Pelton turbine (1)
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Transcript of Pelton turbine (1)
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
cc
n
11 =
⋅⋅=
Circumferential speed:
nu1
1 Hg22
1
2
cu ⋅⋅⋅== 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 losses.We 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
96,0
c2u
u1
n1 =
⋅=
⋅η=
From continuity equation:
u1
2s c
4
dzQ ⋅⋅π⋅=
⇓
u1s cz
Q4d
⋅π⋅⋅=
Where:Z = number of nozzlesQ = flow rateC1u = nHg2 ⋅⋅
The size of the bucket and number of nozzles
4.3d
B1.3
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 diameter Rules 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
5,0u
0,1c
1
u1
==
4
dc
4
dQ
2s
u1
2s ⋅π=⋅⋅π=
D
1
Hg2D
Hg2
Hg2D
u2
Hg2 n
n
n
1
n
=⋅⋅⋅
⋅⋅=
⋅⋅⋅⋅=
⋅⋅ω=ω
4
z
D
ds ⋅π=Ω
4
zd
D
1zQ
2s ⋅⋅π⋅=⋅⋅ω=Ω
For the diameter: D = 10 · ds and one nozzle: z = 1
09,04
1
10
1
4
z
D
ds =⋅π=⋅π=Ω
For the diameter: D = 10 · ds and six nozzle: z = 6
22,04
6
10
1
4
z
D
ds =⋅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/s
u1 = 0,48 ⇒ u1 = 71 m/s
3. Choose the number of nozzlesz = 5
4. Calculate ds from continuity for one nozzle
m22,0cz
Q4d
u1s =
⋅π⋅⋅=
6. Find the diameter by interpolation
D/ds
Hn [m]
10
15
400 1400
m0,3d65,13D
65,138H005,0d
D
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:
rpm452D
60un
2
D
60
n2
2
Du
1
1
=⋅Π⋅=
⇓
⋅⋅Π⋅=⋅ω=
]rpm[Z
3000n
p
=
764,6n
3000Zp ===
]rpm[6,428Z
3000n
p
==
m16,3n
60uD
2
D
60
n2
2
Du 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 = 0,48
z = 5B = 0,73 mz = 22Zp = 7
Calculated values:
ds = 0,22 m
n = 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/s
u1 = 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 width B = 3,2 · ds= 0, 5 m
mcz
Qd
us 15,0
4
1
=⋅⋅
⋅=π
6. Find the diameter by interpolation
D/ds
Hn [m]
10
15
400 1400
mdD
Hd
D
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
260
2
2
1
1
=⋅Π⋅=
⇓
⋅⋅Π⋅=⋅= ω
]rpm[Z
3000n
p
=
59,43000 ===n
Z p