What is a regenerative heat exchanger? How does a static type of regenerative heat exchanger differ from a dynamic type?

Short Answer

Expert verified
Answer: A regenerative heat exchanger, also known as a regenerator, is a type of heat exchanger where heat from a hot fluid is stored and later transferred to a cold fluid. The main purpose is to improve the thermal efficiency of a system by reducing heat loss or waste heat. The key differences between static and dynamic types are the motion of the heat storage matrix and their efficiency. In static types, the matrix remains stationary, and the fluids flow alternately through it, while in dynamic types, the matrix is in continuous motion, transferring heat between the fluids as it moves. Static regenerative heat exchangers typically have lower capital cost and easier maintenance but may have lower efficiency compared to dynamic ones, which have higher capital costs and may need more maintenance but can achieve higher heat transfer efficiency.

Step by step solution

01

Regenerative Heat Exchanger

A regenerative heat exchanger, also known as a regenerator, is a type of heat exchanger where heat from a hot fluid is stored and later transferred to a cold fluid. The main purpose of a regenerative heat exchanger is to improve the thermal efficiency of a system by reducing the heat loss or waste heat. Regenerative heat exchangers are often used in industries and power plants where energy saving and efficiency are essential.
02

Static Type of Regenerative Heat Exchanger

A static type of regenerative heat exchanger is a system where the heat storage matrix remains stationary during the heat exchange process. In this type, the hot and cold fluids flow alternately through the heat storage matrix, which is typically made of a solid material like metal or ceramic. The heat from the hot fluid is absorbed by the matrix, which later releases the stored heat to the cold fluid as it flows through the matrix. An example of a static regenerator is a fixed matrix heat exchanger, in which the matrix does not move and the flow direction alternates between the hot and cold fluids.
03

Dynamic Type of Regenerative Heat Exchanger

A dynamic type of regenerative heat exchanger, on the other hand, is a system where the heat storage matrix is in continuous motion during the heat exchange process. In this type, the matrix rotates between the hot and cold fluid streams, allowing heat transfer between the two as it moves. An example of a dynamic regenerator is a rotary regenerator, in which a rotating matrix wheel transfers heat between the hot and cold fluids.
04

Comparing Static and Dynamic Regenerative Heat Exchangers

The primary difference between static and dynamic regenerative heat exchangers lies in the motion of the heat storage matrix. In a static type, the matrix remains stationary, and the fluids flow alternately through it. In contrast, a dynamic type involves the continuous motion of the matrix, transferring heat between the fluids as it moves. Another significant difference is that static regenerative heat exchangers typically have lower capital cost and easier maintenance due to their simple design. However, they may suffer from lower efficiency and effectiveness compared to dynamic regenerative heat exchangers. Dynamic regenerative heat exchangers can achieve higher heat transfer efficiency due to the continuous heat exchange resulting from the motion of the matrix. However, they generally have higher capital costs, require more robust materials to withstand the motion, and may need more maintenance due to their more complex design.

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 Vaia!

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

Oil in an engine is being cooled by air in a crossflow heat exchanger, where both fluids are unmixed. Oil \(\left(c_{p h}=\right.\) \(2047 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K})\) flowing with a flow rate of \(0.026 \mathrm{~kg} / \mathrm{s}\) enters the tube side at \(75^{\circ} \mathrm{C}\), while air \(\left(c_{p c}=1007 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters the shell side at \(30^{\circ} \mathrm{C}\) with a flow rate of \(0.21 \mathrm{~kg} / \mathrm{s}\). The overall heat transfer coefficient of the heat exchanger is \(53 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and the total surface area is \(1 \mathrm{~m}^{2}\). If the correction factor is \(F=\) \(0.96\), determine the outlet temperatures of the oil and air.

Consider the flow of saturated steam at \(270.1 \mathrm{kPa}\) that flows through the shell side of a shell-and-tube heat exchanger while the water flows through 4 tubes of diameter \(1.25 \mathrm{~cm}\) at a rate of \(0.25 \mathrm{~kg} / \mathrm{s}\) through each tube. The water enters the tubes of heat exchanger at \(20^{\circ} \mathrm{C}\) and exits at \(60^{\circ} \mathrm{C}\). Due to the heat exchange with the cold fluid, steam is condensed on the tubes external surface. The convection heat transfer coefficient on the steam side is \(1500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), while the fouling resistance for the steam and water may be taken as \(0.00015\) and \(0.0001 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\), respectively. Using the NTU method, determine \((a)\) effectiveness of the heat exchanger, \((b)\) length of the tube, and \((c)\) rate of steam condensation.

Saturated water vapor at \(100^{\circ} \mathrm{C}\) condenses in a 1 -shell and 2-tube heat exchanger with a surface area of \(0.5 \mathrm{~m}^{2}\) and an overall heat transfer coefficient of \(2000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Cold water \(\left(c_{p c}=4179 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) flowing at \(0.5 \mathrm{~kg} / \mathrm{s}\) enters the tube side at \(15^{\circ} \mathrm{C}\), determine \((a)\) the heat transfer effectiveness, \((b)\) the outlet temperature of the cold water, and \((c)\) the heat transfer rate for the heat exchanger.

Consider a heat exchanger that has an NTU of \(0.1\). Someone proposes to triple the size of the heat exchanger and thus triple the NTU to \(0.3\) in order to increase the effectiveness of the heat exchanger and thus save energy. Would you support this proposal?

A shell-and-tube heat exchanger is used for heating \(10 \mathrm{~kg} / \mathrm{s}\) of oil \(\left(c_{p}=2.0 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\) from \(25^{\circ} \mathrm{C}\) to \(46^{\circ} \mathrm{C}\). The heat exchanger has 1 -shell pass and 6-tube passes. Water enters the shell side at \(80^{\circ} \mathrm{C}\) and leaves at \(60^{\circ} \mathrm{C}\). The overall heat transfer coefficient is estimated to be \(1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Calculate the rate of heat transfer and the heat transfer area.

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.

Sign-up for free