dans ledegreacute de reacuteaction frasl119877119877 = (1198791198793 minus 1198791198792) (1198791198793minus1198791198791) est aussi une fonctiondes angles des pales
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
Crsquoest le temps de faireun peu drsquoexercise
4090761
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
Pour un compresseur axial on a les donneacutees suivantes
Neacutetages=5 U= 313 ms T01=293 K p01=01 MpaΨ =0393 R=05 119950=19kgs Rg=287 Jkg Kγ=14 rext =0339m rint =0271m ηp =09
1
bullEacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 du compresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
2
bullCalculez la vitesse axiale et la puissance transmise au fluidebullObtenez les angles α1β1et β2bullCalculez le rendement ηtt (isentropique)bullCalculez les conditions T02 et p02 agrave la sortie 2
3bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette
hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Eacutecrire lrsquoensemble drsquoeacutequations qui vous permettrait de calculer T p et ρ agrave lrsquoentreacutee 1 ducompresseur Est-il possible drsquoutiliser une forme abreacutegeacutee pour obtenir le mecircme reacutesultat
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)120646120646120783120783120587120587(03392 minus 02712
120646120646120783120783 =11990111990111198771198771198791198791
=119953119953120783120783
287 times 119931119931120783120783
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
119940119940120783120783120784120784
2 times 1004
119953119953120783120783 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 106119931119931120783120783
293
1404
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m sρ
==
==
11988811987711987711989211989211987911987901119901119901011198601198601
= 119924119924 120574120574 1 +120574120574 minus 1
2 119924119924120784120784minus 120574120574+1
)2(120574120574minus1
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
12( 1)0 2
0
112
RTm M M
P A
γγγγ+
minusminusminus = +
1
13
1
2840089711
T Kp MPa
kg mρ
==
=
rint
rext
11988811987711987711987911987901119901119901011198601198601
= 119872119872 120574120574 1 +120574120574 minus 1
2 1198721198722minus 120574120574+1
)2(120574120574minus1
119940119940120783120783119961119961 = 1207831207831207851207851207841207841207891207891199501199501199401199401198881198881119909119909 = 1198881198881
119879119879011198791198791
= 1 +120574120574 minus 1
21198721198722
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
Neacutetages=5
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
119934119934119940119940 = 1205951205951198801198802
Le travail speacutecifique
Calculez la puissance transmise au fluide
119934119934119940119940 = 0393 times 313 2
Il y a 5 eacutetages
119934119934119940119940 = 0393 times 313 2 times 120783120783 = 1925091 frasl119888119888 119888119888 2
119934 = 119950119934119934119940119940 = 120785120785120788120788120783120783120789120789119924119924119934119934
La puissance
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120543120543 = 120782120782120786120786120784120784120786120786
119940119940120783120783119961119961 = 120783120783120785120785120784120784120789120789119950119950119940119940
Les angles 120572120572112057312057311205731205732
Calculez les angles 120572120572112057312057311205731205732
120601120601 =1198881198881119909119909119880119880 =
1327313
= 0424
119877119877 = 120601120601(tan1205731205731 + tan1205731205732 frasl) 2
120595120595 = 120601120601(tan1205731205731 minus tan1205731205732)
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09Les angles 120572120572112057312057311205731205732
01 587β =
tan1205731205732 =2119877119877 minus 120595120595
2120601120601 =2 times 05 minus 0393
2 times 0424 = 0715
02 3558β =
tan1205731205731 =2119877119877 + 120595120595
2120601120601=
2 times 05 + 03932 times 0424
= 163
01 2 3558α β= =
119877119877 = 05
α2 β1
C2U
W1UC1UW2U
φR- ψ2
1
1-R-ψ2ψ
β2α1
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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2008
SFX - Humans Voices Male
17763262
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
( )( )
(04)14
(04)(14 09)
5 10875
5 1tt times
minus η = = minus
bull Supposez p02 p01=5 (pour lrsquoensemble des eacutetages) Veacuterifiez cette hypothegravese Comme premiegravere approximation consideacuterez c1x=c1ne0
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783120783120783120783120783
Les rendement total-agrave-total
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120578120578119905119905119905119905 =frasl11990111990102 11990111990101 120574120574minus1 frasl) 120574120574 minus 1
frasl11990111990102 11990111990101 (120574120574minus1 frasl) 120574120574120636120636119953119953 minus 1
