A station raises 2.5 × 10⁶ kg of water through 120 m in 45 min. Pump efficiency is 78%; generation efficiency is 70%. Take g = 9.8 N/kg. (a) Calculate stored GPE. (b) Calculate pump input energy and power. (c) Calculate generated electrical output. (d) Calculate round-trip efficiency. (e) Explain why the system remains useful below 100% efficiency.

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Stored 2.94 × 10⁹ J; pump input 3.77 × 10⁹ J; power 1.40 MW; output 2.06 × 10⁹ J; round-trip efficiency 54.6%. Stored gravitational energy is mgh = 2.5 × 10⁶ × 9.8 × 120 = 2.94 × 10⁹ J. Efficiency is useful output divided by input, so pump input = 2.94 × 10⁹/0.78 = 3.77 × 10⁹ J. Dividing is required because input exceeds stored energy. Forty-five minutes is 2700 s. Average input power = 3.77 × 10⁹/2700 = 1.40 × 10⁶ W = 1.40 MW. Generated output = 0.70 × 2.94 × 10⁹ = 2.06 × 10⁹ J. Round-trip efficiency = output/original input = 0.78 × 0.70 = 0.546 or 54.6%. Losses become thermal energy and sound through turbulence, friction and resistance. Storage remains valuable because surplus electricity can be shifted to high-demand periods and released quickly to stabilise supply.

Pumped-storage energy chain

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