A heat exchanger contains 400 tubes with inner diameter of \(23 \mathrm{~mm}\) and outer diameter of \(25 \mathrm{~mm}\). The length of each tube is \(3.7 \mathrm{~m}\). The corrected log mean temperature difference is \(23^{\circ} \mathrm{C}\), while the inner surface convection heat transfer coefficient is \(3410 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and the outer surface convection heat transfer coefficient is \(6820 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the thermal resistance of the tubes is negligible, determine the heat transfer rate.

Short Answer

Expert verified
Question: Determine the heat transfer rate in a heat exchanger with given dimensions and heat transfer coefficients. Answer: To find the heat transfer rate, follow these steps: 1. Calculate the inner and outer surface area per tube using the formula \(A=2\pi rL\). 2. Convert the surface areas to m² from mm². 3. Calculate the total inner and outer surface areas by multiplying the surface area per tube with the number of tubes (400). 4. Calculate the overall heat transfer coefficient using the equation \(\frac{1}{U_{total}} = \frac{1}{h_i} + \frac{1}{h_o}\). 5. Calculate the heat transfer rate using the equation \(Q = U_{total} \cdot A_{total\_inner} \cdot \Delta T_{lm}\). The heat transfer rate (Q) is the final answer.

Step by step solution

01

Calculate the inner and outer surface area per tube

To calculate the surface area per tube, we need to find the inner and outer surface area for a single tube. The formula for the surface area of a cylinder is \(A=2\pi rL\), where \(r\) is the radius and \(L\) is the length. Let's find the surface area for both surfaces: Inner surface area per tube, \(A_i\): $$A_i = 2 \pi \cdot \frac{23 \,\mathrm{mm}}{2} \cdot 3.7 \,\mathrm{m}$$ Outer surface area per tube, \(A_o\): $$A_o = 2 \pi \cdot \frac{25 \,\mathrm{mm}}{2} \cdot 3.7 \,\mathrm{m}$$
02

Convert the surface areas to m²

As the diameters are provided in millimeters, we should convert the surface areas from \(\mathrm{mm}^2\) to \(\mathrm{m}^2\): $$A_i = \frac{A_i}{1000^2} \,\mathrm{m}^2$$ $$A_o = \frac{A_o}{1000^2} \,\mathrm{m}^2$$
03

Calculate total inner and outer surface areas

Next, we will find the total inner and outer surface area by multiplying the surface area per tube with the number of tubes (400): $$A_{total\_inner} = 400 \cdot A_i \,\mathrm{m}^{2}$$ $$A_{total\_outer} = 400 \cdot A_o \,\mathrm{m}^{2}$$
04

Calculate the overall heat transfer coefficient

As the thermal resistance of the tubes is negligible, we can use the equation for the overall heat transfer coefficient (\(U_{total}\)) based on the inner and outer heat transfer coefficients (\(h_i\) and \(h_o\)) and the inner and outer surface areas (\(A_i\) and \(A_o\)): $$\frac{1}{U_{total}} = \frac{1}{h_i} + \frac{1}{h_o}$$ Now, solving for \(U_{total}\): $$U_{total} = \frac{1}{\left(\frac{1}{h_i} + \frac{1}{h_o}\right)}$$ $$U_{total}=\frac{1}{\left(\frac{1}{3410\,\mathrm{W/m^2K}}+\frac{1}{6820\,\mathrm{W/m^2K}}\right)}$$
05

Calculate the heat transfer rate

Now we can determine the heat transfer rate (\(Q\)) using the overall heat transfer coefficient, total surface area of the heat exchanger, and the corrected log mean temperature difference (\(\Delta T_{lm}\)): $$Q = U_{total} \cdot A_{total\_inner} \cdot \Delta T_{lm}$$ $$Q = U_{total} \cdot 400 \cdot A_i \cdot 23^{\circ} \mathrm{C}$$ The heat transfer rate \(Q\) is the final answer.

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!

