Question

A pump on the ground floor of a building can pump up water to fill a tank of volume $30m^3$ in $15min$. If the tank is $40m$ above the ground, and the efficiency of the pump is $30\%$, how much electric power is consumed by the pump ?[Dencity of water ${10}^{3}kg{m}^{-3}$]$(g=9.8m{s}^{-2})$

Answer 1

Given:

Volume of the tank, $V=30m^3$

Time of operation, $t=15min=15\times 60=900s$

Height of the tank, $h=40m$

Efficiency of the pump, $=30$ %

Density of water, $={10}^{3}kg/m^3$

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

Output power can be obtained as:

$P_0=\frac{Workdone}{Time}=\frac{mgh}{t}$

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

For input power $P_i$, efficiency is given by the relation:

$\frac{P_o}{P_i}=30$%

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

$P_i=43.6kW$