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?

Answer 1

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 × 10^3 × 9.8 × 40 )/ 900`

=  13.067 × 10³ W

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

η = `P_0 /P_i`  =  30%

P_i = `(13.067 × 10^3 × 100  )/ 30`

=  43.56 kW

Answer 2

Here, volume of water = 30 m³ ; t = 15 min = 15 x 60 = 900s

h = 40 m ; n= 30%

As the density of water = p = 10³ kg m^-3

Mass of water pumped, m = volume x density = 30 x 10³ kg

Actual power consumed or output power p₀ = W/t = mgh/t

=>p₀=(30 x 10³ x 9.8 x 40)/900=13070 watt

If pi is input power (required), then as

η=p₀/p_i=> p_i=p₀/η = 13070/(30/100)=43567 W =43.56 KW