A culture of Rhizobium is grown in a chemostat (100 m3 bioreactor).
The feed contains 12 g/L sucrose.
Ks for the organism is 0.2 g/L and
μmax = 0.3 h-1.
Q.48 The flow rate required to result in steady state concentration of sucrose
as 1.5 g/L in the bioreactor will be
Options:
(A) 15 m3 h-1
(B) 26 m3 h-1
(C) 2.6 m3 h-1
(D) 150 m3 h-1
Chemostat Flow Rate Calculation Using Monod Kinetics in Rhizobium Culture
Chemostats are continuous bioreactors where microbial growth is controlled by the
dilution rate. At steady state, microbial growth follows Monod kinetics.
This numerical explains how to calculate the required flow rate to maintain
a given substrate concentration.
Question Overview
- Bioreactor volume (V) = 100 m3
- Feed sucrose concentration (S0) = 12 g/L
- Steady-state sucrose concentration (S) = 1.5 g/L
- Monod constant (Ks) = 0.2 g/L
- Maximum specific growth rate (μmax) = 0.3 h−1
Key Concept
At steady state in a chemostat:
D = μ
where dilution rate D = F / V
According to Monod kinetics:
μ = μmax × S / (Ks + S)
Step 1: Calculate Specific Growth Rate (μ)
μ = 0.3 × (1.5 / (0.2 + 1.5))
μ = 0.3 × (1.5 / 1.7)
μ ≈ 0.265 h−1
Step 2: Calculate Dilution Rate (D)
At steady state:
D = μ = 0.265 h−1
Step 3: Calculate Flow Rate (F)
F = D × V
F = 0.265 × 100
F ≈ 26.5 m3 h−1
Approximation for exams:
F ≈ 26 m3 h−1
Correct Answer
Option (B): 26 m3 h−1
Important Exam Tips
- At steady state, μ = D.
- Always use steady-state substrate concentration in Monod equation.
- High dilution rate leads to washout.
Conclusion
To maintain a steady-state sucrose concentration of 1.5 g/L
in a 100 m3 chemostat, the required flow rate is
approximately 26 m3 h−1. Hence,
Option (B) is the correct answer.
