GATE 2023 ES – Question 58
A BOD bottle contains 25 mL wastewater diluted to 300 mL. Wastewater ultimate BOD75 mg/L, initial diluted DO6.5 mg/L, and day 7 DO1.25 mg/L;day 5 was missed. At equal temperature with clean dilution water, find wastewater BOD₅ in mg/L, to two decimals.
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Correct answer: 54.4 to 54.8
Explanation
Dilution fraction is 25/300=1/12. Seven-day diluted DO depletion is 5.25 mg/L, so wastewater BOD₇ is 5.25(12)=63 mg/L. First-order exertion gives $63=75(1-e^{-7k})$,so $k=-\ln(.16)/7=.2617974$ per day. Then $BOD_5=75(1-e^{-5k})=\mathbf{54.74}$ mg/L. This uses the first-order BOD model implicit in the interpolation.