5 - Pelton Turbine

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    Design of Peltonturbines

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    When to use a Peltonturbine

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    Energy conversion in aPelton turbine

    Outlet Outlet of

    the runner

    Inlet of

    the runner

    Outlet of

    the needle

    Inlet of

    the needle

    2

    2c

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    Main dimensions for thePelton runner

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    The ideal Pelton runnerAbsolute velocity from nozzle:

    n1 Hg2c = 1Hg2

    cc

    n

    11 =

    =

    Circumferential speed:

    nu1

    1 Hg2

    2

    1

    2

    cu == 5.0u1 =

    Euler`s turbine equation:

    )cucu(2 u22u11h =

    1)05,00.15,0(2)(2 2211 === uuh cucu

    1c u1 = 0c 2u =

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    The real Pelton runner For a real Pelton runner there will always be losWe will therefore set the hydraulic efficiency to:

    96.0h =

    The absolute velocity from the nozzle will be:

    995.0c99.0 u1

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    From continuity equation:

    u1

    2

    s c4

    dzQ

    =

    u1

    scz

    Q4d

    =

    Where:

    Z = number of nozzles

    Q = flow rate

    C1u = nHg2

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    The size of the bucket

    and number of nozzles

    4.3

    d

    B1.3

    s

    >

    Rules of thumb:

    B = 3,1 ds 1 nozzle

    B = 3,2 ds 2 nozzles

    B = 3,3 ds

    4-5 nozzles

    B > 3,3 ds 6 nozzles

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    Number of buckets

    17z empirical

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    Number of buckets

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    Runner diameterRules of thumb:

    D = 10 ds

    Hn

    < 500 m

    D = 15 ds Hn = 1300 m

    D < 9,5 ds must be avoided because water

    will be lost

    D > 15 ds is for very high head Pelton

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    Speed number

    zQ =

    5,0u

    0,1c

    1

    u1

    =

    =

    4

    dc

    4

    dQ

    2

    su1

    2

    s =

    =

    D

    1

    Hg2D

    Hg2

    Hg2D

    u2

    Hg2 n

    n

    n

    1

    n=

    =

    =

    =

    4

    z

    D

    ds =

    4

    zd

    D

    1zQ

    2

    s ==

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    For the diameter: D = 10 dsand one nozzle: z = 1

    09,0

    4

    1

    10

    1

    4

    z

    D

    ds =

    =

    =

    For the diameter: D = 10 dsand six nozzle: z = 6

    22,04

    6

    10

    1

    4

    z

    D

    ds =

    =

    =

    The maximum speed number for a Pelton

    turbine today is = 0,22

    The maximum speed number for a Pelton

    turbine with one nozzle is = 0,09

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    Dimensioning of a

    Pelton turbine

    1. The flow rate and head are given*H = 1130 m

    *Q = 28,5 m3/s

    *P = 288 MW

    2. Choose reduced values

    c1u = 1 c1u = 149 m/su1 = 0,48 u1 = 71 m/s

    3. Choose the number of nozzlesz = 5

    4. Calculate ds from continuity for one nozzle

    m22,0

    cz

    Q4d

    u1

    s =

    =

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    5. Choose the bucket width

    B = 3,3 ds= 0,73 m

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    6. Find the diameter by interpolation

    D/ds

    Hn [m]

    10

    15

    400 1400

    m0,3d65,13D

    65,138H005,0d

    D

    s

    n

    s

    ==

    =+=

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    7. Calculate the speed:

    8. Choose the number of poles on the generator:

    The speed of the runner is given by the generator andthe net frequency:

    where Zp=number of poles on the generator

    The number of poles will be:

    rpm452D

    60un

    2

    D

    60

    n2

    2

    Du

    1

    1

    ==

    ==

    ]rpm[Z

    3000n

    p

    =

    764,6n

    3000Zp ===

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    ]rpm[6,428Z

    3000n

    p

    ==

    m16,3n

    60uD

    2

    D

    60

    n2

    2

    Du 11 =

    =

    ==

    9. Recalculate the speed:

    10. Recalculate the diameter:

    11. Choose the number of buckets

    z = 22

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    12. Diameter of the turbine housing (for vertical turbines)

    13. Calculate the height from the runner to the water levelat the outlet (for vertical turbines)

    m4,9BKDD gsinHou =+=

    K

    z

    8

    9

    1 64

    m1,3DB5.3Height =

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    Given values:

    *Q =28,5 m3/s

    *H =1130m

    Chosen values:

    c1u = 1

    u1 = 0,48

    z = 5

    B= 0,73 m

    z = 22

    Zp = 7

    Calculated values:

    ds = 0,22 m

    n =428,6 rpm

    D=3, 16 m

    Height= 3,1 m

    Dhousing= 9,4 m

    *P= 288MW

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    GE Hydro

    *Q=28,5 m3/s

    *H = 1130m

    *P =288MW

    Jostedal, Sogn og Fjordane

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    Jostedal, Sogn og Fjordane

    GE Hydro

    *Q=28,5 m3/s

    *H =1130m

    *P =288MW

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    ExampleKhimti Power Plant

    1. The flow rate and head are given*H = 660 m

    *Q = 2,15 m3/s

    *P = 12 MW

    2. Choose reduced values

    c1u = 1 c1u = 114 m/su1 = 0,48 u1 = 54,6 m/s

    3. Choose the number of nozzles

    z = 1

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    ExampleKhimti Power Plant

    4. Calculate ds from continuity forone nozzle

    5. Choose the bucket widthB = 3,2 ds= 0, 5 m

    mcz

    Qd

    u

    s 15,04

    1

    =

    =

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    6. Find the diameter by interpolation

    D/ds

    Hn [m]

    10

    15

    400 1400

    mdD

    Hd

    D

    s

    n

    s

    7,13,11

    3,118005,0

    ==

    =+=

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    7. Calculate the speed:

    8. Choose the number of poles on thegenerator:

    The speed of the runner is given by

    the generator and the net frequency:

    where Zp=number of poles on the

    generator

    The number of poles will be:

    rpmD

    un

    DnDu

    61360

    260

    2

    2

    1

    1

    =

    =

    ==

    ]rpm[Z

    3000n

    p

    =

    59,43000

    === nZp

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    ][6003000

    rpmZ

    np

    ==

    mn

    uD

    DnDu 74,1

    60

    260

    2

    2

    11 =

    =

    ==

    9. Recalculate the speed:

    10. Recalculate the diameter:

    11. Choose the number of buckets

    z = 22