/*! 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 67 A composite one-dimensional plan... [FREE SOLUTION] | 91Ó°ÊÓ

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A composite one-dimensional plane wall is of overall thickness \(2 L\). Material A spans the domain \(-L \leq x<0\) and experiences an exothermic chemical reaction leading to a uniform volumetric generation rate of \(\dot{q}_{\mathrm{A}}\). Material B spans the domain \(0 \leq x \leq L\) and undergoes an endothermic chemical reaction corresponding to a uniform volumetric generation rate of \(\dot{q}_{\mathrm{B}}=-\dot{q}_{\mathrm{A}}\). The surfaces at \(x=\pm L\) are insulated. Sketch the steady-state temperature and heat flux distributions \(T(x)\) and \(q_{\mathrm{x}}^{\prime \prime}(x)\), respectively, over the domain \(-L \leq x \leq L\) for \(k_{\mathrm{A}}=k_{\mathrm{B}}, k_{\mathrm{A}}=0.5 k_{\mathrm{B}}\), and \(k_{\mathrm{A}}=2 k_{\mathrm{B}}\). Point out the important features of the distributions you have drawn. If \(\dot{q}_{\mathrm{B}}=-2 \dot{q}_{\mathrm{A}}\), can you sketch the steady-state temperature distribution?

Short Answer

Expert verified
In summary, for the three cases we analyzed: 1. For kA = kB, the temperature distribution exhibits symmetry about x=0, due to equal but opposite heat generation rates in the two materials. 2. For kA = 0.5kB, the temperature gradient is steeper in material A than in material B and the heat flux distribution has a steeper gradient in material B. 3. For kA = 2kB, the temperature gradient is flatter in material A than in material B, and the heat flux distribution has a steeper gradient in material A. When \(\dot{q}_{\mathrm{B}}=-2\dot{q}_{\mathrm{A}}\), the temperature distribution is shifted towards material B as it undergoes a greater endothermic reaction.

Step by step solution

01

Case 1: kA = kB

Assuming constant thermal conductivity and constant volumetric generation rates in each material, we can analyze the heat transfer in the wall by considering the important features of these distributions. For this case, since kA = kB, the wall experiences two equal, but opposite, heat generation rates, and we assume no heat generation at the surfaces (x = ± L). Here, the temperature distribution will exhibit symmetry about the x = 0 line.
02

Case 2: kA=0.5kB

In this case, the thermal conductivity of A is half that of B \(k_{\mathrm{A}}=0.5 k_{\mathrm{B}}\), so the resistance to heat transfer in material A is greater than that in material B. As a result, the temperature gradient in material A will be steeper than that in material B. For the heat flux distribution, it will be the opposite, with a steeper gradient in material B (since heat transfer is easier in material B).
03

Case 3: kA=2kB

In this case, the thermal conductivity of A is twice as much as that of B \(k_{\mathrm{A}}=2 k_{\mathrm{B}}\), so the resistance to heat transfer in material A is less than that in material B. This implies a flatter temperature gradient in material A than in material B. For the heat flux distribution, it will be the opposite, with a steeper gradient in material A (as heat transfer is easier in material A).
04

Varied Heat Generation Rates

Now, if \(\dot{q}_{\mathrm{B}}=-2 \dot{q}_{\mathrm{A}}\), the temperature distribution will no longer be symmetric about x=0, as the heat generation rate in material B is twice as much as that in material A but with opposite sign. In this case, the temperature distribution will be shifted toward material B as it undergoes a greater endothermic reaction.

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Key Concepts

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

Thermal Conductivity
Thermal conductivity is a fundamental property of materials that indicates their ability to conduct heat. In the context of a composite wall, thermal conductivity plays a critical role in determining how heat will flow through the different materials. For instance, in our exercise, materials A and B have different thermal conductivities, which affects how the heat generated or absorbed due to the chemical reactions will distribute across the wall.

