Civil Engineering - GATE Exam Questions - Discussion

49. 

The 5-day BOD of a wastewater sample is obtained as 190 mg/l (with k = 0.01 h-1). The ultimate oxygen demand (mg/l) of the sample will be

[A]. 3800
[B]. 475
[C]. 271
[D]. 190

Answer: Option C

Explanation:

No answer description available for this question.

Samala Laxman said: (Dec 17, 2013)  
(BOD)5 = BODu[1- (e^-kt)].

Here give k = 0.01/hour.
t = 5days = 120 hrs.
BOD5= 190mg/lt.

Substitute these values in above formula we will get answer as 272.32.

Roy said: (Dec 12, 2017)  
5-day bod is approximately 68% of the ultimate bod. So, 68%=190, so 100%= (190*100/68) =279.

So we can go with option c.

Shivaprasad Sajjan said: (Feb 14, 2019)  
(BOD)5 = (BOD)u[1-e^-kt].
190 = (BOD)u[1-e^-0.01 * 120],
190 = (BOD)u * 0.70,
(BOD)u = 190/0.70 = 271.42~271.
Ans: C.

Ravi said: (Feb 24, 2019)  
What is e here? Please tell me.

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