Question

A Pelton wheel is running at 2000 r.p.m. and it develops 68.67 kW shaft power. The discharge through a nozzle is 0.02 m³/s and the net head on the Pelton wheel is 500 m. The overall efficiency of the Pelton wheel is
1. 0.45
2. 0.5
3. 0.7
4. 0.9

Answer 1

Correct Answer - Option 3 : 0.7

Concept:

The overall efficiency of the Pelton wheel is given by,

\({η _o} = \frac{{Shaft \;Power}}{{Water\; Power}}\)

Water Power = ρ × g × Q × H 

Calculation:

Given:

Shaft Power = 68.67 kW, Q = 0.02 m³/sec, H = 500 m

Water Power = ρ × g × Q × H = 1000 × 9.81 × 0.02 × 500 = 98.1 kW 

\({η _o} = \frac{{Shaft \;Power}}{{Water\; Power}}= \frac{68.67}{98.1}= 0.7\)

The overall efficiency of the Pelton wheel is 0.7.