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Consider two metal plates pressed against each other. Other things being equal, which of the measures below will cause the thermal contact resistance to increase? (a) Cleaning the surfaces to make them shinier. (b) Pressing the plates against each other with a greater force. (c) Filling the gap with a conducting fluid. (d) Using softer metals. (e) Coating the contact surfaces with a thin layer of soft metal such as tin.

Short Answer

Expert verified
Answer: None of the given actions increase the thermal contact resistance. All measures provided, such as cleaning the surfaces, pressing the plates with greater force, filling the gap with conducting fluid, using softer metals, and coating the surfaces with soft metal, result in a decrease in thermal contact resistance.

Step by step solution

01

Determine the Role of Each Measure

To find the measure that will increase the thermal contact resistance, we need to analyze the role of each measure in the heat transfer process between the metal plates.
02

(a) Cleaning the Surfaces to Make Them Shinier

When the surfaces are shinier, they are also smoother. This will reduce the number of air gaps and micro-voids present at the contact surface, allowing for better conduction of heat. Therefore, making the surfaces shinier will decrease thermal contact resistance.
03

(b) Pressing the Plates Against Each Other with a Greater Force

Increasing the force between the plates will put the plates in better contact, effectively reducing the air gaps and micro-voids at the contact surface. Consequently, better contact between metal surfaces will decrease the thermal contact resistance.
04

(c) Filling the Gap with a Conducting Fluid

Introducing a conducting fluid will replace the air gaps and micro-voids with a better heat-conductive medium, allowing for improved heat conduction between the plates. Therefore, filling the gap with a conducting fluid will decrease the thermal contact resistance.
05

(d) Using Softer Metals

Softer metals can deform more easily and conform to the irregularities of each other's surfaces when pressed together. This conformity will decrease the air gaps and improve contact between the surfaces, leading to a reduction of thermal contact resistance.
06

(e) Coating the Contact Surfaces with Soft Metal

Coating the contact surfaces with a soft metal, such as tin, will allow the coating material to deform and follow the irregularities of the contact surface. This will improve the contact between the metal surfaces and decrease the amount of air gaps and voids, thus reducing the thermal contact resistance.
07

Summary of the Results

Among the given options, none of them leads to an increase in thermal contact resistance. All of the measures - (a) Cleaning the surfaces, (b) Pressing the plates with greater force, (c) Filling the gap with conducting fluid, (d) Using softer metals, and (e) Coating the surfaces with soft metal - result in a decrease in thermal contact resistance.

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Key Concepts

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

Heat Conduction
Heat conduction is a fundamental mode of heat transfer where thermal energy moves through a material from a region of higher temperature to a region of lower temperature. It operates on the principle of kinetic energy transfer between particles within a body.

For example, when two metal plates are pressed against each other, this kinetic energy can be efficiently transferred at the points where the metal atoms are in direct contact. The quality of this contact directly impacts the heat conduction between the plates. Factors like surface roughness and impurities can introduce air gaps or micro-voids, which are poor conductors of heat compared to metals. Reducing these imperfections, as with sanding surfaces to increase smoothness, typically enhances heat conduction and reduces thermal contact resistance.

Key Pointers for Efficient Heat Conduction:

  • Thermal conductivity of the material: Higher conductivity means better heat conduction.
  • Temperature gradient: Larger differences accelerate heat conduction.
  • Surface contact quality: Smoother, cleaner, and more intimate contact enhances conduction.
Heat Transfer Process
The heat transfer process is the movement of thermal energy from one place to another and can occur through conduction, convection, and radiation. In the context of the exercise involving two metal plates, we focus on conduction, but it is important to note that the general principles governing efficient heat transfer are applicable in broader contexts as well.

Efficiency in heat transfer processes is often synonymous with how well a system minimizes resistance to the thermal flow. As an improvement, conducting fluids can sometimes be introduced to further ameliorate this process. For example, when the gap between metal plates is filled with a conducting fluid, it replaces air—a poor conductor—thus enhancing the overall heat transfer capability of the system.

Factors Enhancing the Heat Transfer Process:

  • Surface area: Greater contact area can increase the rate of heat transfer.
  • Pressure applied: Compression can improve interfacial contact, beneficial for conductive heat transfer.
  • Intermediary materials: Using substances with high thermal conductivity can facilitate heat movement.
Conducting Fluids in Heat Transfer
In many heat transfer scenarios, the use of conducting fluids—such as water, oil, or specially formulated coolants—proves invaluable. These fluids can easily fill irregular and narrow spaces between solid surfaces, thereby significantly improving the continuity of the heat transfer path.

In our textbook exercise, adding a conducting fluid between metal plates plays a similar role. The fluid displaces air from the interface and molds to the microstructure of the contact surfaces, which minimizes the resistance that the air gaps would otherwise contribute to the system. This illustrates a key application of conducting fluids: enhancing the conduction path between imperfectly matched solid surfaces.

Advantages of Conducting Fluids in Heat Transfer:

  • Improved contact between surfaces.
  • Greater thermal stability across a range of temperatures.
  • Flexibility in application to irregular geometries.

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Most popular questions from this chapter

A transparent film is to be bonded onto the top surface of a solid plate inside a heated chamber. For the bond to cure properly, a temperature of \(70^{\circ} \mathrm{C}\) is to be maintained at the bond, between the film and the solid plate. The transparent film has a thickness of \(1 \mathrm{~mm}\) and thermal conductivity of \(0.05 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), while the solid plate is \(13 \mathrm{~mm}\) thick and has a thermal conductivity of \(1.2 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Inside the heated chamber, the convection heat transfer coefficient is \(70 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the bottom surface of the solid plate is maintained at \(52^{\circ} \mathrm{C}\), determine the temperature inside the heated chamber and the surface temperature of the transparent film. Assume thermal contact resistance is negligible.

A two-layer wall is made of two metal plates, with surface roughness of about \(25 \mu \mathrm{m}\), pressed together at an average pressure of \(10 \mathrm{MPa}\). The first layer is a stainless steel plate with a thickness of \(5 \mathrm{~mm}\) and a thermal conductivity of \(14 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The second layer is an aluminum plate with a thickness of \(15 \mathrm{~mm}\) and a thermal conductivity of \(237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). On the stainless steel side of the wall, the surface is subjected to a heat flux of \(800 \mathrm{~W} / \mathrm{m}^{2}\). On the aluminum side of the wall, the surface experiences convection heat transfer at an ambient temperature of \(20^{\circ} \mathrm{C}\), where the convection coefficient is \(12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the surface temperature of the stainless steel plate.

A pipe is insulated to reduce the heat loss from it. However, measurements indicate that the rate of heat loss has increased instead of decreasing. Can the measurements be right?

Can the thermal resistance concept be used for a solid cylinder or sphere in steady operation? Explain.

A 50 -m-long section of a steam pipe whose outer (€) diameter is \(10 \mathrm{~cm}\) passes through an open space at \(15^{\circ} \mathrm{C}\). The average temperature of the outer surface of the pipe is measured to be \(150^{\circ} \mathrm{C}\). If the combined heat transfer coefficient on the outer surface of the pipe is \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine (a) the rate of heat loss from the steam pipe; \((b)\) the annual cost of this energy lost if steam is generated in a natural gas furnace that has an efficiency of 75 percent and the price of natural gas is $$\$ 0.52 /$$ therm ( 1 therm \(=105,500 \mathrm{~kJ})\); and \((c)\) the thickness of fiberglass insulation \((k=0.035 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) needed in order to save 90 percent of the heat lost. Assume the pipe temperature to remain constant at \(150^{\circ} \mathrm{C}\).

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