Question 4 of 5: Chemostat with Recycle; Batch vs. Continuous Productivity

Question 4: Chemostat with Recycle; Batch vs. Continuous Productivity (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.

(i) Chemostat-with-recycle: D can exceed µ until washout

In a simple chemostat (no recycle), steady state forces D = µ exactly, and washout occurs the instant D > μ_max (no steady state with cells can exist above that ceiling). Adding cell recycle — concentrating cells from the reactor effluent and returning part of that concentrated stream — changes this constraint: at steady state the reactor's REQUIRED specific growth rate becomes smaller than D itself, which is exactly the "D exceeds µ" result the question asks to prove.

1. Define the recycle system. Sterile feed enters the reactor (volume V, cell concentration X, dilution rate D=F/V) at flow rate F with X₀=0. The reactor effluent passes through a settler that concentrates cells by a factor C (C≥1): the underflow, at flow rate F_r=αF (recycle ratio α=F_r/F), returns to the reactor at concentration C·X; the overflow (waste/bleed), at flow rate F, leaves the system at concentration X (equal to the non-recycled reactor concentration, since the settler is assumed to remove only the concentrated underflow).
2. Steady-state cell balance on the reactor. Accumulation = growth + inflow − outflow = 0: $$\mu XV + F_rCX = (F+F_r)X$$ The reactor receives fresh growth (μXV) plus recycled cells (F_rCX), and loses cells in the combined outflow to the settler, (F+F_r)X.
3. Divide through by XV and simplify. With D=F/V and α=F_r/F (so F_r/V=αD): $$\mu + \alpha D\,C = D+\alpha D \quad\Rightarrow\quad \mu = D + \alpha D - \alpha DC = D\big[1+\alpha(1-C)\big]$$ $$\boxed{\mu = D\big[1-\alpha(C-1)\big]}$$
4. Interpret the result. For a concentrating settler, C>1, so the bracket $[1-\alpha(C-1)]$ is strictly less than 1 (and positive, for a properly designed system). The steady-state relation therefore gives μ < D for ANY dilution rate — the reactor can be run with the dilution rate D exceeding the cells' actual specific growth rate μ, precisely because recycle continuously re-supplies cell mass that growth alone would not sustain at that D. (When C=1, i.e. no concentration benefit, the bracket is exactly 1 and the result collapses to the ordinary chemostat's μ=D — recycle without concentration confers no advantage.)
5. Washout condition. Washout occurs when the REQUIRED μ from the steady-state relation would have to exceed the biological ceiling μ_max (no achievable growth rate can supply enough new cells). Setting μ=μ_max and solving for the corresponding dilution rate: $$D_{washout}=\frac{\mu_{max}}{1-\alpha(C-1)}$$ Since the denominator is <1 for C>1, $D_{washout}>\mu_{max}$: recycle raises the washout dilution rate strictly above μ_max, proving that the system remains at steady state (no washout) for every D in the range $\mu_{max}<D<D_{washout}$ — i.e. it CAN be operated with D exceeding the cells' specific growth rate, right up until D_washout, exactly as the question asks to prove.

(ii) Batch vs. continuous productivity when S₀ ≫ K_s

When S₀ ≫ K_s, Monod kinetics $\mu=\mu_{max}S/(K_s+S)$ reduce to $\mu\approx\mu_{max}$ over essentially the entire batch or continuous run (substrate is never limiting until it is nearly exhausted) — so growth proceeds at (or very near) the maximum specific rate throughout, in both modes. The productivity comparison then comes down entirely to how each mode uses TIME, not to any kinetic difference between them.

Continuous culture at steady state, run at D approaching μ_max (but staying below the washout point), delivers cells continuously at the volumetric rate DX, with $X\approx Y_{X/S}(S_0-K_s)\approx Y_{X/S}S_0$ (since $K_s\ll S_0$). Productivity is essentially $$P_{continuous}\approx \mu_{max}\,Y_{X/S}\,S_0$$ sustained indefinitely with no interruption, once steady state is reached.

Batch culture must instead complete a full cycle each time: a lag phase (no growth, cells adapting), an exponential phase at μ≈μ_max (the only phase functionally equivalent to the continuous case), then substrate exhaustion once S drops toward K_s, followed by non-productive turnaround time (harvesting the vessel, cleaning/sterilizing, and re-inoculating) before the next batch can begin growing. The TIME-AVERAGED productivity is $$P_{batch}=\frac{X_{max}-X_0}{t_{lag}+t_{exp}+t_{turnaround}}$$ which is always lower than the instantaneous exponential-phase rate because the lag and turnaround intervals contribute zero biomass while still consuming clock time.

Consequently, when S₀ ≫ K_s, continuous culture achieves higher time-averaged biomass productivity than batch, because it eliminates the repeated non-productive lag and turnaround intervals and sustains growth at μ_max indefinitely, whereas batch culture pays the lag/turnaround "tax" on every cycle. The advantage of continuous culture shrinks (and can reverse for very short, well-optimized batches with negligible turnaround) only when turnaround time is small relative to the exponential-phase duration — but for typical industrial-scale sterilization/cleaning turnarounds, the continuous mode's advantage is substantial.