/*! 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 84 Ethylene glycol and water, at 60... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 84

Ethylene glycol and water, at 60 and \(10^{\circ} \mathrm{C}\), respectively, enter a shell-and-tube heat exchanger for which the total heat transfer area is \(15 \mathrm{~m}^{2}\). With ethylene glycol and water flow rates of 2 and \(5 \mathrm{~kg} / \mathrm{s}\), respectively, the overall heat transfer coefficient is \(800 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Determine the rate of heat transfer and the fluid outlet temperatures. (b) Assuming all other conditions to remain the same, plot the effectiveness and fluid outlet temperatures as a function of the flow rate of ethylene glycol for \(0.5 \leq \dot{m}_{h} \leq 5 \mathrm{~kg} / \mathrm{s}\).

Short Answer

Expert verified
In summary, to find the heat transfer rate and fluid outlet temperatures in the shell-and-tube heat exchanger, first calculate the log mean temperature difference, then use energy balance equations to relate the heat transfer rate to the flow rates and specific heat capacities of the fluids. Finally, for part (b), repeat the steps for each value of ethylene glycol flow rate and plot effectiveness and outlet temperatures as a function of ethylene glycol flow rate.

Step by step solution

01

Determine the heat transfer rate

To find the heat transfer rate, we will use the formula: $ Q = UA \Delta T_m $ where \(Q\) is the heat transfer rate, \(U\) is the overall heat transfer coefficient, \(A\) is the heat transfer area, and \(\Delta T_m\) is the log mean temperature difference. First, we must find the log mean temperature difference, given by: $ \Delta T_m = \frac{(\Delta T_1 - \Delta T_2)}{\ln(\Delta T_1/\Delta T_2)} $ where \(\Delta T_1\) and \(\Delta T_2\) are the temperature differences at the inlet and outlet. Temperature differences at the inlet: $ \Delta T_1 = T_{h1} - T_{c1} = 60 - 10 = 50^{\circ}\mathrm{C} $ Temperature differences at the outlet: $ \Delta T_2 = T_{h2} - T_{c2} $ Since we don't know the outlet temperatures yet, we will reserve \(\Delta T_2\) for now. Next, we can plug the values we know into the formula for \(\Delta T_m\), as follows: $ \Delta T_m = \frac{(50 - \Delta T_2)}{\ln(50/\Delta T_2)} $ Now, we can find the heat transfer rate: $ Q = UA \Delta T_m = 800 \cdot 15 \cdot \frac{(50 - \Delta T_2)}{\ln(50/\Delta T_2)} = 12,000 \cdot \frac{(50 - \Delta T_2)}{\ln(50/\Delta T_2)} $
02

Determine the fluid outlet temperatures

Using energy balance equations, we can relate the heat transfer rate to the flow rates and specific heat capacities of the fluids. For the ethylene glycol: $ Q = \dot{m}_hC_{ph}(T_{h1} - T_{h2}) $ For the water: $ Q = \dot{m}_cC_{pc}(T_{c2} - T_{c1}) $ Using the relationship between heat transfer rate, flow rates, and specific heat capacities, we can rewrite the equations as: $ T_{h2} = T_{h1} - \frac{Q}{\dot{m}_hC_{ph}} $ $ T_{c2} = T_{c1} + \frac{Q}{\dot{m}_cC_{pc}} $ Now we can update the value of \(\Delta T_2\): $ \Delta T_2 = \left(T_{h1} - \frac{Q}{\dot{m}_hC_{ph}}\right) - \left(T_{c1} + \frac{Q}{\dot{m}_cC_{pc}}\right) $ Plug this value back into the equation for the heat transfer rate and solve for \(Q\). Once we have the value for \(Q\), we can find the outlet temperatures using the equations above.
03

Plot the effectiveness and outlet temperatures as a function of ethylene glycol flow rate

For part (b), we need to plot the following: 1. Effectiveness, given by: $ \varepsilon = \frac{Q}{Q_{\max}} = \frac{Q}{\dot{m}_hC_{ph}(T_{h1} - T_{c1})} $ 2. Fluid outlet temperatures, \(T_{h2}\) and \(T_{c2}\), as a function of ethylene glycol flow rate. We need to repeat steps 1 and 2 for each value of ethylene glycol flow rate in the interval \(0.5 \leq \dot{m}_{h} \leq 5 \mathrm{~kg} / \mathrm{s}\), and then plot effectiveness and outlet temperatures accordingly.

