Question

A pump on the ground floor of a buliding can pump out water 2 a tank of volume 30m cube in 30 minutes.if the tank is 40 m above the ground and efficiency of the pump is 40 percentage .how much electric power is 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 $=\left(30000kg\right)\left(9.8m/s^2\right)\left(40m\right)=1.17\times 107j$
The work is done by the pump in 15 min = 900 s
Power of the pump = $(1.17\times 107j)/\left(900s\right)=13000Watt$
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.