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Chapter 19

Q. 19.13

A dynamic model of a continuous-flow, biological chemostat has the form

\begin{aligned}\dot{X} &=0.063 C-D x \\\dot{C} &=0.9 S[X-C]-0.7 C-D C \\\dot{S} &=-0.9 S[X-C]+D[10-S]\end{aligned}

where X is the biomass concentration, S is the substrate concentration, and C is a metabolic intermediate concentration. The dilution rate, D, is an independent variable, which is defined to be the flow rate divided by the chemostat volume.

Determine the value of D, which maximizes the steady-state production rate of biomass, f, given by

f=D X

Step-by-Step

Verified Solution

Assuming steady state behavior, the optimization problem is,

\max f=D e

Subject to

\begin{aligned}&0.063 c-D e=0              (1)\\&0.9 s e-0.9 s c-0.7 c-D c=0          (2)\end{aligned}

\begin{gathered}-0.9 s e+0.9 s c+10 D-D s=0             (3) \\D, e, s, c \geq 0\end{gathered}

where f=f(D, e, c, s)

Excel-Solver is used to solve this problem,

c D e s
Initial values 1 1 1 1
Final values 0.479031 0.045063 0.669707 2.079784
max f = 0.030179
Constraints
0.063 c –D e 2.08E-09
0.9 s e – 0.9 s c – 0.7 c – Dc -3.10E-07
-0.9 s e + 0.9 s c + 10D – Ds 2.88E-07

Table S19.13. Excel solution

Thus the optimum value of D is equal to 0.045 h ^{-1}