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 \(A=24 \mathrm{~m}^{2}\). The left side of the wall is subjected to a heat flux of \(\dot{q}_{0}=350 \mathrm{~W} / \mathrm{m}^{2}\) while the temperature at that surface is measured to be \(T_{0}=60^{\circ} \mathrm{C}\). Assuming steady one-dimensional heat transfer and taking the nodal spacing to be \(6 \mathrm{~cm},(a)\) obtain the finite difference formulation for the six nodes and (b) determine the temperature of the other surface of the wall by solving those equations.

Short Answer

Expert verified
The purpose of using the finite difference method in this problem is to discretize the heat transfer equation and find the temperature distribution within the wall. This is done by dividing the wall thickness into equally spaced grid points and creating a system of linear equations for the temperatures at each node. Once these equations are solved, we can find the temperature at the other surface of the wall, which is the main objective of the problem.

Step by step solution

01

Understand the problem

We are given a plane wall with thickness L, thermal conductivity k, and surface area A. The heat flux experienced by the wall is given by \(\dot{q}_{0}\). The main objective is to find the temperature of the other surface of the wall (\(T_{L}\)), assuming steady one-dimensional heat transfer. We will also use nodal spacing given by 6 cm.
02

Find the number of grid points

For Finite Difference Method, we will divide the wall thickness into equally spaced grid points. Since the nodal spacing is given to be 6 cm, we can find the number of grid points by dividing the wall thickness by the nodal spacing: Number of grid points = \(\frac{L}{6\mathrm{~cm}} = \frac{0.3\mathrm{~m}}{0.06\mathrm{~m}} = 5\) So, there will be 5 grid points inside the wall (excluding the surfaces).
03

Discretize heat transfer equation

The general one-dimensional heat conduction equation is given by \(\frac{d^{2}T}{dx^{2}} = 0\) Using the central difference formulation, the finite difference method for i-th node can be written as: \(\frac{T_{i+1}-2 T_{i}+T_{i-1}}{(\Delta x)^2} = 0\) Where \(\Delta x = 0.06 \mathrm{~m}\) is the nodal spacing and \(T_{i}\) is the temperature at i-th node.
04

Set up equations for each node

For node 1, we have the following equation: \(\frac{T_{2}-2 T_{1}+T_{0}}{(\Delta x)^2} = 0\) Since we know the values for \(T_{0}\) and \(\dot{q}_{0}\) and \(k\), we can use the conduction heat flux equation to find the extra term: \(\dot{q}_{0} = -kA\frac{T_{1}-T_{0}}{\Delta x}\) Using this equation, we can replace \(T_{0}\) in the above equation: \(T_{0} = T_{1} + \frac{\dot{q}_{0}\Delta x}{kA}\) Now we have a system of five equations for \(T_{1}\), to \(T_{5}\): 1. \(T_{2} - 2T_{1} + T_{1} + \frac{\dot{q}_{0}\Delta x}{kA} = 0\) 2. \(T_{3} - 2T_{2} + T_{1} = 0\) 3. \(T_{4} - 2T_{3} + T_{2} = 0\) 4. \(T_{5} - 2T_{4} + T_{3} = 0\) 5. \(\frac{T_{5} - T_{4}}{\Delta x} = q_{L}\), where \(q_{L}\) is the heat flux at the other surface of the wall.
05

Solve the system of linear equations

Now, we have five linear equations for five unknown temperatures (\(T_{1}\), \(T_{2}\), \(T_{3}\), \(T_{4}\), and \(T_{5}\)). We can solve this system of linear equations using any method, such as substitution, elimination, or a matrix inversion. After solving the linear equations, we will get the values for \(T_{1}\), \(T_{2}\), \(T_{3}\), \(T_{4}\), and \(T_{5}\).
06

Determine the temperature at the other surface

The temperature at the other surface of the wall, \(T_{L}\), can be found by interpolation between \(T_{4}\) and \(T_{5}\). Since we know the nodal spacing, we can find \(T_{L}\) as follows: \(T_{L} = T_{5} + \frac{T_{5} - T_{4}}{\Delta x}(0.12\mathrm{~m})\) Now, calculate the value of \(T_{L}\) using the values of \(T_{4}\) and \(T_{5}\) obtained from the linear equations. This will give us the temperature at the other surface of the wall.

