Maple Solutions to the Chemical Engineering Problem Set

We define yz ( ) as the first derivative term.
> ODE3:=diff(C[A](z),z)=y(z): ODE3;
∂
C ( z ) = yz ( )
∂z
A
The second derivative is obtained on differentiation.
> diff(ODE3,z);
∂
=
∂
2
z 2
∂
C ( z ) ( )
A ∂z
yz
Substitution of the original ODE gives the following expression for the ODE
> ODE4:=subs(ODE,"): ODE4;
The parameter values are specified
> Params :={D[A,B]=1.2e-9,k=1e-3};
k C ( z )
A ∂
= ( )
D ∂z
A, B
yz
Params := { k = .001 ,
D = .12 10 }
A, B -8
and substituted into the set of differential equations.
> Deqns := subs(Params,{ODE3,ODE4});
∂
∂
Deqns := { C ( z ) = yz ( ) ,
833333.3333 C ( z)
= ( ) }
∂z
A A ∂z
yz
The initial conditions are the known initial concentration and a guess for the unknown first derivative
term.
> IC := {C[A](0)=0.2,y(0)=-130};
IC := { y0 ( ) = -130 ,
C ( 0 ) = .2}
A
We combine the sets of ODEs and initial conditions.
> Alleqns := Deqns union IC;
∂
∂
Alleqns := { C ( z ) = yz ( ) , y0 ( ) = -130 , C ( 0 ) = .2 ,
833333.3333 C ( z)
= ( ) }
∂z
A A A ∂z
yz
dsolve is invoked to make a procedure for numerical computation of the result.
> g:=dsolve(Alleqns,{C[A](z),y(z)}, type=numeric, method=rkf45, output=procedurelist);
g := proc( rkf45_x)
... end
We compute the "solution" (subject to the guessed initial value of y) as follows
> g(1e-3);
[ z = .001 , C ( z ) = .1404605642564124 ,
yz ( ) = 2.764382915334039]
A
where the value of 0.001 in the call to g is the final value of z.
We need to find an initial value of y that leads to y.001 ( )
= 0 (more or less). We make a procedure to
automate the preceding computations:
> f := proc(y0)
local IC, Deqns, Params, Alleqns,g, err;
Params :={D[A,B]=1.2e-9,k=1e-3, y[f,0]=0};
Deqns := {833333.3333*C[A](z) = diff(y(z),z), diff(C[A](z),z) = y(z)};
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