/*! 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 35 In the heat transfer relation \(... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 35

In the heat transfer relation \(\dot{Q}=U A_{s} \Delta T_{\operatorname{lm}}\) for a heat exchanger, what is \(\Delta T_{\operatorname{lm}}\) called? How is it calculated for a parallel-flow and counter-flow heat exchanger?

Short Answer

Expert verified
To summarize, the term \(\Delta T_{\operatorname{lm}}\) represents the log mean temperature difference, which is essential for heat transfer calculations in heat exchangers. It helps account for the varying temperature differences between the hot and cold fluids within the exchanger. Two types of heat exchangers, parallel-flow and counter-flow, have different formulas to calculate \(\Delta T_{\operatorname{lm}}\). For both types, the formula is: $$ \Delta T_{\operatorname{lm}} = \frac{\Delta T_1 - \Delta T_2}{\ln{\frac{\Delta T_1}{\Delta T_2}}}, $$ with \(\Delta T_1\) and \(\Delta T_2\) representing temperature differences at different points in the heat exchanger. Finally, the log mean temperature difference (\(\Delta T_{\operatorname{lm}}\)) is used in the heat transfer relation to calculate the heat transfer rate.

Step by step solution

01

Definition of \(\Delta T_{\operatorname{lm}}\)

The term \(\Delta T_{\operatorname{lm}}\) stands for "log mean temperature difference." It is an essential parameter in the heat transfer calculations of a heat exchanger and represents the average temperature difference between the hot and cold fluids across the heat exchanger. It takes into account the varying temperature differences across the exchanger and averages them for more accurate calculations.
02

Parallel-flow heat exchanger

In a parallel-flow heat exchanger, the hot and cold fluids flow in the same direction. To calculate the log mean temperature difference for a parallel-flow heat exchanger, use the following formula: $$ \Delta T_{\operatorname{lm}, \parallel} = \frac{\Delta T_1 - \Delta T_2}{\ln{\frac{\Delta T_1}{\Delta T_2}}}, $$ where \(\Delta T_1\) is the temperature difference between the hot and cold fluids at the inlet (entrance) of the heat exchanger, and \(\Delta T_2\) is the temperature difference between the hot and cold fluids at the outlet (exit) of the heat exchanger.
03

Counter-flow heat exchanger

In a counter-flow heat exchanger, the hot and cold fluids flow in opposite directions. To calculate the log mean temperature difference for a counter-flow heat exchanger, use the following formula: $$ \Delta T_{\operatorname{lm}, \text{counter}} = \frac{\Delta T_1 - \Delta T_2}{\ln{\frac{\Delta T_1}{\Delta T_2}}}, $$ where \(\Delta T_1\) is the temperature difference between the hot and cold fluids at one end of the heat exchanger, and \(\Delta T_2\) is the temperature difference between the hot and cold fluids at the other end of the heat exchanger. In both cases, the log mean temperature difference (\(\Delta T_{\operatorname{lm}}\)) is used in the heat transfer relation, given by: $$ \dot{Q} = U A_s \Delta T_{\operatorname{lm}}, $$ where \(\dot{Q}\) is the heat transfer rate, \(U\) is the overall heat transfer coefficient, and \(A_s\) is the heat transfer surface area. By understanding the differences between parallel-flow and counter-flow heat exchangers and their respective formulas to calculate \(\Delta T_{\operatorname{lm}}\), the student can effectively perform accurate heat transfer calculations for different heat exchanger configurations.

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 91影视!

Key Concepts

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

Heat Exchanger Efficiency
Heat exchanger efficiency is a critical measure of how well a heat exchanger performs its tasks of transferring heat between two or more fluids. This efficiency can be defined as the ratio of the actual heat transfer to the maximum possible heat transfer, given the same inlet and outlet temperatures for the hot and cold fluids. An efficient heat exchanger is one that transfers the maximum amount of heat with the least amount of energy loss.

Key factors impacting the efficiency of a heat exchanger include the design of the unit (parallel-flow or counter-flow), the materials used for construction, the surface area available for heat transfer, and the overall heat transfer coefficient. By optimizing and maintaining these aspects, heat exchangers can operate with greater efficiency, saving energy and reducing costs.
Parallel-Flow Heat Exchanger
In a parallel-flow heat exchanger, the fluids flow in the same direction, entering and exiting the exchanger side by side. This setup means the temperature gradient between the fluids is greatest at the entrance and decreases along the flow path. As a consequence, the efficiency of heat exchange may be lower compared to the counter-flow configuration since the temperature difference, which drives heat transfer, diminishes along the exchanger.

When calculating the efficiency or performance of a parallel-flow heat exchanger, it is essential to consider not only the log mean temperature difference \(\Delta T_{\operatorname{lm}, \parallel}\) but also recognize that this type of exchanger may require more surface area or a higher overall heat transfer coefficient to achieve similar performance to a counter-flow design under the same conditions.
Counter-Flow Heat Exchanger
The counter-flow heat exchanger stands apart due to its fluids moving in opposite directions. This arrangement provides a more consistent temperature gradient between the hot and cold fluids across the length of the exchanger, which typically allows for better heat transfer efficiency than in parallel-flow design. The counter-flow arrangement maintains a higher driving force for heat transfer throughout the device, making it effective for systems where space is limited or where the greatest energy recovery is desired.

