A water pump raises 600\text{ kg} of water through a vertical height of 20\text{ m} in 50\… — JAMB UTME

A water pump raises $600\text{ kg}$ of water through a vertical height of $20\text{ m}$ in $50\text{ s}$ . If the efficiency of the pump is $80\%$ , what is the electrical power input required to operate the pump? (Take $g = 10\text{ m s}^{-2}$ )

$3.0\text{ kW}$ Answer
B
$2.4\text{ kW}$
C
$1.92\text{ kW}$
D
$120.0\text{ kW}$

Answer

3.0\text{ kW}

The correct answer is

3.0\text{ kW}

. Raising

600\text{ kg}

of water by

20\text{ m}

requires

120,000\text{ J}

of gravitational potential energy. Doing this in

50\text{ s}

requires an output power of

2,400\text{ W}

(

2.4\text{ kW}

). Since the pump operates at

80\%

efficiency, the input power must be

\frac{2.4\text{ kW}}{0.80} = 3.0\text{ kW}

Step-by-Step Solution

Calculate the useful work output needed to elevate the water

W_{out} = mgh = 600\text{ kg} \times 10\text{ m s}^{-2} \times 20\text{ m} = 120,000\text{ J}

Work done against gravity equals the gravitational potential energy gained

Calculate the useful output power of the pump

P_{out} = \frac{W_{out}}{t} = \frac{120,000\text{ J}}{50\text{ s}} = 2,400\text{ W} = 2.4\text{ kW}

Power is defined as the rate at which work is performed

Calculate the required electrical power input using efficiency

P_{in} = \frac{P_{out}}{\text{Efficiency}} = \frac{2.4\text{ kW}}{0.80} = 3.0\text{ kW}

Efficiency is the ratio of useful output power to total input power

Key Concept

Work done against gravity, power, and mechanical/electrical efficiency

Estimated Time:1m 30s

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