Civil Engineering - Waste Water Engineering - Discussion
Discussion Forum : Waste Water Engineering - Section 6 (Q.No. 43)
43.
If the depletion of oxygen is found to be 5 ppm after incubating a 2.5% solution of sewage sample for 5 days at 21°C, B.O.D. of the sewage is
Discussion:
5 comments Page 1 of 1.
Talha Hassan said:
4 months ago
BOD= Dissolved Oxygen / Fraction of sample in Mixture.
= 5 ppm /(2.5/100),
= 5 x 100/2.5,
= 200.
= 5 ppm /(2.5/100),
= 5 x 100/2.5,
= 200.
Jitendra Kumar Yadav said:
8 months ago
Why it is multiplied by 100?
Yogesh patil said:
6 years ago
B.O.D = D.O x D.F.
= 5 x 100/2.5.
= 200 ppm.
= 5 x 100/2.5.
= 200 ppm.
(4)
Noushi said:
7 years ago
5/(2.5/100) = 200ppm.
(1)
RANJAN PATRA said:
9 years ago
5 * 100/2.5 = 200 PPM.
(2)
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