a pump on ground floor of a building can pump up water to fill a tank of volume 30 metre cube in 15 minutes. if the tank is 40m above the ground and efficiency of pump is 30℅. how much electric power is consumed by the pump?(take g=9.8m/s)
Answers
Work done = mgh
Power(p) = work/time=mgh/t=V.ρ.g.h/t
(ρ density= maass/Volume)
Power required for pumping = V.ρ.g.h/t×1/η
Given
Volume = 30m3
density of water ρ=1000 kg/m3
g = 9.8m/s2
h = 40m
Efficiency = 30%
∴(30×1000×9.8×40) /(15×60×30/100)
=11760000/270
= 43555.55 watt
Here is your answer
I hope my answer is help you
vinay7860
Hello mate ,
Your answer:-
__________________________
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
___________________________
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
____________________________
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______❤
_____________________________❤