When material A has a lower thermal conductivity (\(k_{\mathrm{A}} = 0.5 k_{\mathrm{B}}\)), heat moves through it more slowly, resulting in a steeper temperature gradient compared to material B. Conversely, if material A has higher conductivity (\(k_{\mathrm{A}} = 2 k_{\mathrm{B}}\)), it transfers heat more quickly, leading to a flatter temperature gradient. Understanding thermal conductivity is essential to predict the temperature distribution and manage the heat transfer effectively.
Volumetric Generation Rate
The volumetric generation rate measures the amount of heat produced or absorbed per unit volume within a material due to a chemical reaction or other process. In our example, material A experiences an exothermic reaction, generating heat at a uniform rate (\(\dot{q}_{\mathrm{A}}\)), while material B undergoes an endothermic reaction, absorbing the same amount of heat (\(\dot{q}_{\mathrm{B}}=-\dot{q}_{\mathrm{A}}\)).

This concept is essential because it significantly impacts the steady-state temperature distribution within the composite wall. If the rates were not equal and opposite, as in the case of one side having twice the rate of the other (\(\dot{q}_{\mathrm{B}}=-2 \dot{q}_{\mathrm{A}}\)), the temperature profile would shift toward the material with the higher endothermic reaction rate. This shows the direct influence of the volumetric generation rate on the thermal behavior of the system.
Steady-State Temperature Distribution
Steady-state temperature distribution refers to the condition where the temperature within a system no longer changes with time. For a composite wall like the one in our exercise, achieving steady state means that each point within the wall has reached a consistent temperature, and there is a balance between the heat entering and leaving any section of the wall.

In the presence of uniform volumetric generation rates and insulated surfaces, such as in our scenarios, the steady-state temperature distribution is a result of the ensuing balance. The distribution can be symmetrical or asymmetrical, depending on the relative thermal conductivities and generation rates of the materials. For example, with equal thermal conductivities and generation rates, the distribution will be symmetrical about the center. However, asymmetry arises with different thermal conductivities and varied generation rates, causing a shift in the temperature profile toward the material with less resistance to heat or greater heat absorption.
Heat Flux Distribution
Heat flux distribution indicates how heat energy travels through a material per unit area and time. It is directly related to the material's thermal conductivity and the temperature gradient across the material. In the context of a composite wall, understanding heat flux distribution helps us understand how heat is shared between the materials with different thermal conductivities and generation rates.

For materials with equal thermal conductivities, heat flux will be uniformly distributed. However, discrepancies in conductivity, as seen with \(k_{\mathrm{A}}\) being half or twice \(k_{\mathrm{B}}\), lead to differences in how heat flows through the wall. A higher thermal conductivity results in a flatter heat flux distribution, while a lower conductivity presents a steeper distribution. In each situation, the total heat flowing through the wall must remain constant due to the conservation of energy, indicating an inverse relationship between thermal conductivity and temperature gradient.

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Most popular questions from this chapter

A plane wall with constant properties is initially at a uniform temperature \(T_{v}\). Suddenly, the surface at \(x=L\) is exposed to a convection process with a fluid at \(T_{\infty}\left(>T_{a}\right)\) having a convection coefficient \(h\). Also, suddenly the wall experiences a uniform internal volumetric heating \(\dot{q}\) that is sufficiently large to induce a maximum steadystate temperature within the wall, which exceeds that of the fluid. The boundary at \(x=0\) remains at \(T_{a}\). (a) On \(T-x\) coordinates, sketch the temperature distributions for the following conditions: initial condition \((t \leq 0)\), steady-state condition \((t \rightarrow \infty)\), and for two intermediate times. Show also the distribution for the special condition when there is no heat flow at the \(x=L\) boundary. (b) On \(q_{x}^{\prime \prime}-t\) coordinates, sketch the heat flux for the locations \(x=0\) and \(x=L\), that is, \(q_{x}^{\prime \prime}(0, t)\) and \(q_{x}^{\prime \prime}(L, t)\), respectively.

