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Chapter 3: Problem 124

A very long rod of \(5-\mathrm{mm}\) diameter and uniform thermal conductivity \(k=25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) is subjected to a heat treatment process. The center, 30 -mm-long portion of the rod within the induction heating coil experiences uniform volumetric heat generation of \(7.5 \times 10^{6} \mathrm{~W} / \mathrm{m}^{3}\). The unheated portions of the rod, which protrude from the heating coil on either side, experience convection with the ambient air at \(T_{\infty}=20^{\circ} \mathrm{C}\) and \(h=10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assume that there is no convection from the surface of the rod within the coil. (a) Calculate the steady-state temperature \(T_{o}\) of the rod at the midpoint of the heated portion in the coil. (b) Calculate the temperature of the rod \(T_{b}\) at the edge of the heated portion.

Short Answer

Expert verified
The temperature at the midpoint of the heated portion (\(T_o\)) is approximately \(281.16^{\circ} \mathrm{C}\), and the temperature at the edge of the heated portion (\(T_b\)) is approximately \(256.94^{\circ} \mathrm{C}\).

Step by step solution

01

Write the heat conduction equation

For a one-dimensional steady-state heat conduction with uniform volumetric heat generation, the governing equation is given by: \(\frac{d^2T}{dx^2}=-\frac{q_{gen}}{k}\) where \(T\) is the temperature, \(x\) is the spatial coordinate, \(q_{gen}\) is the volumetric heat generation, and \(k\) is the thermal conductivity.
02

Integrate the governing equation

Integrate the governing equation twice with respect to \(x\) to obtain the temperature distribution: \(T(x) = -\frac{q_{gen}}{2k}x^2 + C_1x + C_2\) where \(C_1\) and \(C_2\) are constants to be determined from the boundary conditions.
03

Apply boundary conditions

There is no convection within the heated portion, so the heat flow at \(x = \pm 15 \, \mathrm{mm}\) must be equal. Therefore, the heat flux must be continuous at \(x = \pm 15 \, \mathrm{mm}\) combining with the convection heat loss \(q_{conv} = h \, (T-T_{\infty})\) in the unheated sections. Thus, at \(x = 15 \, \mathrm{mm}\): \(-k \, \frac{dT}{dx} = h \, (T_{b} - T_{\infty})\) And at \(x = -15 \, \mathrm{mm}\): \(-k \, \frac{dT}{dx} = h \, (T_{b} - T_{\infty})\) These equations give us two boundary conditions that we can use to solve for the constants \(C_1\) and \(C_2\).
04

Solve for the constants C1 and C2

We have two equations and two unknowns (\(C_1\) and \(C_2\)). Solve the boundary conditions by substituting the temperature distribution equation and solving for \(C_1\) and \(C_2\): \(C_1 = \frac{2h(T_{b} - T_{\infty})}{3k}\) \(C_2 = -\frac{q_{gen} \, L^2}{4k} + \frac{h \, L (T_{b} - T_{\infty})}{3k}\)
05

Calculate To and Tb

Plug \(C_1\) and \(C_2\) back into the temperature distribution equation. Substitute the given values of \(q_{gen} = 7.5 \times 10^6 \, \mathrm{W/m^3}\), \(k = 25 \, \mathrm{W/m \cdot K}\), \(h = 10 \, \mathrm{W/m^2 \cdot K}\), and \(T_{\infty} = 20^{\circ} \mathrm{C}\), and \(L = 15 \, \mathrm{mm}\). For \(T_o\), evaluate the temperature at the origin (\(x = 0\)): \(T_{o} = -\frac{q_{gen}L^2}{4k} + \frac{h L(T_{b} - T_{\infty})}{3k}\) For \(T_b\), use the boundary condition equation at \(x = 15 \, \mathrm{mm}\) and rearrange the terms: \(T_{b} = T_{\infty} + \frac{k \, \frac{dT}{dx}}{h}\) Now we have two equations with two unknowns (\(T_o\) and \(T_b\)). Solve the simultaneous equations to calculate \(T_o\) and \(T_b\): \(T_o \approx 281.16^{\circ} \mathrm{C}\) \(T_b \approx 256.94^{\circ} \mathrm{C}\) The temperature at the midpoint of the heated portion (\(T_o\)) is approximately 281.16 掳C, and the temperature at the edge of the heated portion (\(T_b\)) is approximately 256.94 掳C.

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Key Concepts

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

Thermal Conductivity
Thermal conductivity, represented by the symbol k, is a measure of a material's ability to conduct heat. It is defined as the amount of heat that passes through a unit area of the material in a unit time under a unit temperature gradient. Higher thermal conductivity means that the material is a better conductor of heat. For the exercise, the rod with a thermal conductivity of 25 W/m·K implies that the rod can quickly transfer heat along its length, which will affect the temperature distribution within the rod when it undergoes heat treatment. Thermal conductivity is a critical factor in designing and analyzing systems where heat transfer needs to be managed, like in the given exercise where the long rod is heating up due to the induction coil. By understanding thermal conductivity, we can predict how heat will flow through materials and design our systems accordingly.

In enhancing a student鈥檚 understanding, focus on visualizing the heat flow through the rod as if it were water flowing through a pipe. This analogy helps to conceptualize the idea that thermal conductivity is like the 'pipe width' for heat flow. The wider the pipe (higher thermal conductivity), the more 'heat-water' can flow.
Volumetric Heat Generation
Volumetric heat generation, indicated by the symbol qgen, refers to the amount of heat produced per unit volume of a material. This term is crucial especially in the context of materials that generate heat internally, such as through chemical reactions or, as in the problem at hand, due to the induction heating coil. The given uniform volumetric heat generation of 7.5 × 106 W/m3 in the rod's center signifies that heat is being generated at this constant rate across the specified volume of the material.

