Question

An air-conditioning system is used to maintain a house at a constant temperature of \(20^{\circ} \mathrm{C}\). The house is gaining heat from outdoors at a rate of \(20,000 \mathrm{kJ} / \mathrm{h},\) and the heat generated in the house from the people, lights, and appliances amounts to \(8000 \mathrm{kJ} / \mathrm{h}\). For a COP of \(2.5,\) determine the required power input to this air-conditioning system.

Short answer

Answer: To calculate the power input to the air-conditioning system, you need to follow these steps: 1. Calculate the total heat gain per hour: Total heat gain = \(20,000 \mathrm{kJ/h} + 8,000 \mathrm{kJ/h} = 28,000 \mathrm{kJ/h}\). 2. Convert the total heat gain in kJ/hour to watts: Total heat gain = \(\frac{28,000\,\mathrm{kJ/h}}{3600 \,\mathrm{s/h}}\). 3. Use the COP to find the power input: Power input = \(\frac{\mathrm{Cooling\,capacity}}{\mathrm{COP}}\) 4. Calculate the power input in watts: Power input = \(\frac{(28,000/3600)}{2.5}\) Calculate the power input value in watts after following these steps.

Step-by-step solution

Calculate total heat gain per hour.

Sum the heat gain from outdoors and the heat generated by people, lights, and appliances inside the house. Total heat gain = Heat gain from outdoors + Heat generated inside the house Total heat gain = \(20,000 \mathrm{kJ/h} + 8,000 \mathrm{kJ/h}\).

Calculate the total heat gain in watts.

Convert the total heat gain from kJ/hour to watts by using the conversion factor 1kW = 1kJ/s. Total heat gain = \(\frac{20,000\,\mathrm{kJ/h} + 8,000\,\mathrm{kJ/h}}{3600 \,\mathrm{s/h}}\).

Find the power input using the COP.

The coefficient of performance (COP) relates the cooling capacity to the power input. Use the formula: COP = \(\frac{\mathrm{Cooling\,capacity}}{\mathrm{Power\,input}}\) Power input = \(\frac{\mathrm{Cooling\,capacity}}{\mathrm{COP}}\) Power input = \(\frac{\frac{28,000\,\mathrm{kJ/h}}{3600\,\mathrm{s/h}}}{2.5}\)

Calculate the power input.

Find the final value for the power input in watts by substituting the COP and total heat gain in watts: Power input = \(\frac{(28,000/3600)}{2.5}\)

Key concepts

Heat Transfer

Understanding the concept of heat transfer is essential when studying air-conditioning systems. Heat transfer refers to the movement of heat from one place to another, and it can occur through conduction, convection, and radiation. In the case of an air-conditioning system, the process involves removing heat from inside a house to maintain a comfortable temperature.

The exercise provided outlines a scenario where heat is gained in a house from two sources: the outdoors environment and heat generated by internal sources such as people and appliances. The combined heat gain from these sources needs to be counteracted by the air-conditioning system to keep the house at the desired temperature of 20°C. The total amount of heat that transfers into the house per hour is quantified in kilojoules (kJ), which the air-conditioning system needs to remove efficiently.

Coefficient of Performance (COP)

The Coefficient of Performance (COP) is a measure of the efficiency of a heating or cooling system. The COP is defined as the ratio of useful heating or cooling provided to the work or energy input required to achieve that conditioning. For an air-conditioning system, the COP denotes how much cooling effect it can produce per unit of electrical power consumed.

A higher COP indicates a more efficient system, as it provides more cooling for the same amount of energy input. In the given exercise, the air-conditioning system has a COP of 2.5, meaning that for every unit of energy input into the system, it produces 2.5 units of cooling energy.

Energy Conversion

Energy conversion in this context is about converting different units and forms of energy to assess the power input required for the air-conditioning system. The exercise involves converting the total heat gain, given in kilojoules per hour (kJ/h), to watts (W), since power is typically measured in watts in the context of electrical devices.

One crucial aspect of energy conversion is understanding the relation between different energy units; for example, 1 kW is equivalent to 1 kJ/s. Such conversions are important because they allow us to calculate the power input in familiar terms (watts) which can be easily related to electrical energy consumption.

Power Input Calculation

Power input calculation for an air-conditioning system is the process by which we determine the amount of electrical energy required to remove the unwanted heat from a space. To find this, we use the total heat which needs to be extracted (already converted to watts in the example) and divide it by the Coefficient of Performance (COP).

In the provided exercise, the first step is to calibrate the combined heat gain into watts, and then, using the given COP value, calculate the necessary power input the system needs. This result reflects how much electrical power the air-conditioning system requires to operate effectively and maintain the desired temperature by counteracting the heat gains.