A Pelton wheel is having a mean bucket speed of 10 m/s with a jet of water flowing at the rate of 700 litre/s under a head of 31·25 m. The buckets deflect the jet through an angle of 150°. Assuming co-efficient of velocity as unity, g as 10 m/s², and the density of water as 1000 kg/m³, the approximate power given by water to the runner of the turbine is :
- A196 kWCorrect
- B98 kW
- C49 kW
- D294 kW
Solution
Concept
Power developed by a Pelton runner
Formula
P = ρ Q (V₁ − u)(1 + cos φ) u, φ = 180° − deflection angle
Step-by-step solution
1. V₁ = Cv √(2gH) = √(2 × 10 × 31.25) = 25 m/s. 2. Relative velocity at inlet = 25 − 10 = 15 m/s. Deflection 150° ⇒ φ = 30°. 3. P = 1000 × 0.7 × 15 × (1 + cos 30°) × 10 = 1000 × 0.7 × 15 × 1.866 × 10 = 1.96 × 10⁵ W = 196 kW.
Explanation
The force on the buckets equals the rate of change of momentum of the jet, which depends on the relative velocity and the deflection. Multiplying by the bucket speed gives the power transferred to the runner.
Why the other options are wrong
98 kW and 49 kW are one-half and one-quarter of the answer, and 294 kW is 1.5 times it.
Shortcut
P = ρ Q (V₁ − u)(1 + cos φ) u.
Common mistake
Using 150° itself as φ (cos 150° = −0.866 gives a very small power).
Correct answer: A) 196 kW