Effacer les filtres
Effacer les filtres

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

21 vues (au cours des 30 derniers jours)
Steffen B.
Steffen B. le 20 Juin 2022
Modifié(e) : Torsten le 21 Juin 2022
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
  3 commentaires
Torsten
Torsten le 20 Juin 2022
Modifié(e) : Torsten le 21 Juin 2022
If your boundary condition at the left end is
-lambda*dT/dx = 10
, then
pl = 10, ql = 1
Steffen B.
Steffen B. le 20 Juin 2022
Modifié(e) : Steffen B. le 20 Juin 2022
@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

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