Problems getting answers from ode45

Running this code does not generate any answers, the return matrices only contain NaN values. main file:
clear all;
clc;
%Test Parameters
Kc=0.004;
Dab=0.006;
Dl=0.0088;
cp0=1; %Concentration at particle center
c0=ones(2000,1); %Initial conditions
%C(1) to C(1000)=Cp=1, C(1001) to C(2000)=Cf=0
for i=1001:2000; %Initial Cf=0 throughout
c0(i,1)=0;
end
tspan=linspace(1,200,200);%Time span
[T,C]=ode45(@(t,c) dcdt(t,c,cp0,Kc,Dab,Dl),tspan,c0);
function file:
function dcdt1=dcdt(t,c,cp0,Kc,Dab,Dl)
%Constants
N=2000; %Total Nodes=2000
r=2.5*10^-3; %Particle Radius
delr=r/(N/2); %Node Length for R=1000
z=5; %Bed Length
delz=z/(N/2); %Node Length for Z=1000
ez=0.4; %Bed Voidage
u=2.24*10^-2; %Fluid Velocity (CO2)
dcdt1=zeros(N,1); %Define system of dcdt
%Node R=0
dcdt1(1)=6*Dab/2*(c(2)-cp0);
%Nodes R=2:999
for i=2:999
dcdt1(i)=(Dab/delr^2+Dab/(i*delr^2))*c(i+1)...
-2*(Dab/i*delr^2)*c(i)...
+(Dab/delr^2-Dab/(i*delr^2))*c(i-1);
end
%Node R=1000
dcdt1(1000)=((-2*Dab/(delr^2))*(4*r^2/delr^2-4*r/delr+1)*c(999)+...
(2*Dab/(delr^2)*(4*r^2/delr^2-4*r/delr+1)-Kc*(4*r^2/delr^2))*c(1000)+...
Kc*(4*r^2/delr^2)*c(1001))/(7*r^3/(6*delr^2)-r^2/(2*delr)+r/2-delr/6);
%Nodes Z=1:1000
for i=1001:1999
dcdt1(i)=(Dl/delz^2-u/2*delz)*c(i+1)...
-(2/delz^2+(1-ez/ez))*(3*Kc/r)*c(i)+...
(Dl/delz^2+u/2*delz)*c(i-1)...
+(1-ez/ez)*(3*Kc/r)*c(1000);
end
There are no syntax errors, but the return variable C is empty. Hope to get any suggestions or answers as to why ode45 is returning this.

2 comentarios

Walter Roberson
Walter Roberson el 27 de Sept. de 2012
Considering adding an options structure that has function value checking turned on.
Jan
Jan el 27 de Sept. de 2012
Editada: Jan el 27 de Sept. de 2012
The expression "(1-ez/ez)" in the last FOR-loop is at least confusing. Do you mean "(1-ez)/ez"?
In the first FOR-loop you have "Dab/(i*delr^2)" and "Dab/i*delr^2". Is this correct or did you forget the parenthesis in the second expression?
The code can be simplified substantially by using temporary variables e.g. for "delr^2" and "delz^2". Beside the acceleration, this allows for an easier debugging.
for i=1001:2000; c0(i,1)=0; end
Faster and nicer:
c0(1001:2000) = 0;

Iniciar sesión para comentar.

 Respuesta aceptada

Jan
Jan el 27 de Sept. de 2012
Editada: Jan el 27 de Sept. de 2012
Examples for a simplification and vectorization - assuming that "(1-ez/ez)" is not a typo:
%Nodes Z=1:1000
d2 = delz ^ 2;
c1 = u / 2 * delz;
dcdt1(1001:1999)= (Dl/d2 - c1) * c(1002:2000) ...
-6 * Kc/(d2 * r) * c(1001:1999) ...
+(Dl/d2 + c1) * c(1000:1998);
And:
u = 2.24e-2; % The POWER operator is very expensive!
Note: The leaner and cleaner the code, the easier is finding bugs.

1 comentario

Red
Red el 27 de Sept. de 2012
Thanks for your reply. Will try to optimize my code and get the logical error out.

Iniciar sesión para comentar.

Más respuestas (0)

Categorías

Más información sobre Loops and Conditional Statements en Centro de ayuda y File Exchange.

Preguntada:

Red
el 27 de Sept. de 2012

Community Treasure Hunt

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

Start Hunting!

Translated by