1D Heat equation in Matlab with heat Flux at one side

I’m solving the 1D heat equation with Matlab pdepe function.

The geometry is a steel rod of 1m length, which has a constant temperature on the right side and a constant flux on the other side.

I’m struggeling a little bit with specifying the value of p and q.

Here is the code I use:

function [c,f,s] = heatequation(x,t,u,dudx)
rho=7850; % density
cp=420;   % heat capacity
lambda=40;% conductivity
c = rho*cp;
f = lambda*dudx;
s = 0;
end

function u0 = ic(x)
Tini=300; % initial temperature at t= 0 s
u0 = Tini;
end

function [pl,ql,pr,qr] = bc(xl,ul,xr,ur,t)
pl = 0;      
ql = 10;     % Flux at the left end = 10 W/m2
pr = ur-300; % Temperature at the rigth end = 300 K
qr = 0 
end

m = 0;
sol = pdepe(m,@heatequation,@ic,@bc,x,t);

I'm not sure if I wrote the BC's correct? The notation with p and q is still confusing

[Bjorn Gustavsson]

Good hint (that I should've listened to long time ago): explicitly write in the units of everything in your code, and on paper under the equation. That way you can figure out what should be what. Without that we're more or less relying on luck.

[Torsten]

@Steffen B.

If your boundary condition at the left end is

-lambda*dT/dx = 10

, then

pl = 10, ql = 1

[Steffen B.]

@Torsten thanks for your fast answer. You are rigth, I mixed someting up in the BC function. Now it's working just fine

function [pl,ql,pr,qr] = bc(xl,ul,xr,ur,t)
pl = 10;     % Flux at the left end = 10 W/m2   
ql = 1;      
pr = ur-300; % Temperature at the rigth end = 300 K
qr = 0 
end