Question

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?
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Answer 1

Answer:

Volume of the tank, V = 30 m3

Time of operation, t = 15 min = 15 × 60 = 900 s

Height of the tank, h = 40 m

Efficiency of the pump, η = 30 %

Density of water, ρ = 103 kg/m3

Mass of water, m = ρV = 30 × 103 kg

Output power can be obtained as:

P0 = Work done / Time = mgh / t

= 30 × 103 × 9.8 × 40 / 900 = 13.067 × 103 W

For input power Pi,, efficiency η, is given by the relation:

η = P0 / Pi = 30%

Pi = 13.067 × 100 × 103 / 30

= 0.436 × 105 W = 43.6 kW

Answer 2

Solution

Volume of the tank, V = 30 m3

Time of operation, t = 15 min = 15 × 60 = 900 s

Height of the tank, h = 40 m

Efficiency of the pump, η = 30 %

Density of water, ρ = 103 kg/m3

Mass of water, m = ρV = 30 × 103 kg

Output power can be obtained as:

P0 = Work done / Time = mgh / t

= 30 × 103 × 9.8 × 40 / 900 = 13.067 × 103 W

For input power Pi,, efficiency η, is given by the relation:

η = P0 / Pi = 30%

Pi = 13.067 × 100 × 103 / 30

= 0.436 × 105 W = 43.6 kW