An electric motor of 1hp is used to run the water pump. The water pump takes 10 minutes to fill an overhead tank. The tank has a capacity of 800 liters and a height of 15 m. Find the actual work done by the electric motor to fill the tank. Also, find the efficiency of the system. (Density of water = 1000 $kgm^{-3}$ )
(Mass of 1 liter of water = 1 kg) (447600 J, 26.8%)
Difficulty: Hard
Solution:
Power = P = 1 hp = 746 W
Time = t = 10 min = $10\times 60$ s = 600 s
$\frac{Capacity}{volume}$ = V = 800 liters
Height = h = 15 m
Work done = W = ?
Efficiency = E = ?
Power = $\frac{work}{time}$
P =$\frac{w}{t}$
Or W = $P \times t$
W = $746\times 600$
W = 447600 J
Since the work done by the electric pump to fill the tank is 447600 J. It is equal to the input.
Hence input = actual work doe = W = 447600 J
Output = P.E = mgh
Since 1 litre = 1kg, therefore 800 litres = 800 kg
Output = P.E = $800 \times 10\times 15$ = 120000J
% Efficiency = $\frac{Output}{Input }×100$
% Efficiency = $\frac{120000}{447600}×100$
Efficiency = 26. 8 %
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