Question

A pump installed on ground takes 15 minutes to fill a water tank of 30m³ volume placed at a height of 50 m. If the efficiency of the pump is 30%, the power consumed by the pump will be ?

Answer 1

Energy needed to pump the water up to the tank is the change in the potential energy of water.

Actual Energy needed = m g h = 30m³ * 1000kg/m³ * 9.8m/sec² * 50m
                     E  = 1.47*10^7 Joules

Due to inefficiency of pump, the actual amount consumed = E / efficiency
                     = 1.47 * 10^7 J * 100/30 = 4.9 * 10^7 J

Power consumed by motor = energy utilized / time duration = 4.9*10^7 J/ 15*60 sec
                           = 54.4 kW

In case the efficiency is 33 % instead of 30% the Power consumed will be
           = 49.45 kW

Answer 2

Power consumed by the pump = P 
Power utilized by the pump = 30% of P = (3/10)P 
Volume of water raised = v = 30 m³ 
Density of water = d = 1000 kg/m³ 
Mass of water raised = m = vd = 30 * 1000 kg 
Acceleration due to gravity = g = 9.8 m/s² 
Height to which water is raised = h = 50 m 
Total work done by the pump = W = mgh = 30 * 1000 * 9.8 * 50 J 
Time taken = t = 15 min = 15 * 60 s 
Work done per second by the pump = Power utilized by the pump = W/t 
= (30 * 1000 * 9.8 * 50)/(15 * 60) Watt = 49000/3 watt 
Hence, (3/10) P = 49000/3 
=> P = 490000/9 = 54444.44 watt = 54.44 kW