Monod Plotting Model for Biomass Growth
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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
Torsten
on 28 Jan 2023
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 ?
Regina
on 2 Feb 2023
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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