Consider steady two-dimensional heat transfer in a square cross section \((3 \mathrm{~cm} \times 3 \mathrm{~cm})\) with the prescribed temperatures at the top, right, bottom, and left surfaces to be \(100^{\circ} \mathrm{C}\), \(200^{\circ} \mathrm{C}, 300^{\circ} \mathrm{C}\), and \(500^{\circ} \mathrm{C}\), respectively. Using a uniform mesh size \(\Delta x=\Delta y\), determine \((a)\) the finite difference equations and \((b)\) the nodal temperatures with the Gauss-Seidel iterative method.

Short Answer

Expert verified
Answer: The Gauss-Seidel iterative method is used to solve the finite difference equations for this problem. The updating formula for the Gauss-Seidel method is: $$T_{i,j}^{(k+1)} = \frac{1}{4} (T_{i-1,j}^{(k+1)} + T_{i+1,j}^{(k)} + T_{i,j-1}^{(k+1)} + T_{i,j+1}^{(k)})$$

Step by step solution

01

Set up the model and grid

First, set up the problem by dividing the square cross-section into a uniform grid with mesh size \(\Delta x = \Delta y\). The grid points will be represented as \((i, j)\), where \(i\) is the row index and \(j\) is the column index.
02

Derive the finite difference equations

Next, we need to derive the finite difference equations for the nodal points. Since it is a steady-state heat transfer problem, we can apply the Laplace equation to obtain the finite difference equations: $$\frac{T_{i-1,j} - 2T_{i,j} + T_{i+1,j}}{\Delta x^2} + \frac{T_{i,j-1} - 2T_{i,j} + T_{i,j+1}}{\Delta y^2} = 0$$ Since \(\Delta x = \Delta y\), the finite difference equations become: $$T_{i-1,j} - 2T_{i,j} + T_{i+1,j} + T_{i,j-1} - 2T_{i,j} + T_{i,j+1} = 0$$
03

Implement the Gauss-Seidel iterative method

To solve the finite difference equations, we will use the Gauss-Seidel iterative method which updates the values of the grid points one at a time in a sequential manner. The updating formula for the Gauss-Seidel method is: $$T_{i,j}^{(k+1)} = \frac{1}{4} (T_{i-1,j}^{(k+1)} + T_{i+1,j}^{(k)} + T_{i,j-1}^{(k+1)} + T_{i,j+1}^{(k)})$$
04

Set up initial boundary conditions

Before beginning iteration, set up the initial boundary conditions by assigning the given temperatures: Top surface: \(T_{0,j} = 100\degree\mathrm{C}\) Right surface: \(T_{i,2} = 200\degree\mathrm{C}\) Bottom surface: \(T_{2,j} = 300\degree\mathrm{C}\) Left surface: \(T_{i,0} = 500\degree\mathrm{C}\)
05

Iterate until convergence

Iterate using the Gauss-Seidel method until the nodal temperatures converge to a tolerance level (e.g., 0.01\degree\mathrm{C}). Update the values of the grid points according to the updating formula, and keep track of the previous values to check for convergence.
06

Obtain the nodal temperatures

Once the Gauss-Seidel iterative method converges, the nodal temperatures are obtained. These temperatures represent the steady-state temperature distribution within the square cross-section of the material.

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Key Concepts

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

Steady-state heat transfer
Imagine a material at a fixed temperature, without any changes over time—this scenario is known as steady-state heat transfer. It's fundamental in heat transfer analysis because it simplifies the scenario to just observing heat flow without worrying about the transient effects or time dependency. In the problem at hand, we're examining a square cross-section with different temperatures on each side. The aim is to find a temperature map that doesn't change with time, which is valuable for designing systems that operate under consistent temperatures, like electronic equipment casings or insulated walls.

