Answer
a) ΔEp = mgΔh (1) = 2.0 x 105 kg x 10 N kg-1 x 215 m (1) = 4.30 x 108 J (1)
b) Ek = 1/2mv2 = 4.30 x 108 J (1) v = √(2 x 4.30 x 108 J / 2.0 x 105 kg) (1) = 65.6 m s-1 (1)
c) kinetic energy of water leaving turbines each second = 1/2mv2 = 1/2 x 2.0 x 105 kg x (8 m s-1)2 = 6.4 x 106 J (1). maximum power input = 4.30 x 108 W - 6.4 x 106 W = 4.24 x 108 W (1)
d) Efficiency = useful power out / power in = 2.5 x 108 W / 4.24 x 108 W (1) = 0.59 (1)
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