The thermal contact conductance at the interface of two 1 -cm-thick copper plates is measured to be \(18,000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the thickness of the copper plate whose thermal resistance is equal to the thermal resistance of the interface between the plates.

Short Answer

Expert verified
Answer: The thickness of the copper plate whose thermal resistance is equal to the thermal resistance of the interface is approximately \(2.228 \mathrm{~cm}\).

Step by step solution

01

Write down the given information

We are given the following information: - Thermal contact conductance of the interface: \(h_c = 18,000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) - Thickness of the copper plates: \(d_1 = d_2 = 0.01 \mathrm{~m}\) (1 cm converted to meters) - Thermal conductivity of copper: \(k = 401 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) (from a standard material properties table)
02

Determine the thermal resistance of the interface

Using the formula for thermal resistance of a contact: \(R_{contact} = \frac{1}{A h_c}\), where \(A\) is the area, and \(h_c\) is thermal contact conductance. The problem asks us to find a thickness for which the thermal resistance of a solid is equal to the thermal resistance of the contact, so we need to find \(R_{contact}\) while keeping in mind that the area of the solid and the contact will be the same. So, let's divide by the area to have: \(R_{contact} = \frac{1}{h_c}= \frac{1}{18,000}\).
03

Use the formula for the thermal resistance of a solid

The formula for the thermal resistance of a solid is: \(R_{solid} = \frac{d}{k A}\), where \(d\) is the thickness, \(k\) is the thermal conductivity, and \(A\) is the surface area. In this problem, we are trying to find the thickness \(d\) such that \(R_{solid} = R_{contact}\). So we will set the above equation equal to the thermal resistance of the contact: \(\frac{d}{k A} = \frac{1}{h_c}\).
04

Solve for the thickness \(d\)

To find the thickness \(d\), first multiply both sides of the equation by \(k A\): \(d = \frac{k A}{h_c}\). Now, plug in the known values of thermal conductivity of copper \(k\) and thermal contact conductance \(h_c\) to find the thickness: \(d = \frac{(401)}{(18,000)}\). Evaluating this yields: \(d \approx 0.02228 \mathrm{~m}\). Since we've obtained the thickness in meters, we can now convert it to centimeters: \(d \approx 2.228 \mathrm{~cm}\).
05

State the final answer

The thickness of the copper plate whose thermal resistance is equal to the thermal resistance of the interface between the plates is approximately \(2.228 \mathrm{~cm}\).

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Key Concepts

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

Thermal Contact Conductance
Thermal contact conductance is a property that quantifies the ability of two surfaces in contact to pass heat to one another. It's an important aspect in systems where heat transfer between different components is required, such as in heat exchangers or electronic devices. To calculate the thermal contact conductance, we often make use of a formula involving the area of the interface and the conductance value, like this:
\[\begin{equation}R_{contact} = \frac{1}{A h_c}\end{equation}\]
One must note that this conductance can be influenced by a host of factors, including surface roughness, material types, and how tightly the components are pressed together. The given exercise demonstrates the effort to equate the thermal resistance of the interface (which is reciprocal to the conductance) with that of a material.
Thermal Conductivity
The thermal conductivity of a material, represented by the symbol \[\begin{equation}k\end{equation}\]
is a measure of a material's ability to conduct heat. This intrinsic property is critical when designing systems involving heat transfer. The conductivity value is dependent on the material's structure and bonding; metals typically have high thermal conductivities due to their free electrons which aid in energy transport. In practical terms, if a material has a high thermal conductivity, it will quickly transfer heat from a hot area to a cooler one - copper, the material from the exercise, is well-known for being highly conductive.The calculation for thermal resistance of a solid, which is inversely proportional to conductivity, can be found through:
\[\begin{equation}R_{solid} = \frac{d}{k A}\end{equation}\]
This formula allows us to see the relationship between a material's thickness, its conductivity, and its overall thermal resistance.
Heat Transfer
Heat transfer is a discipline of thermal engineering that concerns the generation, use, conversion, and exchange of thermal energy (heat) between physical systems. The three fundamental modes of heat transfer are conduction, convection, and radiation, with conduction being the primary focus when it comes to discussions of thermal resistance and conductivity. Conduction involves the transfer of heat through a material without any movement of the material itself - it's all about energy moving between atoms and molecules. In exercises like the one we're examining, we often equate the resistance to heat transfer in different scenarios to solve for an unknown value such as thickness or temperature change, which underscores the foundational role heat transfer plays in many engineering problems and solutions.
Material Properties
Material properties are the characteristics that determine how a material reacts to environmental and physical variables, such as heat, force, or electricity. In our context of thermal resistance calculation, the key material property is thermal conductivity, but other properties such as density, specific heat, and tensile strength also play crucial roles in other types of analyses. The inherent properties of materials are determined by their molecular makeup and structure, leading to a wide range of behaviors under similar conditions. When solving for thermal resistance or conductivity, such as in the textbook example provided, understanding the specific material properties allows engineers and scientists to design more efficient thermal systems, predict how materials will behave, and innovate new solutions where existing materials may fall short.

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Most popular questions from this chapter

Consider a stainless steel spoon \(\left(k=8.7 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\right)\) partially immersed in boiling water at \(200^{\circ} \mathrm{F}\) in a kitchen at \(75^{\circ} \mathrm{F}\). The handle of the spoon has a cross section of \(0.08\) in \(\times\) \(0.5\) in, and extends 7 in in the air from the free surface of the water. If the heat transfer coefficient at the exposed surfaces of the spoon handle is \(3 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\), determine the temperature difference across the exposed surface of the spoon handle. State your assumptions. Answer: \(124.6^{\circ} \mathrm{F}\)

What is the reason for the widespread use of fins on surfaces?

Steam at \(235^{\circ} \mathrm{C}\) 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 \(105 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(14 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), 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.

Consider a \(1.5\)-m-high and 2 -m-wide triple pane window. The thickness of each glass layer \((k=0.80 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is \(0.5 \mathrm{~cm}\), and the thickness of each air space \((k=0.025 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is \(1 \mathrm{~cm}\). If the inner and outer surface temperatures of the window are \(10^{\circ} \mathrm{C}\) and \(0^{\circ} \mathrm{C}\), respectively, the rate of heat loss through the window is (a) \(75 \mathrm{~W}\) (b) \(12 \mathrm{~W}\) (c) \(46 \mathrm{~W}\) (d) \(25 \mathrm{~W}\) (e) \(37 \mathrm{~W}\)

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.

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