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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=±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 spherical vessel used as a reactor for producing pharmaceuticals has a 10 -mm-thick stainless steel wall \((k=17 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) and an inner diameter of \(1 \mathrm{~m}\). The exterior surface of the vessel is exposed to ambient air \(\left(T_{\infty}=25^{\circ} \mathrm{C}\right)\) for which a convection coefficient of \(6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) may be assumed. (a) During steady-state operation, an inner surface temperature of \(50^{\circ} \mathrm{C}\) is maintained by energy generated within the reactor. What is the heat loss from the vessel? (b) If a 20 -mm-thick layer of fiberglass insulation \((k=0.040 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is applied to the exterior of the vessel and the rate of thermal energy generation is unchanged, what is the inner surface temperature of the vessel?

Approximately \(10^{6}\) discrete electrical components can be placed on a single integrated circuit (chip), with electrical heat dissipation as high as \(30,000 \mathrm{~W} / \mathrm{m}^{2}\). The chip, which is very thin, is exposed to a dielectric liquid at its outer surface, with \(h_{o}=1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(T_{\infty, 0}=20^{\circ} \mathrm{C}\), and is joined to a circuit board at its inner surface. The thermal contact resistance between the chip and the board is \(10^{-4} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\), and the board thickness and thermal conductivity are \(L_{b}=5 \mathrm{~mm}\) and \(k_{b}=1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), respectively. The other surface of the board is exposed to ambient air for which \(h_{i}=40\) \(\mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(T_{\infty, i}=20^{\circ} \mathrm{C}\). (a) Sketch the equivalent thermal circuit corresponding to steady-state conditions. In variable form, label appropriate resistances, temperatures, and heat fluxes. (b) Under steady-state conditions for which the chip heat dissipation is \(q_{c}^{\prime \prime}=30,000 \mathrm{~W} / \mathrm{m}^{2}\), what is the chip temperature? (c) The maximum allowable heat flux, \(q_{c, m}^{\prime \prime}\), is determined by the constraint that the chip temperature must not exceed \(85^{\circ} \mathrm{C}\). Determine \(q_{c, m}^{\prime \prime}\) for the foregoing conditions. If air is used in lieu of the dielectric liquid, the convection coefficient is reduced by approximately an order of magnitude. What is the value of \(q_{c, m}^{\prime \prime}\) for \(h_{o}=100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) ? With air cooling, can significant improvements be realized by using an aluminum oxide circuit board and/or by using a conductive paste at the chip/board interface for which \(R_{t, c}^{n}=10^{-5} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\) ?

A plane wall of thickness \(0.1 \mathrm{~m}\) and thermal conductivity \(25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) having uniform volumetric heat generation of \(0.3 \mathrm{MW} / \mathrm{m}^{3}\) is insulated on one side, while the other side is exposed to a fluid at \(92^{\circ} \mathrm{C}\). The convection heat transfer coefficient between the wall and the fluid is \(500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the maximum temperature in the wall.

Copper tubing is joined to the absorber of a flat-plate solar collector as shown. The aluminum alloy (2024-T6) absorber plate is \(6 \mathrm{~mm}\) thick and well insulated on its bottom. The top surface of the plate is separated from a transparent cover plate by an evacuated space. The tubes are spaced a distance \(L\) of \(0.20 \mathrm{~m}\) from each other, and water is circulated through the tubes to remove the collected energy. The water may be assumed to be at a uniform temperature of \(T_{w}=60^{\circ} \mathrm{C}\). Under steady-state operating conditions for which the net radiation heat flux to the surface is \(q_{\text {rad }}^{\prime \prime}=\) \(800 \mathrm{~W} / \mathrm{m}^{2}\), what is the maximum temperature on the plate and the heat transfer rate per unit length of tube? Note that \(q_{\text {rad }}^{\prime \prime}\) represents the net effect of solar radiation absorption by the absorber plate and radiation exchange between the absorber and cover plates. You may assume the temperature of the absorber plate directly above a tube to be equal to that of the water.

The energy transferred from the anterior chamber of the eye through the cornea varies considerably depending on whether a contact lens is worn. Treat the eye as a spherical system and assume the system to be at steady state. The convection coefficient \(h_{o}\) is unchanged with and without the contact lens in place. The cornea and the lens cover one-third of the spherical surface area. Values of the parameters representing this situation are as follows: \(\begin{array}{ll}r_{1}=10.2 \mathrm{~mm} & r_{2}=12.7 \mathrm{~mm} \\\ r_{3}=16.5 \mathrm{~mm} & T_{\infty, o}=21^{\circ} \mathrm{C} \\ T_{\infty \infty, i}=37^{\circ} \mathrm{C} & k_{2}=0.80 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} \\ k_{1}=0.35 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} & h_{o}=6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K} \\ h_{i}=12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K} & \end{array}\) (a) Construct the thermal circuits, labeling all potentials and flows for the systems excluding the contact lens and including the contact lens. Write resistance elements in terms of appropriate parameters. (b) Determine the heat loss from the anterior chamber with and without the contact lens in place. (c) Discuss the implication of your results.

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