Question

An electric motor pump kept on the ground floor of AVRS campus can pump up water to fill a tank of volume 50 m3in 25 minutes. If the tank is 27 m above sea level, and the efficiency of the pump is 33%, calculate the electric power consumed by the pump.

Answer 1

Solution:

Density of water is = 1000 kg/m^3

Volume of water filled = 50 m^3

Therefore, mass of water filled in the tank = (1000 kg/m^3)(50m^3) = 50000 kg

Height to which the water is raised = 27 m

Acceleration due to gravity = 9.8 m/s^2

So, rise in potential energy of the water = (50000 kg)(9.8 m/s^2)(27 m) = 1.32 x 10^7 J

The work is done by the pump in 25 min = 1500 s

Power of the pump = (1.32 X 10^7 J)/(1500 s) = 8800 Watt

Let, x be the electrical power given to the pump, since the efficiency of the pump is 33%

Therefore,

x(33/100) = 8800

=> x = 26666.66 Watt

So, the power consumed by the pump is 26666.66 Watt.