Question

Consider a large plane wall of thickness \(L\) and constant thermal conductivity \(k\). The left side of the wall \((x=0)\) is maintained at a constant temperature \(T_{0}\), while the right surface at \(x=L\) is insulated. Heat is generated in the wall at the rate of \(\dot{e}_{\text {gen }}=a x^{2} \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{3}\). Assuming steady one-dimensional heat transfer, \((a)\) express the differential equation and the boundary conditions for heat conduction through the wall, \((b)\) by solving the differential equation, obtain a relation for the variation of temperature in the wall \(T(x)\) in terms of \(x, L, k, a\), and \(T_{0}\), and (c) what is the highest temperature \(\left({ }^{\circ} \mathrm{C}\right)\) in the plane wall when: \(L=1 \mathrm{ft}, k=5 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{\circ}{ }^{\circ} \mathrm{F}, a=1200 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft}^{5}\), and \(T_{0}=700^{\circ} \mathrm{F}\).

Short answer

Answer: The highest temperature in the plane wall is \(700°F\).

Step-by-step solution

Set up the differential equation and boundary conditions

We start by applying Fourier's law of heat conduction: \(q_x = -k\frac{dT}{dx}\). Considering the heat generated within the wall, the heat balance equation in a differential control volume is given by \(\frac{d}{dx}(q_x) = \dot{e}_{gen}\). Substituting the Fourier's law, we get: \(\frac{d}{dx} \left(-k\frac{dT}{dx}\right) = ax^2\) Boundary conditions: 1. At \(x=0\), \(T=T_0\) 2. At \(x=L\), the wall is insulated, so \(\frac{dT}{dx}=0\)

Solving the differential equation for temperature distribution

To solve the differential equation, we will first integrate once to get: \(-k\frac{dT}{dx} = \frac{1}{3}ax^3 + C_1\) Now we will integrate again, keeping in mind that we need to satisfy the boundary conditions: \(T(x) =-\frac{a}{12}x^4 - C_1x + C_2\) Satisfying the boundary conditions: 1. At \(x=0\), \(T=T_0\): \(T_0 = -\frac{a}{12}(0)^4 - C_1(0) + C_2 \Rightarrow C_2 = T_0\) 2. At \(x=L\), \(\frac{dT}{dx}=0\): \(0 = -k\frac{dT}{dx}= -k\left(-\frac{a}{12}(4L^3) + C_1\right) \Rightarrow C_1 = \frac{aL^3}{3}\) The temperature distribution is given by: \(T(x) = -\frac{a}{12}x^4 + \frac{aL^3}{3}x + T_0\)

Find the highest temperature in the wall

Given the values \(L=1\,ft\), \(k=5\,Btu/h\cdot ft\cdot°F\), \(a=1200\,Btu/h\cdot ft^5\), and \(T_0=700°F\), we have the temperature distribution as follows: \(T(x) = -\frac{1200}{12}x^4 + \frac{1200(1)^3}{3}x + 700\) Since the temperature distribution is in terms of \(x^4\), the highest temperature will occur at \(x=0\). Using the given values, let's calculate the highest temperature in the wall: \(T(0) = -\frac{1200}{12}(0)^4 + \frac{1200(1)^3}{3}(0) + 700 = 700°F\) The highest temperature in the plane wall is \(700°F\).

Key concepts

Differential Equation

A differential equation is a mathematical equation that involves derivatives of a function and is used to describe various physical phenomena. In the context of heat conduction through a wall, it helps us determine how temperature changes throughout the wall. Here, we're looking at how heat moves from one side of the wall to the other.

Think of it this way: a differential equation tells us the rate of change of temperature, which is how quickly or slowly the temperature increases or decreases. The differential equation for heat conduction in this problem is:\[\frac{d}{dx} \left(-k\frac{dT}{dx}\right) = ax^2\]

This equation accounts for the heat generated in the wall, represented by \(ax^2\), and the rate at which heat is conducted across the wall, represented by \(-k\frac{dT}{dx}\). With this equation, we can solve for the temperature distribution across the wall.✨

Boundary Conditions

Boundary conditions are constraints necessary for solving differential equations. They specify the values the solution needs to satisfy at specific points. These conditions help us find a unique solution to the equation based on how the system behaves at its boundaries.

In this exercise, we have two boundary conditions:

• At \(x=0\): The temperature is maintained at \(T_0\).
• At \(x=L\): The wall is insulated, meaning no heat passes through, so \(\frac{dT}{dx}=0\).

This setup tells us how the temperature behaves at the edges of the wall and thereby allows us to solve the differential equation for the entire wall.

The first condition ensures that our solution stays true to the given temperature at one side of the wall. The second condition ensures that there is no heat loss (or gain) at the insulated boundary. Based on these, we can identify the constants in the temperature distribution equation.

Fourier's Law

Fourier's Law of heat conduction is a fundamental principle that describes the flow of heat through a material. 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 flows. This can be expressed as:\[q_x = -k \frac{dT}{dx}\]

Here, \(q_x\) is the heat transfer rate per unit area, \(\frac{dT}{dx}\) is the temperature gradient across the material, and \(k\) is the thermal conductivity. A higher thermal conductivity means heat passes through the material more easily.

In our problem, Fourier's Law helps establish the relationship between the thermal conductivity of the wall and the change in temperature. By substituting Fourier's Law into the heat balance equation, we derived our differential equation. This principle illuminates how heat conduction varies with different materials and temperature gradients.

Temperature Distribution

Temperature distribution refers to how temperature varies at different points in a material. In the case of heat conduction in a wall, we're interested in knowing what the temperature is at any point along the wall from one side to another. This understanding helps in predicting thermal behavior under different conditions.

The solution to our differential equation gives us the temperature distribution:\[T(x) = -\frac{a}{12}x^4 + \frac{aL^3}{3}x + T_0\]

This equation tells us how the temperature changes inside the wall as a function of \(x\), the distance through the thickness of the wall. The terms in this equation are influenced by the wall’s material properties and the heat generation rate.

By solving this, we can determine where the highest and lowest temperatures occur. For example, in our exercise, the highest temperature can be found at \(x=0\), which aligns with the boundary condition specified at that point. Calculating actual values gives insight into thermal management in engineering applications.