A bioreactor seeded with Bacillus subtilis is running at the maximum agitation s

Question

A bioreactor seeded with Bacillus subtilis is running at the maximum agitation speed is working under the following conditions. The air sparger working at 0.5 L gas/L reactor volume-min, kia from the system is estimated to be 34 h The critical dissolve oxygen concentration is 0.2 mg L1 and the solubility of oxygen from air to the fermentation broth is 6.9 mg L-1 at 30 oC. (q02 for Bacills subtilis is 11.5 mmol O2/g-dry wt-h). Estimate: (i) Maximum concentration at steady state for Bacillus subtilis in this bioreactor under aerobic conditions. (i) Concentration for Bacillus subtilis in this bioreactor at steady state if pure oxygen is sparged into the reactor. (Air contains approximately 21% oxygen)

Explanation / Answer

(1) From the relationship at steady state where Oxygen uptake rate = Oxygen transfer rate, we have

X*qO2 = kla (C*DO – CDO)

Or X = kla (C*DO – CDO)/qO2

Here qO2 should be in mg units instead of mmoles, so we will convert it in mg-O2/g-cells/h

So our qO2 = 32mg-mmoles -1*11.5mmoles/g-cells-h-1 = 368 mg-O2/g-cells/h (Note mol wt of O2 is 32)

By plugging all the given values in we have

X = 34h-1 (6.9 – 0.2)mgL-1/368 mg-O2/g-cells/h = 0.62 g-cells/L <= this is your maximum concentration at steady state

(2) When we switch pure oxygen (100% oxygen) from air (containing 21% oxygen), the solubility of oxygen into fermentation broth will increase from 6.9 mg/L to 100*6.9/21 = 32.85 mg/L. Rest of the parameters will remain the same.

Therefore, your cell concentration now will be

X = 34h-1 (32.85 – 0.2)mgL-1/368 mg-O2/g-cells/h = 3.02 g-cells/L

Thanks!

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