/*! 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 44 Will the thermal contact resista... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Will the thermal contact resistance be greater for smooth or rough plain surfaces?

Short Answer

Expert verified
Answer: The thermal contact resistance is greater for rough plain surfaces as compared to smooth surfaces.

Step by step solution

01

Understanding Thermal Contact Resistance

Thermal contact resistance occurs at the interface between two materials and is responsible for impeding the transfer of heat between them. It depends on various factors such as surface roughness, material properties, surface temperatures, and the pressure applied to the contact.
02

Effect of Surface Roughness on Thermal Contact Resistance

Surface roughness is an important factor affecting thermal contact resistance. Rough surfaces have more contact points, but less effective contact area for heat transfer as compared to smooth surfaces, which can lead to higher contact resistance. Meanwhile, smooth surfaces have larger effective contact areas for heat transfer, leading to lower contact resistance.
03

Effect of Surface Roughness on Material Properties

Material properties, such as thermal conductivity, can also be affected by surface roughness. Generally, smoother surfaces allow for better contact between the materials, thus enhancing the overall thermal conductivity and leading to lower contact resistance.
04

Conclusion

Based on the discussion on the effect of surface roughness on thermal contact resistance, it can be concluded that the thermal contact resistance is greater for rough plain surfaces as compared to smooth surfaces. This is because rough surfaces have lesser effective contact area for heat transfer and can also affect the material properties unfavorably.

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.

Surface Roughness
Understanding surface roughness is key to grasping how it affects thermal contact resistance. Surface roughness refers to the texture of a surface. It is measured by the deviation of a surface's texture from being perfectly smooth. This texture consists of peaks and valleys. In the context of thermal contact resistance, rough surfaces have more peaks and valleys which result in irregular contact points, reducing the effective area for heat transfer. This leads to increased resistance because heat needs more pathways to travel across the interface. Conversely, smoother surfaces have larger flat areas that make better contact, allowing heat to flow more easily from one material to another. Thus, smoother surfaces are generally preferred in applications requiring efficient heat transfer. In practical terms:
  • Smoother surfaces = lower thermal contact resistance
  • Rougher surfaces = higher thermal contact resistance
Heat Transfer
Heat transfer is the process of thermal energy moving from a hotter object to a cooler one. In contexts involving materials and their surfaces, it is crucial to understand how heat is transferred across interfaces. When two surfaces come into contact, like in many engineering applications, the ease of heat transfer is influenced greatly by how well the surfaces are aligned and the quality of the contact points. These contact points are influenced by surface roughness, surface pressure, and temperature. In situations of poor contact due to rough surfaces, heat transfer is less efficient. This inefficiency can be problematic in many situations, such as in electronic devices where excessive heat could lead to malfunction. Proper management of heat transfer not only involves choosing appropriate materials but also involves preparing surfaces to create as little resistance as possible at contact points.
Material Properties
The properties of the materials in contact significantly affect thermal contact resistance. Material properties such as hardness, elasticity, and thermal expansion play a role in how surfaces conform to each other under pressure, and thus, affect contact resistance. A key property is thermal conductivity, which is a material's ability to conduct heat. Materials with high thermal conductivity transfer heat effectively, reducing thermal contact resistance. Smooth material surfaces can conform more closely, improving contact and thus increasing effective heat transfer. When selecting materials for any application, understanding these properties helps in minimizing thermal contact resistance:
  • Selecting materials with high thermal conductivity
  • Ensuring complementary mechanical properties for better contact under pressure
Thermal Conductivity
Thermal conductivity is a fundamental material property that measures the ability of a material to conduct heat. It is a critical factor in determining how effectively heat can be transferred through a material and across contact interfaces. In the context of thermal contact resistance, materials with higher thermal conductivity are more efficient at transferring heat through their contact surfaces. The smoother the surface, the easier it is for heat to move across the contact area, thus decreasing contact resistance. For optimized heat transfer, especially across interfaces:
  • Choose materials with higher thermal conductivity
  • Prepare surfaces to be as smooth as possible
These steps help in achieving efficient heat management in applications where thermal dissipation is essential.

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

A turbine blade made of a metal alloy \((k=\) \(17 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) has a length of \(5.3 \mathrm{~cm}\), a perimeter of \(11 \mathrm{~cm}\), and a cross-sectional area of \(5.13 \mathrm{~cm}^{2}\). The turbine blade is exposed to hot gas from the combustion chamber at \(973^{\circ} \mathrm{C}\) with a convection heat transfer coefficient of \(538 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The base of the turbine blade maintains a constant temperature of \(450^{\circ} \mathrm{C}\) and the tip is adiabatic. Determine the heat transfer rate to the turbine blade and temperature at the tip.

Consider two identical people each generating \(60 \mathrm{~W}\) of metabolic heat steadily while doing sedentary work, and dissipating it by convection and perspiration. The first person is wearing clothes made of 1 -mm-thick leather \((k=\) \(0.159 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) ) that covers half of the body while the second one is wearing clothes made of 1 -mm-thick synthetic fabric \((k=0.13 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) that covers the body completely. The ambient air is at \(30^{\circ} \mathrm{C}\), the heat transfer coefficient at the outer surface is \(15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and the inner surface temperature of the clothes can be taken to be \(32^{\circ} \mathrm{C}\). Treating the body of each person as a 25 -cm-diameter, \(1.7-\mathrm{m}\)-long cylinder, determine the fractions of heat lost from each person by perspiration.

In a combined heat and power (CHP) generation process, by-product heat is used for domestic or industrial heating purposes. Hot steam is carried from a CHP generation plant by a tube with diameter of \(127 \mathrm{~mm}\) centered at a square crosssection solid bar made of concrete with thermal conductivity of \(1.7 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The surface temperature of the tube is constant at \(120^{\circ} \mathrm{C}\), while the square concrete bar is exposed to air with temperature of \(-5^{\circ} \mathrm{C}\) and convection heat transfer coefficient of \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the temperature difference between the outer surface of the square concrete bar and the ambient air is to be maintained at \(5^{\circ} \mathrm{C}\), determine the width of the square concrete bar and the rate of heat loss per meter length.

The unit thermal resistances ( \(R\)-values) of both 40-mm and 90-mm vertical air spaces are given in Table 3-9 to be \(0.22 \mathrm{~m}^{2} \cdot \mathrm{C} / \mathrm{W}\), which implies that more than doubling the thickness of air space in a wall has no effect on heat transfer through the wall. Do you think this is a typing error? Explain.

Consider an insulated pipe exposed to the atmosphere. Will the critical radius of insulation be greater on calm days or on windy days? Why?

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.