Answer
$P_{e}=48.21MW$
Work Step by Step
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 obtained is:
$P_{p}=1000\frac{kg}{m^3}*65\frac{m^3}{s}*9.81\frac{m}{s^2}*90m=57.39MW$
Taking in count the efficiency, the electric power generated is:
$P_{e}=\eta P_{p}=0.84*57.39MW=48.21MW$
