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BOD

Intermediate Sanitation and Wastewater Engineering Reviewed 24 Oct 2026 2 min read

Biochemical oxygen demand (BOD) measures how much dissolved oxygen microorganisms use to break down the organic matter in a water sample. It is the classic measure of the strength of sewage and the pollution load on a river. When untreated sewage enters a stream, bacteria consume the organic matter and with it the oxygen that fish and other aquatic life need; if oxygen falls too low, the stream turns septic.

The standard test incubates diluted samples in sealed bottles in the dark for a fixed period (three days at 27 °C in Indian practice, five days at 20 °C in many other countries) and measures the drop in dissolved oxygen. Seed microorganisms are added if the sample lacks them. Because dissolved oxygen in a bottle is limited, strong samples are diluted so that oxygen is neither exhausted nor barely used. BOD is calculated from the oxygen drop and the dilution.

Diagram: Indiacon original, not to scale.

BOD exertion follows roughly first-order kinetics: the remaining demand falls exponentially with time, governed by a rate constant k that depends on temperature and waste type. The ultimate BOD is the total carbonaceous demand; the 3-day or 5-day value is a fraction of it. Nitrification adds a later oxygen demand. Chemical oxygen demand (COD), measured by strong chemical oxidation in a few hours, is always higher than BOD and is used for faster control; the BOD/COD ratio indicates how biodegradable a waste is. River models such as Streeter-Phelps combine BOD decay with re-aeration to predict the oxygen sag curve downstream of an outfall.

Key formulas
  • BOD = (D₁ − D₂) / P, P = sample volume fraction
  • BOD_t = L₀ (1 − e^(−k t))
  • Temperature: k_T = k₂₀ × 1.047^(T − 20)
  • Population equivalent = BOD load / per capita BOD
Solved example

Problem. A 300 mL bottle contains 6 mL of sewage. Initial DO is 8.2 mg/L and final DO after 3 days is 3.4 mg/L. Find the 3-day BOD and the ultimate BOD if k = 0.23 per day (base e).

  1. P = 6 / 300 = 0.02.
  2. BOD₃ = (8.2 − 3.4) / 0.02 = 240 mg/L.
  3. Fraction exerted in 3 days = 1 − e^(−0.69) = 0.498.
  4. L₀ = 240 / 0.498 = 482 mg/L.

Answer. BOD₃ = 240 mg/L; ultimate BOD ≈ 480 mg/L

Quick quiz

1What does BOD measure?

Oxygen used by microbes to decompose organic matter.

2Why dilute strong samples?

So oxygen is not used up completely in the bottle.

3Why is COD higher than BOD?

Chemical oxidation also oxidises matter microbes cannot break down quickly.

Code note
  • CPHEEO Manual on Sewerage and Sewage Treatment Systems (2013) (India): Central government manual for sewer networks and sewage treatment plants.

Clauses and documents to read

  • IS 3025 (Part 44):1993 (BOD test, cited by number)
  • IS 3025 (Part 58):2006 (COD test, cited by number)

Cited by number and clause only. We do not reproduce clause text, tables or figures (BIS Act 2016, s.10(5) and s.11). Check the current edition and amendments before use.

References and further reading

Standard texts, listed in no order of merit. We do not copy text, figures, tables or solved problems from any book; our explanations and examples are original.

  • Metcalf and Eddy, Wastewater Engineering: Treatment and Resource Recovery, McGraw Hill, 5th edition.
  • S. K. Garg, Sewage Disposal and Air Pollution Engineering (Environmental Engineering Vol. II), Khanna Publishers, latest edition.

Free and official sources

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