Undefined function or variable using ODE45
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Andrew Knight
on 22 Sep 2019
Commented: Star Strider
on 22 Sep 2019
I'm having trouble with defining an independent variable in matlab. I am getting the error, 'undefined function or variable 'Y'. Y is the variable I am trying to solve for.
%elementary rate law
FA0=10; %mol/min
W=100; %KG
a=.008; %kg^-1
k1C=2; %L/kg cat/min
K1C=.2; %mol/L
k2D=.4; %L/kg cat/min
k3E=5; %L/mol^2/kg cat/min
kc=1; %L/kg cat/min
v=16.67; %L/min
Fa=Y(1);
Fb=Y(2);
Fc=Y(3);
Fd=Y(4);
Fe=Y(5);
y=Y(6);
CT0=FA0;
FT=Fa+Fb+Fc+Fd+Fe;
Ca=CT0*(Fa/FT)*y;
Cb=CT0*(Fb/FT)*y;
Cc=CT0*(Fc/FT)*y;
Cd=CT0*(Fd/FT)*y;
Ce=CT0*(Fe/FT)*y;
r1C=k1C*(Ca-(Cb*Cc/K1C));
r2D=k2D*Cd;
r3E=k3E*Cd*(Ce^2);
ra=-r1C-r2D;
rb=r1C;
rc=r1C-r3E;
rd=r2D-.5*r3E;
re=r3E;
dFadw=ra;
dFbdw=rb;
dFcdw=rc;
dFddw=rd;
dFedw=re;
dydw=-a*FT/(2*FT0*y);
fun=@(w,Y)[dFadw;dFbdw;dFcdw;dFddw;dFedw,dydw];
wspan=0:1:100;
x0=[10;0;0;0;0;1];
[T,X] = ode45(fun,wspan,x0)
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Accepted Answer
Star Strider
on 22 Sep 2019
I believe this is what you actually intend:
function dy = fun(T,Y)
FT0 = 42;
FA0=10; %mol/min
W=100; %KG
a=.008; %kg^-1
k1C=2; %L/kg cat/min
K1C=.2; %mol/L
k2D=.4; %L/kg cat/min
k3E=5; %L/mol^2/kg cat/min
kc=1; %L/kg cat/min
v=16.67; %L/min
Fa=Y(1);
Fb=Y(2);
Fc=Y(3);
Fd=Y(4);
Fe=Y(5);
y=Y(6);
CT0=FA0;
FT=Fa+Fb+Fc+Fd+Fe;
Ca=CT0*(Fa/FT)*y;
Cb=CT0*(Fb/FT)*y;
Cc=CT0*(Fc/FT)*y;
Cd=CT0*(Fd/FT)*y;
Ce=CT0*(Fe/FT)*y;
r1C=k1C*(Ca-(Cb*Cc/K1C));
r2D=k2D*Cd;
r3E=k3E*Cd*(Ce^2);
ra=-r1C-r2D;
rb=r1C;
rc=r1C-r3E;
rd=r2D-.5*r3E;
re=r3E;
dFadw=ra;
dFbdw=rb;
dFcdw=rc;
dFddw=rd;
dFedw=re;
dydw=-a*FT/(2*FT0*y);
dy = [dFadw;dFbdw;dFcdw;dFddw;dFedw;dydw];
end
wspan=0:1:100;
x0=[10;0;0;0;0;1];
[T,X] = ode45(@fun,wspan,x0);
figure
plot(T,X)
grid
The ‘FT0’ value and assignment are missing, so I created one to test the code. (You may want to substitute a different value.) This runs without error.
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