64. Biomass is being produced in a continuous stirred tank bioreactor of 750 L capacity. The sterile feed containing 8 g·L−1 glucose
as substrate was fed at a flow rate of 150 L·h−1. The microbial system follows Monod’s model with μm = 0.4 h−1, Ks = 1.5 g·L−1 and
Yx/s = 0.5 g cell mass·g substrate−1. Determine the cell productivity (g·L−1·h−1) at steady state.
(A) 0.85
(B) 0.65
(C) 0.45
(D) 0.25
Cell Productivity in Continuous Stirred Tank Bioreactor (CSTR) Using Monod Kinetics
Numerical problems based on continuous stirred tank bioreactors (CSTR)
and Monod growth kinetics are very common in biotechnology examinations.
One of the most important performance parameters is cell productivity,
defined as biomass formed per unit volume per unit time.
Given Data
- Bioreactor volume (V) = 750 L
- Feed flow rate (F) = 150 L·h−1
- Feed substrate concentration (S0) = 8 g·L−1
- Maximum specific growth rate (μm) = 0.4 h−1
- Monod constant (Ks) = 1.5 g·L−1
- Yield coefficient (Yx/s) = 0.5 g biomass per g substrate
Step 1: Calculation of Dilution Rate
Dilution rate (D) is given by:
D = F / V
D = 150 / 750 = 0.2 h−1
Step 2: Steady-State Substrate Concentration
At steady state in a CSTR:
μ = D
Using Monod equation:
μ = (μm · S) / (Ks + S)
0.2 = (0.4 · S) / (1.5 + S)
Solving the equation:
S = 1.5 g·L−1
Step 3: Biomass Concentration
Biomass concentration (X) is calculated using:
X = Yx/s (S0 − S)
X = 0.5 (8 − 1.5)
X = 3.25 g·L−1
Step 4: Cell Productivity
Cell productivity in a CSTR is given by:
Productivity = D × X
= 0.2 × 3.25
= 0.65 g·L−1·h−1
Correct Answer
Option (B): 0.65 g·L−1·h−1
Conclusion
By applying Monod kinetics and steady-state mass balance for a continuous stirred tank bioreactor,
the correct cell productivity is 0.65 g·L−1·h−1.
This type of problem is frequently asked in biotechnology competitive examinations.
