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In a Pelton wheel, the bucket peripheral speed is 10 m/s, the water jet velocity is 25 m/s and volumetric flow rate of the jet is 0.1 m3/s. If the jet deflection angle is 120° and the flow is ideal, the power developed is
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- 7.5 kW
- 15.0 kW
- 22.5 kW
- 37.5 kW
- 7.5 kW
Correct Option: C
u = 10 m/s
Vi = 25 m/s
Q = 0.1 m3 /s
Jet deflection angle = 120°
∴ β = 180° – 120° = 60°
P = ρQ (Vi – u)(1 + cosβ)u
P = 1000 × 0.1(25 – 10)(1 + cos60°) × 10 P = 22500 W = 22.5 kW