Question 12.21: When a test was conducted on the model of Francis turbine, i......
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%.
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