How to plot temperatures at nine different points of 2D grid resulting from Finite Difference method?
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</matlabcentral/answers/uploaded_files/47655/2D.HeatEqGridImage.png> I need to plot temperature at 9 points of the grid i.e. T11,T12,T13,T21,T22,T23,T31,T32,T33 So basically temp. vs position in the grid for an appropriately sized grid, let's say 8cmX8cm Results are T11=49.99,T12=71.42,T13=85.71,T21=28.57,T22=49.99,T23=71.427,T31=14.28,T32=28.569,T33=49.99 The code for 2D Heat Eq for steady state HT using Guass Seidel method is: (* Pl. excuse me for not including comments in the code)
clear;clc;format('long','g'); i=1; T12(i)=0;T21(i)=0; T13(i)=0;T22(i)=0; T23(i)=0; T31(i)=0; T32(i)=0; T33(i)=0; error_T11(i)=9999; while error_T11(i)>=0.01 T11(i+1)=(100+T12(i)+T21(i))/4; T12(i+1)=(100+T11(i+1)+T13(i)+T22(i))/4; T13(i+1)=(200+T12(i+1)+T23(i))/4; T21(i+1)=(0+T11(i+1)+T31(i)+T22(i))/4; T22(i+1)=(0+T21(i+1)+T23(i)+T12(i+1)+T32(i))/4; T23(i+1)=(100+T13(i+1)+T22(i+1)+T33(i))/4; T31(i+1)=(0+T21(i+1)+T32(i))/4; T32(i+1)=(0+T31(i+1)+T22(i+1)+T33(i))/4; T33(i+1)=(100+T23(i+1)+T32(i+1))/4;
error_T11(i+1)=abs(T11(i+1)-T11(i))/T11(i+1)*100;
error_T12(i+1)=abs(T12(i+1)-T12(i))/T12(i+1)*100;
error_T13(i+1)=abs(T13(i+1)-T13(i))/T13(i+1)*100;
error_T21(i+1)=abs(T21(i+1)-T21(i))/T21(i+1)*100;
error_T22(i+1)=abs(T22(i+1)-T22(i))/T22(i+1)*100;
error_T23(i+1)=abs(T23(i+1)-T23(i))/T23(i+1)*100;
error_T31(i+1)=abs(T31(i+1)-T31(i))/T31(i+1)*100;
error_T32(i+1)=abs(T32(i+1)-T32(i))/T32(i+1)*100;
error_T33(i+1)=abs(T33(i+1)-T33(i))/T33(i+1)*100;
i=i+1;
end
disp(i);
T11=T11(i);T12=T12(i);T13=T13(i);T21=T21(i);T22=T22(i);T23=T23(i);T31=T31(i);T32=T32(i);T33=T33(i);
disp(' T11 ');
% disp([T11',error_T11']);
disp(T11);disp(T12);disp(T13);disp(T21);disp(T22);disp(T23);disp(T31);disp(T32);disp(T33);
4 commentaires
Réponse acceptée
KSSV
le 16 Mar 2016
It shall be easy....You will be having X,Y,T of size mXn each. Use surf(X,Y,T).
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