In a proposed hydel project, 250 m^3/s of water is available at a head of 18 m. A Francis turbine with a specific speed of 350 was proposed. If the normal running speed is 180 rpm, find out the number of machines needed. If, instead of Francis turbine, Kaplan turbine with a specific speed of 700 is used, what is the number of machines required? Also workout the power output of each turbine. Assume that all machines in one type are of equal power and the efficiency is 90%.

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Given:Q=250 m ^3 / s ; H=18 m ; N=180 rpm ; N_s \text { (Francis) }=350 ; N_s \text { (Kaplan) }=700 ; \eta_o=0.9

Total power available =\eta_o \times \text { Water power }

\begin{aligned}&=0.9 \times w Q H \\&=0.9 \times 9810 \times 250 \times 18 \\&=39730500 W=39730.5 kW\end{aligned}

If Francis turbine is employed,

\begin{aligned}N_s &=\frac{N \sqrt{P}}{H^{5 / 4}}\\350 &=\frac{180 \times \sqrt{P}}{18^{5 / 4}}\\ or P &=5197.235 kW\end{aligned}

Each turbine develops 5197.235 kW power. So, the number of turbines required,

n=\frac{39730.5}{5197.235}=7.65

i.e., 8 turbines are required.

If Kaplan turbine is employed,

\begin{aligned}N_s &=\frac{N \sqrt{p}}{H^{5 / 4}}\\700 &=\frac{180 \times \sqrt{p}}{18^{5 / 4}}\\P &=20789 kW\end{aligned}

Each turbine develops 20789 kW power. So, the number of turbines required,

n=\frac{39730.5}{20789}=1.911

i.e., 2 turbines are required.

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