Civil Engineering - Waste Water Engineering - Discussion
Discussion Forum : Waste Water Engineering - Section 1 (Q.No. 4)
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
Discussion:
16 comments Page 1 of 2.
Noushin said:
8 years ago
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.
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.
(3)
Asaithambi said:
9 years ago
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.
Dilution factor = 100/2 = 50.
B.O.D = 50 * 5 = 250ppm.
Butum said:
4 years ago
BOD means total amount of microorganisms present in the sample. Let it be 'x'.
Given;
2% of 'X' = 5 ppm.
So, x = 5/0.02 ppm.
X = 250 ppm (total BOD).
Given;
2% of 'X' = 5 ppm.
So, x = 5/0.02 ppm.
X = 250 ppm (total BOD).
(7)
Roshan said:
1 decade ago
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.
BOD = Oxygen consumed/P = 5/0.02 = 250 ppm.
Tyson said:
6 years ago
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.
BOD = Oxygen consumed/P = 5/0.02 = 250 ppm.
(6)
Robin said:
6 years ago
Oxygen Depletion = DOi - DOf = 5 ppm (Given).
Decimal fraction = 0.02 (Given).
BOD = (DOi - DOf)/P.
So, BOD = 5/0.02 = 250 mg/L.
Decimal fraction = 0.02 (Given).
BOD = (DOi - DOf)/P.
So, BOD = 5/0.02 = 250 mg/L.
(5)
BUJJIBABU said:
8 years ago
BOD=(DO intial - DO final) *Dilution factor.
Dilution factor=100/% of solution,
So, DF=100/2 = 50,
BOD=5*50 = 250.
Dilution factor=100/% of solution,
So, DF=100/2 = 50,
BOD=5*50 = 250.
(4)
Md seraj said:
1 decade ago
D.O consume = 5 ppm.
Dilution factor = 100/2 = 50.
B.O.D of the sewage = 5*50 = 250 ppm.
Dilution factor = 100/2 = 50.
B.O.D of the sewage = 5*50 = 250 ppm.
Santanu said:
9 years ago
BOD = (DO initial - DO final) x dilution ratio.
= 5 x (100/2).
= 250ppm.
= 5 x (100/2).
= 250ppm.
Uday kumar said:
7 years ago
BOD = D.F * DO consumed.
D.F. = 100/%of solution,
D.F. = 50,
BOD = 5 * 50 = 250 ppm.
D.F. = 100/%of solution,
D.F. = 50,
BOD = 5 * 50 = 250 ppm.
(4)
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