Diapositive 1

14 Ecoulement14 Ecoulement des fluides des fluides visqueuxvisqueux

Fluides visqueux ( frottements ) au repos : Fluides visqueux ( frottements ) au repos : théorème de Bernoulli reste applicablethéorème de Bernoulli reste applicable

Fluides visqueux en mouvement Fluides visqueux en mouvement ⇒⇒ viscositviscositééCollisions intermolCollisions intermolééculairesculaires

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Diapositive 2

14.14.1 1 ViscositéViscosité

Force pour déplacer Force pour déplacer la lame supérieure à la lame supérieure à vitesse constante vitesse constante

vF Ay

η ∆=

∆

A = aire de la surfaceA = aire de la surface

∆∆v = vitesse relativev = vitesse relative

ηη = coefficient de viscosité= coefficient de viscosité

Force qui compense les forces de frottements, telle que :

resF 0 ; a 0 et v const= = =r r r

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Diapositive 3

Plus généralement :

Dimensions :

Unités : kg.m-1.s-1 ou Pa.s

LiquidesLiquides : : ηη diminuediminuequand T augmentequand T augmenteGazGaz : : ηη augmenteaugmenteavec Tavec TTribologieTribologie

[ ]2

1 11

F / A MLT /L² ML Tv / y LT /L

η−

− −−

= = = ∆ ∆

2.182.180.2840.284100100

2.092.090.3570.3578080

2.002.000.4690.4696060

1.901.900.6560.6564040

1.811.811.0051.0052020

1.711.711.7921.79200

ηηAir (10Air (10--55))((PaPa.s).s)

ηηEau (10Eau (10--33))((PaPa.s).s)

Temp (°C)Temp (°C)

η=dvF Ady

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Diapositive 4

Ecoulement laminaireEcoulement laminaire

Distribution parabolique Distribution parabolique des vitessesdes vitessesLoi de Poiseuille : débit Loi de Poiseuille : débit d’un fluide en écoulement d’un fluide en écoulement laminairelaminaire

Vrai pour de faibles vitesses

Poiseuille

1799 - 1869

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Diapositive 5

14.214.2 Ecoulement laminaire dans un Ecoulement laminaire dans un tube : analyse dimensionnelletube : analyse dimensionnelle

maxv v / 2=

maxQ Av Av / 2= =

Expérimentalement

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Diapositive 6

A = constanteA = constante

Q = constante Q = constante ⇒⇒v cons tan te=

mais mais PP varievarie le long du tube (travail nle long du tube (travail néécessaire cessaire pour vaincre les forces de viscositpour vaincre les forces de viscositéé))

Perte de charge = chute de pression : Perte de charge = chute de pression : ∆∆P = PP = P1 1 -- PP2 2

÷ au travail pour vaincre les forces de viscosité P v⇒ ∆ ∝ l

= gradient de v prP/ es on si⇒ ∝ ∆ l

Non visqueux visqueux

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Diapositive 7

Autres facteurs dont dépendAutres facteurs dont dépend

Section du tubeSection du tubeViscosité du liqideViscosité du liqide

v

a b v ( P/ )R β η⇒ = ∆ l

β, a et b = nombres sans dimension

Equation aux dimensions : 2

2 -1 PR v = 1/ 8( P/ )R8

= ηη

∆⇒ ∆ l

l

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Diapositive 8

Loi de PoiseuilleLoi de Poiseuille

2PR =8

vη

∆l

4Q P R =

8πη

∆l

R x 1.19

Q x (1.19)4 = Q x 2

X (π R2)

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Diapositive 9

Dissipation d’énergie mécaniqueDissipation d’énergie mécanique

Puissance dissipée par Puissance dissipée par les les forcesforces de frottement de frottement visqueuxvisqueuxF = (PF = (P11 –– PP22) A = ) A = ∆∆P AP A

Cas particulier :Cas particulier :A = A = ππ RR22

P = Fv = P Av = PQ ∆ ∆

2P = P R vπ∆

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Diapositive 10

14.3 Ecoulement turbulent14.3 Ecoulement turbulent

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Diapositive 11

Dissipation dDissipation d ’énergie mécanique plus ’énergie mécanique plus importante que dans le cas d’un écoulement importante que dans le cas d’un écoulement laminairelaminaireLoi de Poiseuille non valableLoi de Poiseuille non valableTraité au moyen de REGLES empiriquesTraité au moyen de REGLES empiriquesExemple:Exemple:

C’est la valeur du nombre de Reynolds qui détermine C’est la valeur du nombre de Reynolds qui détermine si l’écoulement st turbulent ou laminairesi l’écoulement st turbulent ou laminaire

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Diapositive 12

Nombre de Reynolds pour un Nombre de Reynolds pour un fluide de viscosité fluide de viscosité ηη, de , de masse volumique masse volumique ρρ qui s’écoule dans un qui s’écoule dans un TUBETUBE de de rayon Rrayon Ravec une avec une vitesse moyennevitesse moyenne

RN v2 Rη

ρ=

v

NR < 2000 : écoulement laminaire

NR > 3000 : écoulement turbulent

2000 < NR < 3000 : écoulement instable

Ces résultats ne sont valables que pour des tubes. Pour l’écoulement autour d’une aile d’avion ou de la coque d’un bateau, on définit aussi un nombre de Reynolds, mais donné par une autre expression

