An electric pump is used to fill an overhead tank situated at a height of 15 m . The volume of water pumped varies linearly with time, such that the graph of volume in litres ( y -axis) versus time in seconds ( x -axis) is a straight line passing through the origin and the point (50 s , 400 L ) . If the overall efficiency of the pump is 80 % , what is the electrical power consumed by the pump? (Take g = 10 m s ⁻² and

Step-by-step solution

Slope of the volume-time graph gives the volume flow rate: dV dt = 400 - 0 50 - 0 = 8 L s ⁻¹ = 8 10⁻³ m ^3 s ⁻¹ Mass flow rate of water, dm dt = dV dt = 8 10⁻³ 10^3 = 8 kg s ⁻¹ Useful mechanical power delivered to the water: P_ out = ( dm dt )gh = 8 10 15 = 1200 W = 1.2 kW Since the efficiency of the pump is 80 % , the electrical power consumed (input power) is: P_ in = P_ out 0.80 = 1.2 0.80 = 1.5 kW Answer: 1.50 kW