How to solve a system of parabolic partial differential equations in time and 1 D space

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Lucy
Lucy 2017-11-22
评论： Lucy 2017-11-22
Hi,
I cannot seem to solve the PDF in the figures below. This is the code I have but it produces the wrong graphs:
function New
m=0;
x=0:0.5:15;
t=1:100:28800;
sol = pdepe(m,@pdefun,@icfun,@bcfun,x,t);
u1=sol(:,:,1);
u2=sol(:,:,2);
surf(x,t,u1)
title('u1(x,t)')
xlabel('Distance x')
ylabel('Time t')
figure
surf(x,t,u2)
title('u2(x,t)')
xlabel('Distance x')
ylabel('Time t')
function [c,f,s]=pdefun(x,t,u,DuDx)
P=1200;
R=0.208;
cv=0.313;
rho=7.5;
v=0.3;
A=70;
mdot=1.2;
k=0.016/1000;
alph=0.5;
rhoG=1600;
cG=0.7;
kG=0.7/1000;
c=[((P*cv/R)/u(1));(rhoG*cG)];
f=[((k*DuDx(1))-((P*cv/R)*(mdot*R/(P*A))));(kG*DuDx(2))];
s=[(u(2)-u(1));(u(1)-u(2))]*alph;
function u0=icfun(x)
u0=[300;300];
function[pl,ql,pr,qr]=bcfun(xl,ul,xr,ur,t)
pl = [ul(1)-773; 0];
ql = [0; 1];
pr = [0; 0];
qr = [1; 1];
I think my boundary conditions are incorrect but I don't know how to fit them.
I would appreciate any help with this problem, thank you in advance.
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采纳的回答

Torsten
Torsten 2017-11-22
One mistake in your settings:
f = [k*DuDx(1) ; kG*DuDx(2)];
s = [(u(2)-u(1))*alph-cv*mdot/A*DuDx(1);(u(1)-u(2))*alph];
And maybe in your boundary condition for T_A at x=0, you will need a ramp that linearly increases the temperature from 300 K to 773 K in, say, one minute.
Best wishes
Torsten.
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Torsten
Torsten 2017-11-22
编辑：Torsten 2017-11-22
Increasing the number of discretization points together with the ramp for a=100 solved the problem:
x = linspace(0,15,3000);
Best wishes
Torsten.

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