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Diapositive 13

Exemple : ArtèreExemple : Artère

R = 4 mmR = 4 mmVitesse moyenne : 1,99 cm/sVitesse moyenne : 1,99 cm/sη η = 2,084 10= 2,084 10--33 Pa sPa sρρ = 1,0595 10³ kg/m³= 1,0595 10³ kg/m³

NNRR = 80,9 (écoulement laminaire)= 80,9 (écoulement laminaire)

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Diapositive 14

14.4 Ecoulement du sang dans 14.4 Ecoulement du sang dans le système circulatoirele système circulatoire

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Diapositive 15

Résistance à l’écoulementRésistance à l’écoulement

Aussi appelée résistance vasculaire (qcq)Aussi appelée résistance vasculaire (qcq)

Pa.s.mPa.s.m--33

Si régime laminaireSi régime laminairef

PRQ∆

=

4Q P R =

8πη

∆l f 4

8 R =Rη

π⇒

l

Rf (aorte) = 37,2 kPa.s.m-3 si ∆P = 0,00372 kPa

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Diapositive 16

Résistance vasculaire d’un système Résistance vasculaire d’un système d’artèresd’artères

Rf2

Rf1

Rf3

Débit Q2

Débit Q1

Débit Q3

∆P = différence de pression entre les extrémités des artères du même type

∆P = Rf1 Q1 = Rf2 Q2 = Rf3 Q3

Q = Q1 + Q2 + Q3 = ∆P/ Rf1 + ∆P/ Rf2 + ∆P/ Rf3 = ∆P/ Req

1

1/ Req = 1/ Rf1 + 1/ Rf2 + 1/ Rf3

Résistance équivalente Req :

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Diapositive 17

2

∆P = ∆P 1 + ∆P 2 + ∆P 3

∆P = Q Rf1 + Q Rf2 + Q Rf3 = Q Req

Résistance équivalente Req :

Req = Rf1 + Rf2 + Rf3

Rf1 Rf2 Rf3

Débit Q

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Diapositive 18

Req = 45,3 x 10 5 kPa.s.ms.m--33 Req = 2.22 x 10 5 kPa.s.ms.m--33

4.0 x 10²4.0 x 10²1.55 x101.55 x1011

5.42 x 105.42 x 1011

1.401.402.67 x 102.67 x 10--11

2.35 x 102.35 x 10--22

0.93 x 100.93 x 10--22

0.960.961.461.462.11 x 102.11 x 108.68 x 108.68 x 1022

1.26 x 101.26 x 1044

1.70 x 101.70 x 1044

6.67 x 10³6.67 x 10³

RRf1 f1 (kPa s m(kPa s m--33))

4.0 x 10²4.0 x 10²11Veine mésentériqueVeine mésentérique9.33 x10²9.33 x10²6060Veines secondairesVeines secondaires1.03 x 101.03 x 10551 9001 900Veines tertiairesVeines tertiaires2.53 x 102.53 x 104418 00018 000Veines terminalesVeines terminales4.27 x 104.27 x 1044160 000160 000Rameaux terminauxRameaux terminaux4.93 x 104.93 x 10442 100 0002 100 000VeinulesVeinules4.38 x 104.38 x 105547 300 00047 300 000CapillairesCapillaires1.01 x 101.01 x 10661 050 0001 050 000ArtériolesArtérioles4.79 x 104.79 x 1055328 500328 500Rameaux terminauxRameaux terminaux5.61 x 105.61 x 105526 60026 600Artères terminalesArtères terminales1.65 x 101.65 x 10661 9001 900Rameaux tertiairesRameaux tertiaires

5.69 x 105.69 x 10554545Rameaux secondairesRameaux secondaires2.55 x 102.55 x 10551515Rameaux principauxRameaux principaux

6.67 x 10³6.67 x 10³11Artère mésentériqueArtère mésentérique

Résistance Résistance équivalente équivalente RRf1f1/N (kPa s m/N (kPa s m--33))

Nombre, NNombre, NStructureStructureLit vasculaire mésentérique

d’un petit chien

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Diapositive 19

Req = 61.0 x 10 5 kPa.s.ms.m--33

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Diapositive 20

14.5 Forces de résistance 14.5 Forces de résistance visqueusevisqueuse

Dans le vide, les deux corps Dans le vide, les deux corps tombent avec la même vitesse et tombent avec la même vitesse et atteignent le sol en même tempsatteignent le sol en même tempsDans un fluide, ce n’est pas le Dans un fluide, ce n’est pas le cas!cas!

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Diapositive 21

ModèleModèle

W = poids = 4/3 W = poids = 4/3 ππ R³ R³ ρ ρ ggB = poussée d’Archimède = 4/3 B = poussée d’Archimède = 4/3 ππ R³ R³ ρρ00 ggFFRR = Force de résistance visqueuse = 6 = Force de résistance visqueuse = 6 ππ R v R v η η (Loi de Stokes pour une sphère)(Loi de Stokes pour une sphère)

Faible vitesse

Vitesse limite ?

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Diapositive 22

vitesse limite atteinte quand vitesse limite atteinte quand W = FW = FRR + B+ B c’estc’est--àà--dire quand :dire quand :

4/3 4/3 ππ R³ R³ ρ ρ g = 6 g = 6 ππ R v R v η + η + 4/3 4/3 ππ R³ R³ ρρ00 ggouou

6 6 ππ v R v R η = η = 4/3 4/3 ππ R³ (R³ (ρ ρ −− ρρ0 0 ) ) g g

v = 2/9 (v = 2/9 (R²R²/ / η) η) ((ρ ρ −− ρρ0 0 ) ) gg

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