Civil Engineering - Waste Water Engineering - Discussion

2. 

If the depletion of oxygen is found to be 2.5 mg/litre after incubating 2.5 ml of sewage diluted to 250 ml for 5 days at 20°C, B.O.D. of the sewage is

[A]. 50 mg/l
[B]. 100 mg/l
[C]. 150 mg/l
[D]. 200 mg/l
[E]. 250 mg/l.

Answer: Option E

Explanation:

No answer description available for this question.

Vikash said: (May 25, 2014)  
BOD = DO X(DF).

2.5 X (250/2.5).

250mg/l.

Hem said: (Sep 1, 2015)  
DO and DF stands for?

Md Gulam Gos said: (Nov 24, 2015)  
DO: Dissolved Oxygen.

DF: Dilution Factor.

Yaregal T. said: (Jan 10, 2018)  
BOD =200 because don't forget 5 days.

Alvin A. said: (Sep 6, 2018)  
How did you get 200?

Ram Prasad said: (Sep 24, 2018)  
That is BOD 5 days at 20° temperature.

Shweta said: (Dec 3, 2018)  
Answer is 250.

Swapnil Biraris said: (Jan 5, 2019)  
250 is the correct answer here.

Raisa Paul said: (Feb 25, 2020)  
D.F = sewage dilution/depletion of oxygen.

Shubham Siddharth Nayak said: (May 22, 2020)  
BOD=(DO INITIAL-DO FINAL) * DF.

DO INITIAL - DO FINAL =2.5 mg/l (Given),
DF = (total volume of sewage and distilled water/volume of sewage sample)=250/2.5=100,
BOD = 2.5 * 100 = 250.

Teja said: (Sep 9, 2020)  
Thanks for explaining @Nayak.

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