An electric motor operates a pump with an efficiency of 60 % . The pump lifts 3600 L of water to a height of 15 m in 2 minutes. What is the input power consumed by the electric motor? (Take g = 10 m/s ^2 and density of water = 1 kg/L)

Step-by-step solution

Given: Volume of water, V = 3600 L Mass of water, m = 3600 kg (since density is 1 kg/L) Height, h = 15 m Time, t = 2 minutes = 120 s Efficiency, = 60 % = 0.6 The useful work done by the pump (output work) is: W = mgh = 3600 10 15 = 540000 J The output power of the pump is: P_ out = W t = 540000 120 = 4500 W = 4.5 kW Efficiency is defined as the ratio of output power to input power: = P_ out P_ in Therefore, the input power is: P_ in = P_ out = 4.5 0.6 = 7.5 kW. Answer: 7.5 kW