フィルターのクリア

resolution of PDE

2 ビュー (過去 30 日間)
mouna
mouna 2011 年 5 月 7 日
I have a system of equations:
(?C/?t)+(u/?)?C/?z+((1-?)/?)*(?s/?f)*Ki*a*(q-KéqC)=0
?q/?t=-Ki a(q-KéqC)
the initial condition is:
t=0, q=q0 and C=C0,z>0
the boundary conditions are:
z=0;C=0 and q=q0
z=L; ?C/?z=0 ,?q/?z=0
this is my program: i don't find the error so pleaaase help me. thanks in advance.
function modelediffusion
m=0;
z=linspace(0,1,26);
t=linspace(0,2,200);
sol = pdepe(m,@pdexpde,@pdexic,@pdexbc,z,t);
C = sol(:,:,1);
q = sol(:,:,2);
figure
plot(t,C(:,15))
function[g,f,s]= pdexpde(z,t,C,DCDz)
rhos=0.55;
rhof=0.385;
dp=0.03;
a=6/dp;
epsilon=0.45;
Ki=1.4E-7;
u=0.098;
Keq=16.86;
A=Ki*a*(C(2)-(Keq.*C(1)));
B=((1-epsilon)/epsilon)*(rhos/rhof)*A;
g=[1; 1];
f=[0; 0];
s=[((-u)/epsilon).*DCDz-B; A];
function C0 = pdexic(z)
c0=1.5E-3;
q0=2.53E-2;
C0 = [c0; q0];
% -------------------------------------------------------------------------
function [pl,ql,pr,qr] = pdexbc(zl,Cl,zr,Cr,t)
q0=2.53E-2;
pl = [C1(1);q0-Cl(2)];
ql = [1; 0];
pr = [0; 0];
qr = [0; 1];
  5 件のコメント
Walter Roberson
Walter Roberson 2011 年 5 月 8 日
The question marks indicate derivative. Except the ones that appear isolate right after a "/" -- I don't know what those are.
Is KéqC a complete variable, or is it Kéq * C ?
Walter Roberson
Walter Roberson 2011 年 5 月 8 日
((1-?)/?) is a puzzler alright.

サインインしてコメントする。

回答 (1 件)

Andrew Newell
Andrew Newell 2011 年 5 月 8 日
Here is one error: in this line,
pl = [C1(1);q0-Cl(2)];
the first C1 is C followed by the numeral 1, but it should be C followed by the letter l.
Debugging note: It is a good idea to have just one command per line. That way, when you get an error message, you know which command triggered it (if not which command actually caused it).
  1 件のコメント
mouna
mouna 2011 年 5 月 8 日
thank you very much

サインインしてコメントする。

カテゴリ

Help Center および File ExchangeEigenvalue Problems についてさらに検索

製品

Community Treasure Hunt

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

Start Hunting!

Translated by