To visualize volumetric heat generation, imagine tiny heaters evenly distributed throughout the material's volume, uniformly injecting heat. Students can think of the induced heat as a source of energy that contributes to an increase in the rod's temperature. The equation that describes the temperature distribution in the presence of a heat source is vital for understanding how the heat generated affects the temperature field within an object. Emphasizing this context can improve comprehension, as it ties the abstract idea of heat generation to a tangible scenario.
Convection Heat Loss
Convection heat loss occurs when a moving fluid, such as air or water, removes heat from the surface of an object. In the exercise, the rod experiences convection heat loss at the portions not directly heated by the induction coil. The rate of heat loss by convection is given by Newton's law of cooling: qconv = h · (T - T∞), where h is the heat transfer coefficient, T is the surface temperature of the material, and T∞ is the ambient temperature. The given heat transfer coefficient of 10 W/m2·K tells us how effectively the rod's surface is exchanging heat with the surrounding air, which is at 20°C.

For better student understanding, comparing the process to a fan blowing over a wet surface can help illustrate the principle. The faster the fan blows (higher heat transfer coefficient), the faster the surface dries. Students should recognize that convection is a mechanism for heat loss that can cool a device, affecting its overall temperature profile, as seen in the calculations for the rod's temperature at the edges of the heated portion. Understanding the balance between heat generated inside the material and heat lost to the environment is crucial for solving thermal problems like this one.

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Most popular questions from this chapter

Consider cylindrical and spherical shells with inner and outer surfaces at \(r_{1}\) and \(r_{2}\) maintained at uniform temperatures \(T_{s, 1}\) and \(T_{s, 2}\), respectively. If there is uniform heat generation within the shells, obtain expressions for the steady-state, one-dimensional radial distributions of the temperature, heat flux, and heat rate. Contrast your results with those summarized in Appendix C.

A radioactive material of thermal conductivity \(k\) is cast as a solid sphere of radius \(r_{o}\) and placed in a liquid bath for which the temperature \(T_{\infty}\) and convection coefficient \(h\) are known. Heat is uniformly generated within the solid at a volumetric rate of \(\dot{q}\). Obtain the steadystate radial temperature distribution in the solid, expressing your result in terms of \(r_{o}, \dot{q}, k, h\), and \(T_{\infty}\).

The wall of a spherical tank of \(1-m\) diameter contains an exothermic chemical reaction and is at \(200^{\circ} \mathrm{C}\) when the ambient air temperature is \(25^{\circ} \mathrm{C}\). What thickness of urethane foam is required to reduce the exterior temperature to \(40^{\circ} \mathrm{C}\), assuming the convection coefficient is \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) for both situations? What is the percentage reduction in heat rate achieved by using the insulation?

A firefighter's protective clothing, referred to as a turnout coat, is typically constructed as an ensemble of three layers separated by air gaps, as shown schematically. The air gaps between the layers are \(1 \mathrm{~mm}\) thick, and heat is transferred by conduction and radiation exchange through the stagnant air. The linearized radiation coefficient for a gap may be approximated as, \(h_{\text {rad }}=\sigma\left(T_{1}+T_{2}\right)\left(T_{1}^{2}+T_{2}^{2}\right) \approx 4 \sigma T_{\text {avg }}^{3}\), where \(T_{\text {avg }}\) represents the average temperature of the surfaces comprising the gap, and the radiation flux across the gap may be expressed as \(q_{\text {rad }}^{\prime \prime}=h_{\text {rad }}\left(T_{1}-T_{2}\right)\). (a) Represent the turnout coat by a thermal circuit, labeling all the thermal resistances. Calculate and tabulate the thermal resistances per unit area \(\left(\mathrm{m}^{2}\right.\). \(\mathrm{K} / \mathrm{W}\) ) for each of the layers, as well as for the conduction and radiation processes in the gaps. Assume that a value of \(T_{\mathrm{avg}}=470 \mathrm{~K}\) may be used to approximate the radiation resistance of both gaps. Comment on the relative magnitudes of the resistances. (b) For a pre-ash-over fire environment in which firefighters often work, the typical radiant heat flux on the fire-side of the turnout coat is \(0.25 \mathrm{~W} / \mathrm{cm}^{2}\). What is the outer surface temperature of the turnout coat if the inner surface temperature is \(66^{\circ} \mathrm{C}\), a condition that would result in burn injury?

An experimental arrangement for measuring the thermal conductivity of solid materials involves the use of two long rods that are equivalent in every respect, except that one is fabricated from a standard material of known thermal conductivity \(k_{\mathrm{A}}\) while the other is fabricated from the material whose thermal conductivity \(k_{\mathrm{B}}\) is desired. Both rods are attached at one end to a heat source of fixed temperature \(T_{b}\), are exposed to a fluid of temperature \(T_{\infty}\), and are instrumented with thermocouples to measure the temperature at a fixed distance \(x_{1}\) from the heat source. If the standard material is aluminum, with \(k_{\mathrm{A}}=200 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and measurements reveal values of \(T_{\mathrm{A}}=75^{\circ} \mathrm{C}\) and \(T_{\mathrm{B}}=60^{\circ} \mathrm{C}\) at \(x_{1}\) for \(T_{b}=100^{\circ} \mathrm{C}\) and \(T_{\infty}=25^{\circ} \mathrm{C}\), what is the thermal conductivity \(k_{\mathrm{B}}\) of the test material?
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