Matlab help Ok so for my applied math class, we are writing a matlab script to preform finite differencing on a rectangular heat fin to determine the temperature profile of the heat with with distance. Now before anyone says it I know this is much more easily done in solid works but that apparently not the point according to my prof. The heat fin in question can experience 3 boundry conditions on the outside edge of the fin, one its insulated, 2 its at ambient temperature, 3 convection occurs. The below is what I have so far but the values for y and t are a little funky. Any help and explanations would be greatly appreciated. n=input('type number of intervals: '); bc=input('Boundary conditions: for an insulated system, type 1. if outer edge is at ambient temperature, type 2. if convection occurs at the outer edge, type 3: '); tb=298; %temp of base of fin tsurr=273; %temp of surroundings k=1; %thermal condctivity h=1; %heat transfer thck=1; %thickness of fin l=1; %length of fin w=1; %width of fin m=1; %values derived from above a=w*thck; %crosssectional area of fin dz=l/n; %interval size p=2*w+2*thck; %perimeter of fin % initial equation is T''2hP/ka(TTa)=0 % after transform equation becomes y''(2hpl^2/Ka)y=0 % y = TTsurr/TbTsurr, derivatives of y are with respect to z (z=x/l) alpha=(2*h*p*l*l)/(k*a); % set y and solution vector sizes. set x matrix size z=zeros(n); y=zeros(n,m); g=zeros(n,m); x=zeros(n,m); t=zeros(n,m); z(1,1)=1; g(1)=1; % build matrix between boundaries for i=2:1:n1 z(i,i1)=l/dz^2; z(i,i)=alpha+2/dz^2; z(i,i+1)=l/dz^2; g(i)=0; end % establish boundary conditions for i=n if bc==1 z(i,i1)=2/dz^2; z(i,i)=alpha(2/dz^2); end if bc==2 z(i,i1)=0; z(i,i)=1; end if bc==3 z(i,i1)=2/dz^2; z(i,i)=alpha(2/dz^2)(2*h*l)/(k*dz); end end % Implement Thomas's Algorithm to solve for i=2:1:n z(i,i)= z(i,i)(z(i,i1)/z(i1,i1))*z(i1,i); g(i)=g(i)(z(i,i1)/z(i1,i1))*g(i1); z(i,i1)=0; end y(n)=g(n)/z(n,n); for i=n1:1:1 y(i)=(g(i)z(i,i+1)*y(i+1))/z(i,i); end x(1)=0; for i=2:1:n+1 x(i)=x(i1)+dz; end 
Haha, I just spent my morning writing matlab code, I've still got a bit to go so I feel your pain... What exactly do you mean by funky y and t values, do you have the equations for which you are trying solve? Also be sure to include the following code at the top of you code... clc; clear all; This will clear any previous calculations done so they don't interfere with your code as it runs. 
What is you code supposed to output, when I run it, I enter a number of intervals and a condition (1,2 or 3) and the program runs but I get nothing out of it. 
I would give necessary body parts to be fluent in matlab. In half of my classes we're expected to know how to use it even though we've never been taught. 
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By funky i mean the values alternate between a positive and negative values, I am aware of the clear and clc part. The number of intervals could be as small as five and as large as 10k all it does is generate a better approximation. The issue with the alternating values is that the graphs that result when plotted either converge on a value or form a hyperbola, they should however form a some thing similiar to an expential decay. I am attaching a link to a similiar problem that only incorporates the second boundry condition. the example is halfway down in the link. http://profjrwhite.com/matlab_course...x/fd_intro.pdf Duncan 
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http://i1141.photobucket.com/albums/...Untitled1.png and for further help, writing any kind of code always requires the proper material to assist in programming... http://i1141.photobucket.com/albums/...2/Photo030.jpg 
Could you just insert in absolute value command when collecting y value data to fix the issue? so where ever you calculate y in you code instead of the following... y = "code"; do y = abs("code"); 
That might jst work, I will try it and let you know. That would make the graph come out correctly but I am not sure what that would ddo for the integrity of the data? Duncan 
Ok so my friend and I tried that and it still results in a data set that jumps around whereas it should supposedly result in a nice curve. Duncan 
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