Monod Plotting Model for Biomass Growth

Hello! I am trying to plot graphs biomass vs time from these kinetic equations below:
with So=3 g/L (initial substrate concentration)
μmax = 0.00527
Ks = 0.00159
KI = 22.0755
I want to plot these parameters to the kinetic model to get plot data like these data:
Time(day)| 1 2 3 4 5 6 7 8 9 10
Biomass | 0.136808 0.182631 0.2066696 0.302823 0.362919 0.4132496 0.414752 0.467336 0.569499 0.844438
I have tried the code like this, but it is still error. Any little suggestion or solutions or comments are very welcomed! Thank you so much!
function VariasiFerro_Monod
c0=[0.136808 3];
Miumax= 0.005272;
Ks=0.00159;
Ki=22.0755;
tspan=0:1:10;
[t c]=ode45('monodkinetic',tspan,c0,Miumax,Ks,Ki)
result=[t c]
plot(t,c)
xlabel('Waktu(hari)')
ylabel('Biomassa(g/L)')
end
function dcdt=monodkinetic(t,c,Miumax,Ks,Ki)
mu = (Miumax*c(2))/(Ks+c(2)+(c(2)^2/Ki));
dcdt(1,:) = mu*c(1);
end
and also I tried to code like this
function VariasiFerro_Monod
Rentang = linspace(0, 10, 50);
C0 = [0.136808 3];
Miumax= 0.005272;
Ks=0.00159;
Ki=22.0755;
[t,C]=ode45(@(t,C)VariasiFe(t,C,Miumax,Ks,Ki),Rentang,C0);
figure
plot(t,C(:,1));
ylabel('Biomassa (g/L)')
xlabel('Waktu (hari)')
grid
legend('Biomassa','Location','best')
end
function dCdt=VariasiFe(t,C,Miumax,Ks,Ki)
mu = (Miumax*C(2))/(Ks+C(2)+(C(2)^2/Ki));
dCdt(1,:) = mu*C(1);
end

3 Comments

I guess you want to solve
dS/dt = mu
?
Then why does your vector c0 has 2 instead of only 1 element, namely the initial value for S ?
Hi, thank you for your comment, I really appreciate it.
I actually want to plot biomass to time, where biomass growth rate (dX/dt) is the specific growth rate times biomass concentration (dX/dt = mu.X). The mu itself is following the monod equation (mu = mumax . S/(Ks+S)). Therefore, there are two variables (X and S) which means it needs two initial values as well.
Please let me know if you need more explanation on this. Thank you for reaching out!
Please let me know if you need more explanation on this.
You have two equations
dX/dt = mu*X
mu = mumax * S/(Ks+S)
with three unknowns mu, S and X.
So you need a third (maybe differential) equation for S (your C(2) in the function "VariasiFe")

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Asked:

on 28 Jan 2023

Edited:

on 2 Feb 2023

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