## Q. 12.5

A Pelton turbine having four number of jets with jet diameter 200 mm each develops 44 MW power at 400 rpm consuming 11.5 $m^3/s$ of water. The effective head of water available at the nozzle is 455 m. The mean diameter of the runner is 2 m. The mechanical efficiency of the turbine is 97%. Determine the coefficient of velocity of the nozzle and the blade friction coefficient if the outlet angle of the blade is 15°. Also find the hydraulic power loss in the nozzle and the buckets.

## Verified Solution

Given: $n=4 ; d=0.2 m ; S P=44 \times 10^6 W ; N=400 rpm ; Q=11.5 m ^3 / s ; H=455 m ; D=2 m ; \eta_{\text {mech }}=0.97 ; \phi=15^{\circ}$

Using the definition of mechanical efficiency,

Power developed by runner $=\frac{S P}{\eta_{\text {mech }}}=\frac{44 \times 10^6}{0.97}=45.36 \times 10^6 W$

Flow rate per jet $q=\frac{Q}{4}=\frac{11.5}{4}=2.875 m ^3 / s$

Velocity of the jet  $V_1=\frac{4 q}{\pi d^2}=\frac{4 \times 2.875}{\pi \times 0.2^2}=91.514 m / s$

Peripheral speed of runner  $u=\frac{\pi D N}{60}=\frac{\pi \times 2 \times 400}{60}=41.89 m / s$

Also, from the equation for power developed per jet, we get

\begin{aligned}P &=\rho Q\left(V_1-u\right)(1+K \cos \phi) u \\11.34 \times 10^6 &=1000 \times 2.875 \times(91.54-41.89)\left(1+K \cos 15^{\circ}\right) \times 41.89\end{aligned}

Solving, we get, blade velocity coefficient, $K=0.928$

Relative velocity at inlet  $V_{r 1}=\left(V_1-u\right)=91.514-41.89=49.624 m / s$

Coefficient of velocity  $C_ν=\frac{V_1}{\sqrt{2 g H}}=\frac{91.54}{\sqrt{2 \times 9.81 \times 455}}=0.9886$

\begin{aligned}&=H-C_ν^2 H=\left(1-C_ν^2\right) H \\&=\left(1-0.9886^2\right) \times 455=10.32 m\end{aligned}
\begin{aligned}&=\frac{V_{r 1}^2-V_{r 2}^2}{2 g}=\frac{V_{r 1}^2-K V_{r 1}^2}{2 g}\\&=\frac{V_{r 1}^2\left(1-K^2\right)}{2 g}=\frac{49.624^2\left(1-0.928^2\right)}{2 \times 9.81}=17.42 m\end{aligned}
Power loss in the nozzle $=w Q h_n=9810 \times 11 \times 10.32=1164251 W =1164.25 kW$
Power loss in the buckets  $=9810 \times 11.5 \times 17.42=1965237 W =1965.237 kW$.