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
Download Sound Effects - SoundDogs - Voice Box
httpwwwSounddogscom track 0
2008
SFX - Humans Voices Male
17763262
eng - Royalty Free Sound Effects - Sounddogscom
Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
Le valeur du rendement isentropique total-agrave-total 120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783 aeacuteteacute calculeacutee sur la base drsquoun rapport de pression hypotheacutetiquesupposeacute eacutegal agrave 5
On doit alors veacuterifier si ce reacutesultat est coheacuterent par rapport agravelrsquoensemble de donneacutees et aux calculs reacutealiseacutes
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
02 01 02 01
02 01 02 01
( )s stt p
h h T T c cteh h T T
minus minusη = = =
minus minus
( 1)
02
01
( 1)
02
01
1
1poltt
pp
pp
γminus γ
γminus γη
minus
η =
minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
120636120636119953119953 = 120782120782120783120783
119882119882119890119890 = 119888119888119901119901(11987911987902 minus 11987911987901
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574120578120578119901119901
11987911987902s11987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 120574120574
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
02 01( )e pW c T T= minus
02 01
02 01
stt
T TT T
minusη = minus
Neacutetages=5 U= 313 msT01=293 K p01=01 MpaΨ =0393 R=05119950=19kgs Rg=287 Jkg Kγ=14 rext =0339mrint =0271m ηp =09
119934119934119940119940 = 120783120783120783120783120784120784120783120783120782120782120783120783120783120783 frasl119950119950 119940119940 120784120784
120636120636119940119940119940119940 = 120782120782120790120790120789120789120783120783
119931119931120782120782120784120784119940119940 = 120786120786120788120788120782120782120783120783120784120784119922119922
1199011199010211990111990101
=11993111993112078212078212078412078411994011994011987911987901
frasl120574120574 (120574120574minus1)
=46012
293
frasl14 04 1199011199010211990111990101
= 484
1198791198790211987911987901
=1199011199010211990111990101
(120574120574minus1 frasl) 12057412057412057812057811990111990111987911987902 = 4833 K(veacuterification)
119931119931120782120782120784120784 = 120786120786120790120790120786120786120789120789119922119922
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
Lrsquohypothegravese 119940119940120783120783119961119961 = 119940119940120783120783 a permis drsquoeffectuer un premier calcul desvariables On doit cependant recalculer ces quantiteacutes en fonctiondes reacutesultats obtenus Notamment
01 2 3558α β= =
1198881198881u = 11988811988811199091199091198881198881199051199051198881198881205721205721 = 120783120783120786120786120783120783120785120785119950119950119940119940
1198881198881 = 11988811988811199061199062 + 11988811988811199091199092 = 16316 frasl119888119888 119888119888
1198881198881119909119909 = 1327119888119888119888119888
119931119931120783120783 = 11987911987901 minus11988811988812
2119888119888119901119901= 293 minus
11988811988812
2 times 1004 = 120784120784120789120789120783120783120789120789120786120786119922119922
119940119940120783120783119961119961 = 119940119940120783120783Hypothegravese
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
hellipetc
1199011199011 = 11990111990101119879119879111987911987901
120574120574120574120574minus1
= 01 times 1061198791198791
293
1404
= 008503119872119872119872119872119905119905
1205881205881 =11990111990111198771198771198791198791
=1199011199011
287 times 1198791198791= 1059 frasl119896119896119896119896 1198881198883
119940119940120783120783119961119961 =11988812058812058811198601198601
=19
)1205881205881120587120587(03392 minus 02712 = 120783120783120788120788120783120783120783120783120789120789 frasl119950119950 119940119940
120601120601 =1198881198881119909119909119880119880 =
16157313 = 0516 tan1205731205732 =
2119877119877 minus 1205951205952120601120601 1205731205732 = 30460
1
13
1
1
279740085031059 16316 x
T Kp MPa
kg mc m sρ
==
==
119920119920 119920119920119920119920
1
13
1
1
2840089711 1327 x
T Kp MPa
kg mc m s
==
==
ρ
(1205731205732= 3558 1egravere fois)
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
Download Sound Effects - SoundDogs - Voice Box
httpwwwSounddogscom track 0
2008
SFX - Humans Voices Male
17763262
eng - Royalty Free Sound Effects - Sounddogscom
Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
Pour une turbine axiale on a les donneacutees suivantesΦ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03=1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1 bullCalculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
2
bullCalculez les surfaces annulaires (normales agrave cx ) aux sections 1 2 et 3bullAgrave lrsquoaide de ces surfaces annulaires estimez la hauteur des aubes aux
