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 : 
\(\eta\) = \(\frac{P_0}{P_i}\) = \(30\) %

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

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