BOD Calculator
Work out the five day biochemical oxygen demand of a water sample from its dissolved oxygen readings and dilution.
What it does
Biochemical oxygen demand measures how much oxygen living microorganisms need to break down organic matter in a water. The standard test is a five day version called BOD5: a sample is put in a bottle with dilution water, the dissolved oxygen is measured, the bottle is closed and left in the dark for five days, and the oxygen is measured again.
The difference between the two readings is the oxygen the sample used, but that difference is not the demand of the original sample. A bottle holds 300 mL and a small part of it is the sample, so the reading has to be scaled back up by the dilution factor. The EPA writes the formula as the oxygen loss multiplied by the dilution factor, where the dilution factor is the total volume divided by the sample volume.
The same reasoning turns a concentration into a load. Once you have the demand in milligrams per liter, multiply by the flow and you get the mass of oxygen the stream demands per day, which is the number a permit is written against.
The method has acceptance rules. The USGS field manual asks for at least 2 mg/L of depletion so the number is measurable, and at least 1 mg/L of dissolved oxygen left at the end so the test was not oxygen starved. This page flags a result that misses either target.
A proper BOD test also needs seed correction, a dilution water blank and a check on the incubation temperature, and laboratories are expected to run more than one dilution. The calculation here handles the arithmetic on the readings you supply and leaves the quality work to the protocol.
How to use
- Enter the dissolved oxygen of the diluted sample before incubation.
- Enter the dissolved oxygen of the same bottle after five days.
- Enter the volume of sample in the bottle.
- Enter the dilution water added to fill the bottle.
- Enter the flow if you want the daily load.
- Check the quality line: a valid bottle depletes at least 2 mg/L and keeps at least 1 mg/L of oxygen.
Formula
- D1
- Dissolved oxygen in the diluted sample before incubation, in mg/L.
- D2
- Dissolved oxygen in the same bottle after five days, in mg/L.
- S
- Sample volume divided by total volume in the bottle.
- Q
- Flow in million gallons per day.
- 8.34
- Conversion factor that turns mg/L and MGD into pounds per day.
The 8.34 factor assumes a gallon of water weighs 8.34 pounds, so it is a close estimate for dilute streams. The page does not apply a seed correction, which is needed when the sample does not carry enough microorganisms on its own, and it does not subtract a dilution water blank. A seeded test also uses the adjusted formula from the standard method.
Worked example
Example: 60 mL of sample in a 300 mL bottle at 2 MGD
Defaults: 7.8 mg/L initial oxygen, 2.4 mg/L final oxygen, 60 mL of sample with 240 mL of dilution water, and a flow of 2 million gallons per day.
- Oxygen used in the bottle: 7.8 - 2.4 = 5.4 mg/L.
- Dilution factor: 300 mL total / 60 mL sample = 5.
- BOD5: 5.4 x 5 = 27 mg/L.
- Daily load: 27 mg/L x 2 MGD x 8.34 = 450.36 lb per day.
- In metric the same flow is 7,570.8 m3 per day and the load is 204.41 kg per day.
- Quality check: depletion is 5.4 mg/L and 2.4 mg/L of oxygen remains, so the result sits inside the usual range.
- The tool shows 27 mg/L, 5.4 mg/L, 1 to 5 (60 mL sample in 300 mL), 450.36 lb per day and within the usual range.
Reference table
| Expected BOD5 | Sample | Dilution water |
|---|---|---|
| 0 to 7 mg/L | 300 mL | 0 mL |
| 6 to 21 mg/L | 100 mL | 200 mL |
| 12 to 42 mg/L | 50 mL | 250 mL |
| 30 to 105 mg/L | 20 mL | 280 mL |
| 60 to 210 mg/L | 10 mL | 290 mL |
| 120 to 420 mg/L | 5 mL | 295 mL |
| 300 to 1,050 mg/L | 2 mL | 298 mL |
| 600 to 2,100 mg/L | 1 mL | 299 mL |
Limits
- No seed correction, no blank subtraction and no check of the incubation temperature, all of which a laboratory handles.
- Uses one dilution, while good practice runs several and averages the ones that pass the quality rules.
- The load factor of 8.34 assumes the weight of water, so it is approximate for unusual streams.
- Reads a five day result only, so a nitrification inhibitor test for carbonaceous demand is outside its scope.
Common mistakes
- Using the initial and final readings of the dilution water rather than of the diluted sample.
- Forgetting to divide by the dilution fraction, which reads the oxygen loss in the bottle as the demand of the sample.
- Entering the sample volume as a percentage of the bottle instead of milliliters.
- Ignoring the quality line, since a bottle that ran out of oxygen gives a number that looks low but is not valid.
- Mixing up the order of the readings, so the final oxygen is larger than the initial one.
Try it yourself
Quick checks. Answers sit tucked under each question.
A bottle starts at 8.5 mg/L and ends at 3.5 mg/L with 50 mL of sample in 300 mL. What is the BOD5?
Oxygen used is 5.0 mg/L and the dilution factor is 6, so BOD5 = 5.0 x 6 = 30 mg/L. At 5 MGD the load would be 30 x 5 x 8.34 = 1,251.00 lb per day.
A sample starts at 9.0 mg/L and ends at 6.0 mg/L with 100 mL of sample in 300 mL. What is the BOD5 and the load at 1 MGD?
Oxygen used is 3.0 mg/L, the dilution factor is 3, so BOD5 = 9.00 mg/L. At 1 MGD the load is 9 x 1 x 8.34 = 75.06 lb per day.
A dilute sample starts at 6.5 mg/L and ends at 0.9 mg/L with 20 mL of sample in 300 mL. What is the BOD5, and does the result pass the quality check?
Oxygen used is 5.6 mg/L and the dilution factor is 15, so BOD5 = 84.00 mg/L. The depletion passes the 2 mg/L test, but only 0.9 mg/L of oxygen remains, so the quality line asks you to check the dilution.
Questions
What does BOD5 measure?
The amount of dissolved oxygen that microorganisms need to break down the organic matter in a water sample over five days, expressed in milligrams per liter.
Why is the sample diluted?
A 300 mL bottle holds about 8 mg of oxygen. A strong wastewater sample can use that in a day, so it has to be diluted for the reading to be meaningful. The dilution factor then scales the result back to the original sample.
What is a good BOD5 value?
Municipal sewage treatment plants usually aim for 200 mg/L or less in the effluent. Streams with very low flow can be far lower, and the number that matters is the one written into the permit for the water body receiving the discharge.
How do I turn BOD into a load?
Multiply the concentration in mg/L by the flow in million gallons a day and by 8.34. That gives pounds per day, which is how permit limits are written.
What does COD do that BOD does not?
Chemical oxygen demand measures the same kind of load with a chemical reaction instead of living organisms, so it gives an answer in about two hours. BOD5 takes five days and counts only the biodegradable part.
Can this tool handle a CBOD5 sample?
Not directly, because a nitrification inhibitor changes the result and the seeded calculation is different. The arithmetic on the oxygen readings still applies once you have the right readings.
Sources
- US EPA: NPDES Reporting Requirements Handbook, Appendix A math formulas
- US Geological Survey: Five Day Biochemical Oxygen Demand, National Field Manual
- Electronic Code of Federal Regulations: 40 CFR 136.3, approved analytical methods