sections 1 2 et 3
3
bull Consideacuterez c p=cte= 1148 Jkg Kbull Consideacuterez les variables thermodynamiques de lsquolsquolrsquoenvironnementrdquo comme
eacutetant des quantiteacutes totales ou drsquoarrecirct
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
Download Sound Effects - SoundDogs - Voice Box
httpwwwSounddogscom track 0
2008
SFX - Humans Voices Male
17763262
eng - Royalty Free Sound Effects - Sounddogscom
Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
Marcelo Reggio
Image pour fins drsquoillustration Lrsquoaubage correspond agrave celui drsquoun compresseur
❷
❸
❷❶
Stator 1-2
Rotor 2-3
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
Tenv = 1100K
Penv = 4 bar
U = 340 ms
= 20 m kg s
= 20m kg s
Chambre de combustion
❶
❷
❸
119953119953120782120782120783120783119953119953120782120782120785120785
= 120783120783120790120790120789120789120785120785
120607120607119931119931120782120782120783120783120785120785 = 120783120783120786120786120783120783119922119922
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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2008
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09 c p=cte= 1148 Jkg K
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
120653120653 = 120783120783120786120786120786120786
120595120595 =1198881198881199011199011205491205491198791198791198801198802 =
1148 times 1453402
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
Download Sound Effects - SoundDogs - Voice Box
httpwwwSounddogscom track 0
2008
SFX - Humans Voices Male
17763262
eng - Royalty Free Sound Effects - Sounddogscom
Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
α2
β2
α3
β3
1φ
ψ
C2U
C3U
W3U
ψ2 minusR ψ2 +R-1
W2U
1198881198881199051199051198881198881205731205733 =1
2120601120601 120595120595 + 2119877119877
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
R
2β
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
2α
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 bar p01p03=1873 119950=20kgs Rg=287 Jkg K α3=100 γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1198881198881199051199051198881198881205721205723 = 1198881198881199051199051198881198881205731205733 minus1120601120601
1198881198881199051199051198881198881205721205722 = 1198881198881199051199051198881198881205731205732 +1120601120601
1198881198881199051199051198881198881205731205732 =1
2120601120601 120595120595 minus 2119877119877
1198881198881199051199051198881198881205731205733 =1
2120601120601120595120595 + 2119877119877
120537120537 = 120783120783120786120786120786120786
1198881198881199051199051198881198881205731205733 = 119888119888119905119905119888119888(10) +1
08= 01763 + 125
120631120631120785120785 = 120783120783120786120786120783120783120789120789120782120782
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
Download Sound Effects - SoundDogs - Voice Box
httpwwwSounddogscom track 0
2008
SFX - Humans Voices Male
17763262
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
Hypothegravese IIle processus 1-2 estconsideacutereacute isentropique(sans pertes + adiabatique)
Hypothegravese I T02 = T01
2c
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2| c2 Τ2 p2 ρ2 et A2
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 m=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
119931119931120784120784
119953119953120784120784
1198791198792 = 11987911987902 minus11988811988822
2119888119888119901119901
11990111990101119953119953120784120784
=119879119879011198791198792
)frasl120574120574 (120574120574minus1
=119879119879021198791198792
)frasl120574120574 (120574120574minus1
1198881198882 =119888119888119909119909
119888119888119888119888119888119888 1205721205722=
1206011206011198801198801198881198881198881198881198881198881205721205722
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
Download Sound Effects - SoundDogs - Voice Box
httpwwwSounddogscom track 0
2008
SFX - Humans Voices Male
17763262
eng - Royalty Free Sound Effects - Sounddogscom
Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
22
2
pRT
ρ =
( )21 3xx x xc c c Uc= = = = φ11
1 1
xc Uccos cos
φ= = α α =α α1 3
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Calculez ψ β3 R β 2 α2 c2 Τ2 p2 ρ2 et A2 (normale agrave cx)
Eacutetage reacutepeacutetitif
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
1T1198791198791 = 11987911987901 minus11988811988812
2119888119888119901119901
119912119912120784120784 =119888
12058812058821198881198882119909119909
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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2008
SFX - Humans Voices Male
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
( 1)
01
1 1
01p Tp T
γ γminus
=
11
1
pRT
ρ =
1p
1ρ
φα2
β2α3
β3
Stator Rotor
1
φ
Ψ
C2U
