Question 12.19: A Pelton wheel is designed to operate under a head of 600 m ......

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