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Energy conversion in a Pelton turbine
Outlet Outlet of the runner
Inlet of the runner
Outlet of the needle
Inlet of the needle
2
2c
The ideal Pelton runnerAbsolute velocity from nozzle:
n1 Hg2c ⋅⋅= 1Hg2
ccn
11 =
⋅⋅=
Circumferential speed:
nu1
1 Hg221
2cu ⋅⋅⋅== 5.0u1 =
Euler`s turbine equation:
)cucu(2 u22u11h ⋅−⋅=η
1)05,00.15,0(2)(2 2211 =⋅−⋅⋅=⋅−⋅⋅= uuh cucuη
1c u1 = 0c 2u =
The real Pelton runner• For a real Pelton runner there will always be lossWe will therefore set the hydraulic efficiency to:
96.0h =η
The absolute velocity from the nozzle will be:
995.0c99.0 u1 <≤
C1u can be set to 1,0 when dimensioning the turbine.This gives us:
)cucu(2 u22u11h ⋅−⋅=η
⇓
48,00,12
96,0c2
uu1
n1 =
⋅=
⋅η
=
From continuity equation:
u1
2s c
4dzQ ⋅⋅π
⋅=
⇓
u1s cz
Q4d⋅π⋅⋅
=
Where:Z = number of nozzlesQ = flow rateC1u = nHg2 ⋅⋅
The size of the bucket and number of nozzles
4.3dB1.3
s
≥>
Rules of thumb:B = 3,1 · ds 1 nozzleB = 3,2 · ds 2 nozzlesB = 3,3 · ds 4-5 nozzlesB > 3,3 · ds 6 nozzles
Runner diameterRules of thumb:
D = 10 · ds Hn < 500 mD = 15 · ds Hn = 1300 m
D < 9,5 · ds must be avoided because waterwill be lost
D > 15 · ds is for very high head Pelton
Speed number
zQ ⋅ω=Ω
5,0u0,1c
1
u1
==
4dc
4dQ
2s
u1
2s ⋅π
=⋅⋅π
=
D1
Hg2DHg2
Hg2Du2
Hg2 n
n
n
1
n
=⋅⋅⋅
⋅⋅=
⋅⋅⋅⋅
=⋅⋅
ω=ω
4z
Dds ⋅π
=Ω
4zd
D1zQ
2s ⋅⋅π
⋅=⋅⋅ω=Ω
For the diameter: D = 10 · dsand one nozzle: z = 1
09,041
101
4z
Dds =
⋅π=
⋅π=Ω
For the diameter: D = 10 · dsand six nozzle: z = 6
22,04
6101
4z
Dds =
⋅π=
⋅π=Ω
The maximum speed number for a Pelton turbine today is Ω = 0,22
The maximum speed number for a Pelton turbine with one nozzle is Ω = 0,09
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 valuesc1u = 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
m22,0cz
Q4du1
s =⋅π⋅⋅
=
6. Find the diameter by interpolation
D/ds
Hn [m]
10
15
400 1400
m0,3d65,13D
65,138H005,0dD
s
ns
=⋅=⇓
=+⋅=
7. Calculate the speed:
8. Choose the number of poles on the generator:
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:
rpm452D60un
2D
60n2
2Du
1
1
=⋅Π⋅
=
⇓
⋅⋅Π⋅
=⋅ω=
]rpm[Z
3000np
=
764,6n
3000Zp ===
]rpm[6,428Z
3000np
==
m16,3n60uD
2D
60n2
2Du 1
1 =⋅Π⋅
=⇒⋅⋅Π⋅
=⋅ω=
9. Recalculate the speed:
10. Recalculate the diameter:
11. Choose the number of buckets
z = 22
12. Diameter of the turbine housing (for vertical turbines)
13. Calculate the height from the runner to the water level at the outlet (for vertical turbines)
m4,9BKDD gsinHou =⋅+=
K
z
8
9
1 64
m1,3DB5.3Height =≈⋅≈
Given values:
*Q = 28,5 m3/s*H = 1130 m
Chosen values:
c1u = 1u1 = 0,48z = 5B = 0,73 mz = 22Zp = 7
Calculated values:
ds = 0,22 mn = 428,6 rpmD = 3, 16 mHeight = 3,1 mDhousing= 9,4 m*P = 288 MW
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 valuesc1u = 1 ⇒ c1u = 114 m/su1 = 0,48 ⇒ u1 = 54,6 m/s
3. Choose the number of nozzlesz = 1
ExampleKhimti Power Plant
4. Calculate ds from continuity for one nozzle
5. Choose the bucket widthB = 3,2 · ds= 0, 5 m
mcz
Qdu
s 15,04
1
=⋅⋅⋅
=π
6. Find the diameter by interpolation
D/ds
Hn [m]
10
15
400 1400
mdD
HdD
s
ns
7,13,11
3,118005,0
=⋅=⇓
=+⋅=
7. Calculate the speed:
8. Choose the number of poles on the generator: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
2602
2
1
1
=⋅Π⋅
=
⇓
⋅⋅Π⋅
=⋅=ω
]rpm[Z
3000np
=
59,43000===
nZ p