A pump on the ground floor of a building can pump up water
Question.
A pump on the ground floor of a building can pump up water to fill a tank of volume $30 \mathrm{~m}^{3}$ in $15 \mathrm{~min}$. If the tank is $40 \mathrm{~m}$ above the ground, and the efficiency of the pump is $30 \%$, how much electric power is consumed by the pump?

solution:

Volume of the tank, $V=30 \mathrm{~m}^{3}$

Time of operation, $t=15 \mathrm{~min}=15 \times 60=900 \mathrm{~s}$

Height of the tank, $h=40 \mathrm{~m}$

Density of water, $\rho=10^{3} \mathrm{~kg} / \mathrm{m}^{3}$

Mass of water, $m=\rho V=30 \times 10^{3} \mathrm{~kg}$

Output power can be obtained as:

$P_{0}=\frac{\text { Work done }}{\text { Time }}=\frac{m g h}{t}$

$=\frac{30 \times 10^{3} \times 9.8 \times 40}{900}=13.067 \times 10^{3} \mathrm{~W}$

For input power $P_{i}$, efficiency $\eta$, is given by the relation:

$\eta=\frac{P_{0}}{P_{i}}=30 \%$

$P_{i}=\frac{13.067}{30} \times 100 \times 10^{3}$

$=0.436 \times 10^{5} \mathrm{~W}$

$=43.6 \mathrm{~kW}$
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