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 m^{3} 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 m^{3}

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 = 10^{3} kg m^{-3}

So, Mass of water, m = ρV

⇒ m = 10^{3} m^{3} × 30 kg m^{-3}

⇒ m = 3 × 10^{4} kg

Output power, P_{o} = Work done/Time of operation

∴ P_{o} = mgh/t

⇒ P_{o} = (3×10^{4} kg × 9.8 m s^{-2} × 40 m)/(900 sec)

⇒ P_{o} = 1.307×10^{4} W = 13.07 kW

We know,

Efficiency (η) = Output power (P_{o})/Input power (P_{i})

∴ Input power (P_{i})= Output power (Po)/Efficiency (η)

⇒ P_{i} = (1.307×10^{4} W)/(0.3)

⇒ P_{i} = 4.355×10^{4} W = 43.55 kW

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