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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Answered by
11
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.
mimansh:
but the answer provided is 1.1 * 10^5 sec
Then either you have written the question wrong or the answer is wrong.
are the dimensions of the tank given ?
Answered by
2
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
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.
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