Question

E. coli is cultivated in a chemostat operated at a dilution rate of 0.2 h\(^{-1}\). The values of biomass yield due to oxygen consumption and the steady state biomass concentration are 0.2 g g\(^{-1}\) and 10 g L\(^{-1}\), respectively. The oxygen transfer rate (in g L\(^{-1}\) h\(^{-1}\)) is ____________.

Hint:

In chemostats at steady state, substrate consumption rate = substrate transfer rate.
Always calculate productivity as growth rate × biomass concentration.

Answer 1

Correct Answer: 10

Step 1: In steady state chemostat:
Dilution rate (\(D\)) = growth rate (\(\mu\)) = 0.2 h\(^{-1}\).
Step 2: Biomass concentration = 10 g L\(^{-1}\).
Biomass productivity = \(\mu \times X = 0.2 \times 10 = 2 \, g \, L^{-1} h^{-1}\).
Step 3: Biomass yield on oxygen = 0.2 g biomass / g oxygen.
So, oxygen consumption rate = \(\frac{2}{0.2} = 10 \, g \, L^{-1} h^{-1}\).
Step 4: At steady state, oxygen transfer rate = oxygen consumption rate.
\(\therefore\) Oxygen transfer rate = \(\mathbf{10 \, g \, L^{-1} h^{-1}}\).