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
50 ppm
100 ppm
150 ppm
200 ppm
250 ppm.
Answer: Option
Explanation:
No answer description is available. Let's discuss.
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.

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.
(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)

Post your comments here:

Your comments will be displayed after verification.