A fermentation system has KLa = 3 s^-1 and C* = 5 ppm O2. If the bulk liquid is already saturated (CL = C*), what is the oxygen transfer rate (OTR)?

Difficulty: Easy

Correct Answer: Zero

Explanation:


Introduction / Context:
Oxygen transfer from bubbles into liquid is driven by the concentration difference between saturation (C*) and bulk dissolved oxygen (CL). When CL equals C*, there is no driving force, and the net transfer stops regardless of the KLa magnitude. This scenario frequently occurs at low cell densities or during idle gassing.


Given Data / Assumptions:

  • KLa = 3 s^-1 (system capacity).
  • C* = 5 mg·L^-1; CL = C* = 5 mg·L^-1.
  • Assume steady conditions with no rapid transients.


Concept / Approach:
Use OTR = KLa * (C* − CL). If CL = C*, the term in parentheses is zero; therefore, OTR is zero. Oxygen may still circulate between phases, but net transfer into the bulk does not occur until CL drops below C* due to cellular consumption.


Step-by-Step Solution:

Compute driving force: C* − CL = 5 − 5 = 0.Multiply by KLa: OTR = 3 * 0 = 0 mg·L^-1·s^-1.Interpretation: the system is supply-capable but demand-limited.Conclusion: OTR = 0 until cells consume oxygen.


Verification / Alternative check:
As soon as OUR > 0 lowers CL below C*, the same KLa will yield OTR = KLa * ΔC to match OUR at steady state.


Why Other Options Are Wrong:

  • 3, 5, or 15 mg·L^-1·s^-1: confuse KLa or C* with OTR; ignore ΔC = 0.
  • 0.2 mg·L^-1·s^-1: arbitrary nonzero value without basis.


Common Pitfalls:
Assuming large KLa always means high OTR; without driving force, capacity is unused.


Final Answer:
Zero

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