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A cross-flow heat exchanger with both fluids unmixed has an overall heat transfer coefficient of \(200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and a heat transfer surface area of \(400 \mathrm{~m}^{2}\). The hot fluid has a heat capacity of \(40,000 \mathrm{~W} / \mathrm{K}\), while the cold fluid has a heat capacity of \(80,000 \mathrm{~W} / \mathrm{K}\). If the inlet temperatures of both hot and cold fluids are \(80^{\circ} \mathrm{C}\) and \(20^{\circ} \mathrm{C}\), respectively, determine the exit temperature of the cold fluid.

Short Answer

Expert verified
Based on the given information, calculate the exit temperature of the cold fluid in the heat exchanger. Solution: Step 1: Calculate the total heat exchanged in the heat exchanger: \(Q = U \cdot A \cdot \Delta T = 200 \frac{\mathrm{W}}{\mathrm{m}^2 \cdot \mathrm{K}} \times 400 \mathrm{~m}^{2} \times (80 \mathrm{~^\circ C} - 20 \mathrm{~^\circ C})\) Step 2: Use the heat exchange formula to find the exit temperature of the cold fluid: \(T_{cold, exit} = \frac{Q}{80000 \frac{\mathrm{W}}{\mathrm{K}}} + 20 \mathrm{~^\circ C}\) The exit temperature of the cold fluid is \(T_{cold, exit}\).

Step by step solution

01

Calculate the heat exchanged in the heat exchanger

First, we need to compute the total amount of heat exchanged by both fluids in the heat exchanger since the heat exchange is equal on both sides. Use the given overall heat transfer coefficient, surface area for heat transfer, and the temperature difference between the inlet temperatures of the hot and cold fluids: \(Q = U \cdot A \cdot \Delta T\) \(Q = 200 \frac{\mathrm{W}}{\mathrm{m}^2 \cdot \mathrm{K}} \times 400 \mathrm{~m}^{2} \times (80 \mathrm{~^\circ C} - 20 \mathrm{~^\circ C})\) \(Q = W\)
02

Use the heat exchange formula to find the exit temperature of the cold fluid

Now we can use the heat exchange formula, using the heat capacity rate of the cold fluid and the inlet temperature of the cold fluid, and the known total heat exchange (\(Q\)) value from Step 1: \(Q = C_{cold}(T_{cold, exit} - T_{cold, inlet})\) \(W = 80000 \frac{\mathrm{W}}{\mathrm{K}} \times (T_{cold, exit} - 20 \mathrm{~^\circ C})\) Now rearrange the equation to solve for \(T_{cold, exit}\): \(T_{cold, exit} = \frac{W}{80000 \frac{\mathrm{W}}{\mathrm{K}}} + 20 \mathrm{~^\circ C}\) \(T_{cold, exit} = T\) The exit temperature of the cold fluid is \(T\).

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Key Concepts

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

Cross-flow Heat Exchanger
In the field of thermal engineering, heat exchangers are vital components used to transfer heat between two or more fluids. A cross-flow heat exchanger is a specific type in which the fluids move perpendicular to each other. This configuration is common in air conditioning systems and automotive radiators. Here, neither fluid mixes, which is critical for maintaining separate flow paths and ensuring efficient heat transfer.
Cross-flow heat exchangers are popular because they can achieve a better balance between heat transfer and pressure drop. This means they can effectively manage the flow resistance introduced by the fluid. They often consist of a series of ducts and fins arranged to maximize the surface area available for heat exchange. This arrangement enhances the effectiveness of heat transfer between the two fluids.
Understanding how these heat exchangers function helps us apply them in various applications, ensuring optimal energy efficiency and system functionality.
Overall Heat Transfer Coefficient
The overall heat transfer coefficient (\(U\)) plays a crucial role in determining the efficiency of a heat exchanger. It's a measure of how effectively heat is transferred from one fluid to another across the exchanger surface. This coefficient considers several variables:
  • Properties of the fluids involved
  • Type and thickness of the material used in the exchanger
  • The cleanliness of the surfaces

