Question

In a dam generating hydro electric power, water falls at a rate of 1000 kg s-1 from a height of 100 m.
(a) Calculate the initial potential energy of the water.
(b) Assuming that 60% of the energy of the falling water is converted to electrical energy, calculate the power generated (g=9.8 m s-2)

Can u show how to do it?

Answer 1

Answer:

given,

mass of water =1000kg  

height =100m

gravity =9.8m s-2

thus ,

initial potential energy of water =>

U =mgh

U =1000×9.8×100

U =980000J

Answer 2

Explanation:

It is given that,

In a dam generating hydroelectric power, water falls at a rate of 1000 kg s-1 from a height of 100 m.

(a) The initial potential energy of the water is equal to the energy when it was at ground level. It is given by :

$P=mgh\\\\P=1000\times 9.8\times 100\\\\P=980000\ J$

(b) If 60% of the energy of the falling water is converted to electrical energy, then the power generated is given by work done divided by time i.e.

$P=60\%\ \text{of}\ P\\\\P=\dfrac{60}{100} \times 980000\\\\P=588000\ W\\\\P=588\ W$

Hence, this is the required solution.