Figure 5.14:

Fractional error in the QSSA molar flowrate of C_2H_4 versus reactor volume.

Code for Figure 5.14

Text of the GNU GPL.

main.m


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% Copyright (C) 2001, James B. Rawlings and John G. Ekerdt
%
% This program is free software; you can redistribute it and/or
% modify it under the terms of the GNU General Public License as
% published by the Free Software Foundation; either version 2, or (at
% your option) any later version.
%
% This program is distributed in the hope that it will be useful, but
% WITHOUT ANY WARRANTY; without even the implied warranty of
% MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the GNU
% General Public License for more details.
%
% You should have received a copy of the GNU General Public License
% along with this program; see the file COPYING.  If not, write to
% the Free Software Foundation, 59 Temple Place - Suite 330, Boston,
% MA 02111-1307, USA.

% This program solves part of ethane pyrolysis example that is in the
% text.  It generates the error plot for the ethane example
% It is titled ethane_error.m
%
% It was last edited 2/5/97
%
% Revised 7/24/2018

%E in Joules, mass in grams, T in Kelvin, time in sec, volume in cm3

Components_1 = [' C2H6 = 1,',' CH3 = 2,',' CH4 = 3,',' C2H5 = 4'];
Components_2 = [' H = 5,',' C2H4 = 6,', ' H2 = 7,',' H2O = 8'];

Ao = [1e17,2e11, 3e14, 3.4e12, 1.6e13]';
Ea = [356000, 44000, 165000, 28000, 0]';
nu = [-1,2,0,0,0,0,0,0
      -1,-1,1,1,0,0,0,0
       0,0,0,-1,1,1,0,0
      -1,0,0,1,-1,0,1,0
       1,0,0,-1,-1,0,0,0];

R = 8.3144;  % Gas Constant (J/gmole-K)
p.R1 = 82.057; % Gas Constant (cc-atm/gmole-K)
p.T = 925; % Temp (K)
EXP = exp(-Ea/(R*p.T));
p.k = Ao.*EXP;
p.kp = (p.k(1)/(2*p.k(3)) + ((p.k(1)/(2*p.k(3)))^2 + ...
     ((p.k(1)*p.k(4))/(p.k(3)*p.k(5))))^0.5);
C1o = (50/760)/(82.057*p.T);  %gmole/cm3
C8o = (710/760)/(82.057*p.T);
Qf = 35.0;	%cc/sec
N1o = C1o*Qf;	%gmole/sec
N8o = C8o*Qf;
p.P = 1.0;	%atm
Initial = [N1o,0,0,0,0,0,0,N8o]';


v = [0:1:100]';

opts = odeset ('AbsTol', sqrt (eps), 'RelTol', sqrt (eps));
[tsolver, solution] = ode15s(@(v,x) rxrate(v,x,p),v,Initial,opts);
answer = [v solution];

Initial_q = [N1o, 0, 0, 0, N8o]';
% The total mole balance neglects H, CH3 and C2H5 because at 925K they
% are of order 10^-6 the concentration(molar flowrate) of ethane and
% ethane will only change by one order of magnitude.


[tsolver, solution_q] = ode15s(@(v,x) rate_q(v,x,p),v,Initial_q,opts);
answer_q = [v solution_q];

% Delete first row to avoid floating point exception.

Ethylene_exact = solution(2:end,6);
Ethylene_qss = solution_q(2:end,3);
err = abs((Ethylene_exact - Ethylene_qss)./Ethylene_exact);
erro = err*100;
%error = [v(2:length(v)), erro];
% JBR, 2/22/98
error = [v(2:length(v)), err];
save -ascii ethane_error.dat error;

if (~ strcmp (getenv ('OMIT_PLOTS'), 'true')) % PLOTTING
    semilogy (error(:,1),error(:,2))
    % TITLE
end % PLOTTING

rxrate.m


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function dNdv = rxrate(v,x,p)
    N1 = x(1);
    N2 = x(2);
    N3 = x(3);
    N4 = x(4);
    N5 = x(5);
    N6 = x(6);
    N7 = x(7);
    N8 = x(8);

    Ntot = N1 + N2 + N3 + N4 + N5 + N6 + N7 + N8;
    Ctot = p.P/(p.R1*p.T);
    C1 = (N1/Ntot)*Ctot;
    C2 = (N2/Ntot)*Ctot;
    C4 = (N4/Ntot)*Ctot;
    C5 = (N5/Ntot)*Ctot;

    r1 = p.k(1)*C1;
    r2 = p.k(2)*C1*C2;
    r3 = p.k(3)*C4;
    r4 = p.k(4)*C1*C5;
    r5 = p.k(5)*C4*C5;

    dNdv = zeros (8, 1);

    dNdv(1) = -r1 - r2 - r4 + r5;
    dNdv(2) = 2*r1 - r2;
    dNdv(3) = r2;
    dNdv(4) = r2 - r3 + r4 - r5;
    dNdv(5) = r3  - r4 - r5;
    dNdv(6) = r3;
    dNdv(7) = r4;
    dNdv(8) = 0;

rate_q.m


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function dNdv = rate_q(v,x,p)

    N1 = x(1);
    N3 = x(2);
    N6 = x(3);
    N7 = x(4);
    N8 = x(5);

    Ntot = N1 + N3 + N6 + N7 + N8;
    Ctot = p.P/(p.R1*p.T);
    C1 = (N1/Ntot)*Ctot;
    C2 = (2*p.k(1))/p.k(2);
    C4 = p.kp*C1;
    C5 = p.k(1)/(p.k(5)*p.kp);

    r1 = p.k(1)*C1;
    r2 = 2*p.k(1)*C1;
    r3 = p.k(3)*p.kp*C1;
    r4 = p.k(4)*p.k(1)/(p.k(5)*p.kp)*C1;
    r5 = p.k(1)*C1;

    dNdv = zeros (5, 1);

    %  dNdv(1) = -r1 - r2 - r4 + r5;
    dNdv(1) = - r2 - r4;
    dNdv(2) = r2;
    dNdv(3) = r3;
    dNdv(4) = r4;
    dNdV(5) = 0;