Question

A pump lifts water at a rate of 283 liters per second from a lake and force it into a tank 8 meters above the level of the water at a pressure of 137kPa. What is the power required in kilowatts if the suction is 5 cm larger in diameter than a discharge pipe.

Answer 1

The power required to lift water at a certain rate from a lake into a tank can be calculated using the formula Power = Force x Velocity. By finding the force and velocity, we can determine the power required. In this case, the power required is 37.18 kW.

Step-by-step explanation:

To calculate the power required, we can use the formula:

Power = Force x Velocity

In this case, the force is the pressure multiplied by the area of the cross-section of the pipe. The velocity is the rate at which water is being lifted.

First, we need to find the area of the cross-section of the pipe. The diameter of the discharge pipe is 3.5 cm, so the radius is 1.75 cm. Hence, the area is:

Area = π x radius^2 = 3.14 x (1.75 cm)^2 = 9.62 cm^2

Now, we can calculate the force:

Force = Pressure x Area = 137 kPa x 10^3 Pa/kPa x 9.62 cm^2 x 10^-4 m^2/cm^2 = 131.4 N

Finally, we can calculate the power:

Power = Force x Velocity = 131.4 N x 283 L/s x 10^-3 m^3/L = 37.18 kW

Power required: 37.18 kW