Applying the Nonlinear Least Squares Method to Minimize the Objective Function to Find the Parameters of the Equation
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The liquid-liquid equilibrium data were fitted using the NRTL equation. The equation parameters of NRTL are derived from the experimental data.
below is my code:
clc,clear
options=optimset('MaxIter',4000,'MaxFunEvals',2000000,'algorithm','levenberg-marquardt');
b0=[0,0,0,0,0,0];
[b,gamma1,gamma2,gamma3,gamma4,gamma5,gamma6]=lsqnonlin('NRTL',b0,[],[],options);
Although the code gives the result, the result is not what I want. And the format of the output gamma is also wrong. It should be a matrix of the same order as x1, but it has become something I don't understand. Can someone please give a reason? It would be best to give some solutions. Thanks!
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Matt J
2022 年 7 月 4 日
編集済み: Matt J
2022 年 7 月 4 日
Your code makes strange assumptions about the output syntax of lsqnonlin.
Why do you think lsqnonlin will return values for the gamma variables in NRTL?
6 件のコメント
Torsten
2022 年 7 月 4 日
編集済み: Torsten
2022 年 7 月 4 日
If you want
x(:,1)+x(:,2)+x(:,3)-1.0 = 0;
x(:,4)+x(:,5)+x66-1.0 = 0
to be satisfied exactly and if you want x >= 0, use
options=optimset('MaxIter',40000,'MaxFunEvals',2000000,'algorithm','levenberg-marquardt');
x0=0.1.*ones(5,5);
lb = zeros(5,5);
ub = ones(5,5);
x=lsqnonlin('NRTLyan',x0,lb,ub,options);
together with the x(:,:) squared in your equations in function NRTLyan.
If you want x>= 0 and if it suffices if the relations
x(:,1)+x(:,2)+x(:,3)-1.0 = 0;
x(:,4)+x(:,5)+x66-1.0 = 0
are only approximately satisfied, use
options=optimset('MaxIter',40000,'MaxFunEvals',2000000,'algorithm','levenberg-marquardt');
x0=0.1.*ones(5,5);
lb = zeros(5,5);
ub = ones(5,5);
x=lsqnonlin('NRTLyan',x0,lb,ub,options);
together with
f=[x(:,1).*gamma1-gamma4.*x(:,4);x(:,2).*gamma2-gamma5.*x(:,5);x(:,3).*gamma3-gamma6.*x66;x(:,1)+x(:,2)+x(:,3)-1.0;x(:,4)+x(:,5)+x66-1.0];
If you want the relations
x(:,1)+x(:,2)+x(:,3)-1.0 = 0;
x(:,4)+x(:,5)+x66-1.0 = 0
to be satisfied exactly, but x>= 0 does not matter, only solve for x(:,i) for i=1,2,4 and calculate x(:,j) for j=3,5 from the relation.
Matt J
2022 年 7 月 4 日
Do you have any other suggestions?
@yu zhang I suggest you Acccept-click this answer, since your original question appears to have been resolved.
Since you have a new question and since it is related to code different from this question, I suggest you post that in a new thread.
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