Question

A strain of Azotobacter vinelandii is cultured in a 15 m^3 stirred fermenter for alginate production. Under current operating conditions kLa is 0.17 s^-1. Oxygen solubility in the broth is approximately 8 x 10^-3 kg m^-3. The specific rate of oxygen uptake is 12.5 mmol g^-1 h^-1. What is the maximum possible cell concentration?

(a) 10 gl^-1

(b) 15 gl^-1

(c) 20 gl^-1

(d) 12 gl^-1

I got this question in an interview for internship.

Origin of the question is Oxygen Uptake in Cell Cultures topic in section Mass Transfer of Bioprocess Engineering

Answer 1

To calculate the maximum possible cell concentration, we need to determine the oxygen transfer rate in the fermenter and how much oxygen is required for the cells.

Given Data:

• kLa = 0.17 s⁻¹ (volumetric mass transfer coefficient).
• Oxygen solubility = 8×10−38 \times 10^{-3}8×10−3 kg/m³.
• Specific rate of oxygen uptake (qO2) = 12.5 mmol/g/h.
• Volume of fermenter = 15 m³.

Step-by-Step Calculation:

1. Calculate the maximum oxygen transfer rate (OTR):

   The oxygen transfer rate (OTR) is given by the equation:

   OTR=kLa×C∗\text{OTR} = k_L a \times C^* OTR=kL​a×C∗

   where:

   • kLak_L akL​a is the volumetric mass transfer coefficient (0.17 s⁻¹),
   • C∗C^*C∗ is the oxygen solubility in the broth (8 × 10⁻³ kg/m³).

   First, convert the oxygen solubility to the appropriate units (g/m³):

   C∗=8×10−3 kg/m3=8 g/m3C^* = 8 \times 10^{-3} \, \text{kg/m}^3 = 8 \, \text{g/m}^3C∗=8×10−3kg/m3=8g/m3

   Now, calculate the oxygen transfer rate:

   OTR=0.17×8=1.36 g O2/m3s\text{OTR} = 0.17 \times 8 = 1.36 \, \text{g O}_2/\text{m}^3\text{s}OTR=0.17×8=1.36g O2​/m3s

2. Calculate the total oxygen available in the fermenter:

   The total oxygen transfer rate for the fermenter, considering its volume (15 m³), is:

   Total OTR=OTR×Volume of fermenter\text{Total OTR} = \text{OTR} \times \text{Volume of fermenter}Total OTR=OTR×Volume of fermenter Total OTR=1.36 g/m3s×15 m3=20.4 g/s\text{Total OTR} = 1.36 \, \text{g/m}^3\text{s} \times 15 \, \text{m}^3 = 20.4 \, \text{g/s}Total OTR=1.36g/m3s×15m3=20.4g/s

   Now, convert to g/h:

   Total OTR=20.4×3600 s/h=73,440 g/h\text{Total OTR} = 20.4 \times 3600 \, \text{s/h} = 73,440 \, \text{g/h}Total OTR=20.4×3600s/h=73,440g/h

3. Calculate the maximum cell concentration:

   The oxygen required for cell growth is related to the specific rate of oxygen uptake (qO2qO2qO2):

   qO2=12.5 mmol/g/h=12.5×10−3 mol/g/hqO2 = 12.5 \, \text{mmol/g/h} = 12.5 \times 10^{-3} \, \text{mol/g/h}qO2=12.5mmol/g/h=12.5×10−3mol/g/h

   Convert oxygen uptake from moles to grams:

   qO2=12.5×10−3 mol/g/h×32 g/mol=0.4 g/g/hqO2 = 12.5 \times 10^{-3} \, \text{mol/g/h} \times 32 \, \text{g/mol} = 0.4 \, \text{g/g/h}qO2=12.5×10−3mol/g/h×32g/mol=0.4g/g/h

   The maximum possible cell concentration is found by dividing the total oxygen available by the specific rate of oxygen uptake:

   Maximum cell concentration=Total OTRqO2\text{Maximum cell concentration} = \frac{\text{Total OTR}}{qO2}Maximum cell concentration=qO2Total OTR​ Maximum cell concentration=73,440 g/h0.4 g/g/h=183,600 g/h\text{Maximum cell concentration} = \frac{73,440 \, \text{g/h}}{0.4 \, \text{g/g/h}} = 183,600 \, \text{g/h}Maximum cell concentration=0.4g/g/h73,440g/h​=183,600g/h

   Convert this to g/L:

   Maximum cell concentration=183,60015=12.24 g/L\text{Maximum cell concentration} = \frac{183,600}{15} = 12.24 \, \text{g/L}Maximum cell concentration=15183,600​=12.24g/L

Answer:

(d) 12 g/L

This is the maximum possible cell concentration, considering the oxygen transfer rate and the oxygen uptake rate.