Question

An electric pump raises 10.4 metre cube of water from a reservoir whose water level is 4m below ground level to a storage tank above ground.If the discharge pipe outlet is 32 m above ground level and the operation takes 60 minutes,find the minimum power rating of the pump if its efficiency is 80%.​

Answer 1

Given info : An electric pump raises 10.4 m³ of water from a reservoir whose water level is 4m below ground level to a storage tank above ground.If the discharge pipe outlet is 32 m above ground level and the operation takes 60 minutes.

To find : the minimum power rating of tyhe pump if its efficiency is 80% is..

solution : total height covered by water to reach reservoir , h = 4m + 32 m = 36 m

volume of water, V = 10.4 m³

∵ density of water , ρ = 1000 kg/m³

∴ the mass of water , m = Vρ = 10.4 × 1000 = 10400 kg

now potential energy of water , U = mgh

= 10400 kg × 10 m/s² × 36 m

= 3744000 J = 3744 kJ

now the power of electric pump , P = potential energy of water/time taken

= $\bf{\frac{3744kJ}{60min}}$

= $\frac{3744\times10^3J}{60\times60s}$

= 1040 watt

but it has given that the efficiency of the electric pump is 80%.

∴ minimum power rating of the pump = 1040 × 100/80 = 1300 watt.

therefore the minimum power rating of the pump if its efficiency is 80%.