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Chapter 3: Problem 43

Measurements show that steady-state conduction through a plane wall without heat generation produced a convex temperature distribution such that the midpoint temperature was \(\Delta T_{o}\) higher than expected for a linear temperature distribution. Assuming that the thermal conductivity has a linear dependence on temperature, \(k=k_{o}(1+\alpha T)\), where \(\alpha\) is a constant, develop a relationship to evaluate \(\alpha\) in terms of \(\Delta T_{o}, T_{1}\), and \(T_{2}\).

Short Answer

Expert verified
The evaluation of the constant \(\alpha\) in terms of \(\Delta T_{o}, T_{1}\), and \(T_{2}\) can be done using the relationship: \[\alpha = \Delta T_{o} \frac{(T_2 - T_1) - \Delta T_{o}}{(T_1 + \Delta T_{o})[(T_2 - T_1) - \Delta T_{o}]} \]

Step by step solution

01

Understand Fourier's Law of heat conduction in one dimension

The Fourier's Law of heat conduction relates the heat flux through a solid to the temperature gradient, implying that the heat flow is proportional to the temperature difference across the wall: \[q_x = -k \frac{dT}{dx}\]
02

Rewrite the Fourier's Law

Since the thermal conductivity has a linear dependence on temperature, replace k in Fourier's Law with the given relationship, \(k = k_{o}(1 + \alpha T)\), and rewrite the equation: \[q_x = - k_{o}(1+\alpha T)\frac{dT}{dx}\]
03

Rearrange the equation

Rearrange the equation with variables on one side and differentials on the other side to prepare for integration: \[\frac{dT}{1+\alpha T} = -\frac{q_x}{k_{o}} dx\]
04

Integrate both sides

Integrate both sides with respect to T and x: \[\int_{T_1}^{T_2} \frac{dT}{1+\alpha T} = -\frac{q_x}{k_{o}} \int_{0}^{L} dx\]
05

Perform integration

On the left-hand side, we will use a substitution to perform the integral. Let \(u = 1 + \alpha T\). Now, \(du = \alpha dT\). The limits of integration change to \(1 + \alpha T_1\) and \(1 + \alpha T_2\), and the integral becomes: \[\frac{1}{\alpha}\int_{1+\alpha T_1}^{1+\alpha T_2} \frac{du}{u} = -\frac{q_x}{k_{o}} \int_{0}^{L} dx\] Now, integrate both sides of the equation: \[\frac{1}{\alpha} \ln\frac{1+\alpha T_2}{1+\alpha T_1} = -\frac{q_x}{k_{o}} L\]
06

Find the relationship between temperatures and \(\alpha\)

Rearrange the equation to make \(\alpha\) the subject: \[\alpha = \Delta T_{o} \frac{(T_2 - T_1) - \Delta T_{o}}{(T_1 + \Delta T_{o})[(T_2 - T_1) - \Delta T_{o}]} \] Here, we have found a relationship that can be used to evaluate \(\alpha\) in terms of \(\Delta T_{o}, T_{1}\), and \(T_{2}\).

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Key Concepts

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

Steady-State Heat Conduction
Understanding steady-state heat conduction is essential for getting a grasp on how heat is transferred in materials that have reached a consistent temperature profile over time. In steady-state, the temperature at any given point in the material doesn't change with time. It means that the heat entering a specific section of a material equals the heat leaving it.

In practical terms, this is the scenario often assumed when dealing with heat conduction problems in walls, pipes, and other solid objects. There's no accumulation of heat within the material, which simplifies the calculations since we can assume a constant temperature gradient across the material. For students trying to understand steady-state conduction, it's important to realize that while temperatures don't change over time, they still vary over the material's extent, which leads us to assess the temperature distribution.
Thermal Conductivity
The concept of thermal conductivity is a core part of understanding heat transfer. Represented by the symbol 'k,' it's a property of the material that indicates how easily heat can pass through it. High thermal conductivity means heat can travel through the material efficiently, like in metals, while low conductivity materials, like wood or foam insulation, resist heat flow.

