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 165.0 minute long period of low usage, 71.0 MW is used to pump water up an elevation of 74.1 m in a series of 1.00 m diameter pipes. The friction loss in the pipe going to the upper reservoir is 39.5 m2/s2. a) What is the mass flow rate, M* , of water during the low usage period? b) What is the total mass of water raised 74.1 m during this 165.0 minute period? c) How long can the turbines produce 380.0 MW of power from the release of the mass of stored water calculated above? Assume the water sustains a friction loss of 39.5 m2/s2 on the way down to the turbine. d) What is the efficiency of this process? (Assume that the pump and tubines operate at 100% efficiency, and the only losses are due to frictional losses within the pipes.)

Explanation / Answer

see here:- 71*10*6=DM/DT(g*h+39.5) dm/dt=92.72*10^3kg/sec total mass=165*60*dm/dt total mass=9.18*10^8Kg 380*10*6=dm/dt(74.1*9.8-39.5) 5.53*10^5kg/sec=dm/dt total mass/5.53*10^5=1658.8sec=27.64min Efficiency=(74.1*9.8-39.5)/(74.1*9.8+39.5)=0.89

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