フィルターのクリア

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

1 回表示 (過去 30 日間)
Lucy
Lucy 2017 年 11 月 22 日
コメント済み: Lucy 2017 年 11 月 22 日
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.
<<
>>
%

採用された回答

Torsten
Torsten 2017 年 11 月 22 日
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.
  7 件のコメント
Torsten
Torsten 2017 年 11 月 22 日
編集済み: Torsten 2017 年 11 月 22 日
Increasing the number of discretization points together with the ramp for a=100 solved the problem:
x = linspace(0,15,3000);
Best wishes
Torsten.
Lucy
Lucy 2017 年 11 月 22 日
Hi Torsten,
Thank you so much for your help. It is now working how I believe it should.
Lucy

サインインしてコメントする。

その他の回答 (0 件)

カテゴリ

Help Center および File ExchangeOrdinary Differential Equations についてさらに検索

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by