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.

Understanding Shell-and-Tube Heat Exchangers
Shell-and-tube heat exchangers are widely used in various industries for heating or cooling fluids. They consist of a bundle of tubes enclosed by a shell. One fluid flows through the tubes, referred to as the tube-side fluid, while another fluid flows outside the tubes but within the shell, known as the shell-side fluid.

These heat exchangers are designed to maximize heat transfer between the two fluids. The key variables in calculating the performance of a shell-and-tube heat exchanger include the flow rates of the fluids, the specific heat capacities of the fluids, the temperature differences between the fluids, and the heat transfer area of the tubes.

Ease of Maintenance and Versatility

One of the reasons shell-and-tube heat exchangers are so prevalent is their ease of maintenance. The tubes can often be cleaned, replaced, or repaired without needing to dismantle the entire unit. Moreover, they are versatile, capable of handling high pressures and a wide range of temperatures.

The heat transfer rate calculation in a shell-and-tube heat exchanger takes the specific characteristics of the exchange鈥攕uch as the thermal conductivities of the tube material鈥攊nto account, along with the aforementioned key variables.
Log Mean Temperature Difference (LMTD) Explained
The Log Mean Temperature Difference (LMTD) is a critical factor in the thermal performance calculation of a heat exchanger. It signifies the driving force behind the heat transfer process. In simple terms, LMTD represents an 'average' temperature difference between the hot and cold streams over the length of the exchanger.

To calculate LMTD, you need to know the inlet and outlet temperatures of both the hot and cold fluids. The formula for LMTD is as follows: \[\Delta T_m = \frac{(\Delta T_1 - \Delta T_2)}{\ln(\Delta T_1/\Delta T_2)}\]where \(\Delta T_1\) and \(\Delta T_2\) are the temperature differences between the hot and cold fluids at the inlet and outlet, respectively.

Importance in Heat Exchanger Design

Understanding and accurately determining LMTD is vital. Engineers use it to size heat exchangers and to estimate how effective a heat exchanger will be under certain operating conditions. It's a standard way of representing temperature difference when the temperatures at either end of the heat exchanger vary.
Overall Heat Transfer Coefficient (U-Factor)
The overall heat transfer coefficient, commonly referred to as U-factor, is a measure of the total thermal resistance between the two fluids in a heat exchanger. It incorporates the conductive, convective, and sometimes radiative heat transfer mechanisms. The U-factor is critical for determining the rate of heat transfer through the unit's surface area.

Mathematically, the heat transfer rate \(Q\) can be expressed using U-factor as:\[Q = UA\Delta T_m\]where \(A\) is the heat transfer area and \(\Delta T_m\) is the log mean temperature difference. The higher the U-factor, the more efficient the heat exchanger, as it implies lower thermal resistance and thus a greater capability to transfer heat.

Factors Affecting U-Factor

The U-factor can be influenced by several parameters, including the types of fluids involved, the velocity of the fluids, the nature of the flow (laminar or turbulent), and the material properties of the tubes. It's also affected by the cleanliness of the heat transfer surfaces鈥攆ouling can significantly reduce the U-factor over time.
Energy Balance Equations in Heat Exchangers
In the context of heat exchangers, energy balance equations are fundamental. They state that the rate of energy loss by the hot fluid is equal to the rate of energy gain by the cold fluid. This principle adheres to the law of conservation of energy and is fundamental in solving for unknown variables in heat exchanger calculations.

The general form of the energy balance equation for a fluid in a heat exchanger is given by:\[Q = \dot{m}C_p(T_{in} - T_{out})\]where \(Q\) is the rate of heat transfer, \(\dot{m}\) is the mass flow rate, \(C_p\) is the specific heat at constant pressure, and \((T_{in} - T_{out})\) are the inlet and outlet temperatures of the fluid.

