Chapter 2 - Energy, Energy Transfer, and General Energy Analysis - Problems - Page 107: 2-130

b) $38MW$

Here we are working with potential energy: The potential work is: $W_{p}=mgh$ And the potential power is: $P_{p}=\frac{W_{p}}{t}=\frac{mgh}{t}=\rho Qgh$ The potential power is: $P_{p}=1000\frac{kg}{m^3}*70\frac{m^3}{s}*9.81\frac{m}{s^2}*65m=44.64MW$ Taking in count the efficiency, the electric power to be generated is: $P_{e}=\eta P_{p}=0.85*44.64MW=37.94MW$