An electric water pump with an efficiency of 80\% lifts 60\text{ kg} of water vertically t… — JAMB UTME

An electric water pump with an efficiency of $80\%$ lifts $60\text{ kg}$ of water vertically through a height of $10\text{ m}$ in $1\text{ minute}$ . What is the input power of the pump in watts? (Take $g = 10\text{ m s}^{-2}$ )

Answer: 125 W

Answer

The input power of the pump is

125\text{ W}

The total gravitational potential energy gained by

60\text{ kg}

of water lifted

10\text{ m}

W = mgh = 60 \times 10 \times 10 = 6000\text{ J}

. Performed over

60\text{ seconds}

, the useful power output is

100\text{ W}

. Accounting for an efficiency of

80\%

(

0.80

), the required input power is

P_{\text{in}} = \frac{100\text{ W}}{0.80} = 125\text{ W}

Step-by-Step Solution

Calculate the useful work done to lift the water

W = mgh = 60\text{ kg} \times 10\text{ m s}^{-2} \times 10\text{ m} = 6000\text{ J}

The useful work done equals the gravitational potential energy gained by the lifted mass of water.

Determine the useful output power of the pump

P_{\text{out}} = \frac{W}{t} = \frac{6000\text{ J}}{60\text{ s}} = 100\text{ W}

Power is the rate at which work is done, and

1\text{ minute}

must be converted to

60\text{ seconds}

Calculate the required input power using efficiency

P_{\text{in}} = \frac{P_{\text{out}}}{\text{Efficiency}} = \frac{100\text{ W}}{0.80} = 125\text{ W}

Efficiency is defined as

\text{Efficiency} = \frac{P_{\text{out}}}{P_{\text{in}}}

, so

P_{\text{in}} = \frac{P_{\text{out}}}{\text{Efficiency}}

Key Concept

Work, Power, and Efficiency of a Pump System

Estimated Time:1m 30s

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