Physics

A pump is used to lift 500kg of water from a depth of 80m in 10s.
Calculate:
(a) The work done by the pump,
(b) The power at which the pump works, and
(c) The power rating of the pump if its efficiency is 40%. (Take g = 10m s^-2).

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Given,

Mass of water = 500kg

Height to which the water has to be raised = 80 m

Time = 10 s

(a) Work done by the pump = mgh

Substituting the values we get,

$W = 500 \times 10 \times 80 \\[0.5em] \Rightarrow W = 4 \times 10^5 J \\[0.5em]$

(b) Power of the pump = work done / time taken

Power = mgh / t

Substituting the values we get,

$\text{Power} = \dfrac{500 \times 10 \times 80 }{10} = 4000 W\\[0.5em] \Rightarrow \text{Power} = 40 kW \\[0.5em]$

(c)

$\text{Efficiency} = \dfrac{\text{useful power}}{\text{power input}}$

Given,

Efficiency = 40 %

$0.4 = \dfrac{40}{\text{input power}} \\[0.5em] \text{input power} = \dfrac{40}{0.4} \\[0.5em] = 100 kW \\[0.5em]$

∴ Power rating of pump = 100 kW.

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