Real-life applications range from HVAC system design to thermal stress analysis in construction materials. Understanding this concept helps in optimizing thermal management and improving energy efficiency by ensuring that the material can conduct or insulate heat according to the specific needs of an application.
Gauss-Seidel iterative method
The Gauss-Seidel iterative method is a popular algorithm in numerical analysis used to solve systems of linear equations. It belongs to the family of iterative methods for solving equations, which means it repeatedly refines the solution until it converges on an answer. The beauty of Gauss-Seidel lies in its simplicity and efficiency, especially when dealing with large systems, like in our heat transfer problem. Here's how it works in simple terms:
  • Start with an initial guess for the solution.
  • Proceed through each equation in the system, updating the solution as you go.
  • Keep iterating until changes between consecutive iterations are below a pre-defined threshold.
It's akin to solving a puzzle by gradually tweaking one piece after the other, circling back to adjust earlier pieces as the picture becomes clearer.

For students, it's essential to understand this method since it is widely used in engineering fields, such as civil, mechanical, and aerospace engineering, where modelling and simulations are critical. It shows practical approaches for tackling large-scale computational problems.
Laplace equation
Central to our heat transfer problem is the Laplace equation. Without getting too deep into the mathematical jungle, the Laplace equation is a second-order partial differential equation with a broad array of applications, from electromagnetism to fluid dynamics. In the context of heat transfer, it describes the distribution of temperature in a region that is steady-state—meaning, again, time-independent. Especially, \[\begin{equation}abla^2 T = \frac{\partial^2 T}{\partial x^2} + \frac{\partial^2 T}{\partial y^2} = 0\end{equation}\]where \( T \) is the temperature. Think of it as ensuring that the temperature at any point in the material is the average of the temperatures around it. No peaks or dips, just a smooth landscape of thermal equilibrium. This equation is key to making sure engineers can design systems that maintain consistent temperatures, crucial for processes that need precise thermal conditions.
Nodal temperature analysis
Imagine you're trying to understand how heat spreads through a material, like a metal bar or a section of a wall. Nodal temperature analysis is a technique that helps make sense of this. In this context, 'nodes' are specific points within the material mesh or grid. To conduct this analysis, the material's cross-section is divided into a grid, and temperatures at each node are calculated. The goal is to find the temperature at each of these points (nodes), giving us a detailed temperature profile.

This method seems a bit like piecing together a temperature map—it's a snapshot of how warm or cool each section of the material is. And why should students wrap their heads around this concept? Because it's wildly applicable—from designing car engines that don't overheat to crafting spacesuits that can handle the harsh conditions of space. Nodal temperature analysis plays a crucial role in the safety and efficiency of countless engineering projects.

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Most popular questions from this chapter

Consider a large plane wall of thickness \(L=0.4 \mathrm{~m}\), thermal conductivity \(k=2.3 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and surface area \(A=20 \mathrm{~m}^{2}\). The left side of the wall is maintained at a constant temperature of \(95^{\circ} \mathrm{C}\), while the right side loses heat by convection to the surrounding air at \(T_{\infty}=15^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(h=18 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assuming steady one-dimensional heat transfer and taking the nodal spacing to be \(10 \mathrm{~cm},(a)\) obtain the finite difference formulation for all nodes, \((b)\) determine the nodal temperatures by solving those equations, and (c) evaluate the rate of heat transfer through the wall.

Consider a long concrete dam \((k=0.6 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), (es \(\alpha_{s}=0.7\) ) of triangular cross section whose exposed surface is subjected to solar heat flux of \(\dot{q}_{s}=\) \(800 \mathrm{~W} / \mathrm{m}^{2}\) and to convection and radiation to the environment at \(25^{\circ} \mathrm{C}\) with a combined heat transfer coefficient of \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The \(2-\mathrm{m}\)-high vertical section of the dam is subjected to convection by water at \(15^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and heat transfer through the 2-m-long base is considered to be negligible. Using the finite difference method with a mesh size of \(\Delta x=\Delta y=1 \mathrm{~m}\) and assuming steady two-dimensional heat transfer, determine the temperature of the top, middle, and bottom of the exposed surface of the dam.

What are the limitations of the analytical solution methods?

Starting with an energy balance on the volume element, obtain the three- dimensional transient explicit finite difference equation for a general interior node in rectangular coordinates for \(T(x, y, z, t)\) for the case of constant thermal conductivity and no heat generation.

Suggest some practical ways of reducing the roundoff error.

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