C3U
W3U
ψ2 minusR
W2U
ψ2 +R-1
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
mecircme endroit
119912119912120783120783 =119888
12058812058811198881198881119909119909
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
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httpwwwSounddogscom track 0
2008
SFX - Humans Voices Male
17763262
eng - Royalty Free Sound Effects - Sounddogscom
Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
23
3 03 2 p
cT Tc
= minus
) ( 1)
03 03
3 3
p Tp T
γ γminus
=
33
3
pRT
ρ =
01
03
1873pp
=3p
013( 145 )T K∆ =03 01 013T T T= minus ∆
Φ=08 n= 250 rps Tenv=1100 K penv=4 barp01p03= 1873 119950=20kgs Rg=287 Jkg K α3=100
γ=133 U=340 ms ΔT01-03=145 K ηtt =09
Sortie 3 Calculez les surfaces annulaires aux sections 1 2 et 3 3
1198881198883 = 1198881198881
119912119912120785120785 =1198881205881205883119888119888119909119909
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
Download Sound Effects - SoundDogs - Voice Box
httpwwwSounddogscom track 0
2008
SFX - Humans Voices Male
17763262
eng - Royalty Free Sound Effects - Sounddogscom
Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
2
2
asymp π
= π
m
m
A r h
U r n
Calculez la hauteur des aubes aux sections 1 2 et 3
ℎ =119888119888119860119860119880119880
Download Sound Effects - SoundDogs - Voice Box
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Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
Dans le domaine des turbomachines on a introduit un terme quelrsquoon peut associer agrave lrsquoenthalpie totale deacutecrite en fonction de lavitesse relative Cette nouvelle variable deacutesigneacutee par rothalpiese conserve agrave travers le rotor
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
Eacutenergie Euler02 01 2 2 1 1( )u uh h c U c Uminus = minus
02 2 2 01 1 1u uh c U h c Uminus = minus
202 2 2 1 1 1 1
12u uh c U h c c Uminus = + minus
rothalpie Rthℎ0 minus 119888119888119906119906119880119880
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
2 2 21 1 12 2 2th uR h c c U U U= + minus + minus
( )2 2 21 122 2uh c c U U U= + minus + minus
2 21 12 2
h W U+ minus
20
12w
enthalpie de stagantion relative
h h W= +
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
La quantiteacute Rth se conserve Elle est deacutesigneacutee par rothalpie
La rothalpie correspond agrave la geacuteneacuteralisation du casstationnaire(119880119880 = 0119882119882 = 119862119862) pour lequel lrsquoenthalpie de stagnationest conserveacutee
119929119929119940119940119945119945 = ℎ 0minus119888119888119906119906119880119880
= ℎ +12119882119882
2 minus12119880119880
2
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
En geacuteneacuteral lrsquoenthalpie drsquoarrecirct relative dans un repegravere mobile119945119945 + frasl119934119934120784120784 120784120784 nrsquoest pas une constante Cependant pour une
machine axiale 119932119932 = cnste Dans ce cas 119945119945 + frasl119934119934120784120784 120784120784=cnste
119929119929119940119940119945119945 = ℎ +12119882119882
2 minus12119880119880
2 = 119888119888119888119888119888119888119888119888119888119888
119945119945120784120784 +120783120783120784120784119934119934120784120784
120784120784 = 119945119945120785120785 +120783120783120784120784119934119934120785120785
120784120784
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
Trois valeurs particuliers du degreacute de reacuteaction R sont agrave noter
R=0 toute variation de pression (drsquoenthalpie) a lieu dans lestator Le rotor simplement deacutevie lrsquoeacutecoulement
R=05 megravene agrave un triangle de vitesse symeacutetrique Lavariation de pression (drsquoenthalpie) est eacutequireacutepartie entre lerotor et le stator
R=10 toute variation de pression (drsquoenthalpie) a lieu dansle rotor
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
α2 β1
C2U
W1UC1U
W2UΦ
R-ψ2
1
1-R-ψ2ψ
β2 α1
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
α2
β2α3
β3
1Φ
ψ
C2U
C3U
W3U
ψ2 -R ψ2 +R-1
W2U
119929119929 =119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
119929119929 = 120783120783 minus119940119940120785120785119958119958 + 119940119940120784120784119958119958
120784120784119932119932120509120509 =
119940119940119961119961119932119932 120511120511 =
119934119934119940119940119932119932120784120784
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
La valeur du degreacute de reacuteaction 119825119825 deacutetermine des relations entreles angles drsquoentreacutee et de sortie du rotor (120514120514120515120515) ainsi que desrelations particuliegraveres entre le coefficient de charge 120511120511 et de deacutebit120509120509 afin drsquooptimiser le transfert drsquoeacutenergie entre le fluide enmouvement et le rotorDans la suite nous allons regarder trois cas speacutecifiques 119825119825 = 120782120782 119825119825 = 120783120783120784120784et 119825119825 = 120783120783 pour le cas drsquoune turbine
