## Q. 12.21

When a test was conducted on the model of Francis turbine, it developed 5 kW at 280 rpm under a head of 2 m. The scale ratio of the model was 1:4. Determine the speed and power of full size turbine operating under a head of 6 m, if the efficiency of the full size turbine and the model turbine are the same. Also, find the above parameters if scale effect is considered. Take the efficiency of the model as 72%.

## Verified Solution

Given: $P_m=5 kW ; N_m=280 rpm ; H_m=2 m ; H_p=6 m ; \eta_m=0.72 \text {; Scale ratio }$$D_{m m} / D_p=1 / 4$

If the efficiency is same, then for geometrically similar turbines, the head coefficient and power coefficient are the same. From the equation for head coefficient $C_H$,

$\left(\frac{H}{N^2 D^2}\right)_m=\left(\frac{H}{N^2 D^2}\right)_p$

Speed of prototype  $N_p=N_m \times\left(\frac{D_m}{D_p}\right)\left(\frac{H_p}{H_m}\right)^{1 / 2}=280 \times\left(\frac{1}{4}\right)\left(\frac{6}{2}\right)^{1 / 2}=121.24 rpm$

From the equation for power coefficient $C_p$,

$\left(\frac{P}{N^3 D^5}\right)_m=\left(\frac{P}{N^3 D^5}\right)_p$

Power delivery by prototype

\begin{aligned}P_p &=P_m\left(\frac{N_p}{N_m}\right)^2\left(\frac{D_p}{D_m}\right)^5 \\&=5 \times\left(\frac{121.24}{280}\right)^3(4)^5=415.66 kW\end{aligned}

If the efficiency of model turbine is 0.72, then from Moody’s equation,

\begin{aligned}& \frac{1-\eta_p}{1-\eta_m}=\left(\frac{D_m}{D_p}\right)^{0.2}\\& \frac{1-\eta_p}{1-0.72}=\left(\frac{1}{4}\right)^{0.2}\end{aligned}

Efficiency of prototype = 0.788 = 78.8%

From the equation for head coefficient,

$\left(\frac{\eta H}{N^2 D^2}\right)_m=\left(\frac{\eta H}{N^2 D^2}\right)_p$

$N_p=N_m\left(\frac{D_m}{D_p}\right)\left(\frac{H_p}{H_m}\right)^{1 / 2}\left(\frac{\eta_p}{\eta_m}\right)^{1 / 2}=121.24 \times\left(\frac{0.788}{0.72}\right)^{1 / 2}=126.82 rpm$

From the equation for power coefficient,

$P_p=P_m\left(\frac{N_p}{N_m}\right)^3\left(\frac{D_p}{D_m}\right)^5=5 \times\left(\frac{126.82}{280}\right)^2(4)^5=475.74 kW$