Question

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

1. The work done by the pump,
2. the power at which the pump works, and
3. The power rating of the pump if its efficiency is 40%. (Take g = 10m s^-2).

`["Hint" : "Efficiency" = "useful power"/"power input"]`

Answer 1

Mass of water = 600 kg

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

Time = 10 s

(a) Work done by the pump = mgh

W = mgh

W = 600 × 10 × 75

W = 450000

W = 4.5 × 10⁵ J

(b) Power at which pump works `= "work done"/"Time taken"`

`= "mgh"/"t"`

`= (600 xx 10 xx 75 "J")/"10 s"`

= 45000 W

= 45 kW   ...[1 kW = 1000 W]

(c) `"Efficiency" = "useful power"/"power input"`

Efficiency = 40 % = 0.4

`=> 0.4 = "45  kW"/"Power input"`

Power input = `"45 kW"/0.4`

Power input = 112.5  kW