Got to do this program about heat transer and finite equations and although the program is done I have litle errors dont know how to fix. The issuesare remarked down. THANKS

>> %ANA BATRES MATLAB PROGRAM FINITE EQUATIONS MODELING.

clear;

B=24; %(mm) distance between heat wires

D=4; %(mm)distance from upper plate to the heat wire

a=6; %(mm) distance from upper plate to inferior plate

T_inf=30; %(centigrades), we dont change to kelvin the rest of the parameters are in degrees.

k=2; %(w/m2C), conductive parameter

h=100; %(W/m2C), convective parameter

q_perunit=50; %(W/m)

mesh=14; %this is choosed and can be changed as we want

delta_y=(B+4)/mesh; %sidelength of each element, same in both directions

delta_y=(delta_y)*(10^3); %I pass it to m.

>> Bi=h*delta_y/k; %biot number

>> n=mesh+1;

>> m=mesh+1;

>> k=n*m;

>> X=0:1/mesh:1;

>> Y=0:1/mesh:1;

>>

>> Con=ones(n,m);

>> A=zeros(k,k);

>> C=zeros(k,1);

>>

>> for i=1:n;

Con(i,1)=2;

Con(i,m)=4;

Con(n,i)=2;

end

>>

>> for i=2:m-1;

Con(1,i)=3;

end

>>

>> counter=1;

>> for i=1:m;

for j=1:n;

switch Con(i,j)

case 1

%internal node

A(counter,counter+1)=1;

A(counter,counter-1)=1;

A(counter,counter+n)=1;

A(counter,counter-n)=1;

A(counter,counter)=-4;

case 2

A(counter,counter+1)=1;

A(counter,counter-1)=1;

A(counter,counter+n)=1;

A(counter,counter-n)=1;

A(counter,counter)=-4;

C(counter,1)=-q_perunit*delta_y;

%this is an interior node with generation

case 3

%corner node with convection

A(counter,counter+1)=2;

A(counter,counter+n)=1;

A(counter,counter)=-4-Bi;

C(counter,1)=-Bi*T_inf;

case 4

%corner node with no convection (insulated surface)

A(counter,counter+1)=2;

A(counter,counter+n)=1;

A(counter,counter)=-4;

case 5

%convection and conduction

A(counter,counter-1)=1;

A(counter,counter+1)=1;

A(counter,counter+n)=2;

A(counter,counter)=-4-Bi;

C(counter,1)=-2*Bi*T_inf;

end;

counter=counter+1;

end;

end;

Index in position 2 is invalid. Array indices must

be positive integers or logical values.

>> T2=inv(A)*C;

Warning: Matrix is singular to working precision.

>> counter2=1;

>> for i=1:m;

for j=1:n;

T(i,j)=T2(counter2);

counter2=counter2+1;

end

end

>>

>> figure(1)

>> [C1,h1]=contour(X,Y,T,15);

Warning: Contour not rendered for non-finite ZData

> In contour (line 51)

>> clabel(C1,h1)

>> grid on

>>

>> figure (2)

>> surf(X,Y,T)

>>

>>

>> figure(1)

>> surf(T)

[Walter Roberson]

A(counter,counter-1)=1;

Which array location is being used when counter is 1?

Answer 1

[Neelanshu]

Hi Ana,

It looks like you're trying to fix the "Array indices must be positive integers or logical values" error that's popping up in your MATLAB code.

As Walter pointed out, using A(counter, counter-1) when counter = 1 results in counter-1 being equal to 0. Since MATLAB indexing starts at 1, attempting to access A(1, 0) is invalid. Similarly, when i=1 and j=1, the left boundary case statement activates with counter = 1, but A(counter, counter-n) would again result in negative index values as illustreated in the image, which are not permitted.

To fix this error, please ensure that you are using positive integers for array indexing. When counter is 1, you must avoid using expressions that result in zero or negative indices. You can implement conditional checks to handle boundary cases or adjust your loop ranges to start from a higher index when necessary.

For more information on array indexing in MATLAB and to avoid such errors, you can refer to the following documentation:

https://www.mathworks.com/help/matlab/learn_matlab/array-indexing.html

Hope this helps.