Question

A batch cultivation of E. coli follows zeroth order Monod kinetics. Growth stops when dissolved oxygen reaches 10% of saturation. The oxygen mass transfer coefficient is \(k_La = 80\ \text{h}^{-1}\). Saturation dissolved oxygen = 0.007 kg m\(^{-3}\). If maximum specific growth rate = 0.2 h\(^{-1}\) and yield coefficient \(Y_{X/O_2} = 1.5\ (\text{kg cells})(\text{kg O}_2)^{-1\), then the final biomass concentration (kg m\(^{-3}\)) is \(\underline{\hspace{1cm}}\).}

Hint:

For oxygen-limited batch cultures, set OTR = OCR at the point of growth termination to compute biomass concentration.

Answer 1

Correct Answer: 3.7

The residual dissolved oxygen at termination is 10% of saturation:
\[ C_{L,\min} = 0.1 \times 0.007 = 0.0007\ \text{kg m}^{-3} \]
Oxygen transfer rate under steady gas–liquid conditions:
\[ \text{OTR} = k_La (C^_L - C_L) \]
Substitute values:
\[ \text{OTR} = 80(0.007 - 0.0007) = 80(0.0063) = 0.504\ \text{kg O}_2 \text{ m}^{-3}\text{ h}^{-1} \]
For zeroth-order kinetics, oxygen consumption rate is constant:
\[ \text{OCR} = \mu_{\max} X / Y_{X/O_2} \]
At termination: \( \text{OTR} = \text{OCR} \):
\[ 0.504 = \frac{0.2 X}{1.5} \]
Solve for biomass \(X\):
\[ X = \frac{0.504 \times 1.5}{0.2} = \frac{0.756}{0.2} = 3.78\ \text{kg m}^{-3} \]
Rounded to two decimal places:
\[ X = 3.78\ \text{kg m}^{-3} \]