Civil Engineering - UPSC Civil Service Exam Questions - Discussion
Discussion Forum : UPSC Civil Service Exam Questions - Section 20 (Q.No. 11)
11.
What is the BOD5 at 20°C of a waste that yields an oxygen consumption of 2 mg/l from a 0.5% diluted sample ?
Discussion:
2 comments Page 1 of 1.
Chethan said:
8 years ago
Thank you @Deepak.
Deepak said:
8 years ago
BOD5= (Initial oxygen level - Final Oxygen level) x Dilution Factor.
Dilution Factor = Volume of raw sewage/Volume of treated(or Diluted) Sample.
ie BOD5 = 2 x (100 /0.5).
BOD5 = 400 mg/l.
Dilution Factor = Volume of raw sewage/Volume of treated(or Diluted) Sample.
ie BOD5 = 2 x (100 /0.5).
BOD5 = 400 mg/l.
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