1. Design of Pelton
turbines

2. When to use a Pelton
turbine

3. Energy conversion in a
Pelton turbine
Outlet Outlet of Inlet of Outlet of Inlet of
the runner the runner the needle the needle
c2
2

4. Main dimensions for the
Pelton runner

6. The ideal Pelton runner
Absolute velocity from nozzle:
c1
c1 = 2 ⋅ g ⋅ H n c1 = =1
2 ⋅ g ⋅ Hn
Circumferential speed:
c1u 1
u1 = = ⋅ 2 ⋅ g ⋅ Hn u1 = 0.5
2 2
Euler`s turbine equation:
ηh = 2(u1 ⋅ c1u − u 2 ⋅ c 2 u )
c1u = 1 cu 2 = 0
η h = 2 ⋅ (u1 ⋅ c1u − u 2 ⋅ c 2u ) = 2 ⋅ (0,5 ⋅1.0 − 0,5 ⋅ 0) = 1

7. The real Pelton runner
• For a real Pelton runner there will always be losses
We will therefore set the hydraulic efficiency to:
ηh = 0.96
The absolute velocity from the nozzle will be:
0.99 ≤ c1u < 0.995
C1u can be set to 1,0 when dimensioning the turbine.
This gives us:
ηh = 2(u1 ⋅ c1u − u 2 ⋅ c 2 u )
⇓
ηn 0,96
u1 = = = 0,48
2 ⋅ c1u 2 ⋅1,0

8. From continuity equation:
π⋅d 2
Q = z⋅ ⋅ c1u
s
4
⇓
4⋅Q
ds =
z ⋅ π ⋅ c1u
Where:
Z = number of nozzles
Q = flow rate
C1u = 2 ⋅ g ⋅ Hn

9. The size of the bucket
and number of nozzles
B
3.1 > ≥ 3.4
ds
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

10. Number of buckets
z ≥ 17 empirical

11. Number of buckets

12. 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

13. Speed number
Ω = ω Q⋅z
c1u = 1,0
π ⋅ ds
2
π ⋅ ds
2
Q= ⋅ c1u = u1 = 0,5
4 4
ω 2 ⋅ u1 2 ⋅ g ⋅ Hn 1
ω= = = =
2 ⋅ g ⋅ Hn D ⋅ 2 ⋅ g ⋅ Hn D ⋅ 2 ⋅ g ⋅ Hn D
1 π ⋅ ds ⋅ z
2
Ω = ω⋅ Q ⋅ z = ⋅
D 4
ds π⋅z
Ω=
D 4

14. For the diameter: D = 10 · ds
and one nozzle: z=1
ds π ⋅ z 1 π ⋅1
Ω= = = 0,09
D 4 10 4
The maximum speed number for a Pelton
turbine with one nozzle is Ω = 0,09
For the diameter: D = 10 · ds
and six nozzle: z=6
ds π⋅z 1 π⋅6
Ω= = = 0,22
D 4 10 4
The maximum speed number for a Pelton
turbine today is Ω = 0,22

15. 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
4⋅Q
ds = = 0,22 m
z ⋅ π ⋅ c1u

16. 5. Choose the bucket width
B = 3,3 · ds= 0,73 m

17. 6. Find the diameter by interpolation
D
= 0,005 ⋅ H n + 8 = 13,65
ds
⇓
D = 13,65 ⋅ d s = 3,0 m
D/ds
15
10
400 1400 Hn [m]

18. 7. Calculate the speed:
D 2⋅Π ⋅n D
u1 = ω ⋅ = ⋅
2 60 2
⇓
u ⋅ 60
n= 1 = 452 rpm
Π⋅D
8. Choose the number of poles on the generator:
The speed of the runner is given by the generator and
the net frequency:
3000
n= [rpm]
Zp
where Zp=number of poles on the generator
The number of poles will be:
3000
Zp = = 6,64 = 7
n

19. 9. Recalculate the speed:
3000
n= = 428,6 [rpm]
Zp
10. Recalculate the diameter:
D 2⋅Π ⋅n D u ⋅ 60
u1 = ω ⋅ = ⋅ ⇒ D= 1 = 3,16 m
2 60 2 Π⋅n
11. Choose the number of buckets
z = 22

20. 12. Diameter of the turbine housing (for vertical turbines)
D Hou sin g = D + K ⋅ B = 9,4 m
K
9
8
1 4 6 z
13. Calculate the height from the runner to the water level
at the outlet (for vertical turbines)
Height ≈ 3.5 ⋅ B ≈ D = 3,1 m

21. Given values: Chosen values: Calculated values:
*Q = 28,5 m3/s c1u = 1 ds = 0,22 m
*H = 1130 m u1 = 0,48 n = 428,6 rpm
z=5 D = 3, 16 m
B = 0,73 m Height = 3,1 m
z = 22 Dhousing= 9,4 m
Zp = 7 *P = 288 MW

22. *Q = 28,5 m3/s
*H = 1130 m
*P = 288 MW
GE Hydro
Jostedal, Sogn og Fjordane

23. *Q = 28,5 m3/s
*H = 1130 m
*P = 288 MW
GE Hydro
Jostedal, Sogn og Fjordane

24. 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

25. Example
Khimti Power Plant
4. Calculate ds from continuity for
one nozzle
4⋅Q
ds = = 0,15 m
z ⋅ π ⋅ c1u
5. Choose the bucket width
B = 3,2 · ds= 0, 5 m

26. 6. Find the diameter by interpolation
D
= 0,005 ⋅ H n + 8 = 11,3
ds
⇓
D = 11,3 ⋅ d s = 1,7 m
D/ds
15
10
400 1400 Hn [m]

27. 7. Calculate the speed:
D 2⋅Π ⋅n D
u1 = ω ⋅ = ⋅
2 60 2
⇓
u1 ⋅ 60
n= = 613 rpm
Π⋅D
8. Choose the number of poles on the
generator:
The speed of the runner is given by
the generator and the net frequency:
3000
n= [rpm]
Zp
where Zp=number of poles on the
generator
The number of poles will be:
3000
Zp = = 4,9 = 5
n

28. 9. Recalculate the speed:
3000
n= = 600 [rpm]
Zp
10. Recalculate the diameter:
D 2⋅Π ⋅n D u1 ⋅ 60
u1 = ω ⋅ = ⋅ ⇒ D= = 1,74 m
2 60 2 Π⋅n
11. Choose the number of buckets
z = 22