/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 48 Consider two surfaces pressed ag... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider a tube for transporting steam that is not centered properly in a cylindrical insulation material \((k=\) \(0.73 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The tube diameter is \(D_{1}=20 \mathrm{~cm}\) and the insulation diameter is \(D_{2}=40 \mathrm{~cm}\). The distance between the center of the tube and the center of the insulation is \(z=5 \mathrm{~mm}\). If the surface of the tube maintains a temperature of \(100^{\circ} \mathrm{C}\) and the outer surface temperature of the insulation is constant at \(30^{\circ} \mathrm{C}\), determine the rate of heat transfer per unit length of the tube through the insulation.

Consider a house with a flat roof whose outer dimensions are \(12 \mathrm{~m} \times 12 \mathrm{~m}\). The outer walls of the house are \(6 \mathrm{~m}\) high. The walls and the roof of the house are made of \(20-\mathrm{cm}-\) thick concrete \((k=0.75 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The temperatures of the inner and outer surfaces of the house are \(15^{\circ} \mathrm{C}\) and \(3^{\circ} \mathrm{C}\), respectively. Accounting for the effects of the edges of adjoining surfaces, determine the rate of heat loss from the house through its walls and the roof. What is the error involved in ignoring the effects of the edges and corners and treating the roof as a \(12 \mathrm{~m} \times 12 \mathrm{~m}\) surface and the walls as \(6 \mathrm{~m} \times 12 \mathrm{~m}\) surfaces for simplicity?

Consider two cold canned drinks, one wrapped in a blanket and the other placed on a table in the same room. Which drink will warm up faster?

The wall of a refrigerator is constructed of fiberglass insulation \((k=0.035 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) sandwiched between two layers of 1 -mm-thick sheet metal \((k=15.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The refrigerated space is maintained at \(3^{\circ} \mathrm{C}\), and the average heat transfer coefficients at the inner and outer surfaces of the wall are \(4 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(9 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. The kitchen temperature averages \(25^{\circ} \mathrm{C}\). It is observed that condensation occurs on the outer surfaces of the refrigerator when the temperature of the outer surface drops to \(20^{\circ} \mathrm{C}\). Determine the minimum thickness of fiberglass insulation that needs to be used in the wall in order to avoid condensation on the outer surfaces.

In the United States, building insulation is specified by the \(R\)-value (thermal resistance in \(\mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F} /\) Btu units). A homeowner decides to save on the cost of heating the home by adding additional insulation in the attic. If the total \(R\)-value is increased from 15 to 25 , the homeowner can expect the heat loss through the ceiling to be reduced by (a) \(25 \%\) (b) \(40 \%\) (c) \(50 \%\) (d) \(60 \%\) (e) \(75 \%\)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.