Matlab中矩阵中的二维热传播答案

【问题标题】：2D-Heat propagation in a matrix in MatlabMatlab中矩阵中的二维热传播
【发布时间】：2016-12-19 02:12:37
【问题描述】：

我想计算和可视化 24 x 24 矩阵中的热传播。矩阵中有两个热点，其温度随时间保持恒定。随着时间的推移，矩阵的外边缘也保持不变。我是 MATLAB 新手，到目前为止我已经编写了以下脚本：

%declaration and initialization of matrix with
%24 x 24 cells, start temp. = 20°C
PlateOrg (1:24, 1:24) = 20;
% size of matrix
[height, width] = size(PlateOrg);
cells = height * width;
% percent that is given from one cell to its neighbouring cells
prop = 0.2;
%These are the hot-spots, they heat up the neighbouring cells
%but do not heat up themselves any higher -> their
%temperature is constant over the simulation
PlateOrg(4,4:20) = 100;
PlateOrg(5:9,20) = 80;
PlateOrg(11:19,4) = 50;
PlateOrg(20,4:20) = 60;
%First Contourf 2D-Diagram with start temperatures
contourf(PlateOrg);

PlateOld = [];
PlateOld = PlateOrg;
PlateNew = [];
% the amount of cycles 
cycles = 50;
% Calculation of the heat propagation
for k = 1:cycles
    %the outer edges stay constant at temp. = 20°C
    %for the whole simulation
    PlateNew(:,1) = 20;
    PlateNew(1,:) = 20;
    PlateNew(24,:) = 20;
    PlateNew(:,24) = 20;
    %every cell gets 20 percent heat of four neighbouring cells
    for i=2:height-1
        for j=2:width-1
            PlateNew(i,j)=...
            (1-4*prop)*PlateOld(i,j)+prop*...
            (PlateOld(i,j-1)+PlateOld(i,j+1)+...
             PlateOld(i-1,j)+PlateOld(i+1,j));
        end
    end
end

pause(0.2);
contourf(PlateNew)
colorbar
title('Heat propagation in a plate');
PlateOld = PlateNew;

我计算出了二维热传播的计算方法，但模拟仍然无法正常工作。有什么问题吗？

【问题讨论】：

标签： matlab matrix propagation

【解决方案1】：

部分：

pause(0.2);
contourf(PlateNew)
colorbar
title('Heat propagation in a plate');
PlateOld = PlateNew;

必须在for k = 1:cycle 循环中。其余的应该没问题。但目前热点并没有保持不变。您必须对边缘执行相同的操作。

问候，迈克尔

【讨论】：

感谢您的回答。我想通了！

【解决方案2】：

至于你的代码，看起来像你的忘记集

PlateOld = PlateNew;

在第 41 行（在“for k = 1:cycles”循环的末尾）。

您的代码似乎在第 k 次迭代时设置了错误的外边缘恒温。在第一次迭代中，您的 PlateNew 矩阵为空，因此设置

PlateNew(:,1) = 20;
PlateNew(1,:) = 20;

只填充矩阵的一个（第一个）元素。您的代码看起来像 C 风格。相反，MATLAB 可以更有效地处理矩阵。因此，您的 for 循环可以更有效地实现为（替换代码中的第 21-41 行）

PlateNew = PlateOld;
% the amount of cycles 
cycles = 50;
% Calculation of the heat propagation
for k = 1:cycles
    %every cell gets 20 percent heat of four neighbouring cells
    PlateNew = filter2(prop * [0 1 0; 1 1/prop-4 1; 0 1 0], PlateNew);

    %the outer edges stay constant at temp. = 20°C
    %for the whole simulation
    PlateNew(:,1) = 20;
    PlateNew(1,:) = 20;
    PlateNew(24,:) = 20;
    PlateNew(:,24) = 20;
end

【讨论】：