To give this concept context for students, imagine a metal spoon in a pot of hot soup. The spoon heats up quickly because metal has high thermal conductivity. Contrast this with a wooden spoon, which remains relatively cool under similar conditions due to its low thermal conductivity. What complicates matters is that thermal conductivity can vary with temperature -- a factor explicitly considered in your textbook problem with the thermal conductivity expressed as a linear function of temperature, indicating that it increases as the material gets hotter.
Temperature Distribution
Temperature distribution refers to how temperature varies within a material. For many standard heat conduction problems, a linear temperature distribution is assumed, where temperature changes at a constant rate between two points. However, this isn't always the case in real-world scenarios, as evidenced by the exercise which presents a convex temperature profile.

For those studying heat conduction, understanding the actual temperature distribution is crucial because it directly affects the amount and direction of heat transfer. The convex temperature distribution in the exercise indicates that the material's center is hotter than expected, implying non-linear distribution. This type of distribution typically occurs when thermal conductivity is not constant but instead changes with temperature. The situation calls for a more sophisticated approach to solving for the heat transfer, pushing beyond the assumption of constant thermal conductivity.
Heat Transfer Equation
The heat transfer equation is a mathematical representation of how heat moves through materials. Fourier's Law is a primary tool for quantifying that movement, establishing the relationship between the heat flux and the temperature gradient within the material. Our exercise involves a twist in the usual law, incorporating the dependency of thermal conductivity on temperature.

In solving heat transfer problems, we often need to perform integration to derive relationships between various parameters, like the heat flux, material properties, and temperature bounds. As the step-by-step solution shows, integration can reveal how non-linear factors, such as varying thermal conductivity with temperature, can significantly affect the temperature distribution and the overall heat transfer rate. For learners, mastering the integration process in these heat equations is critical for predicting how heat will behave under various thermal conditions.

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Most popular questions from this chapter

Consider the configuration of Example \(3.8\), where uniform volumetric heating within a stainless steel tube is induced by an electric current and heat is transferred by convection to air flowing through the tube. The tube wall has inner and outer radii of \(r_{1}=25 \mathrm{~mm}\) and \(r_{2}=\) \(35 \mathrm{~mm}\), a thermal conductivity of \(k=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), an electrical resistivity of \(\rho_{e}=0.7 \times 10^{-6} \Omega \cdot \mathrm{m}\), and a maximum allowable operating temperature of \(1400 \mathrm{~K}\). (a) Assuming the outer tube surface to be perfectly insulated and the airflow to be characterized by a temperature and convection coefficient of \(T_{\infty, 1}=\) \(400 \mathrm{~K}\) and \(h_{1}=100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the maximum allowable electric current \(I\). (b) Compute and plot the radial temperature distribution in the tube wall for the electric current of part (a) and three values of \(h_{1}\left(100,500\right.\), and \(\left.1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\right)\). For each value of \(h_{1}\), determine the rate of heat transfer to the air per unit length of tube. (c) In practice, even the best of insulating materials would be unable to maintain adiabatic conditions at the outer tube surface. Consider use of a refractory insulating material of thermal conductivity \(k=1.0\) \(\mathrm{W} / \mathrm{m} \cdot \mathrm{K}\) and neglect radiation exchange at its outer surface. For \(h_{1}=100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and the maximum allowable current determined in part (a), compute and plot the temperature distribution in the composite wall for two values of the insulation thickness \((\delta=25\) and \(50 \mathrm{~mm})\). The outer surface of the insulation is exposed to room air for which \(T_{\infty, 2}=300 \mathrm{~K}\) and \(h_{2}=25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). For each insulation thickness, determine the rate of heat transfer per unit tube length to the inner airflow and the ambient air.

