Model the exertion of biochemical oxygen demand over time with first-order deoxygenation kinetics, and check the percent BOD removal across a treatment process. Enter the ultimate BOD, the rate constant, and the incubation time to find the BOD exerted by day t.
Biochemical oxygen demand (BOD) is the master variable of wastewater engineering: it measures how much dissolved oxygen microorganisms will consume as they break down the organic matter in a water sample. This calculator does two things — it projects how much BOD is exerted by any incubation day using first-order deoxygenation kinetics, and it computes the percent BOD removal a treatment process achieves between influent and effluent.
BOD is the mass of oxygen (mg O₂ per litre of sample) that aerobic bacteria require to oxidise the biodegradable organics in water. Because the reaction is slow, the standard test incubates a sample in the dark at 20 °C and measures oxygen consumed over five days — this is BOD5, the most-reported value in discharge permits. BOD5 is a proxy for organic pollution strength: raw domestic sewage is typically 100–400 mg/L, while a well-treated secondary effluent is under 30 mg/L.
If you let the test run long enough, oxygen demand approaches a plateau called the ultimate (carbonaceous) BOD, L₀ — the total oxygen needed to fully oxidise the organics. The BOD exerted by any time t follows first-order kinetics: BOD_t = L₀·(1 − e^(−k·t)). With a typical rate constant the 5-day test captures roughly two-thirds of the ultimate demand, which is why BOD5 < L₀. Knowing L₀ and k lets you convert between any two incubation times.
The rate constant k (base e, units 1/day) sets how fast the demand is exerted. For untreated domestic wastewater k is roughly 0.20–0.30 /day; well-treated effluents are slower (0.05–0.15 /day) because the easily degraded fraction is already gone. Beware of conventions: laboratory k is sometimes quoted in base 10 (k₁₀), and k_e = 2.303·k₁₀. This calculator uses the base-e form directly in the exponential.
The percent removal η = (BOD_in − BOD_out) / BOD_in × 100 expresses how effectively a unit process strips organic load. United States secondary-treatment standards require at least 85% BOD removal and a 30-day average effluent BOD5 of 30 mg/L or less. Primary clarification alone removes only 25–40% of BOD; the biological stage (activated sludge, trickling filter, or similar) does the heavy lifting to reach the 85%+ target.
BOD measures only the biodegradable organic fraction, using living microorganisms over days. Chemical oxygen demand (COD) uses a strong chemical oxidant to measure essentially all oxidisable matter in a couple of hours, so COD is always greater than or equal to BOD. The BOD/COD ratio is a quick index of biodegradability — a ratio above ~0.5 indicates the wastewater treats well biologically.
Five days is a practical compromise dating to British river studies: it captures a large, repeatable fraction of the carbonaceous demand before nitrification (oxygen demand from ammonia-oxidising bacteria) typically begins to interfere, and it fits a workweek. The ultimate BOD would need 20+ days of incubation, which is impractical for routine monitoring.
Rearrange the kinetics: L₀ = BOD5 / (1 − e^(−k·5)). With k = 0.23 /day, the factor (1 − e^(−1.15)) ≈ 0.68, so the ultimate BOD is roughly BOD5 / 0.68 ≈ 1.46 × BOD5. Use the actual k for your wastewater for an accurate conversion.
United States Clean Water Act secondary-treatment regulations require at least 85% removal of BOD5 (and TSS) as a monthly average, with effluent BOD5 not exceeding 30 mg/L (30-day average) or 45 mg/L (7-day average). Many discharge permits impose stricter limits where the receiving water is sensitive.
Yes. After several days, ammonia-oxidising bacteria begin exerting nitrogenous oxygen demand (NBOD), inflating the result above the carbonaceous demand. To measure only carbonaceous BOD (CBOD), a nitrification inhibitor is added. The first-order model here represents carbonaceous deoxygenation; if nitrification is uninhibited, measured demand at long t will exceed the model.
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