# Civil Engineering - GATE Exam Questions