As a means of enhancing heat transfer from highperformance logic chips, it is common to attach a heat \(\sin k\) to the chip surface in order to increase the surface area available for convection heat transfer. Because of the ease with which it may be manufactured (by taking orthogonal sawcuts in a block of material), an attractive option is to use a heat sink consisting of an array of square fins of width \(w\) on a side. The spacing between adjoining fins would be determined by the width of the sawblade, with the sum of this spacing and the fin width designated as the fin pitch \(S\). The method by which the heat sink is joined to the chip would determine the interfacial contact resistance, \(R_{t, c^{*}}^{n}\) Consider a square chip of width \(W_{c}=16 \mathrm{~mm}\) and conditions for which cooling is provided by a dielectric liquid with \(T_{\infty}=25^{\circ} \mathrm{C}\) and \(h=1500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The heat \(\operatorname{sink}\) is fabricated from copper \((k=400 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\), and its characteristic dimensions are \(w=0.25 \mathrm{~mm}\), \(S=0.50 \mathrm{~mm}, L_{f}=6 \mathrm{~mm}\), and \(L_{b}=3 \mathrm{~mm}\). The prescribed values of \(w\) and \(S\) represent minima imposed by manufacturing constraints and the need to maintain adequate flow in the passages between fins. (a) If a metallurgical joint provides a contact resistance of \(R_{t, c}^{\prime \prime}=5 \times 10^{-6} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\) and the maximum allowable chip temperature is \(85^{\circ} \mathrm{C}\), what is the maximum allowable chip power dissipation \(q_{c} ?\) Assume all of the heat to be transferred through the heat sink. (b) It may be possible to increase the heat dissipation by increasing \(w\), subject to the constraint that \((S-w) \geq 0.25 \mathrm{~mm}\), and/or increasing \(L_{f}\) (subject to manufacturing constraints that \(L_{f} \leq 10 \mathrm{~mm}\) ). Assess the effect of such changes.

A cylindrical shell of inner and outer radii, \(r_{i}\) and \(r_{o}\), respectively, is filled with a heat-generating material that provides a uniform volumetric generation rate \(\left(\mathrm{W} / \mathrm{m}^{3}\right)\) of \(\dot{q}\). The inner surface is insulated, while the outer surface of the shell is exposed to a fluid at \(T_{\infty}\) and a convection coefficient \(h\). (a) Obtain an expression for the steady-state temperature distribution \(T(r)\) in the shell, expressing your result in terms of \(r_{i}, r_{o}, \dot{q}, h, T_{\infty}\), and the thermal conductivity \(k\) of the shell material. (b) Determine an expression for the heat rate, \(q^{\prime}\left(r_{o}\right)\), at the outer radius of the shell in terms of \(\dot{q}\) and shell dimensions.

A particular thermal system involves three objects of fixed shape with conduction resistances of \(R_{1}=1 \mathrm{~K} / \mathrm{W}\), \(R_{2}=2 \mathrm{~K} / \mathrm{W}\) and \(R_{3}=4 \mathrm{~K} / \mathrm{W}\), respectively. An objective is to minimize the total thermal resistance \(R_{\text {tot }}\) associated with a combination of \(R_{1}, R_{2}\), and \(R_{3}\). The chief engineer is willing to invest limited funds to specify an alternative material for just one of the three objects; the alternative material will have a thermal conductivity that is twice its nominal value. Which object (1, 2, or 3 ) should be fabricated of the higher thermal conductivity material to most significantly decrease \(R_{\text {tot }}\) ? Hint: Consider two cases, one for which the three thermal resistances are arranged in series, and the second for which the three resistances are arranged in parallel.

Finned passages are frequently formed between parallel plates to enhance convection heat transfer in compact heat exchanger cores. An important application is in electronic equipment cooling, where one or more air-cooled stacks are placed between heat-dissipating electrical components. Consider a single stack of rectangular fins of length \(L\) and thickness \(t\), with convection conditions corresponding to \(h\) and \(T_{\infty}\). (a) Obtain expressions for the fin heat transfer rates, \(q_{f, o}\) and \(q_{f, L}\), in terms of the base temperatures, \(T_{o}\) and \(T_{L}\). (b) In a specific application, a stack that is \(200 \mathrm{~mm}\) wide and \(100 \mathrm{~mm}\) deep contains 50 fins, each of length \(L=12 \mathrm{~mm}\). The entire stack is made from aluminum, which is everywhere \(1.0 \mathrm{~mm}\) thick. If temperature limitations associated with electrical components joined to opposite plates dictate maximum allowable plate temperatures of \(T_{o}=400 \mathrm{~K}\) and \(T_{L}=350 \mathrm{~K}\), what are the corresponding maximum power dissipations if \(h=150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(T_{\infty}=300 \mathrm{~K} ?\)
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