Question

Question 6.15 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?

Chapter Work, Energy And Power Page 136

Answer 1

Hi

Volume of the tank, V =30m^3
Time of operation, t = 15 min = 15 ×60 = 900 s
Height of the tank, h = 40 m
Efficiency of the pump, = 30%
Density of water=10^3 kg/m^3

now we need to find out mass
we know density =m/volume
m=density *volume


Mass of water=10^3*30 kg

we know that
power=workdone /time

workdone is equal to energy
here energy=mgh

so
power=mgh/t

=30x10^3 x9.8x40/900

=13x10^3watt

we know efficiency is output power into input power

efficiency =output/input
which is give 30%

output/input=30%

input=output/30%

=13x10^3*100/30
=43.6kw
hope u got ur answet
hope it help

Answer 2

Given,
volume of tank = 30 m³
density of water = 10³ Kg/m³
so, mass of water pump up water to fill the tank = volume of tank × density of water
= 30m³ × 10³ Kg/m³
= 3 × 10⁴ Kg

height of tank = 40 m
hence, work done by pump to fill the tank = mgh
= 3 × 10⁴ × 9.8 × 40 [ g = 9.8 m/s² ]
= 1176× 10⁴ Joule
= 1.176 × 10⁷ joule

A/C to question ,
time taken to fill the tank by pump = 15min
= 15 *60 = 900 sec

power required = workdone/time taken
= 1.176 *10⁷/900
= 13.07 Kw

ELECTRIC power consumed = 30% of total power required for pumping in the tank
= 30/100 * 13.07
= 43.6 Kw