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Consider two surfaces pressed against each other. Now the air at the interface is evacuated. Will the thermal contact resistance at the interface increase or decrease as a result?

Short Answer

Expert verified
Answer: Evacuating the air between two surfaces pressed against each other decreases the thermal contact resistance at the interface. This is because it increases the direct contact points between the surfaces, improving heat transfer through conduction and eliminating the less efficient heat transfer mechanisms through air.

Step by step solution

01

Explain Thermal Contact Resistance

Thermal contact resistance is the resistance to heat flow across the interface between two solid surfaces in contact. When two surfaces are pressed together, contact occurs only at distinct points, creating small air-filled gaps between the surfaces. Heat is transferred across these small gaps through a combination of conduction, convection, and radiation from the solid surfaces to the air and back to the other solid surface.
02

Consider the Role of Air in Heat Transfer

Air is a poor thermal conductor compared to most solids, so the presence of air at the interface of the two solid surfaces can reduce the efficiency of heat transfer between them. Additionally, when air is evacuated from the interface, the number of contact points between the surfaces increases leading to an increase in heat transfer through direct conduction.
03

Analyze the Effect of Evacuating Air on Thermal Contact Resistance

By evacuating the air at the interface, we are essentially eliminating the air-filled gaps and increasing the direct contact points between the two surfaces. This increases the heat transfer by conduction through the direct contact between the surfaces and does not rely on the poor thermal conductivity of air.
04

Determine if Thermal Contact Resistance Increases or Decreases

Since evacuating the air from the interface between the two solid surfaces improves heat transfer through the direct contact points and removes the less efficient heat transfer mechanism through air, the thermal contact resistance at the interface will decrease as a result of evacuating the air.

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Key Concepts

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

Heat Transfer
Heat transfer is a fundamental concept in physics that describes the movement of heat energy from one place to another. This can occur in various ways, such as conduction, convection, and radiation. In conduction, heat is transferred through direct contact between materials, a process that is heavily influenced by their thermal conductivity.

With convection, heat is carried away by flowing fluids, which could be liquids or gases. Lastly, radiation heat transfer occurs through electromagnetic waves and does not require a medium. Understanding the mechanisms of heat transfer is essential in many practical applications, including designing heat sinks, insulating homes, and analyzing thermal systems.
Thermal Conductivity
Thermal conductivity is a property of materials that measures their ability to conduct heat. It's defined as the quantity of heat, typically measured in watts, that passes through a material with a given area and thickness over a time and temperature difference. Materials with high thermal conductivity, such as metals, are efficient at transferring heat, while those with low conductivity, like rubber or air, are good insulators.

In the context of the textbook problem, understanding thermal conductivity is crucial as it helps explain why eliminating air from the interface between two surfaces can improve the overall heat transfer—since air's low thermal conductivity is a bottleneck in the process.
Conduction in Solids
Conduction in solids is a mode of heat transfer occurring within a solid material or between solid surfaces in contact. The thermal motion of particles within a solid increases with temperature, causing neighboring particles to vibrate more vigorously. This motion and subsequent interaction between particles is how energy is passed through the solid.

This process is also described by Fourier's law of heat conduction, which states that the rate of heat transfer through a material is proportional to the negative gradient of temperatures and the area to which the heat is being transferred, with a constant of proportionality known as thermal conductivity. In the instance of two surfaces in contact, increasing the contact area by evacuating air enhances this conductive transfer significantly.
Radiation Heat Transfer
Radiation heat transfer differs from conduction and convection as it involves the transfer of heat through electromagnetic waves, such as infrared radiation. This type of transfer does not require a medium; it can occur through a vacuum. Every object emits, absorbs, and possibly transmits or reflects thermal radiation according to its temperature.

In the process of heat transfer between two surfaces, when there's an air gap, radiation can contribute to the transfer of energy across that gap. However, direct solid contact tends to be more efficient, leading to the conclusion that removing the air gap can significantly reduce the resistance to heat flow via increased conduction, overshadowing the contribution by radiation.
Evacuating Air Effect
Evacuating air from the interface of two contacting surfaces has a profound effect on thermal contact resistance. When the air is present, its low thermal conductivity compared to that of solids limits the efficiency of heat transfer between those surfaces. Upon evacuation of air, the solid surfaces come into closer contact, rapidly increasing the area of direct solid-to-solid conduction.

This increased contact area reduces the paths' resistance, where heat can flow more freely due to higher thermal conductivity of solids—effectively decreasing thermal contact resistance. This concept is essential when designing interfaces for heat transfer in mechanical and electronic systems to ensure efficient thermal management.

