PARAMETER estimation of kinetic and adsorbtion constant of langmuir hinshelwood haugen watson model

please help me calculated parameter for different model atttached in literature. i tried for model E but could not get my parameter values close to anywhere mentioned in literature. Ch2 is consant = 2.47e-3

star strider please help me

Answer 1

[GOVIND BHANDARI]

function model4

% 2016 12 03

% NOTES:

%

% 1. The ‘theta’ (parameter) argument has to be first in your

% ‘kinetics’ funciton,

% 2. You need to return ALL the values from ‘DifEq’ since you are fitting

% all the values

function C=kinetics(theta,t)

c0=[7.00E-02;0.00E+00;0.00E+00;0.00E+00];

[T,Cv]=ode45(@DifEq,t,c0);

%

function dC=DifEq(t,c)

dcdt=zeros(4,1);

dcdt(1)=(-theta(1)*c(1)*0.00273-theta(2)*c(1)*0.00273)/(1+(theta(6)*c(1)+theta(7)*c(2)+theta(8)*c(3)+theta(9)*c(4)+theta(10)*0.00273))^3;

dcdt(2)= (theta(1)*c(1)*0.00273-theta(3).*c(2)*0.00273-theta(5)*c(2))/(1+(theta(6)*c(1)+theta(7)*c(2)+theta(8)*c(3)+theta(9)*c(4)+theta(10)*0.00273))^3;

dcdt(3)= (theta(2)*c(1)*0.00273-theta(4)*c(2)*0.00273+theta(5)*c(2))/(1+(theta(6)*c(1)+theta(7)*c(2)+theta(8)*c(3)+theta(9)*c(4)+theta(10)*0.00273))^3;

dcdt(4)= (theta(3)*c(2)*0.00273+theta(4)*c(3)*0.00273)/(1+(theta(6)*c(1)+theta(7)*c(2)+theta(8)*c(3)+theta(9)*c(4)+theta(10)*0.00273))^3;

dC=dcdt;

end

C=Cv;

end

t=[20

40

60

90

120

180

240];

c=[5.69E-02 1.36E-03 8.25E-03 3.48E-03

4.28E-02 9.79E-03 9.50E-03 7.94E-03

2.93E-02 2.85E-03 1.53E-02 2.26E-02

1.46E-02 4.69E-03 2.00E-02 3.28E-02

1.19E-03 3.52E-03 1.10E-02 5.43E-02

1.75E-03 0.00E+00 0.00E+00 6.10E-02

1.00E-09 1.00E-09 1.00E-09 7.00E-02];

theta0=[1.86e-1,1.70e-1,1.28,1.51e-1,1.0e-1,15.90,2.60e-1,1.87e-2,3.25,2.33e-2];

[theta,Rsdnrm,Rsd,ExFlg,OptmInfo,Lmda,Jmat]=lsqcurvefit(@kinetics,theta0,t,c);

fprintf(1,'\tRate Constants:\n')

for k1 = 1:length(theta)

fprintf(1, '\t\tTheta(%d) = %8.5f\n', k1, theta(k1))

end

tv = linspace(min(t), max(t));

Cfit = kinetics(theta, tv);

figure(1)

plot(t, c, 'p')

hold on

hlp = plot(tv, Cfit);

hold off

grid

xlabel('Time')

ylabel('Concentration')

legend(hlp, 'C_1(t)', 'C_2(t)', 'C_3(t)', 'C_4(t)', 'Location','N')

end

THE PROBLEM IS ORANGE STAR AND RED LINE ARE NOT FITTING THE TOP SHOULD BE AROUND 0.02

Answer 2

[Alex Sha]

if want all parameters to be positive:

Root of Mean Square Error (RMSE): 0.0035717712315911
Sum of Squared Residual: 0.00035721139246301
Correlation Coef. (R): 0.892442040492228
R-Square: 0.796452795637932

Parameter          	Best Estimate
--------------------	-------------
theta1	17.172493498211
theta2	21.5160808499238
theta3	38.9519219861006
theta4	51.8990082731777
theta5	5.17756884421286E-16
theta6	17.526757560822
theta7	6.33562009831634E-21
theta8	1.27030055679559E-14
theta9	9.92523484202955
theta10	11.9701814364323

while, if all parameters are free ranges, the solutions will be not unique, one of them like below:

Root of Mean Square Error (RMSE): 0.00271818383364327
Sum of Squared Residual: 0.000206878653897429
Correlation Coef. (R): 0.894998138423552
R-Square: 0.801021667781624

Parameter          	Best Estimate
--------------------	-------------
theta1	0.306719359089766
theta2	0.808060248987747
theta3	4.89340760623797
theta4	-0.0316673300900705
theta5	-0.00965914856097772
theta6	-79.2660953582632
theta7	-52.8331941956074
theta8	-94.6702336579354
theta9	-83.3482760040795
theta10	1914.31778883528