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Waste Water Engineering - Engineering

Q1:

If a 2% solution of sewage sample is incubated for 5 days at 20°C and the dissolved oxygen depletion was found to be 8 mg/l. The BOD of the sewage is

A 100 mg/l

B 200 mg/l

C 300 mg/l

D 400 mg/l.

ANS:D - 400 mg/l.

Given:

  • Initial dissolved oxygen concentration (D0​) = 8 mg/L
  • Dissolved oxygen concentration after 5 days (D5​) = 0 mg/L
  • Dilution factor = 0.02
  • Temperature = 20°C
Using the BOD formula: BOD=(D0​−D5​)×Dilution×100​ / F Substituting the given values: BOD=(8−0)×0.02×100/1​  BOD=18×0.02×100​ /1 BOD=18×2​/1 BOD=16mg/L The calculated BOD for the sewage sample is 16 mg/L. Since the calculated BOD is 16 mg/L, the option "400 mg/l" is incorrect based on the provided information. It seems there may be an error or discrepancy in the options provided. The accurate BOD value based on the given data is 16 mg/L.