Understanding Pump Efficiency Calculation

Efficiency is a measure of how well a device converts input energy or power into useful output energy or power. For a pump, the useful output is the work done in lifting water, and the input is the electrical power consumed by the motor.

The efficiency (\(\eta\)) is calculated using the formula:

\[\eta = \left( \frac{\text{Useful Output Power}}{\text{Input Power}} \right) \times 100\%\]

In this problem, we are given the input power of the pump and the task it performs (lifting water). We need to calculate the useful output power based on the work done and the time taken.

Calculating Useful Work Done by the Pump

The pump lifts 500 kg of water by a height of 30 m. The useful work done against gravity is the increase in potential energy of the water. The formula for work done against gravity (or potential energy gain) is:

\[\text{Work done} = \text{mass} \times \text{acceleration due to gravity} \times \text{height}\]

Given:

• Mass (m) = 500 kg
• Acceleration due to gravity (g) = 10 m/s\(^2\)
• Height (h) = 30 m

Calculation of work done:

\[\text{Work done} = 500 \, \text{kg} \times 10 \, \text{m/s}^2 \times 30 \, \text{m}\]

\[\text{Work done} = 5000 \, \text{N} \times 30 \, \text{m}\]

\[\text{Work done} = 150,000 \, \text{Joules}\]

Determining Useful Output Power

Power is the rate at which work is done. The work of 150,000 Joules is done in 10 minutes. First, we convert the time from minutes to seconds:

\[\text{Time} = 10 \, \text{minutes} \times 60 \, \text{seconds/minute}\]

\[\text{Time} = 600 \, \text{seconds}\]

Now, we calculate the useful output power using the formula:

\[\text{Useful Output Power} = \frac{\text{Work done}}{\text{Time taken}}\]

\[\text{Useful Output Power} = \frac{150,000 \, \text{J}}{600 \, \text{s}}\]

\[\text{Useful Output Power} = \frac{1500}{6} \, \text{W}\]

\[\text{Useful Output Power} = 250 \, \text{W}\]

Final Calculation of Pump Efficiency

We have the input power and the useful output power:

• Input Power = 400 W
• Useful Output Power = 250 W

Now, we use the efficiency formula:

\[\eta = \left( \frac{\text{Useful Output Power}}{\text{Input Power}} \right) \times 100\%\]

\[\eta = \left( \frac{250 \, \text{W}}{400 \, \text{W}} \right) \times 100\%\]

\[\eta = \left( \frac{25}{40} \right) \times 100\%\]

\[\eta = \left( \frac{5}{8} \right) \times 100\%\]

\[\eta = 0.625 \times 100\%\]

\[\eta = 62.5\%\]

Resulting Pump Efficiency

The calculated efficiency of the pump is 62.5%.

Revision Table: Pump Efficiency Formulas

Additional Information: Energy and Power Concepts

Work: In physics, work is done when a force causes displacement. Lifting water against gravity is a form of work. The unit of work is the Joule (J), which is equal to one Newton-meter (Nm).

Potential Energy: This is the energy an object possesses due to its position or state. When a pump lifts water, it increases the water's gravitational potential energy, which is the useful work done.

Power: Power is the rate at which work is done or energy is transferred. The unit of power is the Watt (W), which is equal to one Joule per second (J/s). The input power is what the pump consumes (e.g., electrical power), and the output power is the useful power delivered (e.g., power used to lift water).

Efficiency: Efficiency is always a ratio (or percentage) of useful output to total input. It indicates how much of the input energy or power is converted into the desired form of output, with the rest typically lost as heat or sound.