Question

You are trying to build the small scale hydroelectric power plant on the hill. Potential energy is the main energy that you can use to generate the electricity. How much of the water (in kg) should be in the source to provide the 7 kW of power to the house? Assume that the efficiency of the electricity generator is 0.7 and g=9.8 m/s2 .

Answer 1

If we want an output power of 7kW that means we want the output energy to be 7000 joules or 7kJ every second.

To produce this much output power I need to give an input power higher than 7kW since the efficiency of the electric generator is not 1.

So we know that efficiency of the generator can be given by the ratio of input and output power.

So P(input) = P(output)/n

So P(input) = 7kW/0.7 = 10kW

So the input power should be 10kW or we can say that the generator must be having an input of 10000 J of energy every second so that it is producing 7000 J of energy every second.

So here the input energy is in the form of gravitational potential energy which can be given by

U = mgh

So the output power must convert 10000 J of energy every second.

Since the height of the reservoir is not given in the question let's assume it to be 'h'.

So for each second the mass of water that should fall down the hill should be such to produce an energy of 1000 J.

So mgh= 10000

That implies m = 10000/gh = 10000/9.8h = 1020.408/h

You just need to put the value of h over here to get your answer.

And the answer would be that 'm' kg of water should be there in the source every second.