Practical Application

In practice, if you know the mass flow rates and specific heat capacities of the fluids, as well as either the inlet or outlet temperatures, you can calculate the heat transfer rate and the other unknown temperatures using these energy balance equations. They are pivotal for design, analysis, and troubleshooting of heat exchangers in any thermal system.

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

The human brain is especially sensitive to elevated temperatures. The cool blood in the veins leaving the face and neck and returning to the heart may contribute to thermal regulation of the brain by cooling the arterial blood flowing to the brain. Consider a vein and artery running between the chest and the base of the skull for a distance \(L=250 \mathrm{~mm}\), with mass flow rates of \(3 \times 10^{-3} \mathrm{~kg} / \mathrm{s}\) in opposite directions in the two vessels. The vessels are of diameter \(D=5 \mathrm{~mm}\) and are separated by a distance \(w=7 \mathrm{~mm}\). The thermal conductivity of the surrounding tissue is \(k_{\mathrm{r}}=0.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). If the arterial blood enters at \(37^{\circ} \mathrm{C}\) and the venous blood enters at \(27^{\circ} \mathrm{C}\), at what temperature will the arterial blood exit? If the arterial blood becomes overheated, and the body responds by halving the blood flow rate, how much hotter can the entering arterial blood be and still maintain its exit temperature below \(37^{\circ} \mathrm{C}\) ? Hint: If we assume that all the heat leaving the artery enters the vein, then heat transfer between the two vessels can be modeled using a relationship found in Table 4.1. Approximate the blood properties as those of water.

As part of a senior project, a student was given the assignment to design a heat exchanger that meets the following specifications: \begin{tabular}{lccc} \hline & \(\dot{m}(\mathrm{~kg} / \mathrm{s})\) & \(T_{m, i}\left({ }^{\circ} \mathrm{C}\right)\) & \(T_{m, \theta}\left({ }^{\circ} \mathrm{C}\right)\) \\ \hline Hot water & 28 & 90 & \(-\) \\ Cold water & 27 & 34 & 60 \\ \hline \end{tabular} Like many real-world situations, the customer hasn't revealed, or doesn't know, additional requirements that would allow you to proceed directly to a final configuration. At the outset, it is helpful to make a first-cut design based upon simplifying assumptions, which can be evaluated to determine what additional requirements and trade-offs should be considered by the customer. (a) Design a heat exchanger to meet the foregoing specifications. List and explain your assumptions. Hint: Begin by finding the required value for \(U A\) and using representative values of \(U\) to determine \(A\). (b) Evaluate your design by identifying what features and configurations could be explored with your customer in order to develop more complete specifications.

A plate-fin heat exchanger is used to condense a saturated refrigerant vapor in an air-conditioning system. The vapor has a saturation temperature of \(45^{\circ} \mathrm{C}\), and a condensation rate of \(0.015 \mathrm{~kg} / \mathrm{s}\) is dictated by system performance requirements. The frontal area of the condenser is fixed at \(A_{\mathrm{fr}}=0.25 \mathrm{~m}^{2}\) by installation requirements, and a value of \(h_{f g}=135 \mathrm{~kJ} / \mathrm{kg}\) may be assumed for the refrigerant. (a) The condenser design is to be based on a nominal air inlet temperature of \(T_{c, i}=30^{\circ} \mathrm{C}\) and nominal air inlet velocity of \(V=2 \mathrm{~m} / \mathrm{s}\) for which the manufacturer of the heat exchanger core indicates an overall coefficient of \(U=50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). What is the corresponding value of the heat transfer surface area required to achieve the prescribed condensation rate? What is the air outlet temperature? (b) From the manufacturer of the heat exchanger core, it is also known that \(U \propto V^{0 . t}\). During daily operation the air inlet temperature is not controllable and may vary from 27 to \(38^{\circ} \mathrm{C}\). If the heat exchanger area is fixed by the result of part (a), what is the range of air velocities needed to maintain the prescribed condensation rate? Plot the velocity as a function of the air inlet temperature.

