Chapter 12

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


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


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