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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

A spherical, cryosurgical probe may be imbedded in diseased tissue for the purpose of freezing, and thereby destroying, the tissue. Consider a probe of \(3-\mathrm{mm}\) diameter whose surface is maintained at \(-30^{\circ} \mathrm{C}\) when imbedded in tissue that is at \(37^{\circ} \mathrm{C}\). A spherical layer of frozen tissue forms around the probe, with a temperature of \(0^{\circ} \mathrm{C}\) existing at the phase front (interface) between the frozen and normal tissue. If the thermal conductivity of frozen tissue is approximately \(1.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and heat transfer at the phase front may be characterized by an effective convection coefficient of \(50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), what is the thickness of the layer of frozen tissue (assuming negligible perfusion)?

A rod of diameter \(D=25 \mathrm{~mm}\) and thermal conductivity \(k=60 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) protrudes normally from a furnace wall that is at \(T_{w}=200^{\circ} \mathrm{C}\) and is covered by insulation of thickness \(L_{\text {ins }}=200 \mathrm{~mm}\). The rod is welded to the furnace wall and is used as a hanger for supporting instrumentation cables. To avoid damaging the cables, the temperature of the rod at its exposed surface, \(T_{o}\), must be maintained below a specified operating limit of \(T_{\max }=100^{\circ} \mathrm{C}\). The ambient air temperature is \(T_{\infty}=\) \(25^{\circ} \mathrm{C}\), and the convection coefficient is \(h=15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Derive an expression for the exposed surface temperature \(T_{o}\) as a function of the prescribed thermal and geometrical parameters. The rod has an exposed length \(L_{o}\), and its tip is well insulated. (b) Will a rod with \(L_{o}=200 \mathrm{~mm}\) meet the specified operating limit? If not, what design parameters would you change? Consider another material, increasing the thickness of the insulation, and increasing the rod length. Also, consider how you might attach the base of the rod to the furnace wall as a means to reduce \(T_{o}\).

A device used to measure the surface temperature of an object to within a spatial resolution of approximately \(50 \mathrm{~nm}\) is shown in the schematic. It consists of an extremely sharp-tipped stylus and an extremely small cantilever that is scanned across the surface. The probe tip is of circular cross section and is fabricated of polycrystalline silicon dioxide. The ambient temperature is measured at the pivoted end of the cantilever as \(T_{\infty}=\) \(25^{\circ} \mathrm{C}\), and the device is equipped with a sensor to measure the temperature at the upper end of the sharp tip, \(T_{\text {sen. }}\). The thermal resistance between the sensing probe and the pivoted end is \(R_{t}=5 \times 10^{6} \mathrm{~K} / \mathrm{W}\). (a) Determine the thermal resistance between the surface temperature and the sensing temperature. (b) If the sensing temperature is \(T_{\text {sen }}=28.5^{\circ} \mathrm{C}\), determine the surface temperature. Hint: Although nanoscale heat transfer effects may be important, assume that the conduction occurring in the air adjacent to the probe tip can be described by Fourier's law and the thermal conductivity found in Table A. \(4 .\)

Consider a tube wall of inner and outer radii \(r_{i}\) and \(r_{o}\), whose temperatures are maintained at \(T_{i}\) and \(T_{o}\), respectively. The thermal conductivity of the cylinder is temperature dependent and may be represented by an expression of the form \(k=k_{o}(1+a T)\), where \(k_{o}\) and \(a\) are constants. Obtain an expression for the heat transfer per unit length of the tube. What is the thermal resistance of the tube wall?

A bonding operation utilizes a laser to provide a constant heat flux, \(q_{o}^{\prime \prime}\), across the top surface of a thin adhesivebacked, plastic film to be affixed to a metal strip as shown in the sketch. The metal strip has a thickness \(d=1.25 \mathrm{~mm}\), and its width is large relative to that of the film. The thermophysical properties of the strip are \(\rho=7850 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=435 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), and \(k=60 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The thermal resistance of the plastic film of width \(w_{1}=40 \mathrm{~mm}\) is negligible. The upper and lower surfaces of the strip (including the plastic film) experience convection with air at \(25^{\circ} \mathrm{C}\) and a convection coefficient of \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The strip and film are very long in the direction normal to the page. Assume the edges of the metal strip are at the air temperature \(\left(T_{\infty}\right)\). (a) Derive an expression for the temperature distribution in the portion of the steel strip with the plastic film \(\left(-w_{1} / 2 \leq x \leq+w_{1} / 2\right)\). (b) If the heat flux provided by the laser is 10,000 \(\mathrm{W} / \mathrm{m}^{2}\), determine the temperature of the plastic film at the center \((x=0)\) and its edges \(\left(x=\pm w_{1} / 2\right)\). (c) Plot the temperature distribution for the entire strip and point out its special features.
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