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- 1. Design of Pelton turbines
- 2. When to use a Pelton turbine
- 3. Energy conversion in a Pelton turbineOutlet 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
- 5. The ideal Pelton runnerAbsolute velocity from nozzle: c1 c1 = 2 ⋅ g ⋅ H n c1 = =1 2 ⋅ g ⋅ Hn Circumferential speed: c1u 1u1 = = ⋅ 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 = 0η h = 2 ⋅ (u1 ⋅ c1u − u 2 ⋅ c 2u ) = 2 ⋅ (0,5 ⋅1.0 − 0,5 ⋅ 0) = 1
- 6. The real Pelton runner• For a real Pelton runner there will always be lossesWe will therefore set the hydraulic efficiency to: ηh = 0.96 The absolute velocity from the nozzle will be: 0.99 ≤ c1u < 0.995C1u 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
- 7. From continuity equation: π⋅d 2 Q = z⋅ ⋅ c1u s 4 ⇓ 4⋅Q ds = z ⋅ π ⋅ c1uWhere: Z = number of nozzles Q = flow rate C1u = 2 ⋅ g ⋅ Hn
- 8. The size of the bucketand 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
- 9. Number of buckets z ≥ 17 empirical
- 10. Number of buckets
- 11. Runner diameterRules of thumb: D = 10 · ds Hn < 500 m D = 15 · ds Hn = 1300 mD < 9,5 · ds must be avoided because water will be lostD > 15 · ds is for very high head Pelton
- 12. Speed number Ω = ω Q⋅z c1u = 1,0 π ⋅ ds 2 π ⋅ ds 2Q= ⋅ c1u = u1 = 0,5 4 4 ω 2 ⋅ u1 2 ⋅ g ⋅ Hn 1ω= = = = 2 ⋅ g ⋅ Hn D ⋅ 2 ⋅ g ⋅ Hn D ⋅ 2 ⋅ g ⋅ Hn D 1 π ⋅ ds ⋅ z 2Ω = ω⋅ Q ⋅ z = ⋅ D 4 ds π⋅z Ω= D 4
- 13. For the diameter: D = 10 · dsand one nozzle: z=1 ds π ⋅ z 1 π ⋅1 Ω= = = 0,09 D 4 10 4The maximum speed number for a Peltonturbine 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
- 14. Dimensioning of a Pelton turbine1. The flow rate and head are given *H = 1130 m *Q = 28,5 m3/s *P = 288 MW2. Choose reduced values c1u = 1 ⇒ c1u = 149 m/s u1 = 0,48 ⇒ u1 = 71 m/s3. Choose the number of nozzles z=54. Calculate ds from continuity for one nozzle 4⋅Q ds = = 0,22 m z ⋅ π ⋅ c1u
- 15. 5. Choose the bucket width B = 3,3 · ds= 0,73 m
- 16. 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]
- 17. 7. Calculate the speed: D 2⋅Π ⋅n D u1 = ω ⋅ = ⋅ 2 60 2 ⇓ u ⋅ 60 n= 1 = 452 rpm Π⋅D8. 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
- 18. 9. Recalculate the speed: 3000 n= = 428,6 [rpm] Zp10. Recalculate the diameter: D 2⋅Π ⋅n D u ⋅ 60 u1 = ω ⋅ = ⋅ ⇒ D= 1 = 3,16 m 2 60 2 Π⋅n11. Choose the number of buckets z = 22
- 19. 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 z13. Calculate the height from the runner to the water level at the outlet (for vertical turbines) Height ≈ 3.5 ⋅ B ≈ D = 3,1 m
- 20. 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
- 21. *Q = 28,5 m3/s *H = 1130 m *P = 288 MW GE HydroJostedal, Sogn og Fjordane
- 22. *Q = 28,5 m3/s *H = 1130 m *P = 288 MW GE HydroJostedal, Sogn og Fjordane
- 23. Example Khimti Power Plant1. The flow rate and head are given *H = 660 m *Q = 2,15 m3/s *P = 12 MW2. Choose reduced values c1u = 1 ⇒ c1u = 114 m/s u1 = 0,48 ⇒ u1 = 54,6 m/s3. Choose the number of nozzles z=1
- 24. Example Khimti Power Plant4. Calculate ds from continuity for one nozzle 4⋅Q ds = = 0,15 m z ⋅ π ⋅ c1u5. Choose the bucket width B = 3,2 · ds= 0, 5 m
- 25. 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]
- 26. 7. Calculate the speed: D 2⋅Π ⋅n D u1 = ω ⋅ = ⋅ 2 60 2 ⇓ u1 ⋅ 60 n= = 613 rpm Π⋅D8. 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
- 27. 9. Recalculate the speed: 3000 n= = 600 [rpm] Zp10. Recalculate the diameter: D 2⋅Π ⋅n D u1 ⋅ 60u1 = ω ⋅ = ⋅ ⇒ D= = 1,74 m 2 60 2 Π⋅n11. Choose the number of buckets z = 22

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what's the liquid used?