Question

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

Answer 1

Volume of the tank, $V$ = $30 \; m^3$
Time of operation, $t$ = $15\, \text {min}$ = $15\times 60$ = $900\; s$
Height of the tank, $h$ = $40\, m$
Efficiency of the pump, $\eta$ = $30$%
Density of water, $\rho$ = $10^3\;kg/m^3$
Mass of water, $m$ = $\rho V$ = $30 \times 10^3\;kg$
Output power can be obtained as:

$P_0$ = $\frac{\text{Work done }}{\text{ Time}}$ = $\frac{\text {mgh}}{\text t}$

= $\frac{30\times 10^3\times 9.8\times 40}{900}$= 1$13.067\times 10^3 \;W$
For input power $P_i$, efficiency $\eta$, is given by the relation :
$\eta$ = $\frac{P_0}{P_i}$ = $30$ %

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

= $0.436\times 10^5 \;\text W$
= $43.6 \;\text{kW}$