/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 156 The condenser of a room air cond... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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

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

Step by step solution

01

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) = \(\dfrac{15,000 \times 1000}{3600} \mathrm{~W} = 4166.67 \mathrm{~W}\)
02

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 (\(\Delta T\)) = \(\dfrac{(40 - 25) + (40 - 35)}{2} = \dfrac{15 + 5}{2} = 10^{\circ} \mathrm{C}\)
03

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: \(A = \dfrac{4166.67}{150 \cdot 10} = \dfrac{4166.67}{1500} = 2.78 \mathrm{~m}^2\)
04

Final answer for heat transfer area

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

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider two double-pipe counter-flow heat exchangers that are identical except that one is twice as long as the other one. Which heat exchanger is more likely to have a higher effectiveness?

Cold water \(\left(c_{p}=4.18 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters a cross-flow heat exchanger at \(14^{\circ} \mathrm{C}\) at a rate of \(0.35 \mathrm{~kg} / \mathrm{s}\) where it is heated by hot air \(\left(c_{p}=1.0 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) that enters the heat exchanger at \(65^{\circ} \mathrm{C}\) at a rate of \(0.8 \mathrm{~kg} / \mathrm{s}\) and leaves at \(25^{\circ} \mathrm{C}\). Determine the maximum outlet temperature of the cold water and the effectiveness of this heat exchanger.

A counterflow double-pipe heat exchanger with \(A_{s}=\) \(9.0 \mathrm{~m}^{2}\) is used for cooling a liquid stream \(\left(c_{p}=3.15 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) at a rate of \(10.0 \mathrm{~kg} / \mathrm{s}\) with an inlet temperature of \(90^{\circ} \mathrm{C}\). The coolant \(\left(c_{p}=4.2 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters the heat exchanger at a rate of \(8.0 \mathrm{~kg} / \mathrm{s}\) with an inlet temperature of \(10^{\circ} \mathrm{C}\). The plant data gave the following equation for the overall heat transfer coefficient in W/m \({ }^{2} \cdot \mathrm{K}: U=600 /\left(1 / \dot{m}_{c}^{0.8}+2 / \dot{m}_{h}^{0.8}\right)\), where \(\dot{m}_{c}\) and \(\dot{m}_{h}\) are the cold-and hot-stream flow rates in kg/s, respectively. (a) Calculate the rate of heat transfer and the outlet stream temperatures for this unit. (b) The existing unit is to be replaced. A vendor is offering a very attractive discount on two identical heat exchangers that are presently stocked in its warehouse, each with \(A_{s}=5 \mathrm{~m}^{2}\). Because the tube diameters in the existing and new units are the same, the above heat transfer coefficient equation is expected to be valid for the new units as well. The vendor is proposing that the two new units could be operated in parallel, such that each unit would process exactly one-half the flow rate of each of the hot and cold streams in a counterflow manner; hence, they together would meet (or exceed) the present plant heat duty. Give your recommendation, with supporting calculations, on this replacement proposal.

Saturated liquid benzene flowing at a rate of \(5 \mathrm{~kg} / \mathrm{s}\) is to be cooled from \(75^{\circ} \mathrm{C}\) to \(45^{\circ} \mathrm{C}\) by using a source of cold water \(\left(c_{p}=4187 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) flowing at \(3.5 \mathrm{~kg} / \mathrm{s}\) and \(15^{\circ} \mathrm{C}\) through a \(20-\mathrm{mm}-\) diameter tube of negligible wall thickness. The overall heat transfer coefficient of the heat exchanger is estimated to be \(750 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the specific heat of the liquid benzene is \(1839 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) and assuming that the capacity ratio and effectiveness remain the same, determine the heat exchanger surface area for the following four heat exchangers: \((a)\) parallel flow, \((b)\) counter flow, \((c)\) shelland-tube heat exchanger with 2 -shell passes and 40-tube passes, and \((d)\) cross-flow heat exchanger with one fluid mixed (liquid benzene) and other fluid unmixed (water).

The cardiovascular counter-current heat exchanger has an overall heat transfer coefficient of \(100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Arterial blood enters at \(37^{\circ} \mathrm{C}\) and exits at \(27^{\circ} \mathrm{C}\). Venous blood enters at \(25^{\circ} \mathrm{C}\) and exits at \(34^{\circ} \mathrm{C}\). Determine the mass flow rates of the arterial blood and venous blood in \(\mathrm{g} / \mathrm{s}\) if the specific heat of both arterial and venous blood is constant and equal to \(3475 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and the surface area of the heat transfer to occur is \(0.15 \mathrm{~cm}^{2}\).

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.