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Chapter 3: Problem 38

A one-dimensional plane wall of thickness \(L\) is constructed of a solid material with a linear, nonuniform porosity distribution described by \(\varepsilon(x)=\varepsilon_{\max }(x / L)\). Plot the steady-state temperature distribution, \(T(x)\), for \(k_{s}=10 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \quad k_{f}=0.1 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, L=1 \mathrm{~m}, \quad \varepsilon_{\max }=\) \(0.25, T(x=0)=30^{\circ} \mathrm{C}\) and \(q_{x}^{\prime \prime}=100 \mathrm{~W} / \mathrm{m}^{2}\) using the expression for the minimum effective thermal conductivity of a porous medium, the expression for the maximum effective thermal conductivity of a porous medium, Maxwell's expression, and for the case where \(k_{\mathrm{efl}}(x)=k_{s}\).

Short Answer

Expert verified
The steady-state temperature distribution $T(x)$ in a porous medium with a linear, nonuniform porosity distribution can be found by calculating the effective thermal conductivity using different expressions, such as minimum effective thermal conductivity, maximum effective thermal conductivity, Maxwell's expression, and the case where \(k_{\mathrm{efl}}(x)=k_{s}\). For each case, we can integrate the temperature gradient obtained from Fourier's law to find the temperature distribution, considering the given boundary condition \(T(x=0)=30^{\circ} \mathrm{C}\). Plotting these temperature distributions as a function of position \(x\) allows us to compare the influence of the different effective thermal conductivity expressions on the temperature profile in the porous medium.

Step by step solution

01

Calculate porosity distribution \(\varepsilon(x)\)

First, we need to find the porosity as a function of the position \(x\). We know that \[ \varepsilon(x)=\varepsilon_{\max }(x / L) \] where \(\varepsilon_{\max}=0.25\) and \(L=1\)m. Thus, \[ \varepsilon(x)=0.25\frac{x}{1} \] So, \(\varepsilon(x)=0.25x\).
02

Calculate the effective thermal conductivity using different expressions

Now we have to find the effective thermal conductivity using the different expressions given: 1. The minimum effective thermal conductivity, \(k_{\min}(x)\), is given by the harmonic mean of \(k_s\) and \(k_f\) \[ k_{\min}(x)=\frac{k_{s} k_{f}}{(\varepsilon \cdot k_f) + k_s(1-\varepsilon)} \] 2. The maximum effective thermal conductivity, \(k_{\max}(x)\), is given by the arithmetic mean of \(k_s\) and \(k_f\) \[ k_{\max}(x)=\frac{k_s+k_f \varepsilon}{1+\varepsilon} \] 3. Maxwell's expression gives \[ k_{M}(x)=k_f \left(1+\frac{2 \varepsilon (k_s-k_f)}{k_f+(2 \varepsilon k_s)}\right) \] 4. The case where \(k_{\mathrm{efl}}(x)=k_{s}\). For each of these expressions, we replace \(k_s=10 \mathrm{~W}/\mathrm{m} \cdot \mathrm{K}\), \(k_f=0.1 \mathrm{~W}/\mathrm{m} \cdot \mathrm{K}\), and \(\varepsilon(x)=0.25x\) to obtain the expressions for effective thermal conductivity as a function of \(x\).
03

Compute the steady-state temperature gradient

In this problem, the heat flux is given by \(q_{x}^{\prime \prime}=100 \mathrm{~W} /\mathrm{m}^2\), so the temperature gradient can be found by using Fourier's law, which is: \[q_{x}^{\prime \prime}=-k_{\mathrm{efl}}\frac{dT}{dx}\] For all four cases, we can solve this equation to find the temperature gradient, \(\frac{dT}{dx}\), as a function of the location \(x\). This preliminary step for each expression would be: 1. Minimum effective thermal conductivity: \[q_{x}^{\prime \prime}=-k_{\min}(x)\frac{dT}{dx}\] 2. Maximum effective thermal conductivity: \[q_{x}^{\prime \prime}=-k_{\max}(x)\frac{dT}{dx}\] 3. Maxwell's expression: \[q_{x}^{\prime \prime}=-k_{M}(x)\frac{dT}{dx}\] 4. The case where \(k_{\mathrm{efl}}(x)=k_{s}\): \[q_{x}^{\prime \prime}=-k_{s}\frac{dT}{dx}\]
04

Integrate the temperature gradient to find the temperature distribution

Now, we need to integrate the temperature gradient for each case to find the actual temperature distribution. For each expression, we have: 1. Minimum effective thermal conductivity: \[T(x)-T(0)=\int_{0}^{x} \frac{q_{x}^{\prime \prime}}{-k_{\min}(x)} dx\] 2. Maximum effective thermal conductivity: \[T(x)-T(0)=\int_{0}^{x} \frac{q_{x}^{\prime \prime}}{-k_{\max}(x)} dx\] 3. Maxwell's expression: \[T(x)-T(0)=\int_{0}^{x} \frac{q_{x}^{\prime \prime}}{-k_{M}(x)} dx\] 4. The case where \(k_{\mathrm{efl}}(x)=k_{s}\): \[T(x)-T(0)=\int_{0}^{x} \frac{q_{x}^{\prime \prime}}{-k_{s}} dx\] Then, we can solve these integrals and add the boundary condition \(T(x=0)=30^{\circ} \mathrm{C}\) to find the temperature distribution \(T(x)\) for each expression of effective thermal conductivity. Now that we have found the temperature distributions, we can plot them as a function of position \(x\). In plotting these distributions, we will be able to compare the effects of each effective thermal conductivity expression on the temperature profile within the porous medium.

