Finite Temperature Variation - Heat Transfer
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I am writing a code that displays temperature variation. There is a triangle section cut off of a square(nxn) that is exposed to convection heat transfer. The part that is cut off is a 90 degree triangle between i and n-.6*i. I need help defining this section for every row because the slope is not in one row. The cut out part is after a certain meter distance.
3 Comments
Image Analyst
on 13 Oct 2013
Edited: Image Analyst
on 13 Oct 2013
If the triangle is formed by two sides and one side is "i" elements long, and the other side is "n-0.6*i" elements long, how is that a 45 degree angle for the hypoteneuse? To get 45 degrees, both sides would have to be the same length.
Answers (3)
Image Analyst
on 13 Oct 2013
Edited: Image Analyst
on 13 Oct 2013
Do you mean like this:
clc;
n = 100
% Create sample random data.
myArray = randi(9, [n, n], 'uint8') + 100;
subplot(2, 2, 1);
imshow(myArray);
title('Original Image', 'FontSize', 20);
% Chop off i by n-6*i triangle
i = 4
xvertices = [0, n-6*i, 0];
yvertices = [0,0, i];
mask = poly2mask(xvertices, yvertices, n, n);
subplot(2,2,2);
imshow(mask);
title('Mask Image', 'FontSize', 20);
out = myArray .* uint8(~mask);
subplot(2,2,3);
imshow(out);
title('Output Image', 'FontSize', 20);
% Enlarge figure to full screen.
set(gcf, 'Units', 'Normalized', 'OuterPosition', [0 0 1 1]);
Note: requires Image Processing Toolbox because of poly2mask.
2 Comments
Image Analyst
on 13 Oct 2013
Kaelyn's comment moved here, since it's not an answer to her original question, but a comment on mine.
Yes but n and i will change according to the size of the mesh. The cut off of the triangle will be identified and used to calculate other temperatures. So I have been trying to use the mod function.
Image Analyst
on 13 Oct 2013
Edited: Image Analyst
on 13 Oct 2013
n and i are variables in my code. You can change them. You can do everything with a for loop, instead of poly2mask(), if you want.
Youssef Khmou
on 13 Oct 2013
Edited: Youssef Khmou
on 13 Oct 2013
Kaelyn,
OK the problem now is clear , here is fast way
N=500; % more resolution better 2D heat conduction resolution .
H=ones(N);
M=flipud(triu(H)); % TOP RIGHT TRIANGLE
surface(M);
shading interp
6 Comments
Youssef Khmou
on 13 Oct 2013
Edited: Youssef Khmou
on 13 Oct 2013
N=100;
M=rot90(triu(ones(N),-60));
surface(M), shading interp
Youssef Khmou
on 13 Oct 2013
N=1000; % 1 meter sampled with 1000 Hz
p=0.6*N; % your 0.6 starting point .
M=zeros(N); % initial matrix
A=(N-p)/N; % the Coefficient for constructing the triangle
for n=1:N
M(n,1:A*n)=1;
end
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