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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 measures (a), (b), (c), (d), or (e) would cause the thermal contact resistance to increase. They all tend to lower the thermal contact resistance by increasing the contact area and improving heat transfer between the metal plates.

Step by step solution

01

Option (a): Cleaning the surfaces to make them shinier

Shiny surfaces indicate reduced surface roughness, which in turn ensures better contact between the two plates. Thus, the thermal contact resistance will be lower due to increased contact area, allowing for better heat flow between the two plates. Therefore, this option will cause the thermal contact resistance to decrease and not increase.
02

Option (b): Pressing the plates against each other with a greater force

Applying a greater pressure on the metal plates will increase the contact area between them, as the surface imperfections will be compressed. This increased contact area will enhance the heat flow between the metal plates, leading to a reduction in thermal contact resistance. So, in this case, thermal contact resistance will not increase.
03

Option (c): Filling the gap with a conducting fluid

Introducing a conducting fluid in the gap between the metal plates will improve the thermal contact by conducting heat in the regions where the plates do not have direct contact. The heat conduction through the fluid enhances the overall heat transfer and thus reduces the thermal contact resistance. Therefore, this measure will cause the thermal contact resistance to decrease as well.
04

Option (d): Using softer metals

Softer metals are more easily compliant or deformable, meaning that they can adapt better to the surface imperfections of the contacting plates and thus increase the contact surface area. This increase in contact area leads to better heat flow and consequently contributes to a reduction in the thermal contact resistance. Therefore, using softer metals will not cause an increase in thermal contact resistance.
05

Option (e): Coating the contact surfaces with a thin layer of soft metal such as tin

Coating the contact surfaces with a thin layer of soft metal, like tin, will have similar effects as using softer metals. The soft metal coating will help in adapting better to the surface imperfections, increasing the contact area between the two metal plates. This leads to improved heat transfer and a reduction in thermal contact resistance. Thus, this measure will not increase the thermal contact resistance. In conclusion, none of the given measures - (a), (b), (c), (d), or (e) - would cause the thermal contact resistance to increase. In fact, they all tend to lower the thermal contact resistance by increasing the contact area and improving heat transfer between the two metal plates.

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Most popular questions from this chapter

Explain how the fins enhance heat transfer from a surface. Also, explain how the addition of fins may actually decrease heat transfer from a surface.

Two finned surfaces with long fins are identical, except that the convection heat transfer coefficient for the first finned surface is twice that of the second one. Which of the following conditions is accurate for the efficiency and effectiveness of the first finned surface relative to the second one? (a) Higher efficiency and higher effectiveness (b) Higher efficiency but lower effectiveness (c) Lower efficiency but higher effectiveness (d) Lower efficiency and lower effectiveness (e) Equal efficiency and equal effectiveness

Steam at \(260^{\circ} \mathrm{C}\) is flowing inside a steel pipe $(k=61 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})$ whose inner and outer diameters are \(10 \mathrm{~cm}\) and \(12 \mathrm{~cm}\), respectively, in an environment at \(20^{\circ} \mathrm{C}\). The heat transfer coefficients inside and outside the pipe are 120 \(\mathrm{W} / \mathrm{m}^{2}, \mathrm{~K}\) and $14 \mathrm{~W} / \mathrm{m}^{2}, \mathrm{~K}\(, respectively. Determine \)(a)$ the thickness of the insulation $(k=0.038 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\( needed to reduce the heat loss by 95 percent and \)(b)$ the thickness of the insulation needed to reduce the exposed surface temperature of insulated pipe to \(40^{\circ} \mathrm{C}\) for safety reasons.

Hot water $\left(c_{p}=4.179 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right)\( flows through an 80 -m-long PVC \)(k=0.092 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\( pipe whose inner diameter is \)2 \mathrm{~cm}$ and outer diameter is \(2.5 \mathrm{~cm}\) at a rate of $1 \mathrm{~kg} / \mathrm{s}\(, entering at \)40^{\circ} \mathrm{C}$. If the entire interior surface of this pipe is maintained at \(35^{\circ} \mathrm{C}\) and the entire exterior surface at \(20^{\circ} \mathrm{C}\), the outlet temperature of water is (a) \(35^{\circ} \mathrm{C}\) (b) \(36^{\circ} \mathrm{C}\) (c) \(37^{\circ} \mathrm{C}\) (d) \(38^{\circ} \mathrm{C}\) (e) \(39^{\circ} \mathrm{C}\)

Using cylindrical samples of the same material, devise an experiment to determine the thermal contact resistance. Cylindrical samples are available at any length, and the thermal conductivity of the material is known.

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