A Pelton wheel is revolving at a speed of 200 r.p.m. and develops 5886 kW S.P. when working under a head of 200 m with an overall efficiency of 80%. Determine unit speed, unit discharge and unit power. The speed ratio for the turbine is given as 0.48. Find the speed, discharge and power when this turbine is working under a head of 150 m.

Pelton Wheel Unit Quantities and Performance

Problem Statement

Given Data for Initial Conditions (H = 200 m)

Speed, $N_1 = 200 \, \text{r.p.m.}$
Shaft Power, $P_1 = 5886 \, \text{kW}$
Head, $H_1 = 200 \, \text{m}$
Overall efficiency, $\eta_o = 80\% = 0.80$
New Head, $H_2 = 150 \, \text{m}$
Density of water, $\rho = 1000 \, \text{kg/m}^3$
Acceleration due to gravity, $g = 9.81 \, \text{m/s}^2$

Solution

Part 1: Determine Unit Quantities

Unit Speed ($N_u$):

$$ N_u = \frac{N_1}{\sqrt{H_1}} = \frac{200}{\sqrt{200}} \approx 14.14 \, \text{r.p.m.} $$

Unit Power ($P_u$):

$$ P_u = \frac{P_1}{H_1^{3/2}} = \frac{5886}{(200)^{1.5}} = \frac{5886}{2828.4} \approx 2.081 \, \text{kW} $$

Unit Discharge ($Q_u$): First, we must find the actual discharge $Q_1$.

$$ \text{Water Power, } P_w = \frac{\text{Shaft Power}}{\eta_o} = \frac{5886000 \, \text{W}}{0.80} = 7357500 \, \text{W} $$ $$ Q_1 = \frac{P_w}{\rho g H_1} = \frac{7357500}{1000 \times 9.81 \times 200} \approx 3.75 \, \text{m}^3/\text{s} $$ $$ Q_u = \frac{Q_1}{\sqrt{H_1}} = \frac{3.75}{\sqrt{200}} \approx 0.265 \, \text{m}^3/\text{s} $$

Part 2: Performance at New Head (H = 150 m)

We use the calculated unit quantities to find the performance at the new head.

New Speed ($N_2$):

$$ N_2 = N_u \times \sqrt{H_2} = 14.14 \times \sqrt{150} \approx 173.2 \, \text{r.p.m.} $$

New Discharge ($Q_2$):

$$ Q_2 = Q_u \times \sqrt{H_2} = 0.265 \times \sqrt{150} \approx 3.246 \, \text{m}^3/\text{s} $$

New Power ($P_2$):

$$ P_2 = P_u \times H_2^{3/2} = 2.081 \times (150)^{1.5} = 2.081 \times 1837.1 \approx 3823 \, \text{kW} $$

Final Results:

Unit Quantities: Unit Speed $ \approx 14.14 $ rpm, Unit Discharge $ \approx 0.265 $ m³/s, Unit Power $ \approx 2.081 $ kW

Performance at 150 m Head: Speed $ \approx 173.2 $ rpm, Discharge $ \approx 3.25 $ m³/s, Power $ \approx 3823 $ kW

Explanation of Unit Quantities

Unit Quantities are performance parameters of a turbine scaled to what they would be if the turbine were operating under a standard head of 1 meter. They are a powerful tool for comparing different turbines and for predicting the performance of a single turbine under varying head conditions, as done in this problem.

Unit Speed ($N_u$): The speed of the turbine under a 1 m head.
Unit Discharge ($Q_u$): The discharge of the turbine under a 1 m head.
Unit Power ($P_u$): The power developed by the turbine under a 1 m head.

By first calculating these base "unit" values from a known operating point, we can then easily scale them to find the speed, discharge, and power at any other head, assuming the efficiency remains constant.