1 Introduction 介绍
在这份报告中,我将解决二维泊松方程为通过直接和迭代法上的长棒的横截面中的热分布的稳态方程。这个方程将在极坐标来解决。In this report, I will solve the 2D Poisson equation as the steady state equation for the heat distribution on the cross section of a long rod through both direct and iterative methods. This equation will be solved in polar coordinates.
1.1 Notations
We will use the following notations.
2 Problem set up: 2D Poisson equation with mixed boundary conditions (BC)
3 Finite difference scheme of Poisson equation in polar coordinates
Notice that uij ≈ u(rj; i). r and θ are grid sizes along r and . The Poisson equation can be expressed in polar coordinates as (where the unit mm in length and the unit K in temperature is ommited.)
4 Question one: draw the contour plots of temperature at steady state.
The contours of temperature at steady state are called isotherms. The isotherms in this Section will be computed by solving the Poisson equation in polar coordinates given in Equation (8) by the direct method of implementing the FD scheme given in Equation (9). Please see Figure 1 for a contour plot of temperature at the steady state.
5 Question two: please calculate the total heat transfer rate per unit length (W/m) from the rod.
6 Question three: use two different computational schemes and comment about the di
erences.
The Poisson solver, after indexing the points on the 2D plane inside the computational domain into a 1D vector for the nodal temperature, can be simpli
ed as a problem of Lx=b (linear problem). For any linear problems, the unknown vector x can be solved by direct Gaussian elimination and by indirect iterative method. Please see the Matlab code poisson equation polar.m as the implementation of the direct method. Please see the Figure 1 as the outcome of implementing the direct method. Please see the Matlab code poisson equation polar iterative.m as the implementation of the Jacobi iterative method (to solve the linear system). Please see the Figure 2 as the outcome of implementing the iterative method. In the iterative method, thenal solution is highly dependent on the initial guess. Therefore, in Figure 2 the steady solution is sensitive to the initial guess of this problem. Figure 2: Contours of temperatures at steady state (isotherms). This is computed by taking 20 points along 4 and 20 points along . The unit of temperature is K here. This is computed by Jacobi iterations with different initial guess of temperature as 200K (left) and 300K (right). #p#分页标题#e#
7 Question four: what are optimum 4r and 4 ? Ideally, to be computational e
cient, the best choice of 4r and 4θ is to have smaller 4θ when r is small (near the origin) and larger 4θ when r is large. Because the domain looks almost singular at the origin, more rened resolution near the origin will help the computational e
ciency and accuracy of the simulation.