## Q. 12.19

A Pelton wheel is designed to operate under a head of 600 m and develops 6 MW while running at 200 rpm. It is desired to test the turbine at a site where the maximum supply head is 100 m. Find out the speed, discharge and power output at the test condition. Assume an overall efficiency of 85% at the best operating point. Also find the unit quantities.

## Verified Solution

$H_1=600 m ; P_1=6 \times 10^6 W ; N_1=200 rpm ; H_2=120 m ; \eta_o=0.85$

From the equation for efficiency,

$\eta_o=\frac{S P}{W P}=\frac{S P}{w Q H}=\frac{6 \times 10^6}{9810 \times Q \times 600}$

Discharge  $Q_1=\frac{6 \times 10^6}{9810 \times 600 \times 0.85}=1.2 m ^3 / s$

Since the same machine has to operate under the new head, unit quantities for the actual and test conditions are to be equated.

Unit speed          $N_u=\frac{N_1}{\sqrt{H_1}}=\frac{200}{\sqrt{600}}=8.165$

Unit discharge  $Q_u=\frac{Q_1}{\sqrt{H_1}}=\frac{1.2}{\sqrt{600}}=0.049$

Unit power    $P_u=\frac{P_1}{H_1^{3 / 2}}=\frac{6 \times 10^3}{600^{3 / 2}}=0.408$

Equating the unit quantities for the actual (suffix 1) and test (suffix 2) conditions, we get

$\frac{N_1}{\sqrt{H_1}}=\frac{N_2}{\sqrt{H_2}}$

Speed during test $N_2=N_1 \times\left(\frac{H_2}{H_1}\right)^{1 / 2}=200 \times\left(\frac{100}{600}\right)^{1 / 2}=81.65 rpm$

$\frac{Q_1}{\sqrt{H_1}}=\frac{Q_2}{\sqrt{H_2}}$

Discharge during test  $Q_2=Q_1 \times\left(\frac{H_2}{H_1}\right)^{1 / 2}=1.2\left(\frac{100}{600}\right)^{1 / 2}=0.49 m ^3 / s$

$\frac{P_1}{H_1^{3 / 2}}=\frac{P_2}{H_2^{3 / 2}}$

Power developed  $P_2=P_1 \times\left(\frac{H_2}{H_1}\right)^{3 / 2}=6 \times 10^3 \times\left(\frac{100}{600}\right)^{3 / 2}=408 kW$