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 2 of 2.
Raaj said:
7 years ago
Thank you all for explaining it.
(1)
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)
Robin said:
7 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)
Bruno said:
6 years ago
(Initial do - final do )*dilution factor.
5*50 = 250.
5*50 = 250.
(2)
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)
Butum said:
5 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)
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers