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BOD5 at 30°C converted to 20°C

Short answer

1-day BOD at 30°C=170 mg/L, K20=0.12/day. Ultimate BOD=479.9 mg/L. At 20°C, K20=0.12/day, so 5-day BOD=359.3 mg/L.

The BOD5 of a sewage incubated for one day at 30°C has been found to be 170 mg/l. What will be the 5 day BOD at 20°C? Assume K = 0.12 per day (base 10) at 20°C.

The data are read as: the 1-day BOD at 30°C is 170 mg/L, K20=0.12K_{20}=0.12 day⁻¹ (base 10), and the 5-day BOD at 20°C is required. θ=1.047\theta=1.047.

Step 1: Rate constant at 30°C

K30=K20(1.047)30−20=0.12×1.04710=0.1900 day−1K_{30}=K_{20}(1.047)^{30-20}=0.12\times1.047^{10}=0.1900\ \text{day}^{-1}

Step 2: Ultimate BOD

L=y11−10−K30t=1701−10−0.1900×1=1701−0.6457=479.9 mg/LL=\frac{y_1}{1-10^{-K_{30}t}}=\frac{170}{1-10^{-0.1900\times1}}=\frac{170}{1-0.6457}=479.9\ \text{mg/L}

Step 3: 5-day BOD at 20°C

y5=L(1−10−K20×5)=479.9(1−10−0.12×5)=479.9(1−0.2512)=359.3 mg/Ly_5=L\left(1-10^{-K_{20}\times5}\right)=479.9\left(1-10^{-0.12\times5}\right)=479.9\left(1-0.2512\right)=359.3\ \text{mg/L}

Answer: The 5-day BOD at 20°C = 359.3 mg/L (ultimate BOD = 479.9 mg/L).

✓ Answer checked by Claude Opus 5.5 (high effort)

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