India Site

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?

Volume of the tank is given as $30 \mathrm{~m}^{3}$

Time taken to fill the tank is given as $15 \mathrm{~min}=15 \times 60=900 \mathrm{~s}$

Height of the tank above the ground is given as $h=40 \mathrm{~m}$

Efficiency of the pump is given as $\eta=30 \%$

Density of water is given as $\rho=10^{3} \mathrm{~kg} \mathrm{~m}^{-3}$

Mass of water pumped, $m=$ volume $\times$ density $=30 \times 10^{3} \mathrm{~kg}$

Power consumed can be calculated as,

$P_{\text {output }}=W / \mathrm{t}=\mathrm{mgh} / \mathrm{t}$

$=\left(30 \times 10^{3} \times 9.8 \times 40\right) / 900=13066$ watt

Efficiency, $\eta=P_{\text {output }} / P_{\text {input }}$

$P_{\text {input }}=P_{\text {output } / n}=13066 /(30 / 100)=1306600 / 30$

$=43553 \mathrm{~W}=43.6 \mathrm{~kW}$