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Most popular questions from this chapter

A \(0.2\)-cm-thick, 10-cm-high, and 15 -cm-long circuit board houses electronic components on one side that dissipate a total of \(15 \mathrm{~W}\) of heat uniformly. The board is impregnated with conducting metal fillings and has an effective thermal conductivity of \(12 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). All the heat generated in the components is conducted across the circuit board and is dissipated from the back side of the board to a medium at \(37^{\circ} \mathrm{C}\), with a heat transfer coefficient of \(45 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Determine the surface temperatures on the two sides of the circuit board. (b) Now a 0.1-cm-thick, 10-cm-high, and 15 -cm-long aluminum plate \((k=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) with \(200.2\)-cm-thick, 2-cm-long, and \(15-\mathrm{cm}\)-wide aluminum fins of rectangular profile are attached to the back side of the circuit board with a \(0.03-\mathrm{cm}-\) thick epoxy adhesive \((k=1.8 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). Determine the new temperatures on the two sides of the circuit board.

A 1-m-inner-diameter liquid-oxygen storage tank at a hospital keeps the liquid oxygen at \(90 \mathrm{~K}\). The tank consists of a \(0.5-\mathrm{cm}\)-thick aluminum ( \(k=170 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) shell whose exterior is covered with a 10 -cm-thick layer of insulation \((k=\) \(0.02 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The insulation is exposed to the ambient air at \(20^{\circ} \mathrm{C}\) and the heat transfer coefficient on the exterior side of the insulation is \(5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The temperature of the exterior surface of the insulation is (a) \(13^{\circ} \mathrm{C}\) (b) \(9^{\circ} \mathrm{C}\) (c) \(2^{\circ} \mathrm{C}\) (d) \(-3^{\circ} \mathrm{C}\) (e) \(-12^{\circ} \mathrm{C}\)

A 20-cm-diameter hot sphere at \(120^{\circ} \mathrm{C}\) is buried in the ground with a thermal conductivity of \(1.2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The distance between the center of the sphere and the ground surface is \(0.8 \mathrm{~m}\) and the ground surface temperature is \(15^{\circ} \mathrm{C}\). The rate of heat loss from the sphere is (a) \(169 \mathrm{~W}\) (b) \(20 \mathrm{~W}\) (c) \(217 \mathrm{~W}\) (d) \(312 \mathrm{~W}\) (e) \(1.8 \mathrm{~W}\)

A 25 -cm-diameter, 2.4-m-long vertical cylinder containing ice at \(0^{\circ} \mathrm{C}\) is buried right under the ground. The cylinder is thin-shelled and is made of a high thermal conductivity material. The surface temperature and the thermal conductivity of the ground are \(18^{\circ} \mathrm{C}\) and \(0.85 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) respectively. The rate of heat transfer to the cylinder is (a) \(37.2 \mathrm{~W}\) (b) \(63.2 \mathrm{~W}\) (c) \(158 \mathrm{~W}\) (d) \(480 \mathrm{~W}\) (e) \(1210 \mathrm{~W}\)

We are interested in steady state heat transfer analysis from a human forearm subjected to certain environmental conditions. For this purpose consider the forearm to be made up of muscle with thickness \(r_{m}\) with a skin/fat layer of thickness \(t_{s f}\) over it, as shown in the Figure P3-138. For simplicity approximate the forearm as a one-dimensional cylinder and ignore the presence of bones. The metabolic heat generation rate \(\left(\dot{e}_{m}\right)\) and perfusion rate \((\dot{p})\) are both constant throughout the muscle. The blood density and specific heat are \(\rho_{b}\) and \(c_{b}\), respectively. The core body temperate \(\left(T_{c}\right)\) and the arterial blood temperature \(\left(T_{a}\right)\) are both assumed to be the same and constant. The muscle and the skin/fat layer thermal conductivities are \(k_{m}\) and \(k_{s f}\), respectively. The skin has an emissivity of \(\varepsilon\) and the forearm is subjected to an air environment with a temperature of \(T_{\infty}\), a convection heat transfer coefficient of \(h_{\text {conv }}\), and a radiation heat transfer coefficient of \(h_{\mathrm{rad}}\). Assuming blood properties and thermal conductivities are all constant, \((a)\) write the bioheat transfer equation in radial coordinates. The boundary conditions for the forearm are specified constant temperature at the outer surface of the muscle \(\left(T_{i}\right)\) and temperature symmetry at the centerline of the forearm. \((b)\) Solve the differential equation and apply the boundary conditions to develop an expression for the temperature distribution in the forearm. (c) Determine the temperature at the outer surface of the muscle \(\left(T_{i}\right)\) and the maximum temperature in the forearm \(\left(T_{\max }\right)\) for the following conditions: $$ \begin{aligned} &r_{m}=0.05 \mathrm{~m}, t_{s f}=0.003 \mathrm{~m}, \dot{e}_{m}=700 \mathrm{~W} / \mathrm{m}^{3}, \dot{p}=0.00051 / \mathrm{s} \\ &T_{a}=37^{\circ} \mathrm{C}, T_{\text {co }}=T_{\text {surr }}=24^{\circ} \mathrm{C}, \varepsilon=0.95 \\ &\rho_{b}=1000 \mathrm{~kg} / \mathrm{m}^{3}, c_{b}=3600 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k_{m}=0.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} \\ &k_{s f}=0.3 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, h_{\text {conv }}=2 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}, h_{\mathrm{rad}}=5.9 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K} \end{aligned} $$

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