Civil Engineering - Waste Water Engineering - Discussion

Discussion Forum : Waste Water Engineering - Section 2 (Q.No. 2)
2.
If the depletion of oxygen is found to be 2.5 mg/litre after incubating 2.5 ml of sewage diluted to 250 ml for 5 days at 20°C, B.O.D. of the sewage is
50 mg/l
100 mg/l
150 mg/l
200 mg/l
250 mg/l.
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
11 comments Page 1 of 2.

Vikash said:   1 decade ago
BOD = DO X(DF).

2.5 X (250/2.5).

250mg/l.

Hem said:   9 years ago
DO and DF stands for?

Md Gulam Gos said:   9 years ago
DO: Dissolved Oxygen.

DF: Dilution Factor.

Yaregal t. said:   7 years ago
BOD =200 because don't forget 5 days.

Alvin A. said:   6 years ago
How did you get 200?

RAM PRASAD said:   6 years ago
That is BOD 5 days at 20° temperature.

Shweta said:   6 years ago
Answer is 250.

Swapnil Biraris said:   6 years ago
250 is the correct answer here.

Raisa Paul said:   4 years ago
D.F = sewage dilution/depletion of oxygen.

SHUBHAM SIDDHARTH NAYAK said:   4 years ago
BOD=(DO INITIAL-DO FINAL) * DF.

DO INITIAL - DO FINAL =2.5 mg/l (Given),
DF = (total volume of sewage and distilled water/volume of sewage sample)=250/2.5=100,
BOD = 2.5 * 100 = 250.
(2)


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