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?

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, P_o = Work done/Time of operation

∴ P_o = mgh/t

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

⇒ P_o = 1.307×10⁴ 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 = (1.307×10⁴ W)/(0.3)

⇒ P_i = 4.355×10⁴ W = 43.55 kW