Question 14.10: A turbine for the dam operates under a hydraulic head of 90 ......
A turbine for the dam operates under a hydraulic head of 90 m, producing a discharge of 50 m³/s. If the reservoir level drops so that the hydraulic head becomes 60 m, determine the discharge from the turbine.
Here h_1 = 90 m and Q_1 = 50 m³/s. Because we also know h_2 = 60 m, then we will eliminate the unknowns \omega_1 and \omega_2 from the flow and head coefficients to obtain a relationship between h and Q. Using Eqs. 14–30 and 14–31, we have
\frac{Q_1}{ω_1D_1^3}=\frac{Q_2}{ω_2D_2^3} (14-30)
Flow coefficient
\frac{h_1}{ω_1^2D_1^2}=\frac{h_2}{ω_2^2D_2^2} (14-31)
Head coefficient
\frac{\omega_{1}}{\omega_{2}}=\frac{Q_{1}D_{2}^{3}}{Q_{2}D_{1}^{3}} (1)
\frac{\omega_{1}^2}{\omega_{2}^2}=\frac{h_{1}D_{2}^{2}}{h_{2}D_{1}^{2}}so that
\frac{Q_{1}^{2}D_{2}^{4}}{Q_{2}^{2}D_{1}^{4}}=\frac{h_{1}}{h_{2}} (2)
Since D_1 = D_2,
Q_{2}=Q_{1}{\sqrt{\frac{h_{2}}{h_{1}}}}\\=(50\,\mathrm{m^{3}/s})\sqrt{\frac{60\,\mathrm{m}}{90\,\mathrm{m}}}=40.8\,\mathrm{m^{3}/s}