Wastewater Treatment – 5-Day BOD vs Ultimate BOD (at 20 °C) In standard wastewater analysis, what percentage of the ultimate biochemical oxygen demand (BODu) is typically exerted in the standard 5-day BOD test (BOD5) conducted at 20 °C?

Introduction / Context:
BOD (biochemical oxygen demand) measures the oxygen required by microorganisms to oxidize biodegradable organic matter. The standard laboratory test BOD5 (5-day BOD at 20 °C) is a practical surrogate for the ultimate BOD (BODu), which would take much longer to measure in full. Understanding the typical relationship between BOD5 and BODu is essential for plant sizing, stream impact analysis, and compliance reporting.

Given Data / Assumptions:

• Standard test temperature: 20 °C.
• Standard incubation period: 5 days.
• Typical municipal wastewater with normal microbial seed and kinetics.

Concept / Approach:

BOD exertion follows first-order kinetics: BODt = BODu * (1 − e^(−k t)), where k is the deoxygenation rate constant at 20 °C. For typical domestic wastewater at 20 °C, k commonly lies around 0.23 day^−1 to 0.3 day^−1, yielding a BOD5 fraction of about two-thirds of BODu. This industry convention underpins many textbook conversions between BOD5 and BODu.

Step-by-Step Solution:

Verification / Alternative check:

Using k = 0.23 day^−1: fraction = 1 − e^(−1.15) ≈ 0.68. This aligns with design manuals and classroom practice for domestic sewage characterization at 20 °C.

Why Other Options Are Wrong:

• 48% and 58%: Underestimate oxygen demand exerted by day 5 and are inconsistent with common k values at 20 °C.
• 78%: Overestimates the day-5 exertion for typical domestic wastes; such a high fraction would imply an unusually large k at 20 °C.

Common Pitfalls:

Assuming the 68% rule holds at other temperatures; in reality, k varies with temperature (Arrhenius relation). Also, seeded dilution and nitrification suppression can affect BOD5 measurements.

Final Answer:

68%