Biology, asked by bhanukumar331, 8 months ago

A wastewater has BOD of 1500 mg/L and COD of 2400 mg/L. Assuming 80% treatment efficiency in an activated sludge bioreactor, the BOD/COD ratio of treated effluent will be? (please explain)

Answers

Answered by pharmashailesh
0

Answer:

ILIAS 3. Typical values for the ratio of BOD/COD for untreated municipal wastewater are in the range from 0.3 to 0.8 (see in table 3). If the BOD/COD ratio for untreated wastewater is 0.5 or greater, the waste is considered to be easily treatable by biological means

Explanation:

Answered by smartbrainz
4

The ratio between biological oxidation demand and COD is equal to 0.5

Explanation:

  • there is 80% efficiency in the active sludge bioreactor. As a result of this there is 80% decrease in the BOD is 0.8 x1500= 1200. The decrease of the biological oxygen demand is 1200.
  • as a result of this the ratio between the biological oxidation demand and COD is equal to 1200/2400= 0.5 or 50%
  • the ratio helps in the determination whether the the treatment is good enough for decrease in pollution

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State one Similarity between digestion and decomposition - Brainly.in

https://brainly.in/question/11169306

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