Pelton turbine (1)

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

2

2c

Main dimensions for the Pelton runner

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

Number of buckets

17≥z empirical

Number of buckets

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⋅⋅=

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

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

GE Hydro

*Q = 28,5 m3/s*H = 1130 m*P = 288 MW

Jostedal, Sogn og Fjordane

Jostedal, Sogn og Fjordane

GE Hydro

*Q = 28,5 m3/s*H = 1130 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

][6003000

rpmZ

np

==

mn

uD

DnDu 74,1

60

260

2

21

1 =⋅Π⋅=⇒⋅⋅Π⋅=⋅= ω

9. Recalculate the speed:

10. Recalculate the diameter:

11. Choose the number of buckets

z = 22