The following data is related to the Pelton wheel : Head at the base of the nozzle = 110 m, Diameter of the jet = 7.5 cm, Discharge of the nozzle = 200 litres/s, Shaft power = 191.295 kW, Power absorbed in mechanical resistance = 3.675 kW. Determine : (i) Power lost in nozzle and, (ii) Power lost due to hydraulic resistance in the runner.

Pelton Wheel Power Loss Analysis

Problem Statement

Given Data & Constants

Head at nozzle base, $H = 110 \, \text{m}$
Jet diameter, $d = 7.5 \, \text{cm} = 0.075 \, \text{m}$
Discharge, $Q = 200 \, \text{L/s} = 0.2 \, \text{m}^3/\text{s}$
Shaft Power, $P_s = 191.295 \, \text{kW} = 191295 \, \text{W}$
Mechanical Losses, $P_m = 3.675 \, \text{kW} = 3675 \, \text{W}$
Density of water, $\rho = 1000 \, \text{kg/m}^3$
Acceleration due to gravity, $g = 9.81 \, \text{m/s}^2$

Solution

1. Calculate Power Input to Nozzle (Water Power)

This is the total potential energy of the water available at the base of the nozzle.

$$ \text{Water Power, } P_w = \rho g Q H $$ $$ P_w = 1000 \times 9.81 \times 0.2 \times 110 = 215820 \, \text{W} = 215.82 \, \text{kW} $$

2. Calculate Power at Nozzle Outlet (Jet Power)

This is the kinetic energy of the water jet. First, we find the velocity of the jet.

$$ \text{Area of jet, } a = \frac{\pi}{4} d^2 = \frac{\pi}{4} (0.075)^2 \approx 0.004418 \, \text{m}^2 $$ $$ \text{Velocity of jet, } V = \frac{Q}{a} = \frac{0.2}{0.004418} \approx 45.27 \, \text{m/s} $$ $$ \text{Jet Power, } P_j = \frac{1}{2} (\rho Q) V^2 = \frac{1}{2} (1000 \times 0.2) \times (45.27)^2 $$ $$ P_j = 100 \times 2049.37 \approx 204937 \, \text{W} = 204.94 \, \text{kW} $$

(i) Power Lost in Nozzle

This is the difference between the power supplied to the nozzle and the kinetic power of the jet it produces.

$$ P_{loss, nozzle} = P_w - P_j $$ $$ P_{loss, nozzle} = 215820 - 204937 = 10883 \, \text{W} \approx 10.88 \, \text{kW} $$

3. Calculate Power Delivered to Runner (Runner Power)

The runner power is the useful shaft power plus the power that was lost to mechanical friction.

$$ \text{Runner Power, } P_r = \text{Shaft Power} + \text{Mechanical Losses} $$ $$ P_r = 191295 + 3675 = 194970 \, \text{W} = 194.97 \, \text{kW} $$

(ii) Power Lost due to Hydraulic Resistance in Runner

This is the difference between the power available in the jet and the power that is successfully transferred to the runner.

$$ P_{loss, runner} = P_j - P_r $$ $$ P_{loss, runner} = 204937 - 194970 = 9967 \, \text{W} \approx 9.97 \, \text{kW} $$

Final Results:

(i) Power lost in nozzle: $ \approx 10.88 \, \text{kW} $

(ii) Power lost due to hydraulic resistance in the runner: $ \approx 9.97 \, \text{kW} $

Explanation of Power Flow and Losses

The energy in a Pelton wheel system flows through several stages, with losses occurring at each step:

Water Power: The initial potential energy available from the head of water.
Nozzle Loss: As water flows through the nozzle, friction converts some energy into heat, reducing the final kinetic energy of the jet.
Jet Power: The kinetic energy of the water jet after it leaves the nozzle. This is the power available to do work on the runner.
Runner Hydraulic Loss: As the jet strikes the buckets, not all of its kinetic energy can be converted. Some is lost to fluid friction on the bucket surface and splash, which is the hydraulic resistance.
Runner Power: The actual mechanical power transferred from the water to the spinning runner.
Mechanical Loss: Friction in the shaft bearings and seals consumes a small amount of the runner power.
Shaft Power: The final, useful power available at the output shaft of the turbine.