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Q.

An electric pump on the ground floor of a building takes 10 min to fill a tank of volume 30 m³ with water. If the tank is 60 m above the ground and the efficiency of the pump is 30% then how much electric power is consumed by the pump in filling the tank? Take g = 10m/ s² (in Kw)

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Detailed Solution

Total mass of water to be lifted up is,
$m = 30 (m^{3}) \times 1000 (kg / m^{3}) = 3 \times 10^{4} kg$
Work done against gravity to lift this much water to a height of 60 m
$W = mg \times h = 3 \times 10^{4} \times 10 \times 60 = 1 .8 \times 10^{7} J$
Work done by pump is, $W_{1} = \frac{W}{0 .3} = 6 \times 10^{7} J$
Required power of pump is,
$P = \frac{W_{1}}{t} = \frac{6 \times 10^{7}}{600} W = 100 KW$

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