/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 93 The exposed surface \((x=0)\) of... [FREE SOLUTION] | 91Ó°ÊÓ

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The exposed surface \((x=0)\) of a plane wall of thermal conductivity \(k\) is subjected to microwave radiation that causes volumetric heating to vary as $$ \dot{q}(x)=\dot{q}_{o}\left(1-\frac{x}{L}\right) $$ where \(\dot{q}_{o}\left(\mathrm{~W} / \mathrm{m}^{3}\right)\) is a constant. The boundary at \(x=L\) is perfectly insulated, while the exposed surface is maintained at a constant temperature \(T_{o}\). Determine the temperature distribution \(T(x)\) in terms of \(x, L, k, \dot{q}_{o}\), and \(T_{o}\).

Short Answer

Expert verified
The temperature distribution \(T(x)\) in the plane wall can be determined by solving the heat conduction equation and applying given boundary conditions. The final temperature distribution is: \[ T(x) = -\frac{\dot{q}_{o}}{6k}x^3 + \frac{\dot{q}_{o}L}{4k}x^2 + \frac{\dot{q}_{o}L}{2k}x + T_{o} \]

Step by step solution

01

Set up 1D heat conduction equation with volumetric heating

The 1D heat conduction equation with volumetric heating can be represented as: \[ \frac{d^2T}{dx^2}+\frac{\dot{q}(x)}{k} = 0 \] where \(T\) is the temperature, \(x\) is the distance along the plane wall, and \(k\) is the thermal conductivity.
02

Substitute given function \(\dot{q}(x)\) into the equation

The volumetric heat generation rate, \(\dot{q}(x)\), is given by: \[ \dot{q}(x) = \dot{q}_{o}\left(1-\frac{x}{L}\right) \] Substitute this function into the heat conduction equation: \[ \frac{d^2T}{dx^2}+\frac{\dot{q}_{o}\left(1-\frac{x}{L}\right)}{k} = 0 \]
03

Integrate the equation for the first time

Integrate the equation with respect to \(x\) once: \[ \frac{dT}{dx} + \frac{\dot{q}_{o}}{k} \left(x - \frac{x^{2}}{2L}\right) = C_{1} \] where \(C_{1}\) is a constant of integration.
04

Apply boundary condition at x=L

The boundary at x=L is perfectly insulated, which means that heat does not flow through this point and the temperature gradient is zero: \[ \frac{dT}{dx}\bigg|_{x=L} = 0 \] Apply this boundary condition and solve for \(C_{1}\): \[ 0 + \frac{\dot{q}_{o}}{k} \left(L - \frac{L^{2}}{2L}\right) = C_{1} \] \[ C_{1} = \frac{\dot{q}_{o}L}{2k} \]
05

Integrate the equation for the second time

Integrate the equation with respect to \(x\) again: \[ T(x) = \int \left( \frac{dT}{dx} + \frac{\dot{q}_{o}}{k} \left(x - \frac{x^{2}}{2L}\right) - \frac{\dot{q}_{o}L}{2k} \right) dx \] \[ T(x) = -\frac{\dot{q}_{o}}{6k}x^3 + \frac{\dot{q}_{o}L}{4k}x^2 + C_{1}x + C_{2} \] where \(C_{2}\) is another constant of integration.
06

Apply boundary condition at x=0

The exposed surface at x=0 is maintained at a constant temperature \(T_{o}\): \[ T(0) = T_{o} \] Apply this boundary condition and solve for \(C_{2}\): \[ T_{o} = -\frac{\dot{q}_{o}}{6k}(0)^3 + \frac{\dot{q}_{o}L}{4k}(0)^2 + C_{1}(0) + C_{2} \] \[ C_{2} = T_{o} \]
07

Write the final temperature distribution equation

Substitute constants \(C_{1}\) and \(C_{2}\) in the equation for \(T(x)\): \[ T(x) = -\frac{\dot{q}_{o}}{6k}x^3 + \frac{\dot{q}_{o}L}{4k}x^2 + \frac{\dot{q}_{o}L}{2k}x + T_{o} \] This is the temperature distribution \(T(x)\) in terms of \(x\), \(L\), \(k\), \(\dot{q}_{o}\), and \(T_{o}\).

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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 material property that measures a substance's ability to conduct heat. It indicates how easily heat passes through a material. In physics and engineering, it is typically expressed in units of watts per meter per kelvin (W/m·K). The higher the thermal conductivity, the more efficient the material is at transferring heat.

In the context of the exercise, it plays a crucial role in the plane wall conduction problem. The value of \(k\) affects the rate at which heat is conducted through the wall and influences the temperature distribution within it. If a wall has high thermal conductivity, heat will spread throughout it more quickly, leading to a more uniform temperature distribution. Conversely, low conductivity will result in a steeper temperature gradient within the wall.
Volumetric Heating
Volumetric heating, often denoted by \(\dot{q}(x)\), refers to the generation of heat within a volume of material, as opposed to heat being transferred to it from another source. This form of heating can occur through various processes, including chemical reactions, electromagnetic absorption (as in the case of microwave radiation), or nuclear reactions.

