Question 5 of 6: Contois-Kinetics Chemostat with Cell Death
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams / EGBC — May 2017 — 04-Bio-A5 Enzyme and Microbial Kinetics. Three-hour open-book examination; any non-communicating calculator is permitted. Six questions constitute a complete paper (10 + 25 + 15 + 20 + 15 + 15 = 100 marks). Content spans quasi-steady-state derivation of a substrate-inhibited enzyme mechanism, diffusion-reaction in an immobilized-enzyme bead (Thiele modulus / effectiveness factor), the rapid-equilibrium integrated batch equation for substrate inhibition, a growth-associated-product batch fermentation, Contois-kinetics chemostat design, and fed-batch quasi-steady-state theory.
Reference texts: Bailey & Ollis, Biochemical Engineering Fundamentals (2nd ed.) — enzyme kinetics, substrate inhibition, immobilized-enzyme diffusion-reaction and effectiveness factors, Monod/Contois growth models, chemostat and fed-batch theory. All quantities are used exactly as printed on the exam (SI/CGS mixed units as given).
Question 5: Contois-Kinetics Chemostat with Cell Death (a. 10 marks; b. 5 marks)
Check: the paper prints the Contois denominator as a product, “$0.1\cdot X\cdot S$”, under which $S$ cancels algebraically and $\mu_g=6/X$ would be independent of substrate concentration entirely. Combined with the reactor's own substrate mass balance at $D=0.2\ \text{h}^{-1}$, that literal reading forces a negative (unphysical) reactor substrate concentration — a clear sign of a printing error in the original exam. The standard Contois form, explicitly named in the question stem and used by every textbook treatment, is additive: $$\mu_g=\mu_{\max}\frac{S}{K_{sx}X+S}, \qquad \mu_{\max}=0.6\ \text{h}^{-1},\ K_{sx}=0.1.$$ This corrected, dimensionally- and physically-consistent form is used below, keeping the printed numeric constants (0.6 and 0.1) unchanged.
Given.
Quantity
Symbol
Value
Contois max. specific growth rate
$\mu_{\max}$
0.6 h$^{-1}$
Contois constant
$K_{sx}$
0.1
Death-rate constant
$k_D$
0.1 h$^{-1}$
True growth yield
$Y_{X/S}^M$
0.4 g cells/g substrate
Feed substrate conc.
$S_f$
10 g/L
Dilution rate (part a)
$D$
0.2 h$^{-1}$
Find. (a) the reactor (= effluent) substrate concentration at $D=0.2\ \text{h}^{-1}$; (b) the maximum dilution rate before washout.
Fig. 3 — chemostat (CSTR, no cell recycle): feed at rate $D=F/V$ and concentration $S_f$; well-mixed effluent leaves at reactor concentrations $S$, $X$.
Approach. Apply the chemostat cell balance (growth must offset both dilution and death at steady state) to fix $\mu_g$, use the Contois relation to link $X$ and $S$, then close with the substrate mass balance; the washout condition is the limit $X\to0$.
(a) Steady-state cell balance. With no recycle and first-order death, $\mu_g X=(D+k_D)X$ at steady state, so $$\mu_g=D+k_D=0.2+0.1=0.3\ \text{h}^{-1}.$$
Contois relation between $X$ and $S$. $0.3=0.6S/(0.1X+S)\Rightarrow 0.1X+S=2S\Rightarrow 0.1X=S$, i.e. $$X=10S.$$
Substrate mass balance. $D(S_f-S)=\mu_g X/Y_{X/S}^M$; substituting $X=10S$: $$0.2(10-S)=\frac{0.3(10S)}{0.4}=7.5S \ \Rightarrow\ 2-0.2S=7.5S \ \Rightarrow\ S=\frac{2}{7.7}=\boxed{0.260\ \text{g/L}}$$ (with $X=10S=2.60$ g/L as a by-product check).
(b) Washout limit. As $X\to0$ the Contois denominator $0.1X+S\to S$, so $\mu_g\to\mu_{\max}S/S=\mu_{\max}$ regardless of $S$ — the biomass-dependence that makes Contois kinetics distinctive vanishes exactly where washout occurs. The steady-state cell balance then gives $$D_{\max}=\mu_{\max}-k_D=0.6-0.1=\boxed{0.5\ \text{h}^{-1}}.$$
