Question

Consider a large plane wall of thickness \(L=0.05 \mathrm{~m}\). The wall surface at \(x=0\) is insulated, while the surface at \(x=L\) is maintained at a temperature of \(30^{\circ} \mathrm{C}\). The thermal conductivity of the wall is \(k=30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and heat is generated in the wall at a rate of \(\dot{e}_{\text {gen }}=\dot{e}_{0} e^{-0.5 x / L} \mathrm{~W} / \mathrm{m}^{3}\) where \(\dot{e}_{0}=8 \times 10^{6} \mathrm{~W} / \mathrm{m}^{3}\). Assuming steady one-dimensional heat transfer, \((a)\) express the differential equation and the boundary conditions for heat conduction through the wall, \((b)\) obtain a relation for the variation of temperature in the wall by solving the differential equation, and (c) determine the temperature of the insulated surface of the wall.

Short answer

#answer# The temperature at the insulated surface (\(x=0\)) is given by: \(T|_{x=0}=30+\frac{8\times10^{6}}{30} \int (\int e^{-0.5 x / L} dx) dx |_{x=L}\)

Step-by-step solution

Derive the heat conduction equation

First, we need to express the problem in the form of the heat conduction differential equation. We know that the general form of the heat diffusion equation for a 1D, steady-state condition with heat generation is: \(\frac{\partial ^{2} T}{\partial x^{2}}=-\frac{\dot{e}_{\text {gen }}}{k}\) In this problem, \(\dot{e}_{\text {gen }}=\dot{e}_{0} e^{-0.5x/L}\) where \(\dot{e}_{0}=8\times10^6 \mathrm{~W} / \mathrm{m}^{3}\), and the thermal conductivity, \(k=30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Substitute these values into the heat diffusion equation: \(\frac{\partial ^{2} T}{\partial x^{2}}=-\frac{8\times10^{6} \mathrm{~W} / \mathrm{m}^{3} e^{-0.5 x / L} }{30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}}\) Now, we have the differential equation and can proceed to solve it.

Determine the general solution

First, we separate variables and integrate the differential equation twice to find a general solution: The differential equation is: \(\frac{\partial ^{2} T}{\partial x^{2}}=-\frac{8\times10^{6}}{30}e^{-0.5 x / L}\) Integrate with respect to x: \(\frac{\partial T}{\partial x}=-\frac{8\times10^{6}}{30} \int e^{-0.5 x / L} dx +C_{1}\) Integrate one more time with respect to x: \(T=-\frac{8\times10^{6}}{30} \int (\int e^{-0.5 x / L} dx) dx + C_{1}x +C_{2}\) Now we need to apply the boundary conditions to find the values of the constants \(C_1\) and \(C_2\).

Apply the Boundary Conditions

We have two boundary conditions for this problem: 1. The wall surface at \(x=0\) is insulated, so \(\frac{\partial T}{\partial x}|_{x=0}=0\) 2. The wall surface at \(x=L\) is maintained at a temperature of \(30^{\circ} \mathrm{C}\), so \(T|_{x=L}=30\) We apply the first boundary condition: \(\frac{\partial T}{\partial x}|_{x=0}=-\frac{8\times10^{6}}{30} \int e^{-0.5 x / L} dx |_{x=0}+C_{1}=0\) From this, we find that \(C_1=0\) We apply the second boundary condition: \(T|_{x=L}=-\frac{8\times10^{6}}{30} \int (\int e^{-0.5 x / L} dx) dx |_{x=L}+C_{2}=30\) Solving for \(C_2\), we get: \(C_2=30+\frac{8\times10^{6}}{30} \int (\int e^{-0.5 x / L} dx) dx |_{x=L}\) Now, we have a relation for the temperature distribution inside the wall: \(T=-\frac{8\times10^{6}}{30} \int (\int e^{-0.5 x / L} dx) dx + 30 +\frac{8\times10^{6}}{30} \int (\int e^{-0.5 x / L} dx) dx |_{x=L}\)

Determine the temperature at the insulated surface

After finding the general solution for the temperature distribution, we can determine the temperature at the insulated surface by plugging \(x=0\) into the previous temperature equation: \(T|_{x=0}=-\frac{8\times10^{6}}{30} \int (\int e^{-0.5 x / L} dx) dx |_{x=0} + 30 +\frac{8\times10^{6}}{30} \int (\int e^{-0.5 x / L} dx) dx |_{x=L}\) Upon solving this equation, we find that: \(T|_{x=0}=30+\frac{8\times10^{6}}{30} \int (\int e^{-0.5 x / L} dx) dx |_{x=L}\)

Key concepts

Thermal Conductivity

Thermal conductivity is a fundamental property of materials that determines how well they can conduct heat. It is represented by the symbol \( k \) and is measured in watts per meter per Kelvin (W/m·K).
In essence, it provides a measure of the efficiency of heat transfer through a material when there is a temperature difference across it.
A high value of thermal conductivity means that the material is efficient at conducting heat, whereas a low value indicates insulating properties.

In our example problem, the wall has a thermal conductivity of \(k=30 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). This means that the wall is reasonably good at transferring heat. One of the main goals in solving heat conduction problems is to understand how heat moves through materials, given their thermal properties like thermal conductivity.

• This property plays a critical role in the steady-state heat conduction equation, as we see it influences the temperature gradient \(\frac{\partial T}{\partial x} \) when heat is generated inside the material.
• Understanding thermal conductivity also helps us evaluate materials best suited for insulation versus those meant for efficient heat transfer.

Boundary Conditions

Boundary conditions are critical in defining the behavior of a system at its limits, helping us solve differential equations uniquely. In the context of heat conduction, they specify temperature or heat flux at the surfaces of the material under consideration.
Our problem has two key boundary conditions that must be used to solve the heat conduction equation accurately:

• At \(x=0\), the wall is insulated. This implies no heat flow through this boundary, which is mathematically expressed as \(\frac{\partial T}{\partial x}|_{x=0}=0\). This condition ensures that the temperature profile in the wall starts flat or has zero slope at \(x=0\).
• At \(x=L\), the wall is maintained at a constant temperature of \(30^{\circ} \mathrm{C}\). This means that regardless of the heat generation inside, the temperature at \(x=L\) is fixed at 30°C, providing a clear reference for solving the system of equations.

Understanding boundary conditions is crucial as they help determine the constants of integration in the solution of differential equations, thereby providing a unique solution to the heat conduction problem.
They are the anchors that fix the temperature profile in place, ensuring a correct and practical solution aligned with physical reality.

Temperature Distribution

The temperature distribution within a material tells us how temperature varies with position inside it. It's crucial in thermal analysis because it allows us to predict which areas get hotter or colder.
For one-dimensional steady-state heat conduction in our insulated wall, the differential equation was given by:\[\frac{\partial ^{2} T}{\partial x^{2}}=-\frac{\dot{e}_{\text {gen }}}{k}\]This equation, once solved with the given boundary conditions, provides us the temperature variation across the wall.Knowing how temperature distributes across a material:

• Enables us to understand the effect of heat generation within the wall, which follows an exponential decay function \(\dot{e}_{\text{gen}}=\dot{e}_{0} e^{-0.5x/L}\).
• Allows for controlling thermal stresses, which are critical in materials subjected to high differential temperatures.

The temperature distribution gained from solving the differential equation gives both the temperature at various depths in the wall and crucial insights into managing the heat flow for safety in engineering designs.