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

Hey buddy,

● Answer - 44.44 kW

● Explaination-
◆ Given -
V = 30 m^3
h = 40 m
ρ = 1000 kg/m^3
η = 30%
t = 15 min = 900 s

◆ Solution-
Here, output power of the pump is calculated by,
Po = Work done / time
Po = mgh/t
Po = ρVgh/t
Po = 1000×30×10×40/900
Po = 13333 W

Now, Input power is calculated by formula
Pi = Po/η
Pi = 13333/30%
Pi = 44444 W
Pi = 44.44 kW

Electric power consumed by the pump is 44.44 kW.

Hope that is useful..

Answer 2

Hello mate ,

Your answer:-

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GIVEN-

Volume of the tank, V = 30 m^3

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 = 10^3 kg/m^3

Mass of water, m = ρV = 30 × 10^3 kg

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Output power can be obtained as:

P (output) = work done/ time

= mgh / t

= 30× 10^39 × 9.8 × 40 / 900

= 13.067× 10^3 W

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For input power P( input) ,efficiency, η is given by the relation:

η = P (output)/ P ( input)= 30%

=> P(input) = 13.067× 10^3× 100/ 30

=> P(input)= 0.436× 10^5 W= 43.6 kW

SO ELECTRIC POWER CONSUMED BY ELECTRIC MOTOR IS 43.6 kW.

I hope, this will help you.☺

Thank you______❤

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