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 hot surface at \(80^{\circ} \mathrm{C}\) in air at \(20^{\circ} \mathrm{C}\) is to be cooled by attaching 10 -cm-long and 1 -cm-diameter cylindrical fins. The combined heat transfer coefficient is \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and heat transfer from the fin tip is negligible. If the fin efficiency is \(0.75\), the rate of heat loss from 100 fins is (a) \(325 \mathrm{~W}\) (b) \(707 \mathrm{~W}\) (c) \(566 \mathrm{~W}\) (d) \(424 \mathrm{~W}\) (e) \(754 \mathrm{~W}\)

Cold conditioned air at \(12^{\circ} \mathrm{C}\) is flowing inside a \(1.5\)-cm- thick square aluminum \((k=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) duct of inner cross section \(22 \mathrm{~cm} \times 22 \mathrm{~cm}\) at a mass flow rate of \(0.8 \mathrm{~kg} / \mathrm{s}\). The duct is exposed to air at \(33^{\circ} \mathrm{C}\) with a combined convection-radiation heat transfer coefficient of \(13 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The convection heat transfer coefficient at the inner surface is \(75 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the air temperature in the duct should not increase by more than \(1^{\circ} \mathrm{C}\) determine the maximum length of the duct.

A room at \(20^{\circ} \mathrm{C}\) air temperature is loosing heat to the outdoor air at \(0^{\circ} \mathrm{C}\) at a rate of \(1000 \mathrm{~W}\) through a \(2.5\)-m-high and 4 -m-long wall. Now the wall is insulated with \(2-\mathrm{cm}\) thick insulation with a conductivity of \(0.02 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Determine the rate of heat loss from the room through this wall after insulation. Assume the heat transfer coefficients on the inner and outer surface of the wall, the room air temperature, and the outdoor air temperature to remain unchanged. Also, disregard radiation. (a) \(20 \mathrm{~W}\) (b) \(561 \mathrm{~W}\) (c) \(388 \mathrm{~W}\) (d) \(167 \mathrm{~W}\) (e) \(200 \mathrm{~W}\)

Heat is generated steadily in a 3-cm-diameter spherical ball. The ball is exposed to ambient air at \(26^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(7.5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The ball is to be covered with a material of thermal conductivity \(0.15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The thickness of the covering material that will maximize heat generation within the ball while maintaining ball surface temperature constant is (a) \(0.5 \mathrm{~cm}\) (b) \(1.0 \mathrm{~cm}\) (c) \(1.5 \mathrm{~cm}\) (d) \(2.0 \mathrm{~cm}\) (e) \(2.5 \mathrm{~cm}\)

Chilled water enters a thin-shelled 5-cm-diameter, 150-mlong pipe at \(7^{\circ} \mathrm{C}\) at a rate of \(0.98 \mathrm{~kg} / \mathrm{s}\) and leaves at \(8^{\circ} \mathrm{C}\). The pipe is exposed to ambient air at \(30^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(9 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the pipe is to be insulated with glass wool insulation \((k=0.05 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) in order to decrease the temperature rise of water to \(0.25^{\circ} \mathrm{C}\), determine the required thickness of the insulation.

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