How to function 𝑎𝐴 + 𝑏𝐵 → 𝑝P in ODE89

2022 3 7
1 个回答
更新时间：2022 3 7
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𝑎𝐴 + 𝑏𝐵 → 𝑝P
𝑑𝐴/𝑑𝑡 = −𝐾 ∗ 𝐴 ∗ 𝐵 𝑑𝐵/𝑑𝑡 = (𝑏/𝑎) ∗ (𝑑𝐴/𝑑𝑡) = −𝑌𝐵 ∗ (𝐾 ∗ 𝐴 ∗ 𝐵) 𝑑𝑃/𝑑𝑡 = −(𝑝/𝑎) ∗ (𝑑𝐴/𝑑𝑡) = 𝑌𝑃 ∗ (𝐾 ∗ 𝐴 ∗ 𝐵)

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Davide Masiello
Davide Masiello 2022-3-7
编辑：Davide Masiello 2022-3-7

0 个投票

This should work:
clear,clc
tspan = [0,10];
y0 = [1,1,0];
[t,y] = ode89(@yourODEsystem,tspan,y0);
plot(t,y)
legend('A','B','P','Location','best')
function out = yourODEsystem(t,y)
% Coefficients
K = 1;
a = 2;
b = 1;
p = 0.5;
% Variables
A = y(1);
B = y(2);
P = y(3);
% Time derivatives
dAdt = -K*A*B;
dBdt = -(b/a)*K*A*B;
dPdt = (p/a)*K*A*B;
% Output
out = [dAdt;dBdt;dPdt];
end
Just replace you actual values of stoichiometric coefficients and kinetic constants.

6 个评论
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Torsten
Torsten 2022-3-7
@Navaneetha Krishnan Murugadoss
Have in mind that chemical kinetics ODEs are usually stiff.
Use ode15s instead of ode89.
Error in TASK>odefcn (line 16)
A = y(1);
Error in TASK (line 6)
[t,y] = ode89(odefcn,tspan,y0);
Davide Masiello
Davide Masiello 2022-3-7
编辑：Davide Masiello 2022-3-7
Please show all your code. Moreover, please indicate what version of Matlab you are using.
clear,clc
tspan = [0,12];
y0=[0 1 3];
[t,y] = ode89(DEdef,tspan,y0);
plot(t,y)
legend('CL','NOM','DBP','Location','best')
function [Ddv_div] = odefcn(t,y)
% Coefficients
K = 5E-5;
YB=1;
YP=0.15;
% Variables
A = y(1);
B = y(2);
P = y(3);
% Time derivatives
dAdt = -K*A*B;
dBdt = -YB*(K*A*B);
dPdt = YP*(K*A*B);
Ddv_div = [-K*A*B;-YB*(K*A*B);YP*(K*A*B)];
% Output
out = [dAdt;dBdt;dPdt];
end
2021b version
I need to plot concentrations A,B, and P against time. I have to show how the concentrations change over time until ether A is completely depleted
Davide Masiello
Davide Masiello 2022-3-7
The function call in ode89 must be equal to the function name. Write this
clear,clc
tspan = [0,12];
y0=[0 1 3];
[t,y] = ode89(@DEdef,tspan,y0);
plot(t,y)
legend('CL','NOM','DBP','Location','best')
function Ddv_div = DEdef(t,y)
% Coefficients
K = 5E-5;
YB=1;
YP=0.15;
% Variables
A = y(1);
B = y(2);
P = y(3);
% Output
Ddv_div = [-K*A*B;-YB*(K*A*B);YP*(K*A*B)];
end
However, let me point out that if the initial concentration of one of the two reactants is zero (like in your case) you won't observe any change in the concentration of any of the compounds, since the reaction cannot occur.

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