Continuous Fermentation with Cell Recycle (Self-Flocculating Yeast): 20-Bio-A5 Systems Analysis & Control, December 2017, Q5

Question 5: Continuous Fermentation with Cell Recycle (Self-Flocculating Yeast) (a. 10 marks; b. 5 marks; c. 5 marks; d. 5 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
Feed flow rate	$F$	400 mL/h
Reactor volume	$V$	1000 mL
Feed sugar concentration	$S_0$	10 g/L
Recycle flow rate	$\alpha F$	200 mL/h ($\alpha=0.5$)
Separator concentration factor	$C$	2 (recycle at $2X_1$)
Max. specific growth rate	$\mu_m$	0.5 h$^{-1}$
Monod constant	$K_s$	0.2 g/L
True yield	$Y_{x/s}^m$	0.4 g DCW/g sugar

Find. (a) substrate concentration $S_1$ exiting the bioreactor; (b) cell concentration $X_1$ within the reactor; (c) cell concentration $X_2$ in the separator's effluent (product) stream; (d) cell concentration in the recycle stream returned to the reactor.

Fig. 1 — bioreactor with cell recycle: fresh medium ($F,X_0,S_0$) and concentrated recycle ($\alpha F, C X_1$) enter the reactor; the combined stream $(1+\alpha)F$ at $X_1$ leaves into the cell separator, which splits it into a product stream ($F,X_2$) and the recycle back to the reactor.

Approach. A cell balance across the REACTOR (fed by fresh sterile medium plus concentrated recycle, no cell death) fixes the required steady-state specific growth rate directly from the flow and concentration ratios — independent of the kinetic form. Monod inversion then gives $S_1$, a substrate balance on the reactor gives $X_1$, and a cell balance across the SEPARATOR ALONE (no reaction occurs there) gives $X_2$ and the recycle concentration.

Check: assumes sterile fresh feed ($X_0=0$, not stated explicitly but standard for "medium") and that the cell separator only concentrates cells — it does not remove or add substrate, so the substrate concentration is the same ($S_1$) in the reactor, the recycle stream, and the product stream.

Flow bookkeeping. Recycle ratio $\alpha=200/400=0.5$; total flow leaving the reactor into the separator is $(1+\alpha)F=600$ mL/h.
Reactor cell balance (steady state, $X_0=0$, no death). $$0=\alpha F\,(CX_1)-(1+\alpha)F\,X_1+\mu X_1V \quad\Rightarrow\quad \mu_{ss}=\frac{F\big[(1+\alpha)-\alpha C\big]}{V}=\frac{400(1.5-1.0)}{1000}=\boxed{0.20\ \text{h}^{-1}}.$$
(a) Monod inversion for $S_1$. $$\mu_{ss}=\frac{\mu_mS_1}{K_s+S_1} \ \Rightarrow\ S_1=\frac{K_s\mu_{ss}}{\mu_m-\mu_{ss}}=\frac{0.2(0.20)}{0.5-0.20}=\boxed{0.133\ \text{g/L}}.$$
(b) Reactor substrate balance for $X_1$. $$F(S_0-S_1)=\frac{\mu_{ss}X_1}{Y_{x/s}^m}V \ \Rightarrow\ X_1=\frac{F(S_0-S_1)Y_{x/s}^m}{\mu_{ss}V}=\frac{400(9.867)(0.4)}{0.20(1000)}=\boxed{7.89\ \text{g/L}}.$$
(c)–(d) Separator cell balance (no reaction inside the separator). $$(1+\alpha)F\,X_1=F\,X_2+\alpha F\,(CX_1) \ \Rightarrow\ X_2=X_1\big[(1+\alpha)-\alpha C\big]=X_1(0.5)=\boxed{3.95\ \text{g/L}\ \text{(effluent)}},$$ $$\text{recycle concentration}=CX_1=2(7.89)=\boxed{15.79\ \text{g/L}}.$$

As a closure check, the total growth rate inside the reactor, $\mu_{ss}X_1V=0.20(7.893)(1.0\ \text{L})=1.579$ g/h, equals the net cell washout in the product stream, $FX_2=0.400(3.947)=1.579$ g/h — the overall cell mass balance closes exactly, confirming the recycle bookkeeping.

Result	Value
Steady-state specific growth rate $\mu_{ss}$	0.20 h$^{-1}$
(a) Substrate exiting reactor $S_1$	0.133 g/L
(b) Cell concentration in reactor $X_1$	7.89 g/L
(c) Cell concentration, separator effluent $X_2$	3.95 g/L
(d) Cell concentration, recycle stream $CX_1$	15.79 g/L

Concept & Theory

Topic. Continuous culture with cell recycle (perfusion-type chemostat).

Key relations. Reactor cell balance with recycle gives $\mu_{ss}=D\big[(1+\alpha)-\alpha C\big]$ where $D=F/V$; this replaces the ordinary chemostat relation $\mu_{ss}=D$.

Why this works. Cell recycle decouples the reactor's REQUIRED specific growth rate from the feed-based dilution rate $D=F/V$: because the separator returns cells at a concentration higher than $X_1$, the reactor can sustain a growth rate BELOW $D$ (here $\mu_{ss}=0.20\ \text{h}^{-1} \lt D=0.40\ \text{h}^{-1}$) without washing out, letting the system run at higher throughput/productivity for a given reactor volume than a once-through chemostat could.

Common pitfall. Assuming $\mu_{ss}=F/V$ as in an ordinary chemostat (ignoring recycle), or forgetting that the SEPARATOR conserves cell MASS (flow×concentration), not concentration itself — concentration necessarily drops across the effluent side and rises across the recycle side.

Source. Bailey & Ollis, Biochemical Engineering Fundamentals, Ch. 7 (continuous culture with cell recycle).

Question 5 of 6: Continuous Fermentation with Cell Recycle (Self-Flocculating Yeast)

Question 5: Continuous Fermentation with Cell Recycle (Self-Flocculating Yeast) (a. 10 marks; b. 5 marks; c. 5 marks; d. 5 marks)