A counter-flow heat exchanger can often achieve a closer approach temperature 鈥� the minimum temperature difference between the outlet temperatures of the fluids 鈥� which is an indicator of a highly efficient heat exchange process. Understanding the calculation of log mean temperature difference \(\Delta T_{\operatorname{lm}, \text{counter}}\) for a counter-flow heat exchanger is fundamental in designing and optimizing these systems.
Overall Heat Transfer Coefficient
The overall heat transfer coefficient, denoted as \(U\), is a composite measure that describes how well a heat exchanger can move heat between two fluids. It incorporates the characteristics of the exchanger's materials, the surface area through which the heat is being transferred, and the nature of the fluids involved (including flow configuration and properties).

The coefficient \(U\) plays a vital role in the equation \(\dot{Q} = U A_s \Delta T_{\operatorname{lm}}\) because it directly affects the heat transfer rate \(\dot{Q}\). A higher value of \(U\) indicates better efficiency and effectiveness in transferring heat. To maximize the performance of a heat exchanger, engineers work to optimize the overall heat transfer coefficient by selecting appropriate materials, maximizing surface area, and designing for the most efficient flow configuration.

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

A shell-and-tube heat exchanger with 2-shell passes and 12 -tube passes is used to heat water \(\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) with ethylene glycol \(\left(c_{p}=2680 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\). Water enters the tubes at \(22^{\circ} \mathrm{C}\) at a rate of \(0.8 \mathrm{~kg} / \mathrm{s}\) and leaves at \(70^{\circ} \mathrm{C}\). Ethylene \(\mathrm{glycol}\) enters the shell at \(110^{\circ} \mathrm{C}\) and leaves at \(60^{\circ} \mathrm{C}\). If the overall heat transfer coefficient based on the tube side is \(280 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the rate of heat transfer and the heat transfer surface area on the tube side.

For a specified fluid pair, inlet temperatures, and mass flow rates, what kind of heat exchanger will have the highest effectiveness: double-pipe parallel- flow, double-pipe counterflow, cross-flow, or multipass shell-and-tube heat exchanger?

Water is boiled at \(150^{\circ} \mathrm{C}\) in a boiler by hot exhaust gases \(\left(c_{p}=1.05 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\right)\) that enter the boiler at \(400^{\circ} \mathrm{C}\) at a rate of \(0.4 \mathrm{~kg} / \mathrm{s}\) and leaves at \(200^{\circ} \mathrm{C}\). The surface area of the heat exchanger is \(0.64 \mathrm{~m}^{2}\). The overall heat transfer coefficient of this heat exchanger is (a) \(940 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (b) \(1056 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (c) \(1145 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (d) \(1230 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) (e) \(1393 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\)

A performance test is being conducted on a double pipe counter flow heat exchanger that carries engine oil and water at a flow rate of \(2.5 \mathrm{~kg} / \mathrm{s}\) and \(1.75 \mathrm{~kg} / \mathrm{s}\), respectively. Since the heat exchanger has been in service over a long period of time it is suspected that the fouling might have developed inside the heat exchanger that might have affected the overall heat transfer coefficient. The test to be carried out is such that, for a designed value of the overall heat transfer coefficient of \(450 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and a surface area of \(7.5 \mathrm{~m}^{2}\), the oil must be heated from \(25^{\circ} \mathrm{C}\) to \(55^{\circ} \mathrm{C}\) by passing hot water at \(100^{\circ} \mathrm{C}\left(c_{p}=4206 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) at the flow rates mentioned above. Determine if the fouling has affected the overall heat transfer coefficient. If yes, then what is the magnitude of the fouling resistance?

A counter-flow heat exchanger is used to cool oil \(\left(c_{p}=\right.\) \(2.20 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})\) from \(110^{\circ} \mathrm{C}\) to \(85^{\circ} \mathrm{C}\) at a rate of \(0.75 \mathrm{~kg} / \mathrm{s}\) by cold water \(\left(c_{p}=4.18 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) that enters the heat exchanger at \(20^{\circ} \mathrm{C}\) at a rate of \(0.6 \mathrm{~kg} / \mathrm{s}\). If the overall heat transfer coefficient is \(800 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), the heat transfer area of the heat exchanger is (a) \(0.745 \mathrm{~m}^{2}\) (b) \(0.760 \mathrm{~m}^{2}\) (c) \(0.775 \mathrm{~m}^{2}\) (d) \(0.790 \mathrm{~m}^{2}\) (e) \(0.805 \mathrm{~m}^{2}\)
See all solutions

Recommended explanations on Physics Textbooks

Waves Physics

Read Explanation

Wave Optics

Read Explanation

Translational Dynamics

Read Explanation

Solid State Physics

Read Explanation

Space Physics

Read Explanation

Geometrical and Physical Optics

Read Explanation
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.