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First and second stage BOD

Short answer

First-stage BOD (carbonaceous) involves oxidation of organic carbon, largely complete in20 days. Second-stage BOD (nitrogenous) starts after ~8-10 days with ammonia oxidation. Assuming standard rate constant k=0.1/day (base10) at20°C, ultimate BOD =400/(1-10^(-0.5)) =585.0 mg/L.

Discuss first-stage and second stage BOD with a suitable figure. Calculate ultimate BOD if 5 day BOD of sewage sample at 20°C is 400 mg/l.

First-stage BOD (carbonaceous)

In the first stage, bacteria oxidise the carbonaceous organic matter into CO₂ and water. It is practically complete in about 20 days at 20°C (with k=0.1k=0.1 day⁻¹, about 68% in 5 days and 99% in 20 days). The first stage is called carbonaceous BOD (CBOD).

Second-stage BOD (nitrogenous)

After about 8-10 days, nitrifying bacteria (Nitrosomonas, Nitrobacter) oxidise ammonia to nitrite and then to nitrate, using more oxygen:

NH4+→NitrosomonasNO2−→NitrobacterNO3−NH_4^+\xrightarrow{\text{Nitrosomonas}}NO_2^-\xrightarrow{\text{Nitrobacter}}NO_3^-

This is the nitrogenous BOD (NBOD). Because the nitrifiers grow slowly, it does not appear in the standard 5 day test at 20°C.

 BOD exerted
   |                  ______ second stage
   |            _____/ (nitrification)
   |      _____/___ 
   |   __/ first stage ends (L)
   |  /
   | /
   +----+----+----+----+----> days
   0    5   10   15   20

The first-stage curve is yt=L(1−10−kt)y_t=L(1-10^{-kt}). The second-stage curve starts at about the 8th-10th day and rises above it.

Ultimate BOD calculation

The rate constant is not given, so the standard value K20=0.1K_{20}=0.1 day⁻¹ (base 10) is assumed.

L=y51−10−kt=4001−10−0.1×5=4001−0.3162=585.0 mg/LL=\frac{y_5}{1-10^{-kt}}=\frac{400}{1-10^{-0.1\times5}}=\frac{400}{1-0.3162}=585.0\ \text{mg/L}

Answer: Ultimate first-stage BOD = 585.0 mg/L (with K=0.1K=0.1/day base 10; with the equivalent K=0.23K=0.23/day base ee the result is nearly the same, 585.3 mg/L).

✓ Answer checked by Claude Opus 5.5 (high effort)

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