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The temperature difference between the hot and cold fluids in a heat exchanger is given to be \(\Delta T_{1}\) at one end and \(\Delta T_{2}\) at the other end. Can the logarithmic temperature difference \(\Delta T_{\operatorname{lm}}\) of this heat exchanger be greater than both \(\Delta T_{1}\) and \(\Delta T_{2}\) ? Explain.

Short Answer

Expert verified
Answer: No.

Step by step solution

01

Identifying the components of the question

The problem gives us two temperature differences, 螖T鈧, and 螖T鈧. We will focus on finding the logarithmic mean temperature difference, 螖T鈧椻倶, using the provided definition and comparing it with 螖T鈧 and 螖T鈧.
02

Compute 螖T鈧椻倶 using the provided definition

Let's find the logarithmic mean temperature difference (螖T鈧椻倶) by plugging in the given values for 螖T鈧 and 螖T鈧 into the equation: $$螖T_{\operatorname{lm}} = \frac{螖T_1 - 螖T_2}{\ln{\frac{螖T_1}{螖T_2}}}$$
03

Comparing 螖T鈧椻倶 with 螖T鈧 and 螖T鈧

To see if 螖T鈧椻倶 can be greater than both 螖T鈧 and 螖T鈧, we need to analyze this equation further. We observe that the denominator is the natural logarithm of a ratio; thus, it will be positive when the ratio is greater than 1 and negative when the ratio is less than 1. Now, we will analyze two cases: 1. If 螖T鈧 > 螖T鈧 (which makes the numerator positive), then the ratio \(\frac{螖T_1}{螖T_2} > 1\) and the denominator becomes positive. In this case, we have a positive number divided by a smaller positive number, making the result greater than the numerator. Hence, $$螖T_{\operatorname{lm}} > 螖T_1 - 螖T_2 \Rightarrow 螖T_{\operatorname{lm}} < 螖T_1$$ 2. If 螖T鈧 < 螖T鈧 (which makes the numerator negative), then the ratio \(\frac{螖T_1}{螖T_2} < 1\) and the denominator becomes negative. In this case, we have a negative number divided by a negative number, making the result positive, but less than the numerator. Hence, $$螖T_{\operatorname{lm}} < 螖T_2 - 螖T_1 \Rightarrow 螖T_{\operatorname{lm}} < 螖T_2$$ From both cases, we see that it is not possible for the logarithmic mean temperature difference (螖T鈧椻倶) to be greater than both 螖T鈧 and 螖T鈧. Therefore, the answer to the question is "No."

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Key Concepts

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

Heat Exchanger
A heat exchanger is a device used to transfer heat between two or more fluids without the fluids coming into direct contact with each other. These fluids can be gases or liquids and are usually separated by a solid wall. Heat exchangers are found in a variety of applications including power plants, refrigeration systems, and even in our home heating systems. The efficiency of a heat exchanger is closely related to the temperature difference between the hot and cold fluids passing through it. This is where the concept of the logarithmic mean temperature difference becomes relevant.

In practice, a heat exchanger's design and performance are influenced by factors such as the properties of the fluids, the flow arrangement (e.g., counterflow or parallel flow), and the surface area over which the heat exchange takes place. Understanding the operation and efficiency of heat exchangers is crucial for engineers and technicians when optimizing these systems for energy conservation and cost-effectiveness.

Types of Heat Exchangers

There are several types of heat exchangers, including shell and tube, plate, and finned-tube exchangers. Each type has its design characteristics and is selected based on the specific requirements of the system, like pressure, temperature, and the physical states of the heat transfer mediums.
Temperature Difference
The concept of temperature difference is fundamental in thermodynamics, especially in the operation of heat exchangers. It's the driving force behind heat transfer, with the general rule being that heat flows from the hotter fluid to the cooler fluid until thermal equilibrium is reached. In the context of the heat exchanger, the difference in temperature between the hot fluid at the inlet and outlet is commonly referred to as \(\Delta T_{1}\) and \(\Delta T_{2}\), respectively.

A larger temperature difference usually means more efficient heat transfer since the rate of heat transfer is proportional to the temperature difference, according to Fourier's law of heat conduction. This is why, when designing heat exchangers, engineers strive for the largest possible temperature difference without compromising the integrity or efficiency of the system.

Importance in Heat Exchanger Design

Having a clear understanding of temperature differences within a heat exchanger allows engineers to calculate the heat transfer rate and size the heat exchanger accordingly. By optimizing the temperature difference, energy consumption can be minimized, leading to more sustainable and cost-effective operations.
Natural Logarithm
The natural logarithm is a mathematical concept that is widely used in various scientific and engineering fields. It is denoted as \(\ln\) and is the logarithm to the base \(^e\), where \(^e\) is Euler's number, approximately equal to 2.71828. This particular logarithm is considered 'natural' because it arises in many natural processes, including growth and decay phenomena.

In the context of the logarithmic mean temperature difference (LMTD) in a heat exchanger, the natural logarithm is used to handle the case where the temperature difference between the two fluids varies along the length of the heat exchanger. LMTD provides a mean temperature difference that accounts for the changing temperature gradient and is more accurate than simply averaging the inlet and outlet temperature differences.

Application in Calculating LMTD

When calculating LMTD for a heat exchanger, engineers utilize the natural logarithm to derive an effective temperature difference that is representative of the entire heat transfer process. This calculation balances the variations in temperature, allowing for the precise design and analysis of heat exchangers. Indeed, the natural logarithm is instrumental in establishing a solid mathematical foundation for thermal engineering and beyond.

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Most popular questions from this chapter

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.

Write an essay on the static and dynamic types of regenerative heat exchangers and compile information about the manufacturers of such heat exchangers. Choose a few models by different manufacturers and compare their costs and performance.

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?

Hot water at \(60^{\circ} \mathrm{C}\) is cooled to \(36^{\circ} \mathrm{C}\) through the tube side of a 1-shell pass and 2-tube passes heat exchanger. The coolant is also a water stream, for which the inlet and outlet temperatures are \(7^{\circ} \mathrm{C}\) and \(31^{\circ} \mathrm{C}\), respectively. The overall heat transfer coefficient and the heat transfer area are \(950 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(15 \mathrm{~m}^{2}\), respectively. Calculate the mass flow rates of hot and cold water streams in steady operation.

In a 1-shell and 2-tube heat exchanger, cold water with inlet temperature of \(20^{\circ} \mathrm{C}\) is heated by hot water supplied at the inlet at \(80^{\circ} \mathrm{C}\). The cold and hot water flow rates are \(5000 \mathrm{~kg} / \mathrm{h}\) and \(10,000 \mathrm{~kg} / \mathrm{h}\), respectively. If the shelland-tube heat exchanger has a \(U A_{s}\) value of \(11,600 \mathrm{~W} / \mathrm{K}\), determine the cold water and hot water outlet temperatures. Assume \(c_{p c}=4178 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) and \(c_{p h}=4188 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\).

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