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 件のコメント

Walter Roberson
Walter Roberson 2012 年 9 月 27 日
Considering adding an options structure that has function value checking turned on.
Jan
Jan 2012 年 9 月 27 日
編集済み: Jan 2012 年 9 月 27 日
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;

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 採用された回答

Jan
Jan 2012 年 9 月 27 日
編集済み: Jan 2012 年 9 月 27 日

0 投票

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 件のコメント

Red
Red 2012 年 9 月 27 日
Thanks for your reply. Will try to optimize my code and get the logical error out.

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2012 年 9 月 27 日

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