Difficulty: Easy
Correct Answer: X and P will be ~0 while S equals the substrate concentration in the feed
Explanation:
Introduction / Context:A chemostat (continuous stirred tank bioreactor) operates at a fixed dilution rate with continuous inflow of fresh medium and outflow of culture. “Washout” occurs when the dilution rate exceeds the culture’s net specific growth rate so that the reactor cannot retain cells. Understanding the fate of biomass (X), limiting substrate (S), and product (P) at washout is fundamental to bioprocess control and design.
Given Data / Assumptions:
Concept / Approach:At steady state, mass balances link production/consumption to dilution. If cells are washed out, there is essentially no biomass to consume substrate or to form product. Therefore, S rises to the feed value Sf, X tends to zero, and P tends to zero for a growth-associated product (or to a low value if it is non-growth associated with negligible formation at X ≈ 0).
Step-by-Step Solution:
Biomass balance at washout: generation by growth ≈ 0 while outflow removes any residual cells → X ≈ 0.Substrate balance: with X ≈ 0, there is negligible consumption; by mixing, S in the reactor approaches S in the feed → S ≈ Sf.Product balance: with no cells, formation rate is ~0 for growth-associated products; dilution removes any residual product → P ≈ 0.Hence the only consistent description is X ≈ 0, P ≈ 0, S ≈ Sf.Verification / Alternative check:Steady-state chemostat equations give X = Yx/s (Sf − S). At washout, X = 0 implies S = Sf. For growth-associated product P = Yp/x X (or Yp/s (Sf − S)); with X = 0 and S = Sf, P → 0.
Why Other Options Are Wrong:
Common Pitfalls:Assuming a non-growth-associated product necessarily accumulates; without biomass, its formation ceases and dilution flushes it out. Confusing transient behavior with steady state is another error.
Final Answer:X and P will be ~0 while S equals the substrate concentration in the feed
Discussion & Comments