Asked by Sean Powers
on 16 Apr 2019

I'm trying to optimize a complex reaction system to have the largest output of desired product C by varying the temperature and pressure of the reaction in an adiabatic PFR. I've created the design equation of the reaction:

Function dYdV = Designequation(v,y,T,P)

Fa = y(1);

Fb = y(2);

Fc = y(3);

Fd = y(4);

% Explicit equations

T = T(1);

P = P(1);

Cto = P / (8.314 * 10^-5) / T;

E1 = 15000;

E2 = 17500;

Ft = Fa + Fb + Fc + Fd;

Ca = Cto * Fa / Ft;

Cb = Cto * Fb/Ft;

Cc = Cto * Fc/Ft;

Cd = Cto * Fd/Ft;

k1 = 0.075 * exp(E1 /1.987 * (1/300 - (1/T)));

k2 = 0.0015 * exp(E2 / 1.987 * (1 / 300 - (1 / T)));

Fao = 50;

ra = -k1*Ca*Cb;

rb = -k2*Cb*Cc;

% Differential equations

dFadV = ra;

dFbdV = (2*ra)+rb;

dFcdV = rb-ra;

dFddV = -rb;

dYdV = [dFadV; dFbdV; dFcdV; dFddV];

The design equation is solved for volume and the initial molar flow rates of A and B using ode113 with an initial temperature of 300K and 1 atm:

Vspan = [0 2]; % Range for the volume of the reactor

y0 = [50; 25; 0; 0]; % Initial values for the dependent variables i.e Fa,Fb,and Fc, Fd

T = 300;

P = 1;

[v, y] =ode113(@(v,y) Designequation(v,y,T,P),Vspan,y0);

dFadT = y(:,1);

dFbdT =y(:,2);

dFcdT =y(:,3);

dFddT = y(:,4);

plot(v,dFcdT); hold on

Now I want to find the largest value of the molar flow rate of C (dFcdT), by varying the temperature and pressure, but I haven't found a way that I can implement that based on searches of the MatLab forum.

I apologize if this has been answered in other forum posts, but I'm desperate.

Answer by Torsten
on 16 Apr 2019

Edited by Torsten
on 16 Apr 2019

Accepted Answer

Although I guess that there are bounds on P and T, here is a code that will give you a start.

In case you want to constrain P and T, use "fmincon" instead of "fminsearch".

function main

format long

T = 300;

P = 1;

x0 = [T P];

options = optimset('MaxFunEvals',100000,'MaxIter',100000)

x = fminsearch(@fun,x0,options)

end

function obj = fun(x)

T = x(1);

P = x(2);

Vspan = [0 2]; % Range for the volume of the reactor

y0 = [50; 25; 0; 0]; % Initial values for the dependent variables i.e Fa,Fb,and Fc, Fd

[v, y] =ode15s(@(v,y) Designequation(v,y,T,P),Vspan,y0);

obj = -y(end,3)

end

function dYdV = Designequation(v,y,T,P)

Fa = y(1);

Fb = y(2);

Fc = y(3);

Fd = y(4);

% Explicit equations

Cto = P / (8.314 * 10^-5) / T;

E1 = 15000;

E2 = 17500;

Ft = Fa + Fb + Fc + Fd;

Ca = Cto * Fa / Ft;

Cb = Cto * Fb/Ft;

Cc = Cto * Fc/Ft;

Cd = Cto * Fd/Ft;

k1 = 0.075 * exp(E1 /1.987 * (1/300 - (1/T)));

k2 = 0.0015 * exp(E2 / 1.987 * (1 / 300 - (1 / T)));

Fao = 50;

ra = -k1*Ca*Cb;

rb = -k2*Cb*Cc;

% Differential equations

dFadV = ra;

dFbdV = (2*ra)+rb;

dFcdV = rb-ra;

dFddV = -rb;

dYdV = [dFadV; dFbdV; dFcdV; dFddV];

end

Sean Powers
on 22 Apr 2019

Modify the function Designequation given your differential equations and reaction rates and then run iterate given a temperature and pressure range and the amount of steps you want to take in those ranges. i.e.

function dYdV = Designequation(v,y,T,P)

Fa = y(1);

Fb = y(2);

Fc = y(3);

Fd = y(4);

% Explicit equations

T = T(1);

P = P(1);

Cto = P / (8.314 * 10^-5) / T;

E1 = 15000;

E2 = 17500;

Ft = Fa + Fb + Fc + Fd;

Ca = Cto * Fa / Ft;

Cb = Cto * Fb/Ft;

Cc = Cto * Fc/Ft;

Cd = Cto * Fd/Ft;

k1 = 0.075 * exp(E1 /1.987 * (1/300 - (1/T)));

k2 = 0.0015 * exp(E2 / 1.987 * (1 / 300 - (1 / T)));

Fao = 50;

rA = -(rA1 + rA2)

rB = -(rB1 + rB2 + rB3)

rC = rC1 + rC3

rD = rD1 + rD2 + rD3

rE = rE2 + rE3

% Differential equations

d(FA)/d(V) = rA

d(FB)/d(V) = rB

d(FC)/d(V) = rC

d(FD)/d(V) = rD

d(FE)/d(V) = rE

d(FF)/d(V) = rF

dYdV = [d(FC)/d(V)];

end

I didnt fully modify the script for your problem, but its something like that

Torsten
on 23 Apr 2019

@Sean Powers:

And did the exaustive search lead to a different result than the usual optimization with "fminsearch" ?

Bosnian Kingdom
on 24 Apr 2019

Thank you, I'll try.

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## Torsten (view profile)

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## Sean Powers (view profile)

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