Question

An air conditioner removes heat steadily from a house at a rate of \(750 \mathrm{kJ} / \mathrm{min}\) while drawing electric power at a rate of \(6 \mathrm{kW}\) Determine ( \(a\) ) the COP of this air conditioner and ( \(b\) ) the rate of heat transfer to the outside air.

Short answer

Answer: The COP of the air conditioner is 2.08, and the rate of heat transfer to the outside air is 18,500 W.

Step-by-step solution

Convert the given values to consistent units

First, we need to convert the given values into a consistent unit system. Heat transfer rate is given in \(750 \, \mathrm{kJ/min}\). We will convert this value to watts \((\mathrm{W})\) by the following conversion: \(1 \, \mathrm{kJ/min} = \frac{1000 \, \mathrm{J}}{60 \, \mathrm{s}}\) So, \(750 \, \mathrm{kJ/min} = 750 \times \frac{1000}{60} \, \mathrm{W} = 12,500 \, \mathrm{W}\) Now, we have all values in watts.

Determine the COP (Coefficient of Performance)

The COP of a heat engine is defined as the ratio of the desired output (here, cooling effect) to the required input (here, electric power input). Given: Cooling effect \(Q_c = 12,500 \, \mathrm{W}\) Electric power input \(W = 6,000 \, \mathrm{W}\) (as \(6 \, \mathrm{kW} = 6,000 \, \mathrm{W}\)) COP can be calculated as follows: \(COP = \frac{Q_c}{W} = \frac{12,500}{6,000} = 2.08\) Hence, the COP of this air conditioner is \(2.08\).

Determine the heat transfer rate to the outside air

Now, we need to calculate the rate of heat transfer to the outside air. To do this, we will use the energy conservation principle. Energy conservation: $$Q_h = Q_c + W$$ Where: \(Q_h\) is the heat transfer rate to the outside air \(Q_c\) is the cooling effect (heat removed from the house) \(W\) is the electric power input We already have the values of \(Q_c\) and \(W\). So, we can find the value of \(Q_h\): \(Q_h = 12,500 + 6,000 = 18,500 \, \mathrm{W}\) Therefore, the rate of heat transfer to the outside air is \(18,500 \, \mathrm{W}\).

Key concepts

Understanding Heat Transfer Rate

The concept of heat transfer rate is central to many areas of thermodynamics, including the operation of air conditioners. It refers to the amount of thermal energy moved from one place to another over a certain period of time.

In the context of the exercise, the air conditioner is removing heat from the house at a rate of 750 kJ/min. This heat removal is what makes the house cooler. However, to discuss the rate in terms familiar to most systems, we convert kJ/min into watts, which is a more standard unit of power—representing energy per unit time. Think of the power as the speed of energy flow; more watts mean more energy is being moved quickly.

By converting the heat transfer rate to watts, we get a clear picture of the air conditioner's cooling power. This step is crucial for calculating other essential values such as the coefficient of performance (COP) and understanding how efficiently the air conditioner operates.

The Principle of Energy Conservation

The energy conservation principle is fundamental to all of physics and engineering. It simply states that energy cannot be created or destroyed, only transferred or transformed from one form to another.

When applied to air conditioners, we consider the electrical energy consumed (the work, W, that the air conditioner does) and the amount of heat energy it removes from a space (the cooling effect, Qc). According to the principle of energy conservation, the energy input into the system plus the system's output must equal the total change in energy. Mathematically, we express this balance as:

\
Q_h = Q_c + W\
,

where:\
- \(Q_h\) is the heat transfer rate to the outside air (the total energy output by the system)\
- \(Q_c\) is the cooling effect (the useful energy output)\
- W is the electric power input (the energy input).

In our exercise, by knowing \(Q_c\) and W, we applied this principle to find \(Q_h\), which tells us how much heat is actually being ejected into the outside air by the air conditioning unit.

Air Conditioning Thermodynamics

The field of air conditioning thermodynamics involves the study of heat and thermodynamic processes to cool and dehumidify air. An air conditioner's performance is commonly measured by its coefficient of performance (COP), a dimensionless number that indicates the efficiency of the cooling device.

The COP is defined as the ratio of the heat removed from the cooled space (\(Q_c\)) to the work (W) done by the air conditioner. In mathematical terms, the COP is given by:

\
COP = \frac{Q_c}{W}\
.

A higher COP indicates better efficiency, as it means the air conditioner is capable of removing more heat from the house per unit of electrical energy it consumes. In the exercise, we found a COP of 2.08, which means for every unit of electrical power consumed, the air conditioner removes a bit more than two units of heat from the house. Understanding thermodynamics concepts like heat transfer rate, energy conservation, and COP helps students grasp how air conditioning systems work and the principles behind their operation, which is essential for those studying the design and maintenance of HVAC (heating, ventilation, and air conditioning) systems.