I am writing a code using the 4th order runge kutta method, for heat transfer through an adiabatic fin, i am having some problems getting the code to work, any advice?

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Michael Pine
Michael Pine 2022-4-6
编辑： Torsten 2022-4-6
%%4th order runge kutta-uniform cross section fins
%circular pin fin- pure aluminium
%assumptions
%1D steady state heat transfer
%dimensions
L=0.05;
D=0.005;
%parameters
h=50;
k=237;
Tb=60;
Ta=15;
P=pi*D;
Ac=pi*D^2/4;
m=sqrt(h*P/k*Ac);
%Boundary conditions
%BC1 Theta(0)=Tb-Ta
%BC2 dtheta/dx at x=L =0
%inputs
H=0.0004; %step size
Lmax=0.05; %total length
N=Lmax/H; %Number of steps
%Initial conditions
L(1)=0;
T(1)=60;
z(1)=0;
m=1;
z(100)=0;
x(1)=0;
while m>0.00001;
for i=1:N
%equations for k values
k1=z(i);
k2=z(i)+(H*L1/2);
k3=z(i)+(H*L2/2);
k4=z(i)+(H*L3);
%equation for T(i+1)
T(i+1)=T(i)+(H/6)*(k1+(2*k2)+(2*k3)+k4);
%equations for L values
L1=((h*P/k*Ac).*(T(i)));
L2=((h*P/k*Ac).*(T(i)+(H*k1/2)));
L3=((h*P/k*Ac).*(T(i)+(H*k2/2)));
L4=((h*P/k*Ac).*(T(i)+(H*k3)));
%equation for z(i+1)
z(i+1)=z(i)+(H/6)*(L1+(2*L2)+(2*L3)+L4);
%equation for x(i+1)
x(i+1)=x(i)+H;
end
z(1)=z(1)-z(i+1);
m=abs(z(i+1)-z100);
end
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Torsten
Torsten 2022-4-6
编辑：Torsten 2022-4-6
You solve the boundary value problem
d^2T/dx^2 = 0
T(x=0) = 60
dT/dx(x=0.05) = 0
don't you ?
If this is the case, the solution should be clear:
T(x) = a*x + b
where the constants a and b follow from the boundary conditions (in your case T=60 for all x).

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