One method used to store energy during times of low demand is by pumping water u

Question

One method used to store energy during times of low demand is by pumping water uphill into a reservoir. When energy demand exceeds the output produced by power plants, the water is released, moves downhill, and is converted into electricity by a turbine. During a 208.0 minute long period of low usage, 69.0 MW is used to pump water up an elevation of 59.3 m in a series of 1.00 m diameter pipes. The friction loss in the pipe going to the upper reservoir is 31.6 m^2/s^2 What is the mass flow rate, m, of water during the low usage period? What is the total mass of water raised 59.3 m during this 208.0 minute period? How long can the turbines produce 245.0 MW of power from the release of the mass of stored water calculated above? Assume the water sustains a friction loss of 31.6 m^2/s^2 on the way down to the turbine. What is the efficiency of this process? (Assume that the pump and turbines operate at 100% efficiency, and the only loses are due to frictional loses within the pipes.) Efficiency is defined as the total energy recovered/total energy input.

Explanation / Answer

energy = 69 * 106 * 208 * 60 = 8.61 * 1011 J

b) Total mass of water M = E / (gH + v2) = (8.61 * 1011) / (9.81*59.3 + 31.6)

= 1.40 * 109 kg

a) Mass flow rate = M / t = (1.40 * 109) / (208 / 60) = 1.12 * 105 kg/s

c) Energy = m g H - m v2 = (1.40 * 109 * 9.81 * 59.3) - (1.40 * 109 * 31.6)

= 7.70 * 1011 J

P = energy / time = 7.70 * 1011 / 245 * 106 = 3143.6 s

power = 52.4 min

d) efficiency = 7.70 * 1011 / 8.61 * 1011

= 89.4%

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