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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

A 10-in-thick, 30-ft-long, and 10-ft-high wall is to be constructed using 9 -in-long solid bricks \(\left(k=0.40 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\right)\) of cross section 7 in \(\times 7\) in, or identical size bricks with nine square air holes \(\left(k=0.015 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\right)\) that are 9 in long and have a cross section of \(1.5\) in \(\times 1.5 \mathrm{in}\). There is a \(0.5\)-in-thick plaster layer \(\left(k=0.10 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}{ }^{\circ} \mathrm{F}\right)\) between two adjacent bricks on all four sides and on both sides of the wall. The house is maintained at \(80^{\circ} \mathrm{F}\) and the ambient temperature outside is \(30^{\circ} \mathrm{F}\). Taking the heat transfer coefficients at the inner and outer surfaces of the wall to be \(1.5\) and \(4 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{2} \cdot{ }^{\circ} \mathrm{F}\), respectively, determine the rate of heat transfer through the wall constructed of \((a)\) solid bricks and (b) bricks with air holes.

Hot water at an average temperature of \(85^{\circ} \mathrm{C}\) passes through a row of eight parallel pipes that are \(4 \mathrm{~m}\) long and have an outer diameter of \(3 \mathrm{~cm}\), located vertically in the middle of a concrete wall \((k=0.75 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) that is \(4 \mathrm{~m}\) high, \(8 \mathrm{~m}\) long, and \(15 \mathrm{~cm}\) thick. If the surfaces of the concrete walls are exposed to a medium at \(32^{\circ} \mathrm{C}\), with a heat transfer coefficient of \(12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the rate of heat loss from the hot water and the surface temperature of the wall.

A 25 -cm-diameter, 2.4-m-long vertical cylinder containing ice at \(0^{\circ} \mathrm{C}\) is buried right under the ground. The cylinder is thin-shelled and is made of a high thermal conductivity material. The surface temperature and the thermal conductivity of the ground are \(18^{\circ} \mathrm{C}\) and \(0.85 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) respectively. The rate of heat transfer to the cylinder is (a) \(37.2 \mathrm{~W}\) (b) \(63.2 \mathrm{~W}\) (c) \(158 \mathrm{~W}\) (d) \(480 \mathrm{~W}\) (e) \(1210 \mathrm{~W}\)

The outer surface of an engine is situated in a place where oil leakage can occur. When leaked oil comes in contact with a hot surface that has a temperature above its autoignition temperature, the oil can ignite spontaneously. Consider an engine cover that is made of a stainless steel plate with a thickness of \(1 \mathrm{~cm}\) and a thermal conductivity of \(14 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The inner surface of the engine cover is exposed to hot air with a convection heat transfer coefficient of \(7 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) at \(333^{\circ} \mathrm{C}\). The outer surface is exposed to an environment where the ambient air is \(69^{\circ} \mathrm{C}\) with a convection heat transfer coefficient of \(7 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). To prevent fire hazard in the event of oil leak on the engine cover, a layer of thermal barrier coating (TBC) with a thermal conductivity of \(1.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is applied on the engine cover outer surface. Would a TBC layer of \(4 \mathrm{~mm}\) in thickness be sufficient to keep the engine cover surface below autoignition temperature of \(200^{\circ} \mathrm{C}\) to prevent fire hazard?

A 12-m-long and 5-m-high wall is constructed of two layers of \(1-\mathrm{cm}\)-thick sheetrock \((k=0.17 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) spaced \(16 \mathrm{~cm}\) by wood studs \((k=0.11 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) whose cross section is \(16 \mathrm{~cm} \times 5 \mathrm{~cm}\). The studs are placed vertically \(60 \mathrm{~cm}\) apart, and the space between them is filled with fiberglass insulation \((k=0.034 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The house is maintained at \(20^{\circ} \mathrm{C}\) and the ambient temperature outside is \(-9^{\circ} \mathrm{C}\). Taking the heat transfer coefficients at the inner and outer surfaces of the house to be \(8.3\) and \(34 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively, determine \((a)\) the thermal resistance of the wall considering a representative section of it and (b) the rate of heat transfer through the wall.

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