Zibian1. M
% Isobutene (1) Methanol MTBE (2) (3)
The function xdot=zibian1 (t, x, flag)
If isempty (flag),
T=x (4);
% the Universal gas constant (CAL/mol) (K) 8.314472 J/(k. mol)
R=1.987;
P=8 * (1.013 * 1 e + 5);
% Wilson model binary interaction parameters binary interaction parameters of Wilson model
A (1, 1)=0;
A (1, 2)=169.9953;
A (1, 3)=30.2477;
A (2, 1)=2576.8532;
A (2, 2)=0.
A (2, 3)=1483.2478;
A (3, 1)=271.5669;
A (3, 2)=406.3902;
A (3, 3)=0;
% molar volumes (cm3/mol)
V (1)=93.33;
V (2)=44.44;
V (3)=118.8;
A (1, 1)=1;
A (2, 1)=V (1)/V (2) * exp (- A (2, 1)/(R * T));
(1) A (3, 1)=V/V (3) * exp (A (3, 1)/(R * T));
A (2, 2)=1;
(2) A (1, 2)=V/V (1) * exp (- A (1, 2)/(R * T));
(2) A (3, 2)=V/V (3) * exp (A (3, 2)/(R * T));
A (3, 3)=1;
(3) A (2, 3)=V/V (2) * exp (- (2, 3)/A (R * T));
(3) A (1, 3)=V/V (1) * exp (- A (1, 3)/(R * T));
% the Activity coefficients Activity coefficient
G (1)=exp (1 - x (1) * A (1, 1)/(x (1) * A (1, 1) + x (2) * A (1, 2) + x (3) * A (1, 3))...
- x (2) * A (2, 1)/(x (1) * A (2, 1) + x (2) * A (2, 2) + x (3) * A (2, 3))...
- x (3) * A (3, 1)/(x (1) * A (3, 1) + x (2) * A (3, 2) + x (3) * A (3, 3)))...
/(x (1) * A (1, 1) + x (2) * A (1, 2) + x (3) * A (1, 3));
G (2)=exp (1 - x (1) * A (1, 2)/(x (1) * A (1, 1) + x (2) * A (1, 2) + x (3) * A (1, 3))...
- x (2) * A (2, 2)/(x (1) * A (2, 1) + x (2) * A (2, 2) + x (3) * A (2, 3))...
- x (3) * A (3, 2)/(x (1) * A (3, 1) + x (2) * A (3, 2) + x (3) * A (3, 3)))...
/(x (1) * A (2, 1) + x (2) * A (2, 2) + x (3) * A (2, 3));
G (3)=exp (1 - x (1) * A (1, 3)/(x (1) * A (1, 1) + x (2) * A (1, 2) + x (3) * A (1, 3))...
- x (2) * A (2, 3)/(x (1) * A (2, 1) + x (2) * A (2, 2) + x (3) * A (2, 3))...
- x (3) * A (3, 3)/(x (1) * A (3, 1) + x (2) * A (3, 2) + x (3) * A (3, 3)))...
/(x (1) * A (3, 1) + x (2) * A (3, 2) + x (3) * A (3, 3));
% Vapor pressure using Antoine 's Equation
A (1)=20.64559; A (2)=23.49989; A (3)=20.71616;
B (1)=2125.74886; B (2)=3643.31362; B (3)=2571.58460;
C (1)=33.160; C (2)=33.434; C (3)=48.406;
(Psat) (1)=exp (A (1) + B (1)/(T + C (1)));
(Psat), (2)=exp (A (2) + B (2)/(T + C (2)));
(Psat) (3)=exp (A (3) + B (3)/(T + C (3)));
% Equilibrium constant for the reaction of isobutene + methanol==MTBE isobutylene + methanol MTBE reaction Equilibrium constant of the
Keq=exp (10.0982 + 0.2667 + 4254.05/T * log (T));
% Modified Raoults Law
(1)=y (Psat) G (1) (1) * * x (1)/P;
(2)=y (Psat), (2) (2) * G * x (2)/P;
(3)=y (Psat) (3) * G (3) * x (3)/P;
RR=x (1) * G * x (1) (2) * (2)/G (Keq * x (3) * G (3));
V=[1, 1, 1];
Kf=74.4 * exp (- 3187/T);
Kfref=74.4 * exp (3187/333.35);
Da=0;
Xdot (1) (1) - (1) y=x + kf/kfref * (v (1) - the sum (v) * x (1)) * Da * RR.
