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Chapter 3: Problem 46

A wall consists of two layers of insulation pressed against each other. Do we need to be concerned about the thermal contact resistance at the interface in a heat transfer analysis or can we just ignore it?

Short Answer

Expert verified
Based on the given information, the thermal contact resistance at the interface of two layers of insulation in a wall can typically be ignored in a heat transfer analysis due to the low thermal resistance contributed by the insulation materials. However, if precise results or specific conditions suggest a significant level of contact resistance, it may be necessary to include it in the analysis.

Step by step solution

01

Understanding Thermal Contact Resistance

At the interface between two materials, there may be a resistance to heat transfer due to microscopic gaps and voids that create impedance for heat conduction between the materials. This resistance is called thermal contact resistance, and it can occur in any interface where two materials are pressed against each other.
02

Heat Transfer Analysis

In a heat transfer analysis, the objective is to determine the rate of heat transfer through a system, which is influenced by various factors like material properties, thickness, and surface area. The thermal resistance of the entire system is the sum of the individual resistances, including any contact resistance that may be present.
03

Importance of Thermal Contact Resistance

The importance of considering thermal contact resistance in a heat transfer analysis depends on the specific system being analyzed and the magnitudes of the resistances present. If the contact resistance is significantly lower than the thermal resistance of the insulation layers, it may have a negligible impact on the overall heat transfer and can be ignored. Conversely, if the contact resistance is comparable or larger in magnitude, it should be included in the analysis.
04

Insulation Layers in the Wall

In the given problem, we have two layers of insulation pressed against each other in a wall. The primary purpose of insulation is to reduce heat transfer by minimizing conduction. Insulation materials typically have low thermal conductivity, which results in low thermal resistance between the layers of insulation.
05

Decision on Thermal Contact Resistance

Since the insulation layers have low thermal conductivity, the thermal resistance contributed by the insulation materials is relatively low. In most cases, as the thermal contact resistance between insulation layers is often much smaller in comparison, we can ignore it for a simplified heat transfer analysis. However, if precise results are needed and the specific conditions or materials in the interface suggest a significant level of contact resistance, it might be necessary to include it in the analysis.

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Most popular questions from this chapter

A 4-mm-diameter and 10-cm-long aluminum fin \((k=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is attached to a surface. If the heat transfer coefficient is \(12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the percent error in the rate of heat transfer from the fin when the infinitely long fin assumption is used instead of the adiabatic fin tip assumption.

What is the value of conduction shape factors in engineering?

Two 3-m-long and \(0.4-\mathrm{cm}\)-thick cast iron \((k=\) \(52 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) steam pipes of outer diameter \(10 \mathrm{~cm}\) are connected to each other through two 1 -cm-thick flanges of outer diameter \(20 \mathrm{~cm}\). The steam flows inside the pipe at an average temperature of \(200^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(180 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The outer surface of the pipe is exposed to an ambient at \(12^{\circ} \mathrm{C}\), with a heat transfer coefficient of \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Disregarding the flanges, determine the average outer surface temperature of the pipe. (b) Using this temperature for the base of the flange and treating the flanges as the fins, determine the fin efficiency and the rate of heat transfer from the flanges. (c) What length of pipe is the flange section equivalent to for heat transfer purposes?

A 1-cm-diameter, 30-cm-long fin made of aluminum \((k=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is attached to a surface at \(80^{\circ} \mathrm{C}\). The surface is exposed to ambient air at \(22^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(11 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the fin can be assumed to bery long, its efficiency is (a) \(0.60\) (b) \(0.67\) (c) \(0.72\) (d) \(0.77\) (e) \(0.88\)

A total of 10 rectangular aluminum fins \((k=\) \(203 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) are placed on the outside flat surface of an electronic device. Each fin is \(100 \mathrm{~mm}\) wide, \(20 \mathrm{~mm}\) high and \(4 \mathrm{~mm}\) thick. The fins are located parallel to each other at a center- tocenter distance of \(8 \mathrm{~mm}\). The temperature at the outside surface of the electronic device is \(60^{\circ} \mathrm{C}\). The air is at \(20^{\circ} \mathrm{C}\), and the heat transfer coefficient is \(100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine \((a)\) the rate of heat loss from the electronic device to the surrounding air and \((b)\) the fin effectiveness.
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