design of pelton turbines -...
TRANSCRIPT
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Design of Pelton
turbines
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When to use a Pelton
turbine
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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
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Main dimensions for the
Pelton runner
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The ideal Pelton runner
Absolute velocity from nozzle:
n1 Hg2c ⋅⋅= 1Hg2
cc
n
11 =
⋅⋅=
Circumferential speed:
nu1
1 Hg22
1
2
cu ⋅⋅⋅== 5.0u
1=
Euler`s turbine equation:
)cucu(2 u22u11h ⋅−⋅=η
1)05,00.15,0(2)(22211
=⋅−⋅⋅=⋅−⋅⋅=uuh
cucuη
1c u1 = 0c 2u =
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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(2u22u11h
⋅−⋅=η
⇓
48,00,12
96,0
c2u
u1
n1
=⋅
=⋅η
=
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From continuity equation:
u1
2
s c4
dzQ ⋅
⋅π⋅=
⇓
u1
scz
Q4d
⋅π⋅
⋅=
Where:
Z = number of nozzles
Q = flow rate
C1u = nHg2 ⋅⋅
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The size of the bucket
and number of nozzles
4.3d
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
17≥z empirical
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Number of buckets
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Runner diameter
Rules 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 ⋅ω=Ω
5,0u
0,1c
1
u1
=
=
4
dc
4
dQ
2
su1
2
s ⋅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
2
s ⋅⋅π⋅=⋅⋅ω=Ω
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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
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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/s
u1 = 0,48 ⇒ u1 = 71 m/s
3. Choose the number of nozzles
z = 5
4. Calculate ds from continuity for one nozzle
m22,0cz
Q4d
u1
s =⋅π⋅
⋅=
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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 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 ===
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]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
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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 6 4
m1,3DB5.3Height =≈⋅≈
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GE Hydro
Jostedal, Sogn og Fjordane
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Jostedal, Sogn og Fjordane
GE Hydro
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Example Khimti 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/s
u1 = 0,48 ⇒ u1 = 54,6 m/s
3. Choose the number of nozzles
z = 1
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Example Khimti Power Plant
4. Calculate ds from continuity for one nozzle
5. Choose the bucket width B = 3,2 · ds= 0, 5 m
mcz
Qd
u
s15,0
4
1
=⋅⋅
⋅=
π
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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 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
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][6003000
rpmZ
np
==
mn
uD
DnDu 74,1
60
260
2
2
11 =
⋅Π
⋅=⇒⋅
⋅Π⋅=⋅=ω
9. Recalculate the speed:
10. Recalculate the diameter:
11. Choose the number of buckets
z = 22