Ultimate BOD from a Diluted Sample
Mediumbodtestingwastewater
A 10 mL sample of wastewater is diluted to 300 mL in a BOD bottle. The dissolved oxygen falls from 9.0 mg/L to 3.2 mg/L over 5 days, and the reaction rate constant is k = 0.23 day-1 (base e). The ultimate BOD is most nearly:
- A5.8 mg/L
- B174 mg/L
- C255 mg/L
- D348 mg/L
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Answer: C. 255 mg/L
The DO depletion is 9.0 − 3.2 = 5.8 mg/L, and the dilution factor is 300/10 = 30, so BOD₅ = 5.8 × 30 = 174 mg/L. First-order kinetics give BOD₅ = L(1 − e^(−kt)), so the ultimate demand is L = 174/(1 − e^(−0.23×5)) = 174/0.683 ≈ 255 mg/L. Reporting 174 mg/L stops at the 5-day value, and 5.8 mg/L omits the dilution entirely.
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