Question

A pump is used to lift 500 Kg o water from a depth of 80 m in 10 s. Calculate : a) The work done by the pump b) The power at which the pump works c) The power rating of the pump if its efficiency is 40%. (Take g = 10 m/s) (Efficiency = useful power/power input)

Answer 1

a) workdone by pump=mgh
where m is mass of water , g is acceleration due to gravity and h is depth
now ,
workdone=500 x 10 x 80=400000 j
=400 kj
b) power =workdone/time
=400000/10=40000 w=40 kw

c) efficiency =useful power/input power
40/100= 40 kw/input power
input power=100 kw

Answer 2

a) W=FS

W=mg×S

500×10×80

4,00,000J

b) P=W/t

=400000/10=40000W=40kW

c) efficiency=useful power/power input

40%=40/Power input

Power input = 40 × 100/40

Power input = 40kW