Question

A pump is used to lift 600 kg of water from a depth of 75 m in 10 s. 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).

April 9, 2024

Answer

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,

Work = 600 × 10 × 75 = 450000 J

(b) Power = $work\over time$

= $450000\over 10$

= 45000 W or 45 kW

(c) Efficiency = $P_output\over P_Input$

Efficiency = 40 % = 0.4

0.4 = $45\over P_Input$

⇒ P_input = $45\over 0.4$ = 112.5 kW

∴ Power rating of pump = 112.5 kW.

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