Biochemical Oxygen Demand Calculator
Calculate the 5-day biochemical oxygen demand (BOD₅) of a diluted water sample from initial and final dissolved oxygen, dilution fraction, and optional seed correction, following APHA Standard Methods 5210B.
Find BOD₅ from a Diluted Sample
Biochemical Oxygen Demand (BOD₅)
Where:
- \(D_1\) — dissolved oxygen in the diluted sample at day 0 (mg/L)
- \(D_2\) — dissolved oxygen remaining after 5 days of incubation at 20 °C (mg/L)
- \(B_1\) — day-0 dissolved oxygen in the seed control bottle (mg/L)
- \(B_2\) — dissolved oxygen in the seed control bottle after 5 days (mg/L)
- \(f\) — seed fraction: ratio of seed volume in the sample bottle to seed volume in the control bottle
- \(P\) — dilution fraction, the portion of the final bottle volume that is the original sample (entered as a percentage, e.g. 5% = 0.05)
The numerator is the oxygen actually consumed by the organic matter in the sample (after subtracting the seed's own demand); dividing by \(P\) scales this depletion back up to the demand of the undiluted water.
Biochemical Oxygen Demand – Calculation Example
Example: a wastewater sample is diluted to 5% (1 part sample in 20 parts total) with seeded dilution water. The diluted bottle shows \(D_1 = 9.0\) mg/L initially and \(D_2 = 4.2\) mg/L after 5 days. The seed control bottle drops from \(B_1 = 9.0\) to \(B_2 = 6.6\) mg/L, and the seed fraction is \(f = 0.05\).
\[ D_1 - D_2 = 9.0 - 4.2 = 4.8 \text{ mg/L} \]
\[ (B_1 - B_2)\,f = (9.0 - 6.6) \times 0.05 = 0.12 \text{ mg/L} \]
\[ \mathrm{BOD}_5 = \frac{4.8 - 0.12}{0.05} = \frac{4.68}{0.05} = 93.6 \text{ mg/L} \]
Result: BOD₅ = 93.6 mg/L — a moderately polluted wastewater, well above the 20 mg/L validity threshold of the standard method.
How to Use the BOD Calculator
- Enter the Initial DO (D₁) — the dissolved oxygen of the diluted sample measured immediately after preparing the bottle (day 0), in mg/L.
- Enter the Final DO after 5 Days (D₂) — the dissolved oxygen in the same bottle after 5 days of incubation at 20 °C in the dark.
- Enter the Dilution Fraction (P) as a percentage of the bottle volume that is original sample. A 1:20 dilution means 5%; an undiluted bottle is 100%.
- If the sample was seeded, tick Apply seed correction and fill in Seed Control Initial DO (B₁), Seed Control Final DO (B₂), and the Seed Fraction (f). If no seed was used, leave the box unticked.
- Read the BOD₅ result in mg/L, along with the step-by-step calculation details and any validity warnings from the standard method.
What Is BOD₅ and Why It Matters
The five-day biochemical oxygen demand (BOD₅) is the mass of dissolved oxygen that aerobic microorganisms consume while breaking down the organic matter in a water sample over 5 days at 20 °C. It is the most widely used measure of the biodegradable organic pollution load of a water: the higher the BOD₅, the more oxygen the discharge will strip from the receiving river or lake as it decomposes.
Because most wastewaters would use up all the oxygen in a full bottle long before day 5, the sample is diluted with oxygen-saturated dilution water. The measured depletion then reflects only the diluted mixture, so the formula divides by the dilution fraction \(P\) to recover the demand of the original, undiluted water.
When the water itself contains too few bacteria — typical for treated effluents or industrial wastes — the dilution water is seeded with microorganisms, usually settled domestic sewage or a commercial seed preparation. The seed consumes oxygen too, so its own demand is measured in a parallel seed-control bottle and subtracted, scaled by the seed fraction \(f\), before dividing by \(P\).
BOD₅ is routinely used for treatment-plant design and compliance monitoring, and it complements the Chemical Oxygen Demand (COD), which measures the total oxidizable load (biodegradable and non-biodegradable) in a much faster test. The BOD/COD ratio is a common indicator of how biologically treatable a wastewater is. For the kinetics of the microbial reactions involved, see also the Michaelis–Menten calculator.
