Civil Engineering - Waste Water Engineering - Discussion

4. 

If 2% solution of a sewage sample is incubated for 5 days at 20°C and depletion of oxygen was found to be 5 ppm, B.O.D. of the sewage is

[A]. 200 ppm
[B]. 225 ppm
[C]. 250 ppm
[D]. None of these.

Answer: Option C

Explanation:

No answer description available for this question.

Md Seraj said: (Feb 26, 2014)  
D.O consume = 5 ppm.

Dilution factor = 100/2 = 50.

B.O.D of the sewage = 5*50 = 250 ppm.

Satya said: (Dec 24, 2014)  
2% solution bod 5 ppm.

100% solution 100/2*5 = 250 ppm.

Nilima said: (Jan 3, 2015)  
Dilution ratio = 100/% of solution.
= 100/2.
= 50.
B.O.D = 5 x 50 = 250 ppm.

Dinesh Sharma said: (Jan 26, 2015)  
BOD = Oxygen consumed x dilution factor. So, BOD = 5x100/2 = 250ppm.

Roshan said: (Mar 2, 2015)  
P = Dilution factor = Fraction of waste water in the sample of 300 ml BOD bottle = 2/100 = 0.02.

BOD = Oxygen consumed/P = 5/0.02 = 250 ppm.

Asaithambi said: (Sep 22, 2016)  
Considering total quantity of sewage =100%, solution of sewage = 2%(sample), depletion of oxygen = 5ppm.

Dilution factor = 100/2 = 50.
B.O.D = 50 * 5 = 250ppm.

Santanu said: (Mar 5, 2017)  
BOD = (DO initial - DO final) x dilution ratio.
= 5 x (100/2).
= 250ppm.

Nahid Farjana said: (Sep 26, 2017)  
DF= 2%= 0.02.
DO= 5ppm,
BOD= consumed DO/DF.
= 5/0.02= 250.

Noushin said: (Mar 1, 2018)  
Given,

Depletion of oxygen=5ppm
Solution of sewage=2%

Consider,
Total quantity of sewage= 100%

We have,
BOD = (DO initial -DO final) * Dilution factor.
Dilution factor =(Vol of diluted sample/Vol of in diluted sewage sample).
DO initial-DO final means depleted oxygen content.

Therefore,
BOD = 5 * (100/2) = 250ppm.

Bujjibabu said: (Mar 19, 2018)  
BOD=(DO intial - DO final) *Dilution factor.
Dilution factor=100/% of solution,
So, DF=100/2 = 50,
BOD=5*50 = 250.

Raaj said: (Aug 19, 2018)  
Thank you all for explaining it.

Uday Kumar said: (Nov 2, 2018)  
BOD = D.F * DO consumed.
D.F. = 100/%of solution,
D.F. = 50,
BOD = 5 * 50 = 250 ppm.

Robin said: (May 3, 2019)  
Oxygen Depletion = DOi - DOf = 5 ppm (Given).
Decimal fraction = 0.02 (Given).
BOD = (DOi - DOf)/P.
So, BOD = 5/0.02 = 250 mg/L.

Bruno said: (Dec 30, 2019)  
(Initial do - final do )*dilution factor.
5*50 = 250.

Tyson said: (Jan 6, 2020)  
P = Dilution factor = Fraction of waste water in the sample of 300 ml BOD bottle = 2/100 = 0.02.

BOD = Oxygen consumed/P = 5/0.02 = 250 ppm.

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