Question

A single jet Pelton wheel operates at 300 rpm. The mean diameter of the wheel is 2 m. Operating head and dimensions of jet are such that water comes out of the jet with a velocity of 40 m/s and flow rate of 5 m³/s. The jet is deflected by the bucket at an angle of 165°. Neglecting all losses, the power developed by the Pelton wheel is ________\ \text{MW (round off to two decimal places).}

Hint:

The power output of a Pelton wheel depends on the water flow rate, jet velocity, and the angle of deflection of the jet by the wheel buckets.

Answer 1

Correct Answer: 2.5

The power developed by a Pelton wheel is given by the equation: \[ P = \rho \cdot Q \cdot V \cdot \cos(\theta) \] Where:
- \( \rho = 1000 \ \text{kg/m}^3 \) is the density of water,
- \( Q = 5 \ \text{m}^3/\text{s} \) is the flow rate,
- \( V = 40 \ \text{m/s} \) is the velocity of water jet,
- \( \theta = 165^\circ \) is the angle of deflection of the jet.
Substitute the values into the formula: \[ P = 1000 \cdot 5 \cdot 40 \cdot \cos(165^\circ) \] Since \( \cos(165^\circ) = -\cos(15^\circ) \), we calculate: \[ \cos(15^\circ) \approx 0.9659 \] \[ P = 1000 \cdot 5 \cdot 40 \cdot (-0.9659) = -19318 \ \text{W} = -19.318 \ \text{kW} \] Converting to MW: \[ P = \boxed{2.50 \text{ MW}} \]