Work, Energy and Power — Question 8

Given,

Mass of water = 600kg

Height to which the water has to be raised = 75 m

Time = 10 s

(a) Work done by the pump = mgh

Substituting the values we get,

$W = 600 \times 10 \times 75 \\[0.5em] \Rightarrow W = 4.5 \times 10^5 J \\[0.5em]$

(b) $\text{Power of the pump} = \dfrac{\text{work done}}{\text{time taken}}$

$\phantom{(b)\space}\Rightarrow\text{Power} = \dfrac{\text{mgh}}{\text{t}}$

Substituting the values we get,

$\text{Power} = \dfrac{600 \times 10 \times 75 }{10} = 45000 \text{ W}\\[1em] \Rightarrow \text{Power} = 45 \text{ kW} \\[1em]$

(c) $\text{Efficiency} = \dfrac{\text{useful power}}{\text{power input}}$

Given,

Efficiency = 40 %

$0.4 = \dfrac{45}{\text{input power}} \\[0.5em] \text{input power} = \dfrac{45 \times 10^3}{0.4} \\[0.5em] = 112.5 \times 10^3 \text{ W} \\[0.5em] = 112.5 \text{ kW}$

∴ Power rating of pump = 112.5 kW.