If the 5-day BOD (text BOD _5) of a wastewater sample is 150,text mg/L at 20°C with a rate constant K = 0.23,text day ^ -1 , the Ultimate BOD (L_0) is approximately:

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Civil Engineering QUESTION #6581

Question 1

If the 5-day BOD (\(\text{BOD}_5\)) of a wastewater sample is \(150\,\text{mg/L}\) at \(20°C\) with a rate constant \(K = 0.23\,\text{day}^{-1}\), the Ultimate BOD (\(L_0\)) is approximately:

102.5 mg/L✔️
473.7 mg/L
219.5 mg/L
47.5 mg/L

Correct Answer Explanation

Using \(\text{BOD}_5 = L_0(1 - e^{-Kt})\): \(150 = L_0(1 - e^{-0.23 \times 5}) = L_0(1 - e^{-1.15}) = L_0 \times 0.683\). Therefore \(L_0 = \dfrac{150}{0.683} \approx 219.6\,\text{mg/L}\). Note: Answer key states (a) — verify with base-10 form \(K_{10} = 0.1/\text{day}\): \(L_0 = \dfrac{150}{1-10^{-0.1\times5}} \approx 219\,\text{mg/L}\).

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