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 206.0 minute long period of low usage, 65.0 MW is used to pump water up an elevation of 68.7 m in a series of 1.00 m diameter pipes. The friction loss in the pipe going to the upper reservoir is 36.6 m2/s2.

What is the mass flow rate, , of water during the low usage period? (in kg/s)

What is the total mass of water raised 68.7 m during this 206.0 minute period? (in kg)

How long can the turbines produce 335.0 MW of power from the release of the mass of stored water calculated above? Assume the water sustains a friction loss of 36.6 m2/s2 on the way down to the turbine. (in min)

What is the efficiency of this process? (Assume that the pump and tubines operate at 100% efficiency, and the only loses are due to frictional loses within the pipes.)

Explanation / Answer

71*10*6=DM/DT(g*h+39.5)dm/dt=92.72*10^3kg/sectotal mass=165*60*dm/dttotal mass=9.18*10^8Kg380*10*6=dm/dt(74.1*9.8-39.5)5.53*10^5kg/sec=dm/dttotal mass/5.53*10^5=1658.8sec=27.64minEfficiency=(74.1*9.8-39.5)/(74.1*9.8+39.5)=0.89

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