Q.62 A 2 L bioreactor is being operated as a chemostat, at a flow rate of 0.8 L/h and
sterile feed of 10 g/L substrate. The bacterial growth follows Monod kinetics at
a maximum specific growth rate of 0.6 h-1 with a Monod constant of 0.5 g/L and
a biomass yield coefficient of 0.4 g/g.
The exit biomass concentration is _______ g/L.
(Round off to one decimal place)

Correct Answer: 3.2 g/L

Problem Overview

A chemostat operates at steady state following Monod kinetics. At steady state, the
dilution rate (D) equals the specific growth rate (μ). The objective
is to determine the exit biomass concentration.

Given Parameters

• Bioreactor volume, V = 2 L
• Volumetric flow rate, F = 0.8 L h^-1
• Feed substrate concentration, S₀ = 10 g L^-1
• Maximum specific growth rate, μ_max = 0.6 h^-1
• Monod constant, K_s = 0.5 g L^-1
• Biomass yield coefficient, Y_X/S = 0.4 g g^-1

Step 1: Calculate Dilution Rate

The dilution rate is calculated using:

D = F / V

D = 0.8 / 2 = 0.4 h^-1

Step 2: Apply Monod Equation

At steady state:

μ = D = μ_max × ( S / (K_s + S) )

Solving for substrate concentration (S):

S = (K_s × D) / (μ_max − D)

S = (0.5 × 0.4) / (0.6 − 0.4) = 0.2 / 0.2 = 1 g L^-1

Step 3: Calculate Biomass Concentration

Biomass concentration is obtained from the substrate balance:

X = Y_X/S × (S₀ − S)

X = 0.4 × (10 − 1) = 0.4 × 8 = 3.2 g L^-1

Final Answer

The exit biomass concentration in the chemostat is:

✔ 3.2 g L^-1

Key Takeaway

In chemostat operation with Monod kinetics, the steady-state biomass concentration
depends on dilution rate, kinetic constants, and yield coefficient. Correct application
of mass balance equations leads to the accurate biomass concentration.