How to function 𝑎𝐴 + 𝑏𝐵 → 𝑝P in ODE89
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Navaneetha Krishnan Murugadoss
am 7 Mär. 2022
Kommentiert: Davide Masiello
am 7 Mär. 2022
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
am 7 Mär. 2022
Bearbeitet: Davide Masiello
am 7 Mär. 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 Kommentare
Davide Masiello
am 7 Mär. 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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