A pumped hydro system has a storage reservoir of 300 m^2 in area and an effectiv

Question

A pumped hydro system has a storage reservoir of 300 m^2 in area and an effective head of 200 m. (a) What would be the average depth of the pumped storage reservoir to store 6000 MWh of hydro energy? (Density of water is 1000 kg/m^3). (b) How much power could be generated from this stored energy during high demand assuming the turbo-generator efficiency is 88%? (c) How much coal would be required in a coal-burning power plant to generate the equivalent amount of energy? The energy content of coal is 8, 142 kWh/tonne (1 tonne = 2, 206 pounds). The energy efficiency of a typical coal-burning power plant is about 38%

Explanation / Answer

a.      Let the depth of the reservoir be d meters

The potential energy stored in the water = mgh

m = mass = density * volume = density * area * depth = 1000*300*d

g = acceleration due to gravity = 10m/s*s

Therefore the potential energy stored in the water = 1000*300*d*10*200

But the reservoir has to store energy = 6000MWh ( given ) = 6000MW*3600sec

Equating both, we get

1000*300*d*10*200 = 6000 x10e6*3600

d = 36x10e3 m

b.      Since the efficiency of the generator is 88%, the power that would be generated=

0.88*total power = 0.88*6000MW = 5280MW

c.      Let the amount of coal that would be required be x tonnes. The energy generated by x tonnes of coal would be = x*8142*0.38 kWh ( since efficiency is 38% )

Therefore x*8142*0.38*10e3 = 6000*10e6

x=1939.26 tonnes = 4,278,007.56 pounds ( since 1 tonne = 2206 pounds )

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