Question

a pump on the ground floor of a building can pump up water to fill a tank of volume 30m3 in 15 min if the tank is 40m above the ground, and the efficiency of the pump is 30% how much electric power is consumed by the pump?

Answer 1

see the below image  file  for more clarification 
         
     density of liquid =1000
      volume of water filled=30
      

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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