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BOD1 at 22°C to BOD5 at 30°C

Short answer

1-day BOD at 22°C=310 mg/L, K20=0.12/day. Ultimate BOD=1186.3 mg/L. At 30°C, K30=0.1900/day, so 5-day BOD=1053.1 mg/L.

BOD1 at 22°C of a sewage sample is 310 mg/l. What will be its BOD5 at 30°C? Assume reaction rate K20 = 0.12 per day.

Given: y1y_1 at 22°C = 310 mg/L, K20=0.12K_{20}=0.12 day⁻¹ (base 10). Temperature coefficient θ=1.047\theta=1.047. The ultimate BOD LL is independent of temperature.

Step 1: Rate constant at 22°C and ultimate BOD

K22=0.12×1.0472=0.1315 day−1L=y11−10−K22×1=3101−10−0.1315=1186.3 mg/L\begin{aligned} K_{22}&=0.12\times1.047^{2}=0.1315\ \text{day}^{-1}\\ L&=\frac{y_1}{1-10^{-K_{22}\times1}}=\frac{310}{1-10^{-0.1315}}=1186.3\ \text{mg/L} \end{aligned}

Step 2: Rate constant at 30°C

K30=0.12×1.04710=0.1900 day−1K_{30}=0.12\times1.047^{10}=0.1900\ \text{day}^{-1}

Step 3: 5-day BOD at 30°C

y5=L(1−10−K30×5)=1186.3(1−10−0.9498)=1053.1 mg/Ly_5=L\left(1-10^{-K_{30}\times5}\right)=1186.3\left(1-10^{-0.9498}\right)=1053.1\ \text{mg/L}

Answer: BOD₅ at 30°C = 1053.1 mg/L.

✓ Answer checked by Claude Opus 5.5 (high effort)

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