Maple Solutions to the Chemical Engineering Problem Set

C C C C C ⎫
C D X Y
Z ⎪
C = C + e + e , C = C − e , C = C − e , K = , K = , K = ⎪⎬
A, 0 A R1 R3 Y, 0 Y R2 Z, 0 Z R3 R1 C C R2 C C R3 C C ⎪
A B B C A X ⎭
and a set of unknown variables:
> Vars := indets(Eqns,name);
Vars :=
{ C , C , C , C , C , C , C , C , C , C , e , e , e , C , C , C , C , K , K ,
K }
C, 0 C D, 0 D X, 0 X B B, 0 A, 0 A R1 R2 R3 Y, 0 Y Z, 0 Z R1 R2 R3
The number of degrees of freedom follows directly
> DegFree := nops(Vars)-nops(Eqns);
Case 1
> Specs[1] := C[A,0]=1.5;
> Specs[2] := C[B,0]=1.5;
> Specs[3] := seq(C[i,0]=0,i=[C,D,X,Y,Z]);
DegFree := 10
Specs := C =
1 A, 0
Specs := C =
2 B, 0
1.5
1.5
Specs := C = 0 , C = 0 , C = 0 , C = 0 ,
C = 0
3 C, 0 D, 0 X, 0 Y, 0 Z, 0
At the conditions of interest the reaction equilibrium coefficient has the following values
> Kvals := {K[R1] = 1, K[R2]=2.63, K[R3]=5};
Kvals := { K = 1 , K = 2.63 ,
K = 5}
R1 R2 R3
We now augment the set of equations with the set of specifications
> AllEqns:=Eqns union {seq(Specs[k],k=1..3),op(Kvals)};
⎧
⎪
AllEqns := ⎪⎨
C = 1.5 , C = C − e + e , C = C − e , C = C − e + e ,
⎪ B, 0 C, 0 C R1 R2 D, 0 D R1 X, 0 X R2 R3
⎩
C =
B, 0
+ +
C B e R1
C C
C D
e , C = C + e + e , C = C − e , C = C − e , K = , K = 1
,
R2 A, 0 A R1 R3 Y, 0 Y R2 Z, 0 Z R3 R1 C C R1
A B
C C C
X Y
Z
K = 2.63 , K = 5 , K = , K = , C = 1.5 , C = 0 , C = 0 , C = 0 , C = 0
,
R2 R3 R2 C C R3 C C A, 0 C, 0 D, 0 X, 0 Y, 0
B C A X
⎫
⎪
C = 0⎪⎬
Z, 0 ⎪
⎭
and ask Maple to solve the problem
> result:=solve(AllEqns,Vars);
result := { C = 0 , C = 0 , C = 0 , C = 0 , C = 0 , C = -.3879574470 , C = -.3207318471
,
C, 0 D, 0 X, 0 Y, 0 Z, 0 B X
C = 1.072193162 , C = 1.136496132 , e = 1.072193162 , %3 , C = .2564288777 , K = 5. , C = .8157642847
,
D Z R1
C R3 Y
e = 1.136496132 , e = .8157642847 , C = -.7086892941 , %2 , %1 , K = 1.
} , {
R3
R2
A R1
e = .9256703707 + .1402769617 I , C = 0 , C = 0 , C = 0 , C = 0 , C = 0 , %3 , K = 5.
,
R2
C, 0 D, 0 X, 0 Y, 0 Z, 0 R3
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