In our given example, the overall heat transfer coefficient is \(200 \, \mathrm{W/m^2 \cdot K}\), implying a moderate level of efficiency for transferring heat. This parameter is essential because a higher coefficient signifies that the exchanger can transfer heat more efficiently. It's a critical factor in the design and analysis of heat exchange systems, impacting cost-effectiveness and energy usage.
Keeping the heat exchange surfaces clean and maintaining materials in good condition can positively influence the values of the overall heat transfer coefficient.
Heat Capacity
Heat capacity is a vital component when examining the performance of heat exchangers. It's defined as the amount of heat required to change the temperature of a given amount of substance by 1°C. In the context of a cross-flow heat exchanger, both the hot and cold fluids have specific heat capacities, affecting how heat is absorbed or released as they pass through the exchanger.
In the exercise example, the hot fluid has a heat capacity of \(40,000 \, \mathrm{W/K}\), while the cold fluid is twice that at \(80,000 \, \mathrm{W/K}\). This difference indicates that the cold fluid can absorb more heat per degree of temperature change compared to the hot fluid. Knowing the heat capacities of the fluids involved is crucial for calculating the exit temperatures after heat exchange has occurred.
This concept helps engineers design systems that maximize heat exchange efficiency by considering the specific heat capacities of the fluids used.
Temperature Difference
Temperature difference (\(\Delta T\)) is a fundamental factor in the operation of a heat exchanger. In essence, it's the driving force for heat transfer, with a greater temperature difference leading to higher rates of heat transfer. In our scenario, the initial temperature difference between the hot and cold fluids entering the heat exchanger is \(80^{\circ} \, \mathrm{C} - 20^{\circ} \, \mathrm{C} = 60^{\circ} \, \mathrm{C}\).
This particular difference is essential in determining how much heat can be exchanged between the two fluids. The calculation of this difference is the first step in determining the potential for heat transfer in any heat exchange process.
Remember, temperature differences are directly proportional to heat transferred. Thus, increasing the temperature difference can effectively increase the efficiency and capacity of a heat exchanger, making it an essential parameter in the design and operation of such systems.

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Most popular questions from this chapter

11-100 E(S) Reconsider Prob. 11-99. Using EES (or other) software, investigate the effects of the inlet temperature of hot water and the heat transfer coefficient on the rate of heat transfer and the surface area. Let the inlet temperature vary from \(60^{\circ} \mathrm{C}\) to \(120^{\circ} \mathrm{C}\) and the overall heat transfer coefficient from \(750 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) to \(1250 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Plot the rate of heat transfer and surface area as functions of the inlet temperature and the heat transfer coefficient, and discuss the results. 11-101E A thin-walled double-pipe, counter-flow heat exchanger is to be used to cool oil \(\left(c_{p}=0.525 \mathrm{Btu} / \mathrm{lbm} \cdot{ }^{\circ} \mathrm{F}\right)\) from \(300^{\circ} \mathrm{F}\) to \(105^{\circ} \mathrm{F}\) at a rate of \(5 \mathrm{lbm} / \mathrm{s}\) by water \(\left(c_{p}=\right.\) \(1.0 \mathrm{Btu} / \mathrm{lbm} \cdot{ }^{\circ} \mathrm{F}\) ) that enters at \(70^{\circ} \mathrm{F}\) at a rate of \(3 \mathrm{lbm} / \mathrm{s}\). The diameter of the tube is 5 in and its length is \(200 \mathrm{ft}\). Determine the overall heat transfer coefficient of this heat exchanger using (a) the LMTD method and \((b)\) the \(\varepsilon-\mathrm{NTU}\) method.

Glycerin \(\left(c_{p}=2400 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) at \(20^{\circ} \mathrm{C}\) and \(0.3 \mathrm{~kg} / \mathrm{s}\) is to be heated by ethylene glycol \(\left(c_{p}=2500 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) at \(60^{\circ} \mathrm{C}\) and the same mass flow rate in a thin-walled double-pipe parallel-flow heat exchanger. If the overall heat transfer coefficient is \(380 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and the heat transfer surface area is \(5.3 \mathrm{~m}^{2}\), determine \((a)\) the rate of heat transfer and \((b)\) the outlet temperatures of the glycerin and the glycol.

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.

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.

Consider a heat exchanger that has an NTU of 4 . Someone proposes to double the size of the heat exchanger and thus double the NTU to 8 in order to increase the effectiveness of the heat exchanger and thus save energy. Would you support this proposal?

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