water pump belt

A water pump driven by petrol raises water at a rate of 0.5 \ma...

Question

A water pump driven by petrol raises water at a rate of 0.5 m 3 min−1 from a depth of 30 m . If the pump is 70% efficient, what power is developed by the engine?

1750 W 2450 W 3500 W 7000 W

Explanation

To find the power developed by the engine, we need to calculate the work done per second to raise the water and then account for the efficiency of the pump.

Step by Step Solution

First, convert the rate of water flow from m 3/min to m 3/s. Given rate is 0.5 m 3/min.

Since there are 60 seconds in a minute, the rate in m 3/s is 0.5 m 3/min÷60=60 0.5​m 3/s=0.00833 m 3/s.

Calculate the work done per second (power) to raise the water. The work done to raise a volume V of water to a height h is given by W=ρ V g h, where ρ is the density of water (1000 kg/m 3), g is the acceleration due to gravity (9.8 m/s 2), and h is the height (30 m).

Substitute the values: W=1000×0.00833×9.8×30=2450 W. This is the power required to raise the water without considering efficiency.

Since the pump is 70% efficient, the actual power developed by the engine is 0.7 2450​=3500 W.

Final Answer

The power developed by the engine is 3500 W.