Question 4 of 6: Chemostat Steady State and Batch vs. Continuous Productivity

Question 4: Chemostat Steady State and 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) Proof that μ = D at chemostat steady state

Given. A continuous stirred-tank bioreactor (chemostat) of constant working volume V, fed sterile medium at volumetric flow rate F, with exit stream at the same flow rate F (constant liquid volume). Biomass concentration in the vessel (= exit stream, well-mixed) is X; specific growth rate is μ (a function of substrate concentration, e.g. Monod).

Find. Show that at steady state, μ = D ≡ F/V, up to the point of washout.

1. Unsteady biomass balance on the vessel. Accumulation = growth − outflow (sterile feed carries no biomass in): $$V\frac{dX}{dt}=\underbrace{\mu X V}_{\text{growth}}-\underbrace{FX}_{\text{outflow}}$$ Dividing through by V and defining the dilution rate $D\equiv F/V$ (inverse residence time): $$\frac{dX}{dt}=\mu X-DX=(\mu-D)X$$
2. Impose steady state. At steady state biomass concentration no longer changes with time, $dX/dt=0$, and (below washout) $X\neq0$: $$0=(\mu-D)X\ \Rightarrow\ \boxed{\mu=D}$$ This holds for any growth kinetics (Monod or otherwise) — the only requirement is that a non-trivial steady state ($X>0$) exists.
3. Why this can only hold up to washout. Because μ itself is bounded above by $\mu_{max}$ (e.g. Monod: $\mu=\mu_{max}S/(K_S+S)\le\mu_{max}$ for any S), the relation $\mu=D$ can only be satisfied by a nonzero biomass steady state while $D\le\mu_{max}$. If the operator raises D beyond $\mu_{max}$, cells are washed out of the vessel faster than they can grow to replace themselves: the balance $dX/dt=(\mu-D)X$ is now negative for every $X>0$ (since $\mu\le\mu_{max}<D$), so X decays monotonically to the trivial steady state $X=0$ — washout. At washout the exit substrate rises to equal the feed value, $S=S_{in}$, since nothing is left to consume it. The proven identity $\mu=D$ is therefore valid exactly on $0

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

Approach. When the substrate concentration is always far above K_S (either S₀ throughout a batch run, or S_in in a well-designed chemostat operated well below washout), the Monod term $S/(K_S+S)\to1$, so growth is essentially zero-order in substrate and proceeds at the maximum specific rate, $\mu\approx\mu_{max}$, for as long as substrate remains far in excess — i.e. both cultivation modes are kinetically in the same (fastest-possible) growth regime, and the comparison becomes about time-averaged, whole-cycle productivity rather than instantaneous rate.

1. Batch productivity. In batch, biomass grows exponentially at $\mu\approx\mu_{max}$ from $X_0$ to a final $X_f$ (once S depletes toward K_S, growth slows and eventually stops): $$X_f=X_0e^{\mu_{max}t_g}$$ But the vessel is not producing biomass continuously: each batch cycle also includes a lag phase (t_lag, cells adapting to the medium, μ≈0), plus non-productive turnaround time t_d (harvest, cleaning, sterilization, refilling). The time-averaged volumetric productivity is $$P_{batch}=\frac{X_f-X_0}{t_{lag}+t_g+t_d}$$ — the full cycle time appears in the denominator even though growth only happens during t_g.
2. Continuous (chemostat) productivity. At steady state the volumetric rate of biomass removal in the exit stream is $DX$, and by part (i) $\mu=D$ at every steady operating point, so the continuous volumetric productivity is $$P_{cont}=DX$$ with X fixed by the sterile-feed steady-state balance $X=Y_{X/S}(S_{in}-S)$. Since $S_0\gg K_S$ keeps $\mu\approx\mu_{max}$ available at essentially every D up to (just below) washout, the operator can push D → $\mu_{max}$ and hold $X\approx Y_{X/S}S_{in}$ (S drops to a small residual), giving a productivity that approaches $$P_{cont}\to \mu_{max}\,Y_{X/S}S_{in}$$ continuously, with no lag phase and no turnaround downtime once the steady state is established — every unit of reactor volume-time is producing biomass at (near) the fastest rate the organism can sustain.
3. Comparison. Because $S_0\gg K_S$ removes the kinetic disadvantage batch might otherwise have near depletion (growth stays near $\mu_{max}$ for most of the batch), the productivity gap between the two modes narrows to a purely logistics question: batch pays a fixed overhead (t_lag+t_d) every cycle that continuous operation pays only once (at start-up), so $P_{cont}>P_{batch}$ whenever that overhead is a non-negligible fraction of the batch cycle time — which is the typical industrial case. The trade-off is operational risk: continuous culture must be held strictly below $D=\mu_{max}$ (part i) or it washes out entirely, while batch culture, though less productive on average, cannot wash out and tolerates upsets that simply extend t_g.