Understanding Pump Power and Work

This problem asks us to find the power of a pump. Power is defined as the rate at which work is done or energy is transferred. In this case, the pump does work to lift water against gravity. The work done is converted into the potential energy of the water.

Key Concepts

• Power (P): The rate of doing work, calculated as Work done divided by the time taken (\(P = \frac{W}{t}\)). The unit of power is Watts (W).
• Work (W): When a force moves an object over a distance, work is done. In this problem, the pump exerts a force to lift the water against gravity. The work done against gravity to lift an object is given by the potential energy formula (\(W = mgh\)), where 'm' is mass, 'g' is acceleration due to gravity, and 'h' is the height lifted. The unit of work is Joules (J).
• Potential Energy (PE): The energy stored in an object due to its position or state. For an object lifted against gravity, the potential energy gained is \(PE = mgh\). Assuming 100% efficiency, the work done by the pump equals the potential energy gained by the water.

Analyzing the Given Information

We are given the following information:

• Mass of water (m) = 1 tonne
• Height the water is lifted (h) = 90 m
• Time taken (t) = 30 minutes
• Acceleration due to gravity (g) = 10 m/s²
• Efficiency of the pump = 100%

Step-by-Step Power Calculation

To find the power of the pump, we first need to calculate the total work done in lifting the water and then divide it by the time taken.

Step 1: Convert Units

We need to ensure all units are in the standard SI system (meters, kilograms, seconds).

• Mass: 1 tonne = 1000 kilograms (kg)
• Time: 30 minutes. Since 1 minute = 60 seconds, 30 minutes = \(30 \times 60\) seconds = 1800 seconds (s).

Step 2: Calculate the Work Done

The work done (W) in lifting the water is equal to the potential energy gained by the water. We use the formula \(W = mgh\).

\(W = \text{mass} \times \text{gravity} \times \text{height}\)

Substitute the values:

\(W = (1000 \text{ kg}) \times (10 \text{ m/s}^2) \times (90 \text{ m})\)

\(W = 1000 \times 10 \times 90 \text{ J}\)

\(W = 10000 \times 90 \text{ J}\)

\(W = 900000 \text{ J}\)

The work done by the pump is 900,000 Joules.

Step 3: Calculate the Power

Now we can calculate the power (P) using the formula \(P = \frac{W}{t}\).

\(P = \frac{\text{Work done}}{\text{Time taken}}\)

Substitute the calculated work and the time in seconds:

\(P = \frac{900000 \text{ J}}{1800 \text{ s}}\)

To simplify the division, we can cancel out zeros:

\(P = \frac{9000}{18} \text{ W}\)

We can divide 9000 by 18. \(90 \div 18 = 5\). So, \(9000 \div 18 = 500\).

\(P = 500 \text{ W}\)

The power of the pump is 500 Watts.

Comparing with Options

We found the power to be 500 W. Let's compare this with the given options:

Our calculated value of 500 W matches Option 4.

Revision Table: Pump Power Calculation

Additional Information on Power and Efficiency

The problem assumes 100% efficiency. In reality, pumps are not 100% efficient. Actual efficiency (\(\eta\)) is less than 100%.

Actual Power Output = Work Done / Time Taken (This is the useful power output)

Input Power = Actual Power Output / Efficiency (\(\eta\))

If the efficiency was, say, 80% (or 0.8), the required input power from the motor driving the pump would be higher than the calculated output power.

Input Power \(= \frac{500 \text{ W}}{0.8} = 625 \text{ W}\)

This means the motor would need to supply 625 W of power for the pump to deliver 500 W of useful power in lifting the water. The remaining power is typically lost as heat or sound.

Understanding the relationship between work, energy, power, and efficiency is crucial in solving problems related to machinery and energy transfer.