Question

A pump on the ground floor of a building can pump up water to fill a tank of volume 30m cube in 15 minutes.if the tank is 40 m above the ground,and efficiency of pump is 30%,how much electric power is consumed by the pump?Take the density of water to be 103kg/mcube and g=9.8m/s squared.

Answer 1

Volume of water pumped up in the 15 min is = 30 m³

So, mass of water pumped = Vρ = 30ρ kg

Weight of water pumped = 30ρg N

So, work done in pumping water to a height 40 m above ground = weight × height

= (30ρg)(40)

= 1200ρg

So, power required to lift this water in 15 min or 900 s is = 1200ρg/900 = 4ρg/3

Let, P be the power consumed by the pump. 30% of this power is only used to lift the water.

So, 30% of P = 4ρg/3

=> 0.30P = 4(1000)(9.8)/3

=> P = 43555.5 W