A chemostat is operated with an organism following Monod kinetics (μ_m = 0.5 h^-1, K_s = 0.1 g·L^-1). If the feed substrate S_f = 10 g·L^-1 and the dilution rate D = 0.4 h^-1, what steady-state substrate concentration S in the effluent is expected?

Introduction:
In a chemostat at steady state, the specific growth rate equals the dilution rate. For Monod growth, this constraint determines the residual substrate concentration S, which is crucial for productivity and yield predictions in continuous culture.

Given Data / Assumptions:

• μ_m = 0.5 h^-1 (maximum specific growth rate).
• K_s = 0.1 g·L^-1 (half-velocity constant).
• D = 0.4 h^-1 (dilution rate), S_f = 10 g·L^-1 (feed substrate).
• Steady state with μ = D and no substrate inhibition.

Concept / Approach:
Monod model: μ = μ_m * S / (K_s + S). At steady state, set μ = D and solve for S. Because D is less than μ_m, a finite residual S results, typically close to K_s when D is moderate to high relative to μ_m.

Step-by-Step Solution:

Verification / Alternative check:
Check washout: D < μ_m, so steady state is feasible. Because S_f ≫ K_s, the effluent S is governed by μ = D and not by feed depletion; the 0.4 g·L^-1 value is consistent with Monod behavior near K_s when D/μ_m = 0.8.

Why Other Options Are Wrong:

• 0 g.l-1: Would imply infinite μ, impossible.
• 1.0 g.l-1: Too high for D/μ_m = 0.8 with K_s = 0.1; solving gives 0.4.
• 10 g.l-1: Effluent equaling feed occurs at washout or no consumption, not the case here.

Common Pitfalls:
Forgetting to solve μ = D directly for S; using overall substrate balance without Monod relation leads to incorrect estimates. Also, mixing units (g.l^-1 vs g·L^-1) can cause calculation errors.

Final Answer:
0.5 g.l-1