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Chapter 3: Problem 116

A thin metallic wire of thermal conductivity \(k\), diameter \(D\), and length \(2 L\) is annealed by passing an electrical current through the wire to induce a uniform volumetric heat generation \(\dot{q}\). The ambient air around the wire is at a temperature \(T_{\infty}\), while the ends of the wire at \(x=\pm L\) are also maintained at \(T_{\infty}\). Heat transfer from the wire to the air is characterized by the convection coefficient \(h\). Obtain expressions for the following: (a) The steady-state temperature distribution \(T(x)\) along the wire, (b) The maximum wire temperature. (c) The average wire temperature.

Short Answer

Expert verified
(a) \(T(x)=\frac{\dot{q}}{2k}x^2+T_{\infty}-\frac{\dot{q}}{2k}L^2\) (b) \(T_{max}=T_{\infty}-\frac{\dot{q}}{2k}L^2\) (c) \(T_{avg}=T_{\infty}-\frac{\dot{q}L^2}{6k}\)

Step by step solution

01

Apply the energy conservation equation

For a thin wire with a uniform heat generation \(\dot{q}\), the energy conservation equation in steady state can be represented by the differential equation: \[\frac{d^2T}{dx^2}=\frac{\dot{q}}{k}\]
02

Solve the differential equation

To find the temperature distribution \(T(x)\), we will solve the second-order differential equation. Integration twice gives us: \[\frac{dT}{dx}=\frac{\dot{q}}{k}x+C_1\] \[T(x)=\frac{\dot{q}}{2k}x^2+C_1x+C_2\]
03

Apply boundary conditions

Now, we apply the boundary conditions given in the problem statement: 1. At x=卤L, T=\(T_{\infty}\) \[T_{\infty}=\frac{\dot{q}}{2k}L^2+C_1L+C_2\] 2. At x=-L, T=\(T_{\infty}\) \[T_{\infty}=\frac{\dot{q}}{2k}L^2-C_1L+C_2\] Next, we solve these two equations simultaneously to determine the constant coefficients C_1 and C_2. Subtracting the second equation from the first equation, we get: \[0=2C_1L\] \[C_1=0\] Substitute C_1=0 into either equation to find C_2: \[C_2=T_{\infty}-\frac{\dot{q}}{2k}L^2\]
04

Calculate the steady-state temperature distribution T(x)

Now we have the values of C_1 and C_2, substitute them back into the equation for \(T(x)\) we obtained earlier: \[T(x)=\frac{\dot{q}}{2k}x^2+T_{\infty}-\frac{\dot{q}}{2k}L^2\]
05

Find the maximum wire temperature

The maximum wire temperature will occur at x=0. Substituting x=0 into the steady-state temperature distribution equation, we find: \[T_{max}=\frac{\dot{q}}{2k}(0)^2+T_{\infty}-\frac{\dot{q}}{2k}L^2\] \[T_{max}=T_{\infty}-\frac{\dot{q}}{2k}L^2\]
06

Calculate the average wire temperature

To find the average wire temperature, we integrate the temperature distribution from -L to L and divide by the total length 2L: \[T_{avg}=\frac{1}{2L}\int_{-L}^{L}T(x)dx\] Plugging in our expression for T(x) and integrating: \[T_{avg}=\frac{1}{2L}\int_{-L}^{L}\left(\frac{\dot{q}}{2k}x^2+T_{\infty}-\frac{\dot{q}}{2k}L^2\right)dx = T_{\infty}-\frac{\dot{q}L^2}{6k}\] Summarizing our results, we have found expressions for the steady-state temperature distribution \(T(x)\), the maximum wire temperature \(T_{max}\), and the average wire temperature \(T_{avg}\): (a) \(T(x)=\frac{\dot{q}}{2k}x^2+T_{\infty}-\frac{\dot{q}}{2k}L^2\) (b) \(T_{max}=T_{\infty}-\frac{\dot{q}}{2k}L^2\) (c) \(T_{avg}=T_{\infty}-\frac{\dot{q}L^2}{6k}\)

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Key Concepts

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

Thermal Conductivity
In the context of the given problem, thermal conductivity, represented as k, plays a critical role in understanding the steady-state temperature distribution. Thermal conductivity is a measure of a material's ability to conduct heat. It quantifies the rate at which heat is transferred through a material due to a temperature gradient. In simpler terms, materials with high thermal conductivity, such as metals, transfer heat quickly, whereas materials with low thermal conductivity, like wood or fiberglass, transfer heat slowly.

