BOD: what it measures and how to calculate it
Biochemical oxygen demand, the first-order BOD curve, the dilution test, and the effect of temperature on the rate constant.
Biochemical oxygen demand (BOD) is the oxygen that bacteria use to break down the organic matter in water. It tells us how much a waste stream will drain the oxygen from a river it is discharged into.
The BOD curve
Organic matter is consumed at a rate proportional to how much remains (first-order reaction). The BOD exerted after t days is:
BODt = L₀ ( 1 − e−k t ) = L₀ ( 1 − 10−K t )
L₀ is the ultimate BOD (first stage), k the rate constant to base e, and K = k / 2.303 to base 10.
Standard test conditions
The classic test runs for 5 days at 20 °C (BOD₅). In India, IS 3025 (Part 44) uses 3 days at 27 °C, which gives about the same result faster and closer to Indian temperatures.
Dilution method
Sewage would use up all the oxygen in the bottle, so it is diluted with aerated water. Dissolved oxygen is measured at the start (D₁) and after incubation (D₂):
BOD = D₁ − D₂P
P is the decimal fraction of sample in the mix.
Temperature correction
kT = k₂₀ θT − 20 θ ≈ 1.047
Worked example
A 5 mL sewage sample is diluted to 300 mL. Initial DO 8.6 mg/L, final DO 3.9 mg/L after 5 days at 20 °C. Take k = 0.23 per day.
P = 5300 = 0.0167 BOD₅ = 8.6 − 3.90.0167 = 282 mg/L
L₀ = 2821 − e−0.23 × 5 = 2820.683 = 413 mg/L
Typical values
| Water | BOD₅ (mg/L) |
|---|---|
| Clean river | below 3 |
| Domestic sewage | 200 to 300 |
| Treated effluent (inland surface water discharge) | 30 or less |
COD (chemical oxygen demand) is always higher than BOD, because it oxidises matter that bacteria can't break down.
Quick check
Standard BOD in India is reported as: