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 minutes. If the tank is 40 m above the ground and the efficiency of the pump is 30 %. How much is electric power consumed by the pump?

Answer 1

Density of water is $=1000kg/m^3$
Volume of water filled $=30m^3$
Therefore, mass of water filled in the tank $=\left(1000kg/m^3\right)\left(30m^3\right)$ $=30000kg$
Height to which the water is raised =40 m
Acceleration due to gravity $=9.8m/s^2$
So, rise in potential energy of the water $=\left(30000kg\right)\left(9.8m/s^2\right)\times 40m)=1.17\times 107J$
The work is done by the pump in15 min =900 s
Power of the pump $=\frac{(1.17\times 107J)}{\left(900s\right)}=13000watt$ Let, x be the electrical power given to the pump, since the efficiency of the pump is $30\%$
Therefore, $x\left(\frac{30}{100}\right)=13000$
$\Rightarrow x=43333.33watt$
So, the power consumed by the pump is 43333.3 watt.