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Key Concepts

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

Porosity Distribution
Understanding porosity distribution is essential in many engineering and environmental applications, especially in fields like material science and geology. Porosity refers to the measure of empty spaces in a material and is usually expressed as a fraction or percentage of the total volume. In our exercise, the porosity of the material varies linearly with position, described by the equation \[\begin{equation}\varepsilon(x) = \varepsilon_{\max} \frac{x}{L}\end{equation}\]where \begin{itemize}\item \( \varepsilon_{\max} \) is the maximum porosity at the wall's far end,\item \( x \) is the position within the wall material, and\item \( L \) is the thickness of the wall.\end{itemize}Porosity affects the physical properties of materials, such as density and thermal conductivity. In the context of our exercise, different levels of porosity influence how the wall conducts heat, leading to variations in effective thermal conductivity throughout the material.
Understanding how to integrate porosity into calculations allows engineers and scientists to predict the behavior of porous materials under different conditions, such as when there is a heat flux or a thermal gradient, as presented in our example.
Fourier's Law
Fourier's Law is a fundamental principle used to understand heat conduction. It states that the rate of heat transfer through a material is proportional to the negative gradient of the temperature and the area through which the heat is passing, mathematically expressed as\[\begin{equation}q_x^{prime prime} = -k \frac{dT}{dx}\end{equation}\]where \begin{itemize}\item \( q_x^{prime prime} \) is the heat flux (the rate of heat transfer per unit area), \item \( k \) is the material's thermal conductivity, and\item \( \frac{dT}{dx} \) is the temperature gradient within the material. \end{itemize}
Fourier's law is particularly useful in predicting how temperature changes within a material over time. The law is applicable for steady-state conditions, which implies that the temperature within the system does not change as time progresses. This is the same condition applied in the exercise to find the steady-state temperature distribution. The negative sign in Fourier's Law indicates that heat flows from higher to lower temperatures.
Steady-State Temperature Distribution
Steady-state temperature distribution refers to a condition in which the temperature in a material does not change over time, even though there may be a heat transfer. Under steady-state conditions, the amount of heat entering a specific section of material is equal to the amount leaving it, resulting in a constant temperature throughout that section. In our problem, we aim to calculate the temperature distribution, \( T(x) \), across a one-dimensional plane wall in steady-state. Achieving this requires us to integrate the temperature gradient, obtained from Fourier's Law, with respect to the position, taking into account the effective thermal conductivity of the wall material which may vary due to porosity. The different expressions for the effective thermal conductivity, such as the harmonic mean for minimum conductivity, the arithmetic mean for maximum conductivity, and Maxwell's expression, are used to see how they influence the temperature profile.
Knowing the steady-state temperature distribution is crucial in designing systems for proper thermal management. It allows for the prediction of the thermal behavior of the system and helps in determining insulation requirements, material selection, and other design decisions that are essential for safety, efficiency, and sustainability.

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Most popular questions from this chapter

A composite cylindrical wall is composed of two materials of thermal conductivity \(k_{\mathrm{A}}\) and \(k_{\mathrm{B}}\), which are separated by a very thin, electric resistance heater for which interfacial contact resistances are negligible. Liquid pumped through the tube is at a temperature \(T_{\infty, i}\) and provides a convection coefficient \(h_{i}\) at the inner surface of the composite. The outer surface is exposed to ambient air, which is at \(T_{\infty, o}\) and provides a convection coefficient of \(h_{o^{*}}\) Under steady-state conditions, a uniform heat flux of \(q_{h}^{n}\) is dissipated by the heater. (a) Sketch the equivalent thermal circuit of the system and express all resistances in terms of relevant variables. (b) Obtain an expression that may be used to determine the heater temperature, \(T_{h+}\). (c) Obtain an expression for the ratio of heat flows to the outer and inner fluids, \(q_{o}^{\prime} / q_{i}^{\prime}\). How might the variables of the problem be adjusted to minimize this ratio?

Work Problem \(3.15\) assuming surfaces parallel to the \(x\)-direction are adiabatic.

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.

Circular copper rods of diameter \(D=1 \mathrm{~mm}\) and length \(L=25 \mathrm{~mm}\) are used to enhance heat transfer from a surface that is maintained at \(T_{s, 1}=100^{\circ} \mathrm{C}\). One end of the rod is attached to this surface (at \(x=0\) ), while the other end \((x=25 \mathrm{~mm})\) is joined to a second surface, which is maintained at \(T_{s, 2}=0^{\circ} \mathrm{C}\). Air flowing between the surfaces (and over the rods) is also at a temperature of \(T_{\infty}=0^{\circ} \mathrm{C}\), and a convection coefficient of \(h=100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) is maintained. (a) What is the rate of heat transfer by convection from a single copper rod to the air? (b) What is the total rate of heat transfer from a \(1 \mathrm{~m} \times 1 \mathrm{~m}\) section of the surface at \(100^{\circ} \mathrm{C}\), if a bundle of the rods is installed on 4 -mm centers?

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?
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