a pump fills a tank of capacity 300 m^3 in 5 hours.the tank is situated at a height of 20 m.if the efficiency of the tank is 65%. then calculate the power of the engine which has driven the pump.
work done=weight of water into tank*20m
power= work done/time
power=1/.65 * 300m^3*densitywater*g*20m/(5hr*3600sec/hr)
To calculate the power of the engine that drives the pump, we need to determine the work done by the pump and divide it by the time taken.
First, let's calculate the work done by the pump. The work done by the pump is the product of the force exerted by the pump (which is the weight of the water lifted) and the distance over which the force is applied.
The weight of the water lifted is given by the formula:
Weight = mass * gravity
The mass of water lifted can be calculated by converting the volume of water in the tank into its mass. The density of water is approximately 1000 kg/m^3.
Mass = volume * density
Substituting the given values, we find:
Mass = 300 m^3 * 1000 kg/m^3 = 300,000 kg
The weight of the water lifted is:
Weight = 300,000 kg * gravity
where gravity is approximately 9.8 m/s^2.
Now, let's calculate the power of the engine. We know that power is the rate at which work is done. In this case, the work is done in 5 hours, so we need to convert the time to seconds:
Time = 5 hours * 60 minutes/hour * 60 seconds/minute = 18,000 seconds
The work done by the pump is the product of weight and the height over which it is lifted:
Work = Weight * height
Substituting the values, we get:
Work = (300,000 kg * 9.8 m/s^2) * 20 m = 58,800,000 Joules
Finally, we can find the power using the formula:
Power = Work / Time
Substituting the values, we have:
Power = 58,800,000 Joules / 18,000 seconds
Calculating this, we find:
Power ≈ 3,267,426.67 Watts
Therefore, the power of the engine that drives the pump is approximately 3,267,426.67 Watts.