Question

Consider a large plane wall of thickness \(L=0.3 \mathrm{~m}\), thermal conductivity \(k=2.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and surface area \(A=\) \(12 \mathrm{~m}^{2}\). The left side of the wall at \(x=0\) is subjected to a net heat flux of \(\dot{q}_{0}=700 \mathrm{~W} / \mathrm{m}^{2}\) while the temperature at that surface is measured to be \(T_{1}=80^{\circ} \mathrm{C}\). Assuming constant thermal conductivity and no heat generation in the wall, \((a)\) express the differential equation and the boundary conditions for steady one-dimensional heat conduction through the wall, \((b)\) obtain a relation for the variation of temperature in the wall by solving the differential equation, and \((c)\) evaluate the temperature of the right surface of the wall at \(x=L\).

Short answer

Question: Calculate the temperature at the right surface of a large plane wall with a steady one-dimensional heat conduction when its properties and conditions are given. Answer: The temperature at the right surface of the wall is 44°C.

Step-by-step solution

Write the governing equation for heat conduction

The governing equation for steady one-dimensional heat conduction with no heat generation is given by the following expression: \[\dfrac{d^2T}{dx^2} = 0\]

Write Fourier's Law of heat conduction

Fourier's law states the relationship between heat flux and the temperature gradient as follows: \[\dot{q} = -k \dfrac{dT}{dx}\]

Integrate the governing equation once

Integration of the governing equation (\(\dfrac{d^2T}{dx^2} = 0\)) with respect to x gives: \[\dfrac{dT}{dx} = C_{1}\] where \(C_{1}\) is the integration constant.

Apply the boundary condition at the left surface (x = 0)

At the left surface (\(x=0\)), the net heat flux \(\dot{q}_{0}\) and surface temperature \(T_{1}\) are given. Using Fourier's law, we can find the integration constant \(C_{1}\). \[\dot{q}_{0} = -k \dfrac{dT}{dx} |_{x=0} = -k C_{1} \Rightarrow C_{1} = -\dfrac{\dot{q}_{0}}{k}\]

Substitute the value of C1 into the equation for the temperature gradient

Substituting the calculated value of \(C_{1}\) into the equation for the temperature gradient, we have: \[\dfrac{dT}{dx} = -\dfrac{\dot{q}_{0}}{k}\]

Integrate the temperature gradient to obtain the temperature variation

Integration of the temperature gradient equation with respect to x gives the temperature variation: \[T(x) - T_{1} = -\dfrac{\dot{q}_{0}}{k}x + C_{2}\] Since the temperature at the left surface is \(T_{1}\), we have the boundary condition \(T(0) = T_{1}\). Applying this condition, we can find the value of \(C_{2}\) to be: \[C_{2} = 0\] So, the temperature variation in the wall can be expressed as: \[T(x) = T_{1} - \dfrac{\dot{q}_{0}}{k}x\]

Calculate the temperature at the right surface (x = L)

Now, we can evaluate the temperature at the right surface of the wall (\(x=L\)) using the temperature variation equation: \[T(L) = T_{1} - \dfrac{\dot{q}_{0}}{k}L\] Substitute the given values for \(T_{1}\), \(\dot{q}_{0}\), \(k\), and \(L\): \[T(L) = 80^{\circ} \mathrm{C} - \dfrac{700 \mathrm{~W} / \mathrm{m}^{2}}{2.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}} \cdot 0.3 \mathrm{~m}\] Solve for \(T(L)\): \[T(L) = 44^{\circ} \mathrm{C}\] The temperature of the right surface of the wall at \(x=L\) is \(44^{\circ} \mathrm{C}\).

Key concepts

Fourier's Law

Fourier's Law is fundamental to understanding heat conduction. It describes how heat transfers through a material due to a temperature gradient. This law is expressed mathematically as: \[ \dot{q} = -k \dfrac{dT}{dx} \] Here, \( \dot{q} \) represents the heat flux, which is the rate of heat transfer per unit area. The minus sign indicates the direction of heat flow is opposite to the temperature gradient. The term \( k \) is the thermal conductivity of the material, indicating how well the material conducts heat, and \( \dfrac{dT}{dx} \) is the temperature gradient across the material. Understanding Fourier's Law is crucial because it forms the basis for determining how temperature changes within a material and helps in solving heat conduction problems.

Differential Equation

In the context of heat conduction, we encounter differential equations representing the mathematical model of how heat is distributed over a medium. For steady one-dimensional heat conduction with no heat generation, the governing differential equation is: \[\dfrac{d^2T}{dx^2} = 0 \] This equation signifies that the rate of change of the temperature gradient is zero, implying a linear temperature distribution within the material. To solve for temperature variation, this second-order differential equation is integrated step-by-step. The first integration gives the temperature gradient as a constant: \[ \dfrac{dT}{dx} = C_1 \] Where \( C_1 \) is an integration constant determined by applying boundary conditions.

Boundary Conditions

Boundary conditions are essential for solving differential equations in physical systems. They provide additional information to find unknown constants and ensure the solution is physically meaningful. In heat conduction problems, boundary conditions can specify temperatures or heat flux at the boundaries of the material. For the given plane wall, the boundary conditions include the net heat flux \( \dot{q}_0 = 700 \mathrm{~W/m}^2 \) and the temperature at the left surface \( x=0 \) is \( T_1 = 80^{\circ}\mathrm{C} \). These conditions help in determining the constants in the integrated equation for temperature distribution, allowing the correct description of the temperature variation within the wall.

Temperature Distribution

The temperature distribution is the expression showing how temperature changes with position within a material. It is derived by integrating the differential equation and applying boundary conditions. For the steady-state one-dimensional heat conduction we are examining, the temperature distribution is given by: \[ T(x) = T_1 - \dfrac{\dot{q}_0}{k} x \] This equation shows a linear relationship between temperature \( T \) and the position \( x \) in the material. The slope of this line (\( -\dfrac{\dot{q}_0}{k} \)) is a result of the constant heat flux and characteristic of the material's thermal conductivity. By evaluating this equation at different positions, such as \( x=L \), we can determine the temperature at specific points within the wall, like at its right surface in the original problem, where it was solved to be \(44^{\circ}\mathrm{C}\).