Xdot (2) (2) - (2) y=x + kf/kfref * (v (2) - the sum (v) * x (2)) * Da * RR.
Xdot (3) (3) - (3) y=x + kf/kfref * (v (3) - the sum (v) * x (3)) * Da * RR.
Xdot (4)=x (1) + x (2) + x (3) - 1;
Xdot % xdot=xdot 'must be a column vector
The else
% Return M the mass matirx Return mass matrix
M=zeros (4, 4);
M (1, 1)=1;
M (2, 2)=1;
M (3, 3)=1;
Xdot=M;
End
Zibian2. M
% Isobutene (1) Methanol MTBE (2) (3)
The function xdot=zibian2 (t, x, flag)
If isempty (flag),
T=x (4);
% the Universal gas constant (CAL/mol) (K) 8.314472 J/(k. mol)
R=1.987;
P=8 * (1.013 * 1 e + 5);
% Wilson model binary interaction parameters binary interaction parameters of Wilson model
A (1, 1)=0;
A (1, 2)=169.9953;
A (1, 3)=30.2477;
A (2, 1)=2576.8532;
A (2, 2)=0.
A (2, 3)=1483.2478;
A (3, 1)=271.5669;
A (3, 2)=406.3902;
A (3, 3)=0;
% molar volumes (cm3/mol)
V (1)=93.33;
V (2)=44.44;
V (3)=118.8;
A (1, 1)=1;
A (2, 1)=V (1)/V (2) * exp (- A (2, 1)/(R * T));
(1) A (3, 1)=V/V (3) * exp (A (3, 1)/(R * T));
A (2, 2)=1;
(2) A (1, 2)=V/V (1) * exp (- A (1, 2)/(R * T));
(2) A (3, 2)=V/V (3) * exp (A (3, 2)/(R * T));
A (3, 3)=1;
(3) A (2, 3)=V/V (2) * exp (- (2, 3)/A (R * T));
(3) A (1, 3)=V/V (1) * exp (- A (1, 3)/(R * T));
% the Activity coefficients Activity coefficient
G (1)=exp (1 - x (1) * A (1, 1)/(x (1) * A (1, 1) + x (2) * A (1, 2) + x (3) * A (1, 3))...
- x (2) * A (2, 1)/(x (1) * A (2, 1) + x (2) * A (2, 2) + x (3) * A (2, 3))...
- x (3) * A (3, 1)/(x (1) * A (3, 1) + x (2) * A (3, 2) + x (3) * A (3, 3)))...
/(x (1) * A (1, 1) + x (2) * A (1, 2) + x (3) * A (1, 3));
G (2)=exp (1 - x (1) * A (1, 2)/(x (1) * A (1, 1) + x (2) * A (1, 2) + x (3) * A (1, 3))...
- x (2) * A (2, 2)/(x (1) * A (2, 1) + x (2) * A (2, 2) + x (3) * A (2, 3))...
- x (3) * A (3, 2)/(x (1) * A (3, 1) + x (2) * A (3, 2) + x (3) * A (3, 3)))...
/(x (1) * A (2, 1) + x (2) * A (2, 2) + x (3) * A (2, 3));
G (3)=exp (1 - x (1) * A (1, 3)/(x (1) * A (1, 1) + x (2) * A (1, 2) + x (3) * A (1, 3))...
- x (2) * A (2, 3)/(x (1) * A (2, 1) + x (2) * A (2, 2) + x (3) * A (2, 3))...
- x (3) * A (3, 3)/(x (1) * A (3, 1) + x (2) * A (3, 2) + x (3) * A (3, 3)))...
/(x (1) * A (3, 1) + x (2) * A (3, 2) + x (3) * A (3, 3));
% Vapor pressure using Antoine 's Equation
A (1)=20.64559; A (2)=23.49989; A (3)=20.71616;
B (1)=2125.74886; B (2)=3643.31362; B (3)=2571.58460;
C (1)=33.160; C (2)=33.434; C (3)=48.406;
(Psat) (1)=exp (A (1) + B (1)/(T + C (1)));
(Psat), (2)=exp (A (2) + B (2)/(T + C (2)));
(Psat) (3)=exp (A (3) + B (3)/(T + C (3)));
% Equilibrium constant for the reaction of isobutene + methanol==MTBE isobutylene + methanol MTBE reaction Equilibrium constant of the
Keq=exp (10.0982 + 0.2667 + 4254.05/T * log (T));
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