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Consider steady two-dimensional heat transfer in a rectangular cross section \((60 \mathrm{~cm} \times 30 \mathrm{~cm})\) with the prescribed temperatures at the left, right, and bottom surfaces to be \(0^{\circ} \mathrm{C}\), and the top surface is given as \(100 \sin (\pi x / 60)\). Using a uniform mesh size \(\Delta x=\Delta y\), determine (a) the finite difference equations and \((b)\) the nodal temperatures.

Short Answer

Expert verified
Answer: The main focus in solving the steady-state heat transfer problem in a rectangular cross section is on developing the finite difference equations and solving them to find the nodal temperatures.

Step by step solution

01

Determine the problem setup and boundary conditions

The rectangular cross-section is given by \(60\) cm x \(30\) cm, and we are given temperature boundary conditions on each of the surfaces, as follows: - Left surface: \(0^{\circ} \mathrm{C}\) - Right surface: \(0^{\circ} \mathrm{C}\) - Bottom surface: \(0^{\circ} \mathrm{C}\) - Top surface: \(100 \sin (\pi x / 60)\)
02

Choose a uniform mesh size

Choose \(\Delta x = \Delta y\) as the uniform mesh size for the given problem. Let's choose \(n\) nodes along both \(x\) and \(y\) axes, resulting in \(\Delta x = \frac{60}{n+1}\) cm and \(\Delta y = \frac{30}{n+1}\) cm.
03

Develop finite difference equations

The steady-state heat transfer equation in two dimensions can be written as: \(\frac{\partial^2 T}{\partial x^2} + \frac{\partial^2 T}{\partial y^2} = 0\) Applying finite difference approximations to the second derivatives, we obtain the following equation for each internal node \((i,j)\): \(\frac{T_{i+1,j} - 2T_{i,j} + T_{i-1,j}}{(\Delta x)^2} + \frac{T_{i,j+1} - 2T_{i,j} + T_{i,j-1}}{(\Delta y)^2} = 0\) Since \(\Delta x = \Delta y\), the equation simplifies to: \(T_{i+1,j} + T_{i-1,j} - 4T_{i,j} + T_{i,j+1} + T_{i,j-1} = 0\)
04

Apply boundary conditions

To find the nodal temperatures, we need to apply the given boundary conditions to the finite difference equation: - Left surface: \(T_{0,j} = 0^{\circ} \mathrm{C}\) - Right surface: \(T_{n+1,j} = 0^{\circ} \mathrm{C}\) - Bottom surface: \(T_{i,0} = 0^{\circ} \mathrm{C}\) - Top surface: \(T_{i,n+1} = 100 \sin (\pi i \Delta x / 60)\)
05

Solve the system of equations

The finite difference equation with applied boundary conditions is a set of \((n-1)^2\) linear equations with \((n-1)^2\) unknown nodal temperatures. We can solve this system of linear equations using any suitable method, such as the Gauss-Seidel method, the Jacobi method, or the Conjugate Gradient method. Once the system of equations is solved, the nodal temperatures will be determined, providing a detailed distribution of temperatures within the rectangular cross section subject to given boundary conditions.

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Key Concepts

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

2D Steady-State Heat Conduction
Understanding the concept of 2D steady-state heat conduction is crucial for several engineering applications. This phenomenon occurs when heat is transferred within a material without any changes in temperature with time, creating a steady temperature field. At the steady state, the amount of heat entering any region of the material is equal to the heat leaving that region. In context to our exercise, we consider a rectangular cross-section where the heat transfer is two-dimensional and time-independent.

For instance, in our solved example, the edges of the rectangle are maintained at certain temperatures. The left, right, and bottom surfaces are kept at a constant temperature of 0°°ä, and the top surface follows a sinusoidal temperature distribution. The steady-state condition implies that the interior of the rectangle eventually reaches a temperature configuration that does not change over time, despite the complex boundary conditions.
Boundary Conditions in Heat Transfer
Boundary conditions in heat transfer define how the heat behaves at the borders of the domain being studied. They are essential in setting up a heat transfer simulation or calculation because they directly affect the temperature distribution within the object.

