BOD₅ Calculator (dilution method)

Understanding the BOD₅ calculator — concept and principle

Biochemical oxygen demand (BOD₅) is the amount of dissolved oxygen that microorganisms consume while breaking down organic matter in a water sample over five days at 20 °C. Because a strong sample would use up all the oxygen in the bottle, it is diluted with oxygen-saturated dilution water first. This calculator converts the measured oxygen drop in the diluted bottle back to the BOD of the original, undiluted sample.

4.5 mg/L of oxygen used in a 2% dilution means the undiluted sample has BOD₅ = 225 mg/L.

How it is calculated

BOD₅ (mg/L) = (D₁ − D₂) ÷ P

D₁: dissolved oxygen of the diluted sample right after preparation (mg/L).
D₂: dissolved oxygen after 5 days of incubation (mg/L).
P: decimal fraction of sample in the bottle, sample volume ÷ bottle volume (for example 6 mL in a 300 mL bottle gives 0.02).

This is the unseeded form. If seed microorganisms are added, the seed's own demand is subtracted: BOD₅ = [(D₁ − D₂) − (B₁ − B₂) × f] ÷ P, where B₁ and B₂ are the seed control readings and f is the ratio of seed in the sample bottle to seed in the control.

The principle

Bacteria oxidise biodegradable organic matter and use oxygen in proportion to how much they break down. In a sealed, dark bottle no oxygen enters and none is produced by photosynthesis, so the fall in DO measures microbial consumption alone. Only the fraction P of the bottle is real sample; the rest is dilution water that ideally consumes nothing. Dividing by P scales the oxygen used in the bottle up to what the full-strength sample would demand.

Worked example

Inputs: D₁ = 8.5 mg/L, D₂ = 4.0 mg/L, P = 0.02.
Oxygen depletion: 8.5 − 4.0 = 4.5 mg/L.
BOD₅ = 4.5 ÷ 0.02 = 225 mg/L.
Validity check: the depletion (4.5 mg/L) is at least 2 mg/L and the residual DO (4.0 mg/L) is at least 1 mg/L, so the dilution is acceptable under Standard Methods criteria.

Each point is the calculator re-run with only “Sample fraction P (sample vol ÷ bottle vol)” changed; the gold point is the default example. Change the inputs above and this chart follows your numbers.

Sample fraction P (sample vol ÷ bottle vol)	BOD₅ (mg/L)
0.005	900
0.01	450
0.015	300
0.02	225
0.025	180
0.03	150
0.04	112.5

How to read the result

Untreated domestic wastewater typically has a BOD₅ of a few hundred mg/L, so 225 mg/L is a normal raw-sewage value. Secondary treatment brings it down to the low tens: the EU Urban Waste Water Treatment Directive sets 25 mg/L, and the US secondary treatment standard 30 mg/L as a 30-day average.

Common mistakes

Entering P as a percentage (2) instead of a fraction (0.02).
Accepting bottles where depletion is below 2 mg/L or residual DO below 1 mg/L.
Forgetting the seed correction when seed was added.

Keep in mind: BOD₅ measures only biodegradable matter, and toxic substances can inhibit the microbes and give falsely low values. Regulatory testing must follow the official method and the limits that apply locally.

Frequently asked questions

Why five days?

It became a convention in early 20th-century Britain, where five days was long enough for substantial oxidation and considered longer than river water took to reach the sea.

What is the difference between BOD and COD?

COD uses a strong chemical oxidant and oxidises most organic matter in a few hours, so COD is usually higher than BOD.

How do I choose P?

Prepare several dilutions so that at least one depletes 2 mg/L and leaves 1 mg/L.

Sources: APHA, AWWA and WEF, Standard Methods for the Examination of Water and Wastewater, Method 5210 B · Council Directive 91/271/EEC concerning urban waste-water treatment, Annex I · US EPA, 40 CFR Part 133, Secondary Treatment Regulation

Checked 8 Nov 2026. Standard textbook formulas in SI units, computed in your browser. For learning and estimates — real designs must follow your local codes and be checked by a qualified engineer.

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