Interpreting Typical BOD₅ Values
| Water type | Typical BOD₅ (mg/L) |
|---|---|
| Clean river / stream | 1–5 |
| Moderately polluted river | 5–20 |
| Treated municipal effluent | 10–30 |
| Raw domestic sewage | 150–400 |
| Strong industrial waste | up to several thousand |
Assumptions, Limitations and Practical Tips
- Standard conditions: the calculator assumes incubation for 5 days at 20 ± 1 °C in the dark, per APHA Standard Methods 5210B. Nitrification is not inhibited unless a TCMP inhibitor was added to the bottles.
- Validity criteria: the standard method is only valid if the oxygen depletion is at least 2 mg/L, the residual DO is at least 1 mg/L, and the resulting BOD₅ is at least 20 mg/L. The calculator flags violations of these criteria — if they appear, repeat the test with a more suitable dilution.
- Choosing the dilution: aim for a depletion of roughly 3–5 mg/L from a starting DO near saturation (~9 mg/L at 20 °C). For an unknown sample, prepare several bottles with different dilutions (e.g. 1%, 2%, 5%, 10%) and use the valid one.
- Seed fraction: when the same seeded dilution water is used in both bottles, \(f\) typically equals the dilution fraction expressed as a decimal (e.g. 0.05 for a 5% dilution). Adjust it if the seed volumes differ.
- Immediate readings: D₁ and B₁ must be measured at day 0, not assumed from saturation tables, especially after sample mixing and temperature equilibration.
- Dilution water blank: a separate blank check on the unseeded dilution water should not consume more than 0.2 mg/L over 5 days; higher blanks contaminate every result.
Frequently Asked Questions
How do I convert a dilution ratio into the percentage P?
P is the sample volume divided by the total bottle volume, expressed as a percent. For a 1:20 dilution (1 part sample plus 19 parts dilution water), \(P = 1/20 = 0.05 = 5\%\). A 1:30 dilution is about 3.3%, and an undiluted bottle is 100%. Entering the ratio instead of the percent is the single most common mistake — it makes the result 100 times too large or too small.
What is a good or acceptable BOD₅ value?
It depends on the water. Clean rivers run 1–5 mg/L, and treated municipal effluent typically 10–30 mg/L. Many discharge permits require a monthly average around 20–30 mg/L for secondary-treated sewage (e.g. 30 mg/L under US secondary treatment rules), so anything at or below that level is generally acceptable. Values above 100 mg/L indicate strong pollution. Always check your specific permit or local regulation, since limits vary by receiving water and jurisdiction.
What is the difference between BOD₅ and CBOD₅?
Standard BOD₅ includes oxygen used by nitrifying bacteria, which oxidize ammonia to nitrate. CBOD₅ (carbonaceous BOD) is measured after adding a nitrification inhibitor such as TCMP, so it reflects only the carbonaceous organic load. For raw sewage the two are similar, but for treated effluents nitrification can contribute a large share of the total demand, making CBOD₅ noticeably lower. Use the CBOD option when your permit specifies it or when nitrifiers distort the result.
Can I convert between BOD₅ and COD?
There is no exact conversion, because COD also oxidizes non-biodegradable compounds. As rough rules of thumb, COD is often about 1.5–2.5 times BOD₅ for typical municipal wastewater, and a BOD/COD ratio above about 0.5 indicates readily biodegradable waste. For a specific waste stream, measure both and establish your own empirical ratio rather than relying on generic factors.
Why does the test run for exactly 5 days at 20 °C?
The 5-day, 20 °C protocol is a historical convention from Britain, chosen so that the value approximates the oxygen demand exerted in a receiving stream over its typical travel time. It captures only part of the ultimate demand: at 20 °C, BOD₅ is roughly 60–70% of the 20-day ultimate BOD because slow reactions like nitrification continue beyond day 5. The fixed conditions simply make results comparable between laboratories.
My result came back as an error or a validity warning — what went wrong?
Two frequent causes: if D₂ exceeds D₁, no oxygen was consumed — recheck the readings or the bottle labels, since mixing up samples is common. If the seed correction exceeds the measured depletion, the seed demand was too high relative to the sample; use a smaller seed fraction. Validity warnings mean the dilution was unsuitable: too little depletion needs a stronger sample fraction, residual DO below 1 mg/L needs a weaker one. Run several dilutions and keep the valid bottle.
When must I apply the seed correction?
Seed the dilution water when the sample itself carries too few microorganisms to degrade its organic matter — typical for biologically treated effluent, disinfectant-containing water, or industrial waste. If you seed, always run a parallel seed control bottle and subtract its demand; skipping this step inflates your BOD₅. For raw domestic sewage or river water, which already contain abundant bacteria, seeding and the correction are usually unnecessary.