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BOD1 at 23°C to BOD5 at 30°C

Short answer

1-day BOD at 23°C=105 mg/L, K20=0.1/day. Ultimate BOD=452.1 mg/L. At 30°C, K30=0.1583/day, so 5-day BOD=379.0 mg/L.

If one day BOD of a sewage sample at 23°C is 105 mg/l. What will be its five day BOD at 30°C? Assume K20 = 0.1 per day.

Given: y1y_1 at 23°C = 105 mg/L, K20=0.1K_{20}=0.1 day⁻¹ (base 10), θ=1.047\theta=1.047. The ultimate BOD does not change with temperature.

Step 1: Ultimate BOD

K23=0.1×1.0473=0.1148 day−1L=1051−10−K23×1=1051−10−0.1148=452.1 mg/L\begin{aligned} K_{23}&=0.1\times1.047^{3}=0.1148\ \text{day}^{-1}\\ L&=\frac{105}{1-10^{-K_{23}\times1}}=\frac{105}{1-10^{-0.1148}}=452.1\ \text{mg/L} \end{aligned}

Step 2: 5-day BOD at 30°C

K30=0.1×1.04710=0.1583 day−1y5=L(1−10−K30×5)=452.1(1−10−0.7915)=379.0 mg/L\begin{aligned} K_{30}&=0.1\times1.047^{10}=0.1583\ \text{day}^{-1}\\ y_5&=L\left(1-10^{-K_{30}\times5}\right)=452.1\left(1-10^{-0.7915}\right)=379.0\ \text{mg/L} \end{aligned}

Answer: Five-day BOD at 30°C = 379.0 mg/L.

✓ Answer checked by Claude Opus 5.5 (high effort)

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