
The power of a pump motor is 2kW. How much water per minute can it raise to a height of 10m?
Answer
587.7k+ views
Hint When a body is raised to a height h, it gains potential energy. If all the energy losses are ignored, this potential energy would be equal to the work done by the pump to raise the height of water. So we will equate the potential energy with the energy used by the pump. Energy required by the pump is power time the time for which it runs.
Complete step by step solution
The power of a pump is defined as the work done by the pump or motor per unit time. This work done is exclusive of the power losses that the motor may undergo. In this question, we don’t need to consider the power losses.
To raise an object of mass m against the gravitational pull of the earth, W work done would be required which is equal to the potential energy gained by the object:
\[W\, = \,mgh\]
When this work is expressed per unit time, it becomes power. So, the work done by the motor in 1s is 2000J. further, work done by the pump in 1 minute is equal to \[2000x60\] J. this work done should be equal to the potential energy of the water. Therefore,
\[
2000 \times 60\, = \,m \times 10 \times 10 \\
m = 120 \\
\]
Therefore the mass of water that can be lifted is 120kg. as the density of water is \[1{\text{ }}kg/{m^3}\] , the volume of water that can be lifted is \[120{m^3}\].
Note in actual practice, there are a lot of losses like the viscosity loss, pressure loss, loss at the bend of a pipe etc. Most questions will ignore these losses. But there might be cases where they will give the efficiency of the motor. Efficiency is the ratio of power output of the pump to the input it requires.
Complete step by step solution
The power of a pump is defined as the work done by the pump or motor per unit time. This work done is exclusive of the power losses that the motor may undergo. In this question, we don’t need to consider the power losses.
To raise an object of mass m against the gravitational pull of the earth, W work done would be required which is equal to the potential energy gained by the object:
\[W\, = \,mgh\]
When this work is expressed per unit time, it becomes power. So, the work done by the motor in 1s is 2000J. further, work done by the pump in 1 minute is equal to \[2000x60\] J. this work done should be equal to the potential energy of the water. Therefore,
\[
2000 \times 60\, = \,m \times 10 \times 10 \\
m = 120 \\
\]
Therefore the mass of water that can be lifted is 120kg. as the density of water is \[1{\text{ }}kg/{m^3}\] , the volume of water that can be lifted is \[120{m^3}\].
Note in actual practice, there are a lot of losses like the viscosity loss, pressure loss, loss at the bend of a pipe etc. Most questions will ignore these losses. But there might be cases where they will give the efficiency of the motor. Efficiency is the ratio of power output of the pump to the input it requires.
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