Find the efficiency of a pump (rated 400 W) that can lift 500 kg of water by 30 m in 10 minutes. (Use g = 10m/s²)

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This question was previously asked in

RRB ALP Fitter 23 Jan 2019 Official Paper (Shift 2).

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Answer 1

Concept:

Pump:

A pump is a mechanical device used to force a fluid (a liquid or a gas) to move forward inside a pipeline or hose.
They are also used to produce pressure by the creation of a suction (partial vacuum), which causes the fluid to rise to a higher altitude.

The efficiency of the pump, \(\eta =\frac{Useful ~power}{Power~rating}\)

Calculation:

Given:

Power rating = 400 W, Mass of water lifted = 500 kg, H = 30 m, time = 10 minutes = 60 × 10 = 600 seconds, g = 10 m/s²

Force, F = m × g = 500 × 10 = 5000 N

Work = Force × displacement (H) = 5000 × 30 = 150 kJ

\(Power=\frac{Work}{Time}\)

\(Power=\frac{150}{600}=0.25 ~kW=250~W\)

⇒ Useful work = 250 W

\(\eta=\frac{Useful ~power}{Power~rating}\)

\(\eta=\frac{250}{400}=0.625\)

⇒ η = 62.5%

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