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BOD5 and ultimate BOD, 5% dilution

Short answer

5% dilution, blank DO=5 mg/L, sample DO=0.5 mg/L. Initial mixture DO=4.775 mg/L, after 5 days DO=1 mg/L. BOD5=75.5 mg/L. Assuming K20=0.1/day, ultimate BOD=110.4 mg/L.

The following observations were made on 5% dilution of a sewage sample. The DO of blank is 5 mg/L. The DO of diluted sample after 5 days and incubation at 20°C is 1 mg/L. Calculate BOD5 and ultimate BOD of sample. Assume the DO of original sample 0.5 mg/L.

Initial DO of the diluted sample

The dilution water (blank) has DO = 5 mg/L, the raw sample has DO = 0.5 mg/L, and the dilution is 5% (P=0.05P=0.05 sample, 0.95 dilution water). Initial DO of the mixture:

D1=5(0.95)+0.5(0.05)=4.775 mg/LD_1=5(0.95)+0.5(0.05)=4.775\ \text{mg/L}

The DO after 5 days at 20°C is D2=1D_2=1 mg/L.

5-day BOD

BOD5=D1−D2P=4.775−10.05=75.5 mg/LBOD_5=\frac{D_1-D_2}{P}=\frac{4.775-1}{0.05}=75.5\ \text{mg/L}

Ultimate BOD

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

L=BOD51−10−0.1×5=75.50.6838=110.4 mg/LL=\frac{BOD_5}{1-10^{-0.1\times5}}=\frac{75.5}{0.6838}=110.4\ \text{mg/L}

Answer: BOD₅ = 75.5 mg/L; ultimate BOD ≈ 110.4 mg/L.

✓ Answer checked by Claude Opus 5.5 (high effort)

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