The cylindrical system illustrated has negligible variation of temperature in the \(r\) - and \(z\)-directions. Assume that \(\Delta r=r_{o}-r_{i}\) is small compared to \(r_{i}\), and denote the length in the z-direction, normal to the page, as \(L\). (a) Beginning with a properly defined control volume and considering energy generation and storage effects, derive the differential equation that prescribes the variation in temperature with the angular coordinate \(\phi\). Compare your result with Equation 2.26. (b) For steady-state conditions with no internal heat generation and constant properties, determine the temperature distribution \(T(\phi)\) in terms of the constants \(T_{1}, T_{2}, r_{i}\), and \(r_{\sigma}\). Is this distribution linear in \(\phi\) ? (c) For the conditions of part (b) write the expression for the heat rate \(q_{\phi}\).

One-dimensional, steady-state conduction with uniform internal energy generation occurs in a plane wall with a thickness of \(50 \mathrm{~mm}\) and a constant thermal conductivity of \(5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). For these conditions, the temperature distribution has the form \(T(x)=a+b x+c x^{2}\). The surface at \(x=0\) has a temperature of \(T(0) \equiv T_{o}=120^{\circ} \mathrm{C}\) and experiences convection with a fluid for which \(T_{\infty}=20^{\circ} \mathrm{C}\) and \(h=500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The surface at \(x=L\) is well insulated. (a) Applying an overall energy balance to the wall, calculate the volumetric energy generation rate \(\dot{q}\). (b) Determine the coefficients \(a, b\), and \(c\) by applying the boundary conditions to the prescribed temperature distribution. Use the results to calculate and plot the temperature distribution. (c) Consider conditions for which the convection coefficient is halved, but the volumetric energy generation rate remains unchanged. Determine the new values of \(a, b\), and \(c\), and use the results to plot the temperature distribution. Hint: recognize that \(T(0)\) is no longer \(120^{\circ} \mathrm{C}\). (d) Under conditions for which the volumetric energy generation rate is doubled, and the convection coefficient remains unchanged \(\left(h=500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\right)\), determine the new values of \(a, b\), and \(c\) and plot the corresponding temperature distribution. Referring to the results of parts (b), (c), and (d) as Cases 1, 2 , and 3, respectively, compare the temperature distributions for the three cases and discuss the effects of \(h\) and \(\dot{q}\) on the distributions.

Typically, air is heated in a hair dryer by blowing it across a coiled wire through which an electric current is passed. Thermal energy is generated by electric resistance heating within the wire and is transferred by convection from the surface of the wire to the air. Consider conditions for which the wire is initially at room temperature, \(T_{i}\), and resistance heating is concurrently initiated with airflow at \(t=0\). (a) For a wire radius \(r_{o}\), an air temperature \(T_{\infty}\), and a convection coefficient \(h\), write the form of the heat equation and the boundary/initial conditions that govern the transient thermal response, \(T(r, t)\), of the wire. (b) If the length and radius of the wire are \(500 \mathrm{~mm}\) and \(1 \mathrm{~mm}\), respectively, what is the volumetric rate of thermal energy generation for a power consumption of \(P_{\text {elec }}=500 \mathrm{~W}\) ? What is the convection heat flux under steady-state conditions? (c) On \(T-r\) coordinates, sketch the temperature distributions for the following conditions: initial condition \((t \leq 0)\), steady-state condition \((t \rightarrow \infty)\), and for two intermediate times. (d) On \(q_{r}^{\prime \prime}-t\) coordinates, sketch the variation of the heat flux with time for locations at \(r=0\) and \(r=r_{o^{*}}\).

Steady-state, one-dimensional conduction occurs in a rod of constant thermal conductivity \(k\) and variable crosssectional area \(A_{x}(x)=A_{o} e^{a x}\), where \(A_{o}\) and \(a\) are constants. The lateral surface of the rod is well insulated. (a) Write an expression for the conduction heat rate, \(q_{x}(x)\). Use this expression to determine the temperature distribution \(T(x)\) and qualitatively sketch the distribution for \(T(0)>T(L)\). (b) Now consider conditions for which thermal energy is generated in the rod at a volumetric rate \(\dot{q}=\dot{q}_{s} \exp (-a x)\), where \(\dot{q}_{o}\) is a constant. Obtain an expression for \(q_{x}(x)\) when the left face \((x=0)\) is well insulated.

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