How to function 𝑎𝐴 + 𝑏𝐵 → 𝑝P in ODE89

8 vues (au cours des 30 derniers jours)
Navaneetha Krishnan Murugadoss
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)

Réponses (1)

Davide Masiello
Davide Masiello le 7 Mar 2022
Modifié(e) : Davide Masiello le 7 Mar 2022
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 commentaires
Navaneetha Krishnan Murugadoss
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 le 7 Mar 2022
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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