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=0 R=0
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
β2 β3
C2
W3 C3W2
UW2u W3u
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
α2
120569120569 =119882119882119890119890
1198801198802
(1205731205732 = 1205731205733)
119888119888119909119909tan1205731205732 = 1198881198882sin1205721205722 minus 119880119880
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880
119882119882119890119890 = 119880119880(1198881198883119906119906 minus 1198881198882119906119906)
= 2119880119880119888119888119909119909tan1205731205732119882119882119890119890 = 119880119880(119888119888119909119909tan1205731205733 + 119888119888119909119909tan1205731205732)
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
α2
Ψ =119882119882119890119890
1198801198802
)119882119882119890119890 = 2119880119880(1198881198882sin1205721205722 minus 119880119880 ψ = 21198881198882119880119880 sin1205721205722 minus 1
119932119932119952119952119953119953119940119940 =119940119940120784120784119852119852119852119852119852119852120630120630120784120784
120784120784 )120537120537 = 120784120784 (120543120543 119940119940119957119957119940119940120630120630120784120784 minus 120783120783
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 4119880119880 = 0
= 2119888119888119909119909119880119880 tan1205721205722 minus 1
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=12 R=12
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
C2 W3
C3W2
UW2u W3u
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
2 2sinoptU c= α
120569120569 =119882119882119890119890
1198801198802
119882119882119890119890 = 1205951205951198801198802 = 119880119880(1198881198882 sin1205721205722 + 1199081199083 sin1205731205733 minus 119880119880
1198881198882 sin1205721205722 = 1199081199083 sin1205731205733 (1205721205722 = 1205731205733 1198881198882 = 1199081199083)
119882119882119890119890 = 1205951205951198801198802 = 119880119880(21198881198882sin1205721205722 minus 119880119880)
part119882119882119890119890part119880119880 = 21198881198882 sin1205721205722 minus 2119880119880 = 0
120653120653 = 2119888119888119909119909119880119880 1198881198881199051199051198881198881205721205722 minus 1 = 120784120784120659120659119940119940119957119957119940119940120630120630120784120784 minus 120783120783
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
α2
β2 α3
β3
1
C2U
C3U
W3U
W2U
R=1R=1
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
α3
C2
W3C3W2
UW2u W3u
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
3 2( )e u uW U c c= minus
3 2 2 2( | |) 2 sinu uU c c Uc= + = α
2 22 22
sin2 2 2e xW c c tan tanU U U
αψ = = = α = φ α
22 tanψ = φ α
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
R=0
R=05
R=1
119929119929 =120607120607119945119945119955119955119952119952119940119940119952119952119955119955120607120607119945119945eacute119940119940119957119957119957119957119940119940
=119945119945120785120785 minus 119945119945120784120784119945119945120785120785 minus 119945119945120783120783
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
Dans une machine drsquoactionlrsquoeacutenergie du fluide transmise aurotor (en mode turbine) provientdrsquoune variation de lrsquoeacutenergiecineacutetique
Lors du passage de lrsquoeacutecoulementdans le rotor la pression(enthalpie) demeure constantePar contre il y a un changementde vitesse
rotor
eacutetage
hRh
∆=
∆
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
Dans une machine agrave reacuteaction lapression (enthalpie) et la vitessevarient dans le rotor Ceschangements provoquent unereacuteaction dans les aubes quiproduissent(turbine) une puissance
Dans une machine hypotheacutetique agravereacuteaction pure (R=1) crsquoestuniquement lrsquoeacutenergie de pressionqui change dans le rotor tandis quela variation drsquoeacutenergie cineacutetiquedemeure nulle
rotor
eacutetage
hRh
∆=
∆
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
Pour un mecircme deacutebit une variation de la charge modifie lrsquoinclinaison de aubes
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
Pour une mecircme charge une variation du deacutebit modifie lrsquoinclinaison de aubes
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
R=0
Ψ=05 Ψ=1
Φ=05 Φ=1 Φ=05 Φ=1
R=05
R=1
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
Lrsquoeacutetage reacutepeacutetitif nrsquoest qursquoune ideacutealisation En reacutealiteacute ρ varie drsquoun eacutetage agrave lrsquoautre
Pour compenser cette variation on construit des roues avec des diamegravetres diffeacuterents
Agrave venirLes machines radiales
Agrave venirLes machines radiales