Q.57 In a chemostat, the feed flow rate and culture volume are 100 mL/h and 1.0 L, respectively.
With glucose as substrate, the values of μ_max and K_s are 0.2 h^-1 and 1 g/L, respectively.
For a glucose concentration of 10 g/L in the feed, the effluent substrate concentration (in g/L) is ________

Chemostat Effluent Substrate Concentration: Solved with Monod Kinetics (0.10 g/L)

In a chemostat with a feed flow rate of 100 mL/h and a culture volume of 1.0 L using glucose as substrate (μ_max = 0.2 h^-1, K_s = 1 g/L, feed concentration 10 g/L), the steady-state effluent substrate concentration equals 0.10 g/L.

Key Calculation Steps

At steady state, dilution rate equals microbial growth rate:

D = μ

First compute dilution rate:

D = F / V = 0.1 L/h ÷ 1.0 L = 0.1 h^-1

Apply the Monod equation:

μ = μ_max × S / (K_s + S)

Since μ = D:

D = μ_max × S / (K_s + S)

Solve for substrate concentration:

S = D × K_s / (μ_max − D)

Substituting values:

S = 0.1 × 1 / (0.2 − 0.1) = 0.10 g/L

Options Analysis

• A) 0.05 g/L: Underestimates S; comes from incorrect D or half-μ_max assumption.
• B) 0.10 g/L (Correct): Exact Monod steady-state result.
• C) 0.20 g/L: Occurs if μ ≈ μ_max is wrongly applied.
• D) 0.50 g/L: Possible inversion or use of K_s in wrong place.

Chemostat Principles Explained

• Steady state requires D < μ_max to avoid washout.
• With excess feed glucose (10 g/L), substrate becomes limiting inside reactor.
• Low effluent substrate means high biomass productivity.
• Mass balance ignores biomass in feed (sterile medium assumption).

Exam Tips for Monod–Chemostat Questions

• Compute dilution rate first using F/V.
• Set D = μ at steady state.
• Ignore yield when asked only for substrate concentration.
• Always check D < μ_max to confirm culture stability.