A cross-flow heat exchanger used in a cardiopulmonary bypass procedure cools blood flowing at \(5 \mathrm{~L} / \mathrm{min}\) from a body temperature of \(37^{\circ} \mathrm{C}\) to \(25^{\circ} \mathrm{C}\) in order to induce body hypothermia, which reduces metabolic and oxygen requirements. The coolant is ice water at \(0^{\circ} \mathrm{C}\), and its flow rate is adjusted to provide an outlet temperature of \(15^{\circ} \mathrm{C}\). The heat exchanger operates with both fluids unmixed, and the overall heat transfer coefficient is \(750 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The density and specific heat of the blood are \(1050 \mathrm{~kg} / \mathrm{m}^{3}\) and \(3740 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), respectively. a) Determine the heat transfer rate for the exchanger. b) Calculate the water flow rate. c) What is the surface area of the heat exchanger? d) Calculate and plot the blood and water outlet temperatures as a function of the water flow rate for the range 2 to \(4 \mathrm{~L} / \mathrm{min}\), assuming all other parameters remain unchanged. Comment on how the changes in the outlet temperatures are affected by changes in the water flow rate. Explain this behavior and why it is an advantage for this application.

A recuperator is a heat exchanger that heats the air used in a combustion process by extracting energy from the products of combustion (the flue gas). Consider using a single-pass, cross-flow heat exchanger as a recuperator. Eighty \((80)\) silicon carbide ceramic tubes \((k=20\) \(\mathrm{W} / \mathrm{m} \cdot \mathrm{K})\) of inner and outer diameters equal to 55 and \(80 \mathrm{~mm}\), respectively, and of length \(L=1.4 \mathrm{~m}\) are arranged as an aligned tube bank of longitudinal and transverse pitches \(S_{L}=100 \mathrm{~mm}\) and \(S_{T}=120 \mathrm{~mm}\), respectively. Cold air is in cross flow over the tube bank with upstream conditions of \(V=1 \mathrm{~m} / \mathrm{s}\) and \(T_{c i}=300 \mathrm{~K}\), while hot flue gases of inlet temperature \(T_{\mathrm{h}, \mathrm{I}}=1400 \mathrm{~K}\) pass through the tubes. The tube outer surface is clean, while the inner surface is characterized by a fouling factor of \(R_{f}^{N \prime}=2 \times 10^{-4} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). The air and flue gas flow rates are \(\dot{m}_{c}=1.0 \mathrm{~kg} / \mathrm{s}\) and \(m_{\mathrm{h}}=1.05 \mathrm{~kg} / \mathrm{s}\), respectively. As first approximations, (1) evaluate all required air properties at \(1 \mathrm{~atm}\) and \(300 \mathrm{~K},(2)\) assume the flue gas to have the properties of air at \(1 \mathrm{~atm}\) and \(1400 \mathrm{~K}\), and (3) assume the tube wall temperature to be at \(800 \mathrm{~K}\) for the purpose of treating the effect of variable properties on convection heat transfer. (a) If there is a \(1 \%\) fuel savings associated with each \(10^{\circ} \mathrm{C}\) increase in the temperature of the combustion air \(\left(T_{c o}\right)\) above \(300 \mathrm{~K}\), what is the percentage fuel savings for the prescribed conditions? (b) The performance of the recuperator is strongly influenced by the product of the overall heat transfer coefficient and the total surface area, UA. Compute and plot \(T_{c, \infty}\) and the percentage fuel savings as a function of UA for \(300 \leq U A \leq 600 \mathrm{~W} / \mathrm{K}\). Without changing the flow rates, what measures may be taken to increase \(U A^{*}\) ?
See all solutions

Recommended explanations on Physics Textbooks

Linear Momentum

Read Explanation

Measurements

Read Explanation

Particle Model of Matter

Read Explanation

Electric Charge Field and Potential

Read Explanation

Space Physics

Read Explanation

Circular Motion and Gravitation

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.