Solving ODE 45 of chemical reaction mole balance dF/dV

Question

Proone 2025-7-18

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编辑： Alan Stevens 2025-7-27

Im trying to solve an ODE of dF/dV derived from a chemical reaction as shown below but the figure plotted does not seem to fit the answer of change in flow rate over the volume of the reactor as expected

close all

clear

clc

%Set Conditions

T=318;

P=15;

R=8.205745e-5;

C=P/(R*T);

%Rate Constant Functions

K=(0.076*T^2)/(exp(189*((1/T)-(1/90))));

k_1f=0.06*(exp(40*((1/502)-(1/T))));

k_2=0.00202*(exp(4030*((1/441)-(1/T))));

k_1b=k_1f/K;

%Set Parameter

para.k1f=k_1f;

para.k1b=k_1b;

para.k2=k_2;

para.T=T;

para.P=P;

para.R=R;

para.C=C;

%Initial Condition

F0=[1.5 3 0 0 2.1];

FT=sum(F0);

para.FT=FT;

%Volume of Reaction

Vspan=[0 10];

%Solve ODE

options = odeset('NonNegative',1:length(F0));

[V,F] = ode45(@myOde,Vspan,F0,options,para);

% Plot results

figure

plot(V,F,'LineWidth',1.5)

xlabel('Volume')

ylabel('Molar Flow Rate')

legend('C2H4','H2O','C2H5OH','C2H52O','N2')

function dFdV = myOde(V,F,para)

% Initialization

n=length(F);

dFdV=zeros(n,1); % initialize dydt as a column vector

% Assign state vectors to each column

C2H4=F(1);

H2O=F(2);

C2H5OH=F(3);

C2H52O=F(4);

N2=F(5);

% Parameters

k1f=para.k1f;

k1b=para.k1b;

k2=para.k2;

t=para.T;

p=para.P;

r=para.R;

c=para.C;

ft=para.FT;

% ODE-related expressions

r1f=(k1f*c^2*C2H4*H2O)/ft^2;

r1b=(k1b*c*C2H5OH)/ft;

r2=(k2*c*C2H5OH)/ft;

% ODEs

dFdV = zeros(5,1);

dFdV(1)=-r1f+r1b;

dFdV(2)=-r1f+r1b+((r2)/2);

dFdV(3)=r1f-r1b-r2;

dFdV(4)=((r2)/2);

dFdV(5)=0;

end

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Sam Chak 2025-7-18

Hi @Proone

What are the expected solutions for the five changes in flow rates over the volume of the reactor? Please show the closed-form solutions using equations or formulas so that we can test them for validating numerical solutions.

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Answer 1

Alan Stevens 2025-7-20

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2178566-solving-ode-45-of-chemical-reaction-mole-balance-df-dv#answer_1568186

编辑：Alan Stevens 2025-7-27

在 MATLAB Online 中打开

You had the units of R in m^3.atm/(mol.K), but should probably be in L.atm/(mol.K) , since other volumes are in L.

Does this look more like what you expect to see?

%Set Conditions

T=318; % K

P=15; % atm

R=8.205745e-2; % L.atm/(mol.K)

C=P/(R*T); % mol/L

%Rate Constant Functions

K=(0.076*T^2)/(exp(189*((1/T)-(1/90)))); % L/mol

k_1f=0.06*(exp(40*((1/502)-(1/T)))); % L/(mol.s)

k_2=0.00202*(exp(4030*((1/441)-(1/T)))); % 1/s

k_1b=k_1f/K; % 1/s

%Set Parameter

para.k1f=k_1f; % L/(mol.s)

para.k1b=k_1b; % 1/s

para.k2=k_2; % 1/s

para.T=T; % K

para.P=P; % atm

para.R=R; % L.atm/(mol.K)

para.C=C; % mol/L

%Initial Condition

F0=[1.5 3 0 0 2.1]; % mol/s

FT=sum(F0); % mol/s

para.FT=FT; % mol/s

%Volume of Reaction

Vspan=[0 10000]; % L

%Solve ODE

options = odeset('NonNegative',1:length(F0));

[V,F] = ode45(@myOde,Vspan,F0,options,para);

C2H4=F(:,1);

H2O=F(:,2);

C2H5OH=F(:,3);

C2H52O=F(:,4);

N2=F(:,5);

% Plot results

figure

subplot(2,2,1)

plot(V,C2H4,'LineWidth',1.5)

xlabel('Volume [L]')

ylabel('C2H4 Molar Flow Rate [mol/s]')

subplot(2,2,2)

plot(V,H2O,'LineWidth',1.5)

xlabel('Volume [L]')

ylabel('H2O Molar Flow Rate [mol/s]')

subplot(2,2,3)

plot(V,C2H5OH,'LineWidth',1.5)

xlabel('Volume [L]')

ylabel('C2H5OH Molar Flow Rate [mol/s]')

subplot(2,2,4)

plot(V,C2H52O,'LineWidth',1.5)

xlabel('Volume [L]')

ylabel('C2H52O Molar Flow Rate [mol/s]')

%legend('C2H4','H2O','C2H5OH','C2H52O','N2')

function dFdV = myOde(~,F,para)

% Initialization

% n=length(F);

% dFdV=zeros(n,1); % initialize dydt as a column vector

% Assign state vectors to each column mol/s

C2H4=F(1);

H2O=F(2);

C2H5OH=F(3);

C2H52O=F(4);

N2=F(5);

% Parameters

k1f=para.k1f; % L/(mol.s)

k1b=para.k1b; % 1/s

k2=para.k2; % 1/s

t=para.T; % K

p=para.P; % atm

r=para.R; % L.atm/(mol.K)

c=para.C; % mol/L

ft=para.FT; % mol/s

% ODE-related expressions [c/ft] = s/L

r1f=(k1f*c^2*C2H4*H2O)/ft^2; % (mol/s)/L

r1b=(k1b*c*C2H5OH)/ft; % (mol/s)/L

r2=(k2*c*C2H5OH)/ft; % (mol/s)/L

% ODEs

dFdV = zeros(5,1);

dFdV(1)=-r1f+r1b; % (mol/s)/L

dFdV(2)=-r1f+r1b+((r2)/2); % (mol/s)/L

dFdV(3)=r1f-r1b-r2; % (mol/s)/L

dFdV(4)=((r2)/2); % (m0l/s)/L

dFdV(5)=0; % (mol/s)/L

end

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Solving ODE 45 of chemical reaction mole balance dF/dV

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Solving ODE 45 of chemical reaction mole balance dF/dV

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