contour plot with a circular boundary

3 views (last 30 days)
Zeyad Zeitoun
Zeyad Zeitoun on 5 Oct 2023
Commented: Torsten on 6 Oct 2023
I have the following code which gives me rectangular contour, I would like to chaneg it to be circular contour (boundaries), Can anyone help me please?
L = [-0.875 -0.75 -0.625 -0.5 -0.375 -0.25 -0.125 0 0.125 0.25 0.375 0.5 0.625 0.75 0.875];
r = [-0.875 -0.75 -0.625 -0.5 -0.375 -0.25 -0.125 0 0.125 0.25 0.375 0.5 0.625 0.75 0.875];
[X,Y]=meshgrid(L,r)
T=xlsread('Tempvalues (horizontal 70 psi)try.xlsx');
Z = [T(:,1),T(:,2),T(:,3),T(:,4),T(:,5),T(:,6),T(:,7),T(:,8),T(:,9),T(:,10),T(:,11),T(:,12),T(:,13),T(:,14),T(:,15)];
contourf(X,Y,Z)
pbaspect([1,1,1])
xlabel('Dimensionless axial position (z/L)')
ylabel('Dimensionless radial position (r/R)')
title('Axial velocity distribution along heated channel')

Answers (1)

Torsten
Torsten on 5 Oct 2023
Edited: Torsten on 5 Oct 2023
If you want to plot it within a circular contour, you have a 3d-plot. For a 3d-plot, you can only plot cuts in length direction with temperature as the contours.
So in short: Plotting all your data with a circular boundary is not possible - only the radial temperature distribution at a fixed length.
And I don't know what negative radial positions mean.
  4 Comments
Zeyad Zeitoun
Zeyad Zeitoun on 6 Oct 2023
The temeprature is different radially because the gas is flowing inside the cylinder by natural convection heat transfer, so there is a radial temperature gradient where the hot gas layers are above the cold layers thats why I want to draw it by contour map much better than normal plotting.
I tried your code but it doesn't represent the temperature gradient in the channel. Thanks

Sign in to comment.

Categories

Find more on Contour Plots in Help Center and File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!