Design of Pelton

turbines

When to use a Pelton

turbine

Energy conversion in a

Pelton turbine

Outlet Outlet of

the runner

Inlet of

the runner

Outlet of

the needle

Inlet of

the needle

2c

2

Main dimensions for the

Pelton runner

The ideal Pelton runner

Absolute velocity from nozzle:

n1 Hg2c ⋅⋅= 1Hg2

cc

n

11 =

⋅⋅=

Circumferential speed:

nu1

1 Hg22

1

2

cu ⋅⋅⋅== 5.0u =n1 Hg2

22u ⋅⋅⋅== 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 <≤ 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

2

s c4

dzQ ⋅

⋅π⋅=

⇓

u1

scz

Q4d

⋅π⋅

⋅=

Where:Where:

Z = number of nozzles

Q = flow rate

C1u = nHg2 ⋅⋅

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 · d 4-5 nozzlesB = 3,3 · ds 4-5 nozzles

B > 3,3 · ds 6 nozzles

Number of buckets

17≥z empirical

Number of buckets

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 · d is for very high head PeltonD > 15 · ds is for very high head Pelton

Speed number

zQ ⋅ω=Ω

5,0u

0,1c

1

u1

=

=

4

dc

4

dQ

2

su1

2

s ⋅π=⋅

⋅π=

1Hg2u2 ⋅⋅⋅ω

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

For the diameter: D = 10 · ds

and one nozzle: z = 1

09,04

1

10

1

4

z

D

ds =⋅π

=⋅π

=Ω

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

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

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⋅

⋅=

5. Choose the bucket width

B = 3,3 · ds= 0,73 m

6. Find the diameter by interpolation

D/ds

15

m0,3d65,13D

65,138H005,0d

D

s

n

s

=⋅=

⇓

=+⋅=

Hn [m]

10

400 1400

7. Calculate the speed:

8. Choose the number of poles on the generator:

rpm452D

60un

2

D

60

n2

2

Du

1

1

=⋅Π

⋅=

⇓

⋅⋅Π⋅

=⋅ω=

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:

]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)

m4,9BKDD gsinHou =⋅+=

K

z

8

9

1 64

13. Calculate the height from the runner to the water level

at the outlet (for vertical turbines)

m1,3DB5.3Height =≈⋅≈

GE Hydro

Jostedal, Sogn og Fjordane

Jostedal, Sogn og Fjordane

GE Hydro

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

c = 1 ⇒ c = 114 m/sc1u = 1 ⇒ c1u = 114 m/s

u1 = 0,48 ⇒ u1 = 54,6 m/s

3. Choose the number of nozzles

z = 1

ExampleKhimti Power Plant

4. Calculate ds from continuity for one nozzle

5. Choose the bucket width

mcz

Qd

u

s 15,04

1

=⋅⋅

⋅=

π

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

6. Find the diameter by interpolation

D/ds

15

mdD

Hd

D

s

n

s

7,13,11

3,118005,0

=⋅=

⇓

=+⋅=

Hn [m]

10

400 1400

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:

rpmD

un

DnDu

61360

260

2

2

1

1

=⋅Π

⋅=

⇓

⋅⋅Π⋅

=⋅= ω

the generator and the net frequency:

where Zp=number of poles on the

generator

The number of poles will be:

]rpm[Z

3000n

p

=

59,43000

===n

Z p

][6003000

rpmZ

np

==

mu

DDnD

u 74,1602 1 =

⋅=⇒⋅

⋅Π⋅=⋅= ω

9. Recalculate the speed:

10. Recalculate the diameter:

mn

uD

DnDu 74,1

60

260

2

2

11 =

⋅Π

⋅=⇒⋅

⋅Π⋅=⋅= ω

11. Choose the number of buckets

z = 22