SSE JE Electrical Engineering 24 March 2021 Shift 1 Paper

overall efficiency = 75%
Reservoir capacity (V) $=3.6\times 106{\mathit{m}}^3$
Water Head (h) = 100 m
Now to calculate the potential energy, we know that
Density of water $=997\mathrm{kg}∕{\mathrm{m}}^3$
$\mathit{g}=9.8\mathrm{m}∕{\mathrm{s}}^2$
Put all these in formula of potential energy, we get
E = (D × V) × 9.8 × 100
E = 997 × 3.6 × 106 × 9.8 × 100
E = 3517416 × 106 J
E $=\left(\frac{3517416}{3600}\right)\times 103\mathrm{kWh}$
E = 977060 kWh
Hence, total output energy can be calculated as
E' = 977060 × 0.75
∴ E' ≈ 732795 Wh ≈ 732.7 kWh