Q. 154.3( 39 Votes )

A pump on the ground floor of a building can pump up water to fill a tank of volume 30 m3 in 15 min. If the tank is 40 m above the ground, and the efficiency of the pump is 30%, how much electric power is consumed by the pump?

Answer :

Given,

Volume of the tank, V = 30 m3


Operation time of pump, t = 15 min = 15×60 sec = 900 sec


Height of the tank from ground, h = 40 m


Efficiency of the pump, η = 30%


We know, density of water, ρ = 103 kg m-3


So, Mass of water, m = ρV


m = 103 m3 × 30 kg m-3


m = 3 × 104 kg


Output power, Po = Work done/Time of operation


Po = mgh/t


Po = (3×104 kg × 9.8 m s-2 × 40 m)/(900 sec)


Po = 1.307×104 W = 13.07 kW


We know,


Efficiency (η) = Output power (Po)/Input power (Pi)


Input power (Pi)= Output power (Po)/Efficiency (η)


Pi = (1.307×104 W)/(0.3)


Pi = 4.355×104 W = 43.55 kW


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