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An overhead water tank is fixed at the height 10 m from the ground and has the capacity of 500 litres.Calculate the time required to fill the tank from a pump of 0.5 HP.whose efficiency is 90 %.

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by mimansh

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by mimansh

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Power = 0.5 HP = 0.5×745.7 J/s = 372.85 J/s

efficiency = 90%

Work done in 1 second = 372.85×(90/100) = 335.5 J

Mass of 1l water = 1Kg

Work done to raise 1l water by 10m = mgh = 1×9.8×10 = 98J

in 1s, the pump can raise = 335.5/98 liter

to raise 500l, time = 500 ÷ (335.5/98) = 500×98/335.5 = 146 second

time required is 146 second.

efficiency = 90%

Work done in 1 second = 372.85×(90/100) = 335.5 J

Mass of 1l water = 1Kg

Work done to raise 1l water by 10m = mgh = 1×9.8×10 = 98J

in 1s, the pump can raise = 335.5/98 liter

to raise 500l, time = 500 ÷ (335.5/98) = 500×98/335.5 = 146 second

time required is 146 second.

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.

Change in potential energy of 500 litres from ground level to tank:

= m g h

= 500 kg * 10 m/sec² * 10 m = 50, 000 J

Effective mechanical power of the pump that is available for pumping water

= 0.5 HP * 90 % = 0.5 HP * 745.7 W/HP * 0.9 = 335.45 W

*Time duration to fill the tank = energy / power *

= 50, 000 / 335.45 sec. =* 149 sec = 2 min and 29 sec.*

If the value of g is taken as 9.81 m/s/s then the answer will be a little different.

= m g h

= 500 kg * 10 m/sec² * 10 m = 50, 000 J

Effective mechanical power of the pump that is available for pumping water

= 0.5 HP * 90 % = 0.5 HP * 745.7 W/HP * 0.9 = 335.45 W

= 50, 000 / 335.45 sec. =

If the value of g is taken as 9.81 m/s/s then the answer will be a little different.