Batch Fermentation with a Growth-Associated Product: 20-Bio-A5 Systems Analysis & Control, May 2017, Q4

Question 4: Batch Fermentation with a Growth-Associated Product (a. 5 marks; b. 5 marks; c. 10 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Quantity	Symbol	Value
Reactor volume	$V$	10 L
Initial substrate conc.	$S_0$	20 g/L
Initial biomass conc.	$X_0$	0.1 g/L
Initial product conc.	$P_0$	0 g/L
Specific growth rate	$\mu_g$	$0.1/(P+1)\ \text{h}^{-1}$
True growth yield	$Y_{X/S}^M$	0.3 g cells/g substrate
Product yield	$Y_{P/X}$	0.2 g product/g cells

Cell death is negligible.

Find. (a) biomass concentration when all substrate is consumed; (b) product concentration at that point; (c) the batch time required.

Approach. Parts (a) and (b) follow directly from the stoichiometric yield coefficients once all 20 g/L of substrate is consumed — no kinetics needed. Part (c) requires substituting the growth-associated product $P(X)=Y_{P/X}(X-X_0)$ into $\mu_g$ so the batch growth equation becomes a separable ODE in $X$ alone, then integrating from $X_0$ to the part-(a) endpoint.

(a) Final biomass from the true growth yield. All substrate is consumed ($\Delta S=S_0=20$ g/L), and with no death the true yield applies directly: $$\Delta X=Y_{X/S}^M\,\Delta S=0.3(20)=6.0\ \text{g/L} \ \Rightarrow\ X_f=X_0+\Delta X=0.1+6.0=\boxed{6.1\ \text{g/L}}.$$
(b) Final product from the product yield. $$P_f=Y_{P/X}(X_f-X_0)=0.2(6.1-0.1)=0.2(6.0)=\boxed{1.2\ \text{g/L}}.$$
(c) Substitute the growth-associated product into $\mu_g$. Because $P$ tracks $X$ directly, $P(t)=Y_{P/X}\big(X(t)-X_0\big)=0.2X-0.02$, so $$\mu_g=\frac{0.1}{P+1}=\frac{0.1}{0.98+0.2X}\ \text{h}^{-1}, \qquad \frac{dX}{dt}=\mu_g X=\frac{0.1X}{0.98+0.2X}.$$
Separate and integrate. $$\frac{(0.98+0.2X)}{0.1X}\,dX=dt \ \Rightarrow\ \left(\frac{9.8}{X}+2\right)dX=dt \ \Rightarrow\ t=9.8\ln\frac{X_f}{X_0}+2(X_f-X_0).$$
Evaluate from $X_0=0.1$ to $X_f=6.1$ g/L. $$t=9.8\ln\frac{6.1}{0.1}+2(6.1-0.1)=9.8(4.111)+2(6.0)=40.29+12.00=\boxed{52.3\ \text{h}}.$$ The long batch time is a direct consequence of $\mu_g$ falling as the growth-associated product accumulates — growth self-decelerates over the run.

Result	Value
(a) Final biomass $X_f$	6.1 g/L
(b) Final product $P_f$	1.2 g/L
(c) Batch time $t$	52.3 h

Concept & Theory

Topic. Batch fermentation with a growth-associated (Luedeking–Piret type) product that feeds back on the specific growth rate.

Key relations. Yield-coefficient endpoints $\Delta X=Y_{X/S}^M\Delta S$, $\Delta P=Y_{P/X}\Delta X$; batch growth ODE $dX/dt=\mu_g(P)X$ with $P$ eliminated via the growth-associated yield relation.

Why this works. Because the product is strictly growth-associated (no maintenance or non-growth production term), $P$ is not an independent state — it is a fixed linear function of $X$ at every instant, which is exactly what collapses the coupled $(X,P)$ system down to a single separable ODE in $X$.

Common pitfall. Trying to integrate $\mu_g$ against $t$ directly without first eliminating $P(X)$ (mixing an implicit unknown into the separation) or reusing $X_0$'s numeric value inside the constant term without re-deriving it (here $1-Y_{P/X}X_0=0.98$, not 1).

Source. Bailey & Ollis, Biochemical Engineering Fundamentals, Ch. 6 (batch fermentation kinetics, growth-associated product formation, yield coefficients).

Question 4 of 6: Batch Fermentation with a Growth-Associated Product

Question 4: Batch Fermentation with a Growth-Associated Product (a. 5 marks; b. 5 marks; c. 10 marks)