Question 3.23: A hydroelectric power plant (Fig. E3.23) takes in 30 m3 /s o...

We neglect viscous work and heat transfer and take section 1 at the reservoir surface (Fig. E3.23), where V_{1} \approx 0, p_{1} = p_{atm} , and z_{1} = 100 m. Section 2 is at the turbine outlet. The steady flow energy equation (3.75) becomes, in head form
\frac{p_{1} }{\gamma }+\frac{\alpha_{1}V^{2}_{1}}{2g}+z_{1}= \frac{p_{2} }{\gamma }+ \frac{\alpha_{2}V^{2}_{2}}{2g}+z_{2}+h_{t}+h_{f}

\frac{p_{a} }{\gamma }+\frac{1.06(0)^2}{2(9.81)}+100 m=\frac{p_{a}}{\gamma}+\frac{1.06(2.0 m/s)^2}{2(9.81 m/s^2)}+0m+h_{t}+20m
The pressure terms cancel, and we may solve for the turbine head (which is positive):
h_{t}=100-20-0.2\approx 79.8m
The turbine extracts about 79.8 percent of the 100-m head available from the dam. The total power extracted may be evaluated from the water mass flow:
P=\dot{m}w_{s}=(\rho Q)(gh_{t})=(998 kg/m^3)(30 m^3/s)(9.81 m/s^2)(79.8 m)

=23.4 E6 kg\cdot m^2/s^3= 23.4 E6 N\cdot m/s= 23.4 MW
The turbine drives an electric generator that probably has losses of about 15 percent, so the net power generated by this hydroelectric plant is about 20 MW