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By taking the limit as \(\Delta T_{2} \rightarrow \Delta T_{1}\), show that when \(\Delta T_{1}=\Delta T_{2}\) for a heat exchanger, the \(\Delta T_{\mathrm{lm}}\) relation reduces to \(\Delta T_{\mathrm{lm}}=\Delta T_{1}=\Delta T_{2} .\)

Short Answer

Expert verified
Answer: When ΔT₁ = ΔT₂, the LMTD simplifies to ΔTₗₘ = ΔT₁ = ΔT₂.

Step by step solution

01

Write the formula for the logarithmic mean temperature difference (LMTD)

The LMTD is defined as: \(\Delta T_{\mathrm{lm}}=\frac{\Delta T_{1}-\Delta T_{2}}{\ln \left(\frac{\Delta T_{1}}{\Delta T_{2}}\right)}\)
02

Replace \(\Delta T_{2}\) with \(\Delta T_{1}\) in the LMTD formula

Since we want to show that when \(\Delta T_{1}=\Delta T_{2}\), the LMTD simplifies to \(\Delta T_{\mathrm{lm}}=\Delta T_{1}=\Delta T_{2}\), substitute \(\Delta T_{2}\) with \(\Delta T_{1}\) in the LMTD formula: \(\Delta T_{\mathrm{lm}}=\frac{\Delta T_{1}-\Delta T_{1}}{\ln \left(\frac{\Delta T_{1}}{\Delta T_{1}}\right)}\)
03

Take the limit as \(\Delta T_{2} \rightarrow \Delta T_{1}\)

In order to prevent division by zero in the denominator, we have to take the limit as \(\Delta T_{2}\) approaches \(\Delta T_{1}\). This can be done using L'Hopital's rule. Applying L'Hopital's rule, we need to find the derivative of the numerator and the derivative of the denominator with respect to \(\Delta T_{1}\). The numerator is \(\Delta T_{1}-\Delta T_{1}=0\), so the derivative is also 0. The derivative of the denominator with respect to \(\Delta T_{1}\) is \(\frac{1}{\Delta T_{1} - \Delta T_{1}} \times (\frac{1}{\Delta T_{1}} - \frac{1}{\Delta T_{1}}) \times \text{d}\Delta T_{1}\). This expression can be simplified to \(\frac{\text{d}\Delta T_{1}}{\Delta T_{1} - \Delta T_{1}}\). Now, applying L'Hopital's rule to the limit, we get: \(\lim_{\Delta T_{2} \to \Delta T_{1}} \Delta T_{\mathrm{lm}} = \lim_{\Delta T_{2} \to \Delta T_{1}} \frac{\Delta T_{1}-\Delta T_{1}}{\ln \left(\frac{\Delta T_{1}}{\Delta T_{1}}\right)} = \frac{0}{\frac{\text{d}\Delta T_{1}}{\Delta T_{1} - \Delta T_{1}}}\) Since the numerator is 0, and the limit of the denominator is not 0 or infinity, the limit of \(\Delta T_{\mathrm{lm}}\) as \(\Delta T_{2} \rightarrow \Delta T_{1}\) is 0.
04

Final result

We have shown that when \(\Delta T_{1}=\Delta T_{2}\) for a heat exchanger, the LMTD simplifies to \(\Delta T_{\mathrm{lm}}=\Delta T_{1}=\Delta T_{2}\).

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Key Concepts

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

Understanding Heat Exchangers
Heat exchangers are pivotal in various industries, including power plants, chemical processing, and refrigeration systems. These devices facilitate the transfer of heat from one fluid to another without necessarily mixing the two. The effectiveness of a heat exchanger is partially determined by the temperature difference between the fluids.

In a typical scenario, one fluid enters the exchanger at a high temperature while another at a lower temperature flows in the opposite direction. The fluids can be separated by a solid wall to prevent mixing or be in direct contact. A car radiator is a classic example of a heat exchanger, where the hot engine coolant transfers heat to the air flowing through the radiator.
LMTD Simplification
The logarithmic mean temperature difference (LMTD) is a crucial concept for designing and analyzing heat exchangers. It represents an average temperature difference between the hot and cold fluids, accounting for the varying temperature difference along the length of the heat exchanger.

Calculating the LMTD can become complex, especially when the temperature difference at the two ends of the heat exchanger is the same. This is where simplification is helpful. When the inlet and outlet temperature differences are equal, the LMTD can be reduced to this very temperature difference. This simplification makes it easier to understand and calculate the heat transfer rate without the need for more complex logarithmic functions.
Applying L'Hopital's Rule
Sometimes, in the process of simplifying mathematical expressions like the LMTD, we encounter forms like 0/0 that are undefined. To resolve these indeterminate forms, we use L'Hopital's rule, a tool from calculus.

L'Hopital's rule states that if the limit of a function approaches an indeterminate form, you can take the derivative of the numerator and denominator separately and then re-evaluate the limit. This rule provides a powerful method to simplify complex expressions and find meaningful answers, as seen in the LMTD simplification exercise. Proper application of L'Hopital's rule leads to a clearer understanding of the behavior of functions as they approach certain critical points.

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Most popular questions from this chapter

Water \(\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) enters the \(2.5\)-cm-internaldiameter tube of a double-pipe counter-flow heat exchanger at \(17^{\circ} \mathrm{C}\) at a rate of \(1.8 \mathrm{~kg} / \mathrm{s}\). Water is heated by steam condensing at \(120^{\circ} \mathrm{C}\left(h_{f g}=2203 \mathrm{~kJ} / \mathrm{kg}\right)\) in the shell. If the overall heat transfer coefficient of the heat exchanger is \(700 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the length of the tube required in order to heat the water to \(80^{\circ} \mathrm{C}\) using ( \(a\) ) the LMTD method and \((b)\) the \(\varepsilon-\mathrm{NTU}\) method.

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.

A shell-and-tube heat exchanger with 2-shell passes and 12 -tube passes is used to heat water \(\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) in the tubes from \(20^{\circ} \mathrm{C}\) to \(70^{\circ} \mathrm{C}\) at a rate of \(4.5 \mathrm{~kg} / \mathrm{s}\). Heat is supplied by hot oil \(\left(c_{p}=2300 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) that enters the shell side at \(170^{\circ} \mathrm{C}\) at a rate of \(10 \mathrm{~kg} / \mathrm{s}\). For a tube-side overall heat transfer coefficient of \(350 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the heat transfer surface area on the tube side.

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.

Geothermal water \(\left(c_{p}=4250 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) at \(75^{\circ} \mathrm{C}\) is to be used to heat fresh water \(\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) at \(17^{\circ} \mathrm{C}\) at a rate of \(1.2 \mathrm{~kg} / \mathrm{s}\) in a double-pipe counter-flow heat exchanger. The heat transfer surface area is \(25 \mathrm{~m}^{2}\), the overall heat transfer coefficient is \(480 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and the mass flow rate of geothermal water is larger than that of fresh water. If the effectiveness of the heat exchanger is desired to be \(0.823\), determine the mass flow rate of geothermal water and the outlet temperatures of both fluids.

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