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Environmental Engineering DISCUSSION
In the lab we ran a 5-day BOD test on a wastewater sample. 10 mL of sample was diluted to 300 mL in a BOD bottle with unseeded dilution water. The dissolved oxygen was 8.6 mg/L at the start and 4.2 mg/L after 5 days at 20 °C.
How is the BOD5 of the original sample calculated from this, how do I know whether the dilution was suitable, and how is the ultimate BOD estimated from the 5-day value? I also do not understand why the COD of the same sample is so much higher.
For an unseeded test: BOD5 = (D1 - D2) / P, where D1 and D2 are the initial and final dissolved oxygen and P is the decimal fraction of sample in the bottle. P = 10 / 300 = 0.0333, so BOD5 = (8.6 - 4.2) / 0.0333 = 132 mg/L.
The dilution is needed because water holds only about 9 mg/L of oxygen at 20 °C. An undiluted sample with a BOD of 132 mg/L would use up all the oxygen within the first day and the test would tell you nothing beyond that.
The standard acceptance criteria are a depletion of at least 2 mg/L and a residual of at least 1 mg/L. Yours are 4.4 and 4.2 mg/L, so the result is valid. Bottles outside those limits are discarded, which is why several dilutions are normally set up for each sample, chosen from the expected BOD: with 300 mL bottles, a sample volume of roughly 1200 / (expected BOD) mL uses about 4 mg/L of oxygen.
If the sample has no bacteria of its own, as with disinfected or industrial effluents, the dilution water must be seeded and the seed's own oxygen demand subtracted: BOD = ((D1 - D2) - (B1 - B2) × f) / P, where B1 - B2 is the depletion of the seed control and f the ratio of seed in the sample bottle to seed in the control.
Ultimate BOD comes from the first-order model BOD_t = L × (1 - e^(-k × t)), where L is the ultimate carbonaceous BOD and k the rate constant. With k = 0.23 per day (base e), a typical value for domestic wastewater at 20 °C, the 5-day fraction is 1 - e^(-1.15) = 0.68, so L = 132 / 0.68 = about 193 mg/L.
Watch the base. Older references give k in base 10, where the same wastewater has k = 0.10 per day and the formula is 1 - 10^(-k × t). Mixing the two is a common source of wrong answers. The rate also changes with temperature: k_T = k_20 × 1.047^(T - 20), so at 30 °C the same sample would exert its demand noticeably faster.
COD is higher because it measures something different. A strong chemical oxidant under hot acid conditions oxidises nearly all organic matter in a couple of hours, including material that bacteria degrade slowly or not at all, plus some inorganic species. BOD5 counts only what microorganisms consume in five days, which is roughly two thirds of even the biodegradable part.
For untreated domestic wastewater the BOD5/COD ratio is commonly around 0.4 to 0.6. A much lower ratio suggests poorly biodegradable or toxic constituents, and biologically treated effluent has a low ratio because the easily degradable fraction has already been removed. Once the ratio for a given plant is established, the fast COD test is used for daily control and BOD for compliance.
Two interferences deserve a mention because they inflate results. Nitrifying bacteria oxidise ammonia and consume oxygen as well; this is usually minor in raw sewage within five days but significant in treated effluent, so a nitrification inhibitor is added when carbonaceous BOD (CBOD) is what the permit asks for, and the result should be labelled as such.
Residual chlorine, extreme pH and toxic substances do the opposite and suppress the bacteria. A sign of toxicity is a calculated BOD that increases as the sample is diluted more. Each batch should also include a dilution-water blank and a glucose-glutamic acid check standard to show that the seed and the water are behaving.
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