Two jets strike at buckets of a Pelton wheel, which is having shaft power as 14,715 kW. The diameter of each jet is given as 150 mm. If the net head on the turbine is 500 m, find the overall efficiency of the turbine. Take C_v = 1.0.

Leave a Comment / Fluid Mechanics, Numerical (Water) / By
Two-Jet Pelton Wheel Efficiency Calculation

Problem Statement

Two jets strike at buckets of a Pelton wheel, which is having shaft power as 14,715 kW. The diameter of each jet is given as 150 mm. If the net head on the turbine is 500 m, find the overall efficiency of the turbine. Take C_v = 1.0.

Given Data & Constants

  • Number of jets, \(n = 2\)
  • Shaft Power, \(P_s = 14,715 \, \text{kW} = 14,715,000 \, \text{W}\)
  • Diameter of each jet, \(d = 150 \, \text{mm} = 0.15 \, \text{m}\)
  • Net Head, \(H = 500 \, \text{m}\)
  • Co-efficient of velocity, \(C_v = 1.0\)
  • Density of water, \(\rho = 1000 \, \text{kg/m}^3\)
  • Acceleration due to gravity, \(g = 9.81 \, \text{m/s}^2\)

Solution

1. Calculate the Velocity of the Jets (\(V_1\))

The velocity of the water jet is determined by the head. Since \(C_v = 1.0\), there are no nozzle losses considered.

$$ V_1 = C_v \sqrt{2gH} = 1.0 \times \sqrt{2 \times 9.81 \times 500} $$ $$ V_1 = \sqrt{9810} \approx 99.045 \, \text{m/s} $$

2. Calculate the Total Discharge (Q)

First, find the area of a single jet, then the discharge per jet, and finally the total discharge for two jets.

$$ \text{Area of one jet, } a = \frac{\pi}{4} d^2 = \frac{\pi}{4} (0.15)^2 \approx 0.01767 \, \text{m}^2 $$ $$ \text{Discharge of one jet, } q = a \times V_1 = 0.01767 \times 99.045 \approx 1.75 \, \text{m}^3/\text{s} $$ $$ \text{Total Discharge, } Q = n \times q = 2 \times 1.75 = 3.5 \, \text{m}^3/\text{s} $$

3. Calculate the Power Input from Water (Water Power)

This is the total potential energy of the water entering the turbine per second.

$$ \text{Water Power, } P_w = \rho g Q H $$ $$ P_w = 1000 \times 9.81 \times 3.5 \times 500 $$ $$ P_w = 17167500 \, \text{W} = 17167.5 \, \text{kW} $$

4. Find the Overall Efficiency (\(\eta_o\))

The overall efficiency is the ratio of the useful shaft power output to the total water power input.

$$ \eta_o = \frac{\text{Shaft Power}}{\text{Water Power}} = \frac{P_s}{P_w} $$ $$ \eta_o = \frac{14715000}{17167500} \approx 0.8571 $$
Final Result:

The overall efficiency of the turbine is approximately \(85.7\%\).

Explanation of Overall Efficiency

The overall efficiency (\(\eta_o\)) is the most important real-world measure of a turbine's performance. It tells us how much of the potential energy of the water at the source is converted into useful mechanical work at the output shaft. It accounts for all losses in the system:

  • Hydraulic Losses: Energy lost due to friction in the nozzle, and fluid friction and splash as the water interacts with the buckets.
  • Mechanical Losses: Energy lost to friction in the bearings and seals of the rotating shaft.
  • Windage Losses: Energy lost due to the air resistance (drag) on the spinning buckets.

In this problem, the Water Power (17,167.5 kW) is the total energy available in the water jets, while the Shaft Power (14,715 kW) is what's left after all these losses have been accounted for. The ratio of these two gives the overall efficiency.

Related Posts

Leave a Comment

Your email address will not be published. Required fields are marked *

Scroll to Top