Chemostat theory — Washout in steady-state continuous fermentation occurs under what condition?

Introduction:
In a chemostat (continuous stirred-tank bioreactor with feed and effluent), maintaining biomass requires that cells grow fast enough to replace those leaving with the outflow. Washout is the failure mode where cells are flushed from the reactor.

Given Data / Assumptions:

• Single-limiting nutrient controls growth (Monod kinetics).
• Perfect mixing and constant volume.
• Steady state is established unless instability criteria are met.

Concept / Approach:
At steady state for a single-limiting substrate, the specific growth rate equals the dilution rate (mu = D). If D is raised beyond mu_max, cells cannot divide quickly enough to balance outflow, leading to washout (X → 0).

Step-by-Step Solution:
1) Chemostat balance yields mu = D at steady state.2) Maximum feasible mu is mu_max, determined by organism and conditions.3) If D > mu_max, no steady state with positive X exists; cells decline and wash out.

Verification / Alternative check:
Plotting mu(S) = mu_max * S / (Ks + S) vs D shows that when D exceeds mu_max, the intersection disappears, predicting washout. Experiments confirm rapid drop in OD under such conditions.

Why Other Options Are Wrong:

• (a) D < mu_max allows a stable positive biomass.
• (c) Carrying capacity is a batch/closed-system notion, not chemostat washout.
• (d) mu equals D, not necessarily mu_max; only at high S could mu approach mu_max.
• (e) Product rate is unrelated to washout criterion.

Common Pitfalls:
Confusing D with volumetric flow alone; ignoring that temperature or pH shifts change mu_max and can induce washout unexpectedly.

Final Answer:
Dilution rate exceeds the maximum specific growth rate (D > mu_max).