Question 2 of 6: Comparison of Bioreactor Configurations

Question 2: Comparison of Bioreactor Configurations (20 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.

Find. Substrate conversion (S_in−S_out)/S_in for configurations 1a, 1b and 1c, and a comparison of the three.

Approach. Each stage is a well-mixed CSTR; solve the standard sterile-feed chemostat equations wherever a stage is fed X_in=0, use the washout criterion D>μ_max⇒X=0 where it applies, and for stage 2 of configuration 1c — which is fed by a live (non-sterile) mixed stream — use the general CSTR balance for an inoculated feed.

[Figure not reproduced: Fig. 2 — The three bioreactor networks analysed (redrawn from Figs. 1a–1c of the question). See the official exam paper.]

1. Configuration 1a — single CSTR. Dilution rate D_1a=F/V₁=0.015/0.1=0.15 h^-1, which is below μ_max=0.2 h^-1, so a normal (non-washout) steady state exists. Setting μ=D in the Monod equation and solving for S: $$S=\frac{D\,K_S}{\mu_{max}-D}=\frac{0.15(2)}{0.2-0.15}=\boxed{6.00\ \text{kg/m}^3}$$ $$X=Y_{X/S}(S_{in}-S)=0.5(10-6)=2.00\ \text{kg/m}^3\qquad \text{conversion}=\frac{S_{in}-S}{S_{in}}=\frac{10-6}{10}=\boxed{40.0\%}$$
2. Configuration 1b — two stages in series, no bypass. Both stages have the same flow F and volume V₂, so each has D_1b=F/V₂=0.015/0.05=0.30 h^-1, which now EXCEEDS μ_max=0.2 h^-1. Stage 1 is fed sterile medium (X_in=0); since D>μ_max no steady growth is possible there, so the culture washes out: $$X_1=0,\qquad S_1=S_{in}=10\ \text{kg/m}^3$$ Stage 2 then receives exactly the same sterile-equivalent feed (X_in,2=X₁=0, S_in,2=S₁=10 kg/m³) at the same D=0.30 h^-1>μ_max, so it washes out too: $$X_2=0,\qquad S_2=10\ \text{kg/m}^3\qquad\Rightarrow\qquad\text{conversion}=\boxed{0\%}$$ Splitting the SAME total volume (2V₂=V₁) into two smaller reactors in series, with no downstream inoculation, is catastrophic: each stage's individual dilution rate is now too high to sustain growth at all.
3. Configuration 1c, stage 1 — identical to 1a. Only half the feed (F/2) enters stage 1, so its dilution rate is D_1c=(F/2)/V₂=(0.0075)/0.05=0.15 h^-1 — algebraically identical to D_1a, because halving the flow while halving the volume leaves the ratio unchanged. Stage 1 is sterile-fed, so it reproduces 1a's steady state exactly: $$S_1=6.00\ \text{kg/m}^3,\qquad X_1=2.00\ \text{kg/m}^3$$
4. Configuration 1c, mixing point. The remaining F/2 bypasses stage 1 as fresh sterile feed (S_in=10, X=0) and joins stage 1's effluent before stage 2, in equal flow proportion, so the combined stream is the simple flow-weighted average: $$S_{mix}=\frac{S_1+S_{in}}{2}=\frac{6+10}{2}=8.00\ \text{kg/m}^3\qquad X_{mix}=\frac{X_1+0}{2}=\frac{2}{2}=1.00\ \text{kg/m}^3$$
5. Configuration 1c, stage 2 — inoculated CSTR. Stage 2 sees the FULL flow F, so D₂=F/V₂=0.30 h^-1>μ_max again — but it is NOT sterile-fed (X_mix=1>0), so the sterile-feed washout criterion does not apply here. The general steady-state balances for an inoculated CSTR are $$X_2(D_2-\mu_2)=D_2\,X_{mix}\qquad\text{(biomass)}\qquad\qquad Y_{X/S}D_2(S_{mix}-S_2)=\mu_2 X_2\qquad\text{(substrate)}$$ with μ₂=μ_maxS₂/(K_S+S₂). Trying the ansatz S₂=S₁=6 gives μ₂=μ(S₁)=D_1c=0.15 h^-1 (by definition of stage 1's own steady state), and since D₂=2D_1c and X_mix=X₁/2, the biomass balance gives X₂=D₂X_mix/(D₂−μ₂)=2D_1c(X₁/2)/D_1c=X₁=2, and the substrate balance closes using stage 1's own relation X₁=Y_X/S(S_in−S₁). Both equations are satisfied exactly: $$S_2=\boxed{6.00\ \text{kg/m}^3},\qquad X_2=\boxed{2.00\ \text{kg/m}^3},\qquad \text{conversion}=\frac{10-6}{10}=\boxed{40.0\%}$$

Comparison: configurations 1a and 1c achieve identical outlet conversion (40.0%), even though 1c splits the same total working volume across two smaller vessels. Configuration 1b — the same split volume WITHOUT the feed bypass — fails completely (0% conversion, total washout), because each of its individual stages has a dilution rate (0.30 h^-1) that exceeds μ_max and, being sterile-fed, cannot self-sustain. Routing half the fresh feed to bypass stage 1 and rejoin as "inoculum" ahead of stage 2 is precisely what rescues the second stage from washout: stage 2 is diluted too fast for sterile growth, but it is continuously reseeded by live cells from stage 1, so it operates as an inoculated (not sterile) reactor and reaches the same productive steady state as the single well-sized tank.