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A plane wall of thickness \(2 L\) and thermal conductivity \(k\) experiences a uniform volumetric generation rate 4. As shown in the sketch for Case I. the surface at \(x=-L\) is perfectly insulated, while the other surface is maintained at a uniform, constant temperature \(T_{a}\). For Case 2, a very thin dielectric strip is inserted at the midpoint of the wall \((x=0)\) in order to electrically isolate the two sections, \(A\) and B. The thermal resistance of the strip is \(R_{7}^{*}=0.0005 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). The parameters associated with the wall are \(k=50 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, L=20 \mathrm{~mm}\), \(q=5 \times 10^{6} \mathrm{~W} / \mathrm{m}^{3}\), and \(T_{e}=50^{\circ} \mathrm{C}\). (a) Sketch the temperature distribution for case I on \(T-x\) coordinates. Deseribe the key features of this distribution. Identify the location of the maxirmum temperature in the wall and calculate this temperature. (b) Sketch the temperature distribution for Cuse 2 on the same \(T-x\) coordinates. Describe the key features of this distribution. (c) What is the temperature difference between the two walls at \(x=0\) for Case 2? (d) What is the location of the maximum temperature in the composite wall of Case 2? Calculate this temperature.

Step by step solution

01

Understand the problem statement

We need to solve for temperature distributions in a plane wall with uniform heat generation. For Case 1, there's no heat conduction at one boundary, and for Case 2, a thermal resistance is introduced at the midpoint (x=0) of the wall. We need to plot temperature distributions, find maximum temperature, and compute temperature differences between sections in both cases.

02

Sketch the temperature distribution for Case 1

For Case 1, the heat equation for a plane wall with volumetric generation is \[ T(x) = -\frac{q}{2k}x^2 + C_1x + C_2 \]where C_1 and C_2 are integration constants. With the boundary conditions: insulated surface at \(x = -L\) and \(T = T_a\) at \(x = L\), we can determine C_1 and C_2 by applying these conditions.The insulated boundary gives:\[\frac{dT}{dx}\Big|_{x=-L} = 0\]and\[ T(L) = T_a\]

03

Calculate integration constants for Case 1

Using the insulated boundary condition, we find:\[\frac{dT}{dx} = -\frac{q}{k}x + C_1 = 0 \quad \Rightarrow \quad C_1 = \frac{qL}{k}\]Substituting into the temperature profile equation and using \(T(L) = T_a\):\[T(L) = -\frac{q}{2k}L^2 + \frac{qL^2}{k} + C_2 = T_a \quad \Rightarrow \quad C_2 = T_a - \frac{q}{2k}L^2\]

04

Maximum temperature location and value in Case 1

The location of the maximum temperature is at \(x = 0\) due to symmetry and the parabolic nature of the temperature profile. Substituting \(x = 0\) back:\[ T_{max} = C_2 = T_a - \frac{q}{2k}L^2\]Plugging in the given values:\[ T_{max} = 50 - \frac{5 \times 10^6 \times (0.02)^2}{2 \times 50} = 64^\circ C\]

05

Sketch temperature distribution for Case 2

For Case 2, we have an additional thermal resistance at the midpoint. This divides the wall into two sections, each having its heat generation and distinct temperature profiles, influenced by the thermal resistance:\[T_{1}(x) = -\frac{q}{2k}x^2 + C_3x + C_4\]\[T_{2}(x) = -\frac{q}{2k}x^2 + C_5x + C_6\]

06

Boundary conditions to find constants for Case 2

Using the continuity of heat flux and temperature at \(x = 0\):\[-k \frac{dT_{1}}{dx}\Big|_{x=0} = -k \frac{dT_{2}}{dx}\Big|_{x=0}\]and the condition across the thermal resistance:\[T_{1}(0) - T_{2}(0) = q R^{*}\]

07

Maximum temperature and its location for Case 2

Considering two symmetrical sections and continuous insulation effects, maximum temperature still occurs at the center of each section (near \(x = 0\) but influenced by the thermal resistance). The calculations simplify to show that maximum temperature increases slightly due to the resistance:\[ T_{max} = (T_{max \, \text{without} \, R^{*} }) + \frac{q R^{*}}{k}\]Substitute the known values and solve for Case 2 specifics.

08

Calculate temperature difference across the dielectric strip

The temperature difference across the strip at \(x=0\) is given by:\[ \Delta T = q R^{*}\]Substitute the known value for \(R^{*}\):\[ \Delta T = 5 \times 10^6 \times 0.0005 = 2.5^\circ C\]

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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 material property that measures a material's ability to conduct heat. It is essential in calculating how heat spreads through a material. Imagine holding a metal rod in your hand. If you heat one end, you'll soon feel the warmth at the other end due to the rod's thermal conductivity.
In this exercise, thermal conductivity (\(k\)) is crucial for determining the heat's journey through the plane wall. With \(k = 50 \, \text{W/m}\cdot\text{K}\), this tells us that for every meter of wall thickness, and for each degree of temperature difference, 50 watts of heat passes through. It reflects how efficiently the wall passes heat from a hotter region to a cooler one.

• This factor directly influences the temperature distribution within the wall.
• The higher the thermal conductivity, the faster the heat can flow through the material.
• Application of boundary conditions will determine specific heat flow paths.

Understanding thermal conductivity helps predict how long it will take for heat to reach different parts of the wall and how efficiently it moves. Essentially, it determines the relationship between temperature gradients (how fast temperature changes in space) and heat flux (the rate of heat flow).

Temperature Distribution

Temperature distribution describes how temperature varies within a given system, like a wall, over space. In Case 1, temperature changes in the wall follow a particular pattern or profile because the left boundary is insulated, and the right boundary is kept at a constant temperature \(T_a\).
The pattern or distribution often depends on factors like heat generation, thermal conductivity, and other boundary conditions. For Case 1, the temperature along the wall follows a parabolic shape, based on the equation:\[T(x) = -\frac{q}{2k}x^2 + C_1x + C_2\]Understanding the temperature distribution is vital because it helps locate the maximum and minimum temperatures within the wall. This is particularly important for materials that might degrade or fail under extreme temperatures.

• In Case 1, maximum temperature occurs around the center of the wall due to symmetry.
• In Case 2, the introduction of a thermal resistance changes the distribution, causing a slight increase in maximum temperature.
• Profiles are affected by the volumetric heat generation (\(q\)) within the wall.

Temperature distribution is crucial for designing systems that efficiently manage heat to avoid hotspots that can lead to material failure.

Boundary Conditions

Boundary conditions are specific constraints or conditions at the boundaries of a material or system that must be satisfied for an accurate solution to a heat transfer problem.
Think of them as the rules that govern how heat interacts at the edges of the wall. In our exercise, Case 1 has two boundaries:

• At \(x = -L\), the surface is perfectly insulated, meaning no heat passes through this boundary. This condition leads to \(\frac{dT}{dx} = 0\) at \(x = -L\)
• At \(x = L\), the surface is maintained at a constant temperature \(T_a\)

These conditions allow us to solve for the constants in the heat equation, defining how temperature varies throughout the wall.
For Case 2, additional boundary conditions come into play due to the dielectric strip at \(x=0\). Here, continuity of temperature and heat flux is required:

• Temperatures on either side of the strip need to fulfill the condition \(T_{1}(0) - T_{2}(0) = q R^{*}\)
• Heat flux should be consistent across the strip.

Correctly setting these boundary conditions is essential because they ground the mathematical model in physical reality, leading to accurate predictions of how temperature changes across the wall.