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Key Concepts

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

Steady-State Heat Conduction
Understanding steady-state heat conduction is fundamental for analyzing how heat transfers through materials at a constant rate. Steady-state implies that the temperature within the conducting medium does not change with time, meaning neither does the heat flow.

In the context of the exercise, we consider a plane wall of specific thickness, where the temperature at one surface is held constant, as is the heat flux. In steady-state conditions, the temperature gradient within the wall remains constant over time, which is why the finite difference method can be effectively applied to approximate the temperature distribution across the wall.

The finite difference method helps in transforming the continuous heat conduction equation into a set of algebraic equations that can be solved for nodal temperatures. By dividing the wall into discrete nodes and applying the finite difference equations, one can systematically calculate the temperature at each node, including the opposite face of the wall.
One-Dimensional Heat Transfer
One-dimensional heat transfer refers to the simplification of heat transfer problems where variations in temperature are considered in only one spatial dimension. This assumption is valid under certain conditions, such as when the wall is significantly larger in two dimensions compared to the third, or the temperature variations on the surfaces perpendicular to the heat flow are negligible.

In the exercise presented, the one-dimensional assumption allows the problem to be modeled using a single spatial coordinate, simplifying the mathematical analysis. The heat transfer can be represented along the thickness of the wall, ignoring any heat transfer in the other two dimensions. The finite difference method is then used to discretize the wall into nodes, permitting the calculation of temperature at those discrete positions.
Thermal Conductivity
Thermal conductivity (\( k \)) is a property of materials that quantifies how well they can conduct heat. It is pivotal in determining the rate at which heat energy is transferred through a material when subjected to a temperature gradient.

In the context of the finite difference method, thermal conductivity is utilized to calculate the heat flux between nodes. It aids in understanding how easily heat will pass through the chosen material of the wall in the exercise scenario.

The higher the thermal conductivity, the more efficiently heat will transfer through the material. This allows for constructing a relationship between the temperature difference across nodes and the heat flux exchanged, ultimately contributing to the calculation of temperature at various points within the material.
Boundary Conditions Heat Transfer
Boundary conditions in heat transfer specify the thermal behavior at the boundaries of a domain, which, in this case, is the wall. They are crucial for solving any heat transfer problem because they define the limitations that must be satisfied by the temperature distribution.

The exercise in question presents two types of boundary conditions: a specified heat flux on one side of the wall and a measured temperature on the same side. These conditions serve as the starting points for applying the finite difference method. The heat flux boundary condition defines the rate of heat entry or exit at a surface, while the temperature boundary condition sets a definite temperature value at that point.

Applying boundary conditions allows students to determine the 'fixed' nodal temperatures, which are subsequently used as inputs in the finite difference equations. Calculating temperature at internal nodes and ultimately at the unknown opposite surface becomes possible by incorporating these boundary conditions into the finite difference framework.

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Most popular questions from this chapter

Consider transient one-dimensional heat conduction in a plane wall that is to be solved by the explicit method. If both sides of the wall are at specified temperatures, express the stability criterion for this problem in its simplest form.

What is a practical way of checking if the discretization error has been significant in calculations?

Consider steady one-dimensional heat conduction in a composite plane wall consisting of two layers \(A\) and \(B\) in perfect contact at the interface. The wall involves no heat generation. The nodal network of the medium consists of nodes 0,1 (at the interface), and 2 with a uniform nodal spacing of \(\Delta x\). Using the energy balance approach, obtain the finite difference formulation of this problem for the case of insulation at the left boundary (node 0 ) and radiation at the right boundary (node 2) with an emissivity of \(\varepsilon\) and surrounding temperature of \(T_{\text {surr }}\)

Consider steady one-dimensional heat conduction in a plane wall with variable heat generation and constant thermal conductivity. The nodal network of the medium consists of nodes \(0,1,2\), and 3 with a uniform nodal spacing of \(\Delta x\). The temperature at the left boundary (node 0 ) is specified. Using the energy balance approach, obtain the finite difference formulation of boundary node 3 at the right boundary for the case of combined convection and radiation with an emissivity of \(\varepsilon\), convection coefficient of \(h\), ambient temperature of \(T_{\circ}\), and surrounding temperature of \(T_{\text {surr }}\). Also, obtain the finite difference formulation for the rate of heat transfer at the left boundary.

How can a node on an insulated boundary be treated as an interior node in the finite difference formulation of a plane wall? Explain.

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