Calculate the BOD value for a sample of water in a lab test for BOD. If one BOD

Question

Calculate the BOD value for a sample of water in a lab test for BOD. If one BOD test has an initial Dissolved Oxygen of 9.2 mg/L and a final of 2.2 mg/L and uses 10 mls of sample in the standard bottle, then what is the actual BOD of the sample itself? If the dilution was not known and the final DO level was 0.0 mg/L, what would the BOD value or range of values be for above conditions? What test conditions would be changed before it was re-run? Why? If the sample set up used 15 mls of sample and the beginning and ending values of DO were 9.0 and 2.0, what would the BOD be? The a) value is influent and c) value is effluent for a Primary Sedimentation device, so what is the removal efficiency?

Explanation / Answer

BOD = Oxygne demand

change in concentations --> 9.2-2.2 = 7.0 mg/L

V = 10 mL = 10*10^-3 L

BOD = 7*10*10^-3 = 0.07 mg of O2

b)

Impossible to know since no Oxygen is present, therefore no consumption of BOD

c)

V = 15 mL

change in concentration -- >9-2 = 7 mg/L

DEmand = C*V = 7*15*10^-3 = 0.105 mg of O2

d)

Efficiency = (in-out)/in * 100% = (0.105 - 0.07 ) / 0.105 * 100 = 33.3%

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