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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

From Figure \(2.5\) it is evident that, over a wide temperature range, the temperature dependence of the thermal conductivity of many solids may be approximated by a linear expression of the form \(k=k_{o}+a T\), where \(k_{o}\) is a positive constant and \(a\) is a coefficient that may be positive or negative. Obtain an expression for the heat flux across a plane wall whose inner and outer surfaces are maintained at \(T_{0}\) and \(T_{1}\), respectively. Sketch the forms of the temperature distribution corresponding to \(a>0, a=0\), and \(a<0 .\)

Copper tubing is joined to a solar collector plate of thickness \(t\), and the working fluid maintains the temperature of the plate above the tubes at \(T_{o}\). There is a uniform net radiation heat flux \(q_{\text {rad }}^{\prime \prime}\) to the top surface of the plate, while the bottom surface is well insulated. The top surface is also exposed to a fluid at \(T_{\infty}\) that provides for a uniform convection coefficient \(h\). (a) Derive the differential equation that governs the temperature distribution \(T(x)\) in the plate. (b) Obtain a solution to the differential equation for appropriate boundary conditions.

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 \(T_{o}\).

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 \(T_{\infty}=300 \mathrm{~K} ?\)

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?
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