Oxygen balance check in a bioreactor: assuming dissolved oxygen is at steady state, if the oxygen transfer rate (OTR) is measured as 23 kg·h^-1 at a given instant, what must the oxygen uptake rate (OUR) be at that same instant?

Introduction / Context:
Matching oxygen supply with cellular demand is central to aerobic bioprocess control. Two key quantities are the oxygen transfer rate (OTR), driven by aeration and agitation, and the oxygen uptake rate (OUR), dictated by cellular metabolism. At dissolved oxygen (DO) steady state, these two rates are equal by mass balance.

Given Data / Assumptions:

• OTR = 23 kg·h^-1 at the time of interest.
• DO concentration is momentarily steady (dC_L/dt = 0).
• No other oxygen sources/sinks besides gas–liquid transfer and cellular uptake.

Concept / Approach:
The dissolved oxygen mass balance is dC_L/dt = k_La * (C* − C_L) − OUR/V. Multiplying by V gives d(C_L V)/dt = OTR − OUR. At steady state, d(C_L V)/dt = 0, hence OTR = OUR. Therefore, when OTR is 23 kg·h^-1, OUR must also be 23 kg·h^-1 at that instant.

Step-by-Step Solution:
Write oxygen balance: accumulation = in − out = transfer − uptake.Set accumulation to zero (steady DO): 0 = OTR − OUR.Rearrange: OUR = OTR.Substitute values: OUR = 23 kg·h^-1.

Verification / Alternative check:
If DO were rising (positive accumulation), OTR > OUR; if falling, OTR < OUR. Only at steady DO are the two exactly equal. Controllers often adjust agitation or airflow to maintain this balance.

Why Other Options Are Wrong:
Less than or greater than 23 kg·h^-1 would imply changing DO levels, contradicting the steady-state assumption.2.3 kg·h^-1 is a simple scaling distraction with no basis.

Common Pitfalls:
Confusing OTR capacity with actual transfer; forgetting to check if DO is at steady state before asserting OTR = OUR.

Final Answer:
23 kg·h^-1