Question 14.11: The Francis turbine in Fig. 14–23 is rotating at 75 rev/min ......
The Francis turbine in Fig. 14–23 is rotating at 75 rev/min under a hydraulic head of 10 m and develops 85 kW with a discharge of 0.10 m³/s. If the guide vanes remain in their fixed position, what is the rotation of this turbine when the hydraulic head is 3 m? Also, what is the corresponding discharge and the power of the turbine?
Here \omega_1 = 75 rev/ min , h_1 = 10 m, \dot W_1 = 85 kW, and Q_1 = 0.10 m³/s. For h_2 = 3 m, the rotation \omega_2 can be determined from the head coefficient similitude, Eq. 14–31,
\frac{h_{1}}{\omega_{1}^{2}D_{1}^{2}}=\frac{h_{2}}{\omega_{2}^{2}D_{2}^{2}} (14-31)
Head coefficient
Since D_1 = D_2,
\omega_{2}=\omega_{1}{\sqrt{\frac{h_{2}}{h_{1}}}}\\=(75\;\mathrm{rev/min})\sqrt{\frac{3\;\mathrm{m}}{10\;\mathrm{m}}}=41.08\;\mathrm{rev/min}=41.1\;\mathrm{rev/min}To obtain Q_2 we use the flow coefficient similitude, Eq. 14–30, with D_1 = D_2; that is,
\frac{Q_{1}}{\omega_{1}D_{1}^{3}}=\frac{Q_{2}}{\omega_{2}D_{2}^{3}} (14-30)
Flow coefficient
Finally, \dot W_2 is determined from the power coefficient similitude, Eq. 14–32, with D_1 = D_2. We have
\frac{\dot W_{1}}{\omega_{1}^{3}D_{1}^{5}}=\frac{\dot W_{2}}{\omega_{2}^{3}D_{2}^{5}} (14-32)
Power coefficient
η_1=η_2
Efficiency