Key Concepts

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

Heat Exchanger Analysis
Understanding how heat exchangers work is crucial when dealing with systems designed to transfer heat from one fluid to another. Heat exchangers are used in a variety of applications, including refrigeration, air conditioning systems, and power plants. The analysis begins by determining the surface area available for heat transfer, taking into account the number of tubes and their dimensions. With the surface areas in hand, the next steps focus on the qualities of the surfaces that affect heat transfer, such as the material properties and surface convection heat transfer coefficients.
Through the use of a meticulous step by step method, which covers calculating the individual and total surface area of the tubes, converting these areas to appropriate units, and understanding the role of the heat transfer coefficients, we ensure the accuracy of our findings. This methodical approach is vital when instructing students how to analyze heat exchangers effectively.
Convection Heat Transfer Coefficient
The convection heat transfer coefficient is a measure of the convective heat transfer between a solid surface and a fluid in contact with it. It's a critical aspect of evaluating how efficiently heat is transferred in a heat exchanger. For example, an inner surface convection heat transfer coefficient refers to the transfer on the fluid side that flows inside the tubes, while an outer surface coefficient refers to the transfer where the fluid flows on the outside.
To make studies easier to understand, it’s valuable to showcase real-life examples where these coefficients are used, such as in this exercise, where they directly influence the overall heat transfer coefficient calculation. Clarifying how these individual coefficients impact the overall efficiency drives home the significance of accurate measurement of the convection heat transfer coefficient.
Log Mean Temperature Difference
The log mean temperature difference (LMTD) is a pivotal concept that expresses the driving force behind the overall heat transfer in heat exchangers. It’s a theoretical temperature difference that averages the temperature gradients between the hot and cold sides over the length of the exchanger. Calculating the LMTD corrects for the variations in temperature difference that occur throughout the exchanger, providing a consistent value to use in heat transfer rate equations.
When explaining to students, use visuals or analogies, such as comparing the temperature difference to a car engine that drives the heat from the hot to the cold fluid. The greater the LMTD, the more 'horsepower' the exchanger has to transfer heat. Students should see the LMTD as a critical 'thermometer' that gauges the efficiency of their heat exchanger.
Overall Heat Transfer Coefficient
The overall heat transfer coefficient is an encompassing measure that integrates the different modes and paths of heat transfer into one coefficient. It reflects the total resistance to heat flow through the various layers in a heat exchanger, including conduction through the tube material and convection on both the inside and outside of the tubes.
In simpler terms, it's like a summary of how good an insulator or conductor the entire system is. A higher overall coefficient indicates a more effective heat exchanger. When guiding students, it's essential to emphasize the relationship between the overall coefficient and the individual convection heat transfer coefficients. They must understand that diminishing these individual resistances enhances overall heat transfer, analogous to widening a bottleneck to increase water flow. Breaking down complex equations into real-world applications can significantly improve student comprehension.

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 a water-to-water counter-flow heat exchanger with these specifications. Hot water enters at \(95^{\circ} \mathrm{C}\) while cold water enters at \(20^{\circ} \mathrm{C}\). The exit temperature of hot water is \(15^{\circ} \mathrm{C}\) greater than that of cold water, and the mass flow rate of hot water is 50 percent greater than that of cold water. The product of heat transfer surface area and the overall heat transfer coefficient is \(1400 \mathrm{~W} / \mathrm{K}\). Taking the specific heat of both cold and hot water to be \(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), determine (a) the outlet temperature of the cold water, \((b)\) the effectiveness of the heat exchanger, \((c)\) the mass flow rate of the cold water, and \((d)\) the heat transfer rate.

Hot water \(\left(c_{p h}=4188 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) with mass flow rate of \(2.5 \mathrm{~kg} / \mathrm{s}\) at \(100^{\circ} \mathrm{C}\) enters a thin-walled concentric tube counterflow heat exchanger with a surface area of \(23 \mathrm{~m}^{2}\) and an overall heat transfer coefficient of \(1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Cold water \(\left(c_{p c}=4178 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) with mass flow rate of \(5 \mathrm{~kg} / \mathrm{s}\) enters the heat exchanger at \(20^{\circ} \mathrm{C}\), determine \((a)\) the heat transfer rate for the heat exchanger and \((b)\) the outlet temperatures of the cold and hot fluids. After a period of operation, the overall heat transfer coefficient is reduced to \(500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine (c) the fouling factor that caused the reduction in the overall heat transfer coefficient.

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?

Consider a shell and tube heat exchanger in a milk be heated from \(20^{\circ} \mathrm{C}\) by hot water initially at \(140^{\circ} \mathrm{C}\) and flowing at a rate of \(5 \mathrm{~kg} / \mathrm{s}\). The milk flows through 30 thin-walled tubes with an inside diameter of \(20 \mathrm{~mm}\) with each tube making 10 passes through the shell. The average convective heat transfer coefficients on the milk and water side are \(450 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(1100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. In order to complete the pasteurizing process and hence restrict the microbial growth in the milk, it is required to have the exit temperature of milk attain at least \(70^{\circ} \mathrm{C}\). As a design engineer, your job is to decide upon the shell width (tube length in each pass) so that the milk exit temperature of \(70^{\circ} \mathrm{C}\) can be achieved. One of the design requirements is that the exit temperature of hot water should be at least \(10^{\circ} \mathrm{C}\) higher than the exit temperature of milk.

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.

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