Hydroelectric power plant

In equilibrium, the outflow must correspond to the inlet, so that we can caculate the flow velocity $v$ in the pipe: $A_0 v = \dot V \quad \Rightarrow \quad v = \frac{\dot V}{A_0} = 6 \,\frac{\text{m}}{\text{s}}$ On the other hand, energy conservation must be assured, so that the potential energy of the water volume $\Delta V$ with mass $\Delta m = \rho \Delta V$ is converted into kinetic and electric energy: $\begin{aligned} & & U &= T + W_\text{el} \\ \Rightarrow & & \Delta m g h &= \frac{1}{2} \Delta m v^2 + \eta \Delta m g h \\ \Rightarrow & & h &= \frac{v^2}{2(1-\eta)} = 90\,\text{m} \end{aligned}$ With the height $h$ of the reservoir, we can estimate the electrical power of the plant to $P = \eta \rho \dot V g h = 0.8 \cdot 1000 \cdot 150 \cdot 10 \cdot 90 \,\text{W} = 108 \,\text{MW}$