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Chapter 11: Problem 36

How does the log mean temperature difference for a heat exchanger differ from the arithmetic mean temperature difference? For specified inlet and outlet temperatures, which one of these two quantities is larger?

Short Answer

Expert verified
Answer: The key difference between LMTD and AMTD in a heat exchanger is that LMTD considers the logarithmic nature of the temperature change along the exchanger's length, while AMTD simply averages the inlet and outlet temperature differences. LMTD is usually smaller than AMTD and provides a more accurate representation of the effective driving force for heat transfer in a heat exchanger.

Step by step solution

01

Define LMTD and AMTD for a heat exchanger

In a heat exchanger, the temperature difference between hot and cold fluids varies along the length of the exchanger. The Log Mean Temperature Difference (LMTD) is the logarithmic average of the temperature differences at the inlet and outlet of the exchanger, and it is used to analyze the heat transfer rate in a heat exchanger. LMTD can be calculated using the formula: LMTD = \frac{(ΔT_1 - ΔT_2)}{ln(\frac{ΔT_1}{ΔT_2})} where ΔT_1 is the temperature difference at the inlet, and ΔT_2 is the temperature difference at the outlet. On the other hand, the Arithmetic Mean Temperature Difference (AMTD) is simply the average of the temperature differences at the inlet and outlet: AMTD = \frac{(ΔT_1 + ΔT_2)}{2}
02

Compare LMTD and AMTD

The LMTD and the AMTD differ in the way they account for the varying temperature differences along the length of the heat exchanger. The LMTD takes into account the logarithmic nature of the temperature change, which means it considers both the highest and lowest temperature differences in the heat exchanger. The AMTD, however, averages the inlet and outlet temperature differences without accounting for the logarithmic nature of the temperature change.
03

Determine which quantity is larger

Given the specified inlet and outlet temperatures, we can compare the values of LMTD and AMTD to determine which one is larger. Since the LMTD accounts for the logarithmic nature of the temperature change while the AMTD does not, the LMTD value will typically be smaller than the AMTD value. In some cases, the difference between the two values may not be significant, but in others, it can impact the overall heat transfer rate calculation. Therefore, it is more accurate to use LMTD for heat exchanger design and analysis since it represents the effective driving force for heat transfer more accurately.

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Most popular questions from this chapter

Consider an oil-to-oil double-pipe heat exchanger whose flow arrangement is not known. The temperature measurements indicate that the cold oil enters at \(20^{\circ} \mathrm{C}\) and leaves at \(55^{\circ} \mathrm{C}\), while the hot oil enters at \(80^{\circ} \mathrm{C}\) and leaves at \(45^{\circ} \mathrm{C}\). Do you think this is a parallel-flow or counter-flow heat exchanger? Why? Assuming the mass flow rates of both fluids to be identical, determine the effectiveness of this heat exchanger.

A shell-and-tube heat exchanger with 2-shell passes and 12 -tube passes is used to heat water \(\left(c_{p}=4180 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) with ethylene glycol \(\left(c_{p}=2680 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\). Water enters the tubes at \(22^{\circ} \mathrm{C}\) at a rate of \(0.8 \mathrm{~kg} / \mathrm{s}\) and leaves at \(70^{\circ} \mathrm{C}\). Ethylene \(\mathrm{glycol}\) enters the shell at \(110^{\circ} \mathrm{C}\) and leaves at \(60^{\circ} \mathrm{C}\). If the overall heat transfer coefficient based on the tube side is \(280 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the rate of heat transfer and the heat transfer surface area on the tube side.

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.

Consider a closed loop heat exchanger that carries exit water \(\left(c_{p}=1 \mathrm{Btu} / \mathrm{lbm} \cdot{ }^{\circ} \mathrm{F}\right.\) and \(\left.\rho=62.4 \mathrm{lbm} / \mathrm{ft}^{3}\right)\) of a condenser side initially at \(100^{\circ} \mathrm{F}\). The water flows through a \(500 \mathrm{ft}\) long stainless steel pipe of 1 in inner diameter immersed in a large lake. The temperature of lake water surrounding the heat exchanger is \(45^{\circ} \mathrm{F}\). The overall heat transfer coefficient of the heat exchanger is estimated to be \(250 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2}{ }^{\circ} \mathrm{F}\). What is the exit temperature of the water from the immersed heat exchanger if it flows through the pipe at an average velocity of \(9 \mathrm{ft} / \mathrm{s}\) ? Use \(\varepsilon-\mathrm{NTU}\) method for analysis.

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.
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