Problem Statement: The growth of an organism is limited by the availability of a particular nutrient, which at high concentrations also inhibits cell growth (Andrew's inhibition kinetics). We wish to cultivate this culture in a continuous bioreactor.
Solution:
dx/dt = (μ-D)*x
ds/dt = D*(sf-s) - μ*x/Y
where μ = μm*s/(K+s+Ki*s2)
Structured/Unstructured Segregated/Nonsegregated Stochastic/Deterministic
Solution: Unstructured, nonsegregated, & deterministic.
Solution: D and sf
Solution:
0 = dx/dt = (μ-D)*x
0 = ds/dt = D*(sf-s) - μ*x/Y
There are three distinct regions:
1. For Dmax<D, the only steady-state is washout: x=0 & s=sf
where Dmax=um/(2*sqrt(K+Ki)+1) is the maximum in the specific growth rate
This occurs at smax=sqrt(K/Ki)
2. For D<μsf<Dmax and sf<smax, there are two steady-states.
Washout steady-state x=0 & s=sf
Growth steady-state x=Y*(sf-s) & s=μ-1(D) du(s)/ds>0
3. Otherwise, there are three steady states:
Washout steady-state x=0 & s=sf
Growth steady-state #1 x=Y*(sf-s) & s=μ-1(D) du(s)/ds>0
Growth steady-state #2 x=Y*(sf-s) & s=μ-1(D) du(s)/ds<0
See the following file for detail.
The eventual steady-state is determined by the following three factors:
D, sf, and the initial condition.
Solution:
Step 1. Derive dynamic equations for the system: dx/dt=f(t,x)
Step 2. Find steady-state by setting dx/dt=0
Step 3. Linearize the dynamic equation around the steady-state
Step 4. Find the eigenvalues and eigenvectors
Step 5. Determine localized stability based on the signs of the eigenvalues;
determine trends based on the directions given by the eigenvectors
Solution:
Stable ones:
For D > Dmax, x=0 & s=sf
For D<μsf, x=Y*(sf-s) & s=μ-1(D) dμ(s)/ds>0
Others are unstable.
Solution:
Find the point at which D*x is maximum. Substrate inhibition
is characterized by a point where the specific growth rate
exhibits a maximum as the substate concentration increases.
Let's call this point smax and μmax. When
there is substrate inhibition, the point at which D*x is maximum
coincides with the maximum in the specific growth rate. If
sf<smax, this point is the "hill-top" of
the specific growth rate versus s curve, i.e., μmax.
If sf>smax, this point is
μ(sf). We do not necessariy want to operate at this
point, because it sits on the edge of a "cliff", too close to the
washout condition. The biomass productivity declines very
abruptly once we operate past that point.
Solution: Shift to a lower value because the term -D*sf will pull the original biomass productivity curve D*x downward, more the higher the D. The same curve is pulled downward further the higher the cost of the substrate.
Solution:
Operate at a high sf. In practice, the upper limit
will be limited, among other factors, by the substrate's
solubility. Several files in the class' Handout section deal
with optimization with respect to sf. For example,
see:
--------------------------------------------------------- Mode Consider? Productivity? --------------------------------------------------------- Batch Y/N H/L Fed-batch Y/N H/L Continuous Y/N --- (baseline) Cell recycle Y/N H/L Reactors-in-series Y/N H/L Immobilized cell bioreactor Y/N H/L ---------------------------------------------------------
Solution:
Other than an immobilized cell bioreactor, which is totally
unsuited for biomass production, all other modes of operation,
each with its own advantages and disadvantages, can be
considered. The solution is bold-faced below.
---------------------------------------------------------
Mode Consider? Productivity?
---------------------------------------------------------
Batch Y/N H/L
Fed-batch Y/N H/L
Continuous Y/N --- (baseline)
Cell recycle Y/N H/L
Reactors-in-series Y/N H/L
Immobilized cell bioreactor Y/N H/L (not even a candidate!)
---------------------------------------------------------
Solution: Cell recycle.
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