Question

The condenser of a room air conditioner is designed to reject heat at a rate of \(15,000 \mathrm{~kJ} / \mathrm{h}\) from refrigerant-134a as the refrigerant is condensed at a temperature of \(40^{\circ} \mathrm{C}\). Air \(\left(c_{p}=1005 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) flows across the finned condenser coils, entering at \(25^{\circ} \mathrm{C}\) and leaving at \(35^{\circ} \mathrm{C}\). If the overall heat transfer coefficient based on the refrigerant side is \(150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the heat transfer area on the refrigerant side.

Short answer

Answer: The heat transfer area on the refrigerant side is approximately 2.78 m².

Step-by-step solution

Convert heat rejection rate to Watts

We are given that the heat rejection rate is \(15,000 \mathrm{~kJ} / \mathrm{h}\). To convert this value to Watts, we will use the following conversion factors: 1 kJ = 1000 J 1 h = 3600 s So, the heat rejection rate in Watts can be calculated as: Heat rejection rate (W) = \(\dfrac{15,000 \times 1000}{3600} \mathrm{~W} = 4166.67 \mathrm{~W}\)

Calculate the temperature difference

We are given that the air enters the condenser at a temperature of \(25^{\circ} \mathrm{C}\) and leaves at \(35^{\circ} \mathrm{C}\). Since the refrigerant is at a constant temperature of \(40^{\circ} \mathrm{C}\), the average temperature difference between the refrigerant and the air is: Temperature difference (\(\Delta T\)) = \(\dfrac{(40 - 25) + (40 - 35)}{2} = \dfrac{15 + 5}{2} = 10^{\circ} \mathrm{C}\)

Apply the heat transfer equation

The rate of heat transfer (\(Q\)) can be written as: \(Q = U \cdot A \cdot \Delta T\) where: - \(Q\): heat transfer rate - \(U\): overall heat transfer coefficient - \(A\): heat transfer area - \(\Delta T\): temperature difference between the refrigerant and the air We need to find the heat transfer area \(A\). So, we will rearrange the equation above to get: \(A = \dfrac{Q}{U \cdot \Delta T}\) Now, we can substitute the values we have found in Step 1 and Step 2, and the given overall heat transfer coefficient, into the equation above to calculate the heat transfer area: \(A = \dfrac{4166.67}{150 \cdot 10} = \dfrac{4166.67}{1500} = 2.78 \mathrm{~m}^2\)

Final answer for heat transfer area

The heat transfer area on the refrigerant side is approximately 2.78 square meters.

Key concepts

Refrigerant-134a

Refrigerant-134a is a commonly used refrigerant in many air conditioning and refrigeration systems. Developed as a replacement for older refrigerants, it offers several benefits including low toxicity, non-flammability, and relatively lower Global Warming Potential (GWP) compared to its predecessors. Understanding Refrigerant-134a is key, especially when dealing with heat transfer scenarios in equipment like a condenser. In the exercise, as the refrigerant condenses at a temperature of 40°C, it reaches a phase change. This phase change is crucial for efficient heat rejection, as it allows substantial amounts of heat to be rejected without large changes in temperature, maintaining optimal performance of the cooling system.

Overall Heat Transfer Coefficient

The overall heat transfer coefficient (U) is a measure of a system's ability to conduct heat. For our exercise, U is given as 150 W/m²·K. This coefficient combines all the resistances to heat flow throughout different layers and interfaces. The concept can be likened to how well a thermal "blanket" lets heat pass through, considering all materials involved. The higher the value, the better the system is at transferring heat. This coefficient is vital in determining the necessary heat transfer area in the condenser design, directly influencing efficiency. So, understanding and calculating U helps in designing systems that can effectively handle or reject heat, guiding the design of components like a condenser.

Temperature Difference

Temperature difference (\(\Delta T\)) is a critical factor in the heat transfer process. It represents the driving force for heat flow from the hot refrigerant to the cooler air. In the given problem, \(\Delta T\) is calculated by averaging the difference between the refrigerant temperature (40°C) and the temperatures of air entering (25°C) and leaving (35°C) the system. This average temperature difference, calculated as 10°C, reflects the effective difference that influences the rate of heat transfer. The bigger the temperature difference, the more efficient the heat transfer process becomes. Recognizing and calculating this difference ensures that the condenser operates at its design intent, maximizing heat rejection efficiency.

Condenser Design

Designing a condenser involves carefully balancing various parameters to ensure efficient heat rejection. In the context of this exercise, the design primarily focuses on the heat transfer area required based on the overall heat transfer coefficient and the temperature difference. A well-designed condenser effectively removes unwanted heat from the refrigerant, allowing it to cool and condense. This involves using appropriate materials with high thermal conductivity, optimal fin design for enhanced air flow, and calculating the precise surface area needed. Here, we calculated a heat transfer area of approximately 2.78 m². Such calculations are fundamental in ensuring that the condenser runs efficiently, as they allow for appropriate sizing and selection of materials, enhancing system reliability and performance.