There are generally three types of boundary conditions one may encounter:
  • Dirichlet boundary condition specifies the value of temperature at the boundary.
  • Neumann boundary condition defines the heat flux at the boundary.
  • Robin boundary condition, a combination of Dirichlet and Neumann, where a heat transfer coefficient is specified.
Our example sets Dirichlet conditions by specifying temperatures at all sides of the rectangle. The uniformity of these conditions along the left, right, and bottom surfaces simplifies our calculations, while the varying top condition introduces complexity in the resulting temperature field.
Numerical Solution of Heat Transfer
The numerical solution of heat transfer problems involves solving complex differential equations that describe the behavior of temperature over space and, in transient problems, time. Since analytical solutions are often not feasible for complicated domains and boundary conditions, numerical methods such as the finite difference method (FDM) are employed.

In our example, the application of FDM allows us to approximate the continuous domain with a discrete grid, where the derivatives in the heat equation are replaced with differences. We then get a system of linear equations, which can be solved using computational methods to approximate the temperature field. These methods are quite powerful because they can handle varying material properties, complex geometries, and diverse boundary conditions, which makes them indispensable tools for engineers and scientists solving real-world heat transfer problems.
Nodal Temperature Calculation
Calculating nodal temperatures is the essence of solving a heat transfer problem numerically. In our exercise, a grid is established over the rectangular domain, and we seek to find the temperature at each grid point or node. Using the finite difference approximations to the heat equation, a relationship is established between a node's temperature and its neighbors'.

This leads to a system of linear equations where each equation corresponds to a node inside the grid. The challenge is to resolve this system for temperatures at interior nodes while respecting the given boundary conditions. Various algorithms exist for this, including iterative methods like Gauss-Seidel and Jacobi, often favored for their simplicity and steady convergence characteristics. The resultant temperature at each node gives us a comprehensive look at how heat is distributed throughout the section, paving the way to predicting material behavior and designing efficient thermal systems.

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Most popular questions from this chapter

What are the basic steps involved in solving a system of equations with Gauss- Seidel method?

In many engineering applications variation in thermal properties is significant especially when there are large temperature gradients or the material is not homogeneous. To account for these variations in thermal properties, develop a finite difference formulation for an internal node in case of a three dimensional steady state heat conduction equation with variable thermal conductivity.

Consider transient heat conduction in a plane wall with variable heat generation and constant thermal conductivity. The nodal network of the medium consists of nodes \(0,1,2,3\), and 4 with a uniform nodal spacing of \(\Delta x\). The wall is initially at a specified temperature. The temperature at the right boundary (node 4) is specified. Using the energy balance approach, obtain the explicit finite difference formulation of the boundary node 0 for the case of combined convection, radiation, and heat flux at the left boundary with an emissivity of \(\varepsilon\), convection coefficient of \(h\), ambient temperature of \(T_{\infty}\), surrounding temperature of \(T_{\text {surr }}\), and uniform heat flux of \(\dot{q}_{0}\) toward the wall. Also, obtain the finite difference formulation for the total amount of heat transfer at the right boundary for the first 20 time steps.

Consider an aluminum alloy fin \((k=180 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) of triangular cross section whose length is \(L=5 \mathrm{~cm}\), base thickness is \(b=1 \mathrm{~cm}\), and width \(w\) in the direction normal to the plane of paper is very large. The base of the fin is maintained at a temperature of \(T_{0}=180^{\circ} \mathrm{C}\). The fin is losing heat by convection to the ambient air at \(T_{\infty}=25^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(h=25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and by radiation to the surrounding surfaces at an average temperature of \(T_{\text {surr }}=290 \mathrm{~K}\). Using the finite difference method with six equally spaced nodes along the fin in the \(x\)-direction, determine \((a)\) the temperatures at the nodes and \((b)\) the rate of heat transfer from the fin for \(w=1 \mathrm{~m}\). Take the emissivity of the fin surface to be \(0.9\) and assume steady one-dimensional heat transfer in the fin.

Consider a large uranium plate of thickness \(5 \mathrm{~cm}\) and thermal conductivity \(k=28 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) in which heat is generated uniformly at a constant rate of \(\dot{e}=6 \times 10^{5} \mathrm{~W} / \mathrm{m}^{3}\). One side of the plate is insulated while the other side is subjected to convection to an environment at \(30^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(h=60 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Considering six equally spaced nodes with a nodal spacing of \(1 \mathrm{~cm},(a)\) obtain the finite difference formulation of this problem and \((b)\) determine the nodal temperatures under steady conditions by solving those equations.

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