question_answer

Find the power of the engine deployed to lift water at the rate of 5 litre of water in one second through an average height of 10 m from a well of depth 10 m. Assume 10% of power of engine is wasted.

Answer:

Volume of water lifted in 1 s = 5 litre. \[u=30m/s\]Mass of water = 5 kg The average height = depth + height = 10 + 10 = 20 m \[Power\frac{Work\,\,done}{time}=\frac{5\times 9.8\times 20}{1}=980\,watt\] The useful power = 980 watt, As 10 % of power becomes waste, the useful power = 90% \[power\frac{980\times 100}{90}=1088.W=1.089\,KW\]