In our plane wall example, the volumetric heating rate varies with the position \(x\) within the wall due to the absorption of microwave radiation. The heat generation function \(\dot{q}(x) = \dot{q}_{o}(1-\frac{x}{L})\) captures this variation, showing that the heating is maximum at the surface \(x=0\) and decreases linearly to zero at the insulated boundary \(x=L\). This non-uniform heating creates a temperature gradient across the wall, which is essential for determining the final temperature distribution.
Plane Wall Conduction
Plane wall conduction is a simplified heat transfer model that assumes heat flows in a one-dimensional direction through a flat, uniform wall. This model is particularly useful in situations where the thickness of the wall is much less than the other dimensions, allowing us to disregard heat transfer in the other directions.

In the given exercise, the heat conduction equation \(\frac{d^2T}{dx^2} + \frac{\dot{q}(x)}{k} = 0\) represents the balance of heat within the wall, taking into account the internal heat generation (volumetric heating). This equation gets integrated to find the temperature distribution, which is influenced by both the thermal properties of the wall and the internal heat generation. In reality, walls are rarely perfect conductors or insulators, but this idealized model provides a fundamental understanding of heat transfer in condensed matter.
Boundary Conditions
Boundary conditions in heat transfer are constraints applied to the surfaces of a material through which heat is being transferred. These conditions are essential for solving differential equations that describe thermal systems, as they specify the behavior at the boundaries of the domain.

For the exercise, two boundary conditions are given. First, one surface of the wall is perfectly insulated; mathematically, this translates to \(\frac{dT}{dx}|_{x=L} = 0\), meaning that there is no net heat flow through the boundary at \(x=L\). The second condition maintains that the exposed surface is at a constant temperature \(T(0) = T_{o}\), defining the temperature at \(x=0\). By applying these conditions, we can integrate the heat conduction equation to find two constants of integration \(C_{1}\) and \(C_{2}\), which lead us to the final expression for the temperature distribution within the wall.

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Most popular questions from this chapter

Consider a power transistor encapsulated in an aluminum case that is attached at its base to a square aluminum plate of thermal conductivity \(k=240 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), thickness \(L=6 \mathrm{~mm}\), and width \(W=20 \mathrm{~mm}\). The case is joined to the plate by screws that maintain a contact pressure of 1 bar, and the back surface of the plate transfers heat by natural convection and radiation to ambient air and large surroundings at \(T_{\infty}=T_{\text {sur }}=\) \(25^{\circ} \mathrm{C}\). The surface has an emissivity of \(\varepsilon=0.9\), and the convection coefficient is \(h=4 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The case is completely enclosed such that heat transfer may be assumed to occur exclusively through the base plate. (a) If the air-filled aluminum-to-aluminum interface is characterized by an area of \(A_{c}=2 \times 10^{-4} \mathrm{~m}^{2}\) and a roughness of \(10 \mu \mathrm{m}\), what is the maximum allowable power dissipation if the surface temperature of the case, \(T_{s, c}\), is not to exceed \(85^{\circ} \mathrm{C}\) ? (b) The convection coefficient may be increased by subjecting the plate surface to a forced flow of air. Explore the effect of increasing the coefficient over the range \(4 \leq h \leq 200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

A 40-mm-long, 2-mm-diameter pin fin is fabricated of an aluminum alloy \((k=140 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). (a) Determine the fin heat transfer rate for \(T_{b}=50^{\circ} \mathrm{C}\), \(T_{\infty}=25^{\circ} \mathrm{C}, h=1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and an adiabatic tip condition. (b) An engineer suggests that by holding the fin tip at a low temperature, the fin heat transfer rate can be increased. For \(T(x=L)=0^{\circ} \mathrm{C}\), determine the new fin heat transfer rate. Other conditions are as in part (a). (c) Plot the temperature distribution, \(T(x)\), over the range \(0 \leq x \leq L\) for the adiabatic tip case and the prescribed tip temperature case. Also show the ambient temperature in your graph. Discuss relevant features of the temperature distribution. (d) Plot the fin heat transfer rate over the range \(0 \leq h \leq 1000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) for the adiabatic tip case and the prescribed tip temperature case. For the prescribed tip temperature case, what would the

A long cylindrical rod of diameter \(200 \mathrm{~mm}\) with thermal conductivity of \(0.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) experiences uniform volumetric heat generation of \(24,000 \mathrm{~W} / \mathrm{m}^{3}\). The rod is encapsulated by a circular sleeve having an outer diameter of \(400 \mathrm{~mm}\) and a thermal conductivity of \(4 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The outer surface of the sleeve is exposed to cross flow of air at \(27^{\circ} \mathrm{C}\) with a convection coefficient of \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Find the temperature at the interface between the rod and sleeve and on the outer surface. (b) What is the temperature at the center of the rod?

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 particular thermal system involves three objects of fixed shape with conduction resistances of \(R_{1}=1 \mathrm{~K} / \mathrm{W}\), \(R_{2}=2 \mathrm{~K} / \mathrm{W}\) and \(R_{3}=4 \mathrm{~K} / \mathrm{W}\), respectively. An objective is to minimize the total thermal resistance \(R_{\text {tot }}\) associated with a combination of \(R_{1}, R_{2}\), and \(R_{3}\). The chief engineer is willing to invest limited funds to specify an alternative material for just one of the three objects; the alternative material will have a thermal conductivity that is twice its nominal value. Which object (1, 2, or 3 ) should be fabricated of the higher thermal conductivity material to most significantly decrease \(R_{\text {tot }}\) ? Hint: Consider two cases, one for which the three thermal resistances are arranged in series, and the second for which the three resistances are arranged in parallel.

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