When the metallic wire in the exercise is annealed, its thermal conductivity allows the heat from the electrical current (volumetric heat generation) to be distributed through the wire. The differential equation
\[\frac{d^2T}{dx^2} = \frac{\dot{q}}{k}\]
is central to finding the temperature at various points along the wire. It indicates that the change in temperature is directly proportional to the heat generation rate and inversely proportional to the thermal conductivity. The wire's conductivity is not just a passive property but also dictates how efficiently heat spreads from the source of generation to the surrounding environment.
Volumetric Heat Generation
Volumetric heat generation, denoted by \(\dot{q}\), is another pinnacle concept within this exercise. It refers to the rate at which heat is generated per unit volume of a material. Typically, this occurs due to internal sources, like the electrical resistance encountered by the current flowing through the wire in our scenario. This resistance generates heat throughout the wire's volume, leading to a rise in temperature.

Understanding the uniform volumetric heat generation is essential for solving the problem at hand, as it directly influences the steady-state temperature distribution along the wire. The equation
\[T(x) = \frac{\dot{q}}{2k}x^2 + T_{\infty} - \frac{\dot{q}}{2k}L^2\]
reflects how the generated heat affects the wire's temperature along its length, x. The temperature increase is quadratic in nature, indicating that it varies with the square of the position along the wire's length. The uniform heat generation assumption simplifies the analysis; however, it is crucial to remember that in practical scenarios, heat generation can vary due to multiple factors such as fluctuating current or nonuniform material properties.
Convection Heat Transfer
Lastly, convection heat transfer is the mechanism by which heat energy is carried away from the wire to the surrounding air. It is characterized by the convection coefficient h. Convection is a mode of heat transfer that occurs due to the movement of fluid (air in this case) over the surface of a body with a different temperature. The convection heat transfer plays a defining role in how quickly the wire can cool down and achieve a steady-state temperature.

The ambient temperature, T_\(\infty\), represents the air temperature surrounding the wire. Convection serves as a sink for the heat generated in the wire, pulling away heat at a rate dependent on the temperature difference between the wire and the surrounding air, as well as the convective heat transfer coefficient h. The boundary conditions of the wire being at ambient temperature at x = \pm L inform us that heat is effectively being removed by convection at these points, ensuring that the ends of the wire remain at the ambient temperature.

Understanding how convection works alongside conduction within the wire helps students grasp the full picture of thermal dynamics in play. It underscores the interplay of internal heat generation and heat removal by external means, leading to a comprehensive solution of the steady-state temperature distribution in practical engineering applications.

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Most popular questions from this chapter

A wire of diameter \(D=2 \mathrm{~mm}\) and uniform temperature \(T\) has an electrical resistance of \(0.01 \Omega / \mathrm{m}\) and a current flow of \(20 \mathrm{~A}\). (a) What is the rate at which heat is dissipated per unit length of wire? What is the heat dissipation per unit volume within the wire? (b) If the wire is not insulated and is in ambient air and large surroundings for which \(T_{\infty}=T_{\text {sur }}=20^{\circ} \mathrm{C}\), what is the temperature \(T\) of the wire? The wire has an emissivity of \(0.3\), and the coefficient associated with heat transfer by natural convection may be approximated by an expression of the form, \(h=C\left[\left(T-T_{\infty}\right) / D\right]^{1 / 4}, \quad\) where \(C=1.25\) \(\mathrm{W} / \mathrm{m}^{7 / 4} \cdot \mathrm{K}^{5 / 4}\). (c) If the wire is coated with plastic insulation of 2-mm thickness and a thermal conductivity of \(0.25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), what are the inner and outer surface temperatures of the insulation? The insulation has an emissivity of \(0.9\), and the convection coefficient is given by the expression of part (b). Explore the effect of the insulation thickness on the surface temperatures.

A technique for measuring convection heat transfer coefficients involves bonding one surface of a thin metallic foil to an insulating material and exposing the other surface to the fluid flow conditions of interest. By passing an electric current through the foil, heat is dissipated uniformly within the foil and the corresponding flux, \(P_{\text {elec }}^{\prime \prime}\), may be inferred from related voltage and current measurements. If the insulation thickness \(L\) and thermal conductivity \(k\) are known and the fluid, foil, and insulation temperatures \(\left(T_{\infty}, T_{s}, T_{b}\right)\) are measured, the convection coefficient may be determined. Consider conditions for which \(T_{\infty}=T_{b}=25^{\circ} \mathrm{C}, P_{\text {elec }}^{\prime \prime}=2000\) \(\mathrm{W} / \mathrm{m}^{2}, L=10 \mathrm{~mm}\), and \(k=0.040 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). (a) With water flow over the surface, the foil temperature measurement yields \(T_{s}=27^{\circ} \mathrm{C}\). Determine the convection coefficient. What error would be incurred by assuming all of the dissipated power to be transferred to the water by convection? (b) If, instead, air flows over the surface and the temperature measurement yields \(T_{s}=125^{\circ} \mathrm{C}\), what is the convection coefficient? The foil has an emissivity of \(0.15\) and is exposed to large surroundings at \(25^{\circ} \mathrm{C}\). What error would be incurred by assuming all of the dissipated power to be transferred to the air by convection? (c) Typically, heat flux gages are operated at a fixed temperature \(\left(T_{s}\right)\), in which case the power dissipation provides a direct measure of the convection coefficient. For \(T_{s}=27^{\circ} \mathrm{C}\), plot \(P_{\text {elec }}^{\prime \prime}\) as a function of \(h_{o}\) for \(10 \leq h_{o} \leq 1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). What effect does \(h_{o}\) have on the error associated with neglecting conduction through the insulation?

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. \(4 .\)

When raised to very high temperatures, many conventional liquid fuels dissociate into hydrogen and other components. Thus the advantage of a solid oxide fuel cell is that such a device can internally reform readily available liquid fuels into hydrogen that can then be used to produce electrical power in a manner similar to Example 1.5. Consider a portable solid oxide fuel cell, operating at a temperature of \(T_{\mathrm{fc}}=800^{\circ} \mathrm{C}\). The fuel cell is housed within a cylindrical canister of diameter \(D=\) \(75 \mathrm{~mm}\) and length \(L=120 \mathrm{~mm}\). The outer surface of the canister is insulated with a low-thermal-conductivity material. For a particular application, it is desired that the thermal signature of the canister be small, to avoid its detection by infrared sensors. The degree to which the canister can be detected with an infrared sensor may be estimated by equating the radiation heat flux emitted from the exterior surface of the canister (Equation 1.5; \(E_{s}=\varepsilon_{s} \sigma T_{s}^{4}\) ) to the heat flux emitted from an equivalent black surface, \(\left(E_{b}=\sigma T_{b}^{4}\right)\). If the equivalent black surface temperature \(T_{b}\) is near the surroundings temperature, the thermal signature of the canister is too small to be detected-the canister is indistinguishable from the surroundings. (a) Determine the required thickness of insulation to be applied to the cylindrical wall of the canister to ensure that the canister does not become highly visible to an infrared sensor (i.e., \(T_{b}-T_{\text {sur }}<5 \mathrm{~K}\) ). Consider cases where (i) the outer surface is covered with a very thin layer of \(\operatorname{dirt}\left(\varepsilon_{s}=0.90\right)\) and (ii) the outer surface is comprised of a very thin polished aluminum sheet \(\left(\varepsilon_{s}=0.08\right)\). Calculate the required thicknesses for two types of insulating material, calcium silicate \((k=0.09 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) and aerogel \((k=0.006 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The temperatures of the surroundings and the ambient are \(T_{\text {sur }}=300 \mathrm{~K}\) and \(T_{\infty}=298 \mathrm{~K}\), respectively. The outer surface is characterized by a convective heat transfer coefficient of \(h=12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (b) Calculate the outer surface temperature of the canister for the four cases (high and low thermal conductivity; high and low surface emissivity). (c) Calculate the heat loss from the cylindrical walls of the canister for the four cases.

One modality for destroying malignant tissue involves imbedding a small spherical heat source of radius \(r_{o}\) within the tissue and maintaining local temperatures above a critical value \(T_{c}\) for an extended period. Tissue that is well removed from the source may be assumed to remain at normal body temperature \(\left(T_{b}=37^{\circ} \mathrm{C}\right)\). Obtain a general expression for the radial temperature distribution in the tissue under steady- state conditions for which heat is dissipated at a rate \(q\). If \(r_{o}=0.5 \mathrm{~mm}\), what heat rate must be supplied to maintain a tissue temperature of \(T \geq T_{c}=42^{\circ} \mathrm{C}\) in the domain \(0.5 \leq r \leq\) \(5 \mathrm{~mm}\) ? The tissue thermal conductivity is approximately \(0.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Assume negligible perfusion.
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