A 10 mL sample is added to a 300 mL BOD bottle. Unseeded dilution water is added
ID: 943961 • Letter: A
Question
A 10 mL sample is added to a 300 mL BOD bottle. Unseeded dilution water is added to the sample bottle and the initial dissolved oxygen concentration is measured as 8.5 mg/L. After the sample is sealed, the laboratory incorrectly takes a measurement of DO reading on day 6 of 3 mg/L. The BOD reaction rate coefficient for the sample is 0.30/d.
(a) estimate what the BOD5 should have been.
(b) Estimate what the DO reading should have been on day 5.
(c) Determine the ultimate carbonaceous BOD of the sample.
Explanation / Answer
a) BOD5 = (Initial DO - Final DO ) Xdilution factor
Dilution factor = Volume of bottle / sample volume = 300 / 10
BOD 5 = 8.5 - 3 X 30 = 165
b) DO on day 5 should be
BOD5 = (Initial DO - Final DO ) X Dilution factor
Final DO at day 5 = Initial DO - (BOD5 / Dilution factor )
Final DO at day 5 = 8.5 - (165/30) = 3
c) We know that
Ultimate BOD = BOD5 / (1-e-kt)
Ultimate BOD = 165 / (1-e-0.3X5) = 165 / 1-0.223 = 212.35 mg / L
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