How much electric power could be obtained from a small hydroelectric station if the elevation change was 150 meters and if 10, 000 kg of water passed through the turbines every second, with the overall efficiency being 85 %?
support your answer with equation.
2007-11-07 03:47:35 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Physics
Power P is the rate of energy flow so
P=Pe/t where potential energy Pe gets converted to kinetic energy Ke at a given rate.
P=mgh/t= 10,000 x 9.81x 150/1
P= 14,715,000 W however it is at 100% efficiency so actually
P=0.85 Pmax = .85 x 14,715,000 =
P=12,507,750 W or
P= 12.5 MegaW (3 significant figures)
2007-11-07 03:51:50 · answer #1 · answered by Edward 7 · 0⤊ 0⤋
E(85%) = 0.85mgh = 10000 * 150 * 9.8 * 0.85 = 12.5 Mj
A joule = 2.8E−7 kWh so the above is 3.5 kWh/sec or 12.6 MW
2007-11-07 03:57:28 · answer #2 · answered by gebobs 6 · 0⤊ 0⤋