A fermentation system has KLa = 3 s^-1 and C* = 5 ppm O2. If the bulk liquid contains no dissolved oxygen (CL = 0), what is the oxygen transfer rate (OTR)?

Introduction / Context:In aerated bioreactors, the oxygen transfer rate (OTR) quantifies how fast oxygen crosses from bubbles into the broth. The classic expression OTR = KLa * (C* − CL) connects mass transfer capability (KLa) and driving force (C* − CL). This problem is a direct substitution exercise often used in design and troubleshooting.

Given Data / Assumptions:

• KLa = 3 s^-1 (liquid-phase volumetric mass transfer coefficient).
• C* = 5 ppm O2 (assume ppm ≈ mg·L^-1 for dilute aqueous solutions).
• CL = 0 mg·L^-1 (bulk liquid fully depleted of O2).

Concept / Approach:Use OTR = KLa * (C* − CL). With CL = 0, the driving force is maximized and equals C*. Multiply KLa by C* to obtain the volumetric transfer rate in mg·L^-1·s^-1.

Step-by-Step Solution:

Verification / Alternative check:If CL were half-saturated (2.5 mg·L^-1), OTR would be 3 * (5 − 2.5) = 7.5 mg·L^-1·s^-1, confirming linearity with driving force.

Why Other Options Are Wrong:

• Zero: only true if C* = CL, not when CL = 0.
• 10 or 5 mg·L^-1·s^-1: result from incorrect arithmetic or using KLa or C* alone.
• 3 mg·L^-1·s^-1: equal to KLa, not OTR.

Common Pitfalls:Confusing kL with KLa, or forgetting that ppm ≈ mg·L^-1 for dilute aqueous media at standard conditions.

Final Answer:15 mg·L^-1·s^-1