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1-day BOD 300 mg/l to 5-day at 30°C

Short answer

Ultimate BOD L =300/(1-10^(-0.1)) =1458.6 mg/L. With temperature coefficient θ=1.047, k30 =0.1×1.047^10 =0.1583/day. Then5-day30°C BOD = L(1-10^(-k30×5)) =1222.9 mg/L.

If 1 day BOD of sewage sample at 20°C is 300 mg/l, what will be its 5-day BOD at 30°C? Consider rate constant of 0.1/day (base 10) at 20°C.

Given: y1y_1 at 20°C = 300 mg/L, k20=0.1k_{20}=0.1 day⁻¹ (base 10). Assume θ=1.047\theta=1.047 for the temperature correction. The ultimate BOD LL is independent of temperature.

Step 1: Ultimate BOD

L=y11−10−k20 t=3001−10−0.1×1=3001−0.7943=1458.6 mg/LL=\frac{y_1}{1-10^{-k_{20}\,t}}=\frac{300}{1-10^{-0.1\times1}}=\frac{300}{1-0.7943}=1458.6\ \text{mg/L}

Step 2: Rate constant at 30°C

k30=k20(1.047)30−20=0.1×1.04710=0.1583 day−1k_{30}=k_{20}(1.047)^{30-20}=0.1\times1.047^{10}=0.1583\ \text{day}^{-1}

Step 3: 5-day BOD at 30°C

y5=L(1−10−k30×5)=1458.6(1−10−0.7915)=1222.9 mg/Ly_5=L\left(1-10^{-k_{30}\times5}\right)=1458.6\left(1-10^{-0.7915}\right)=1222.9\ \text{mg/L}

Answer: 5-day BOD at 30°C = 1222.9 mg/L.

✓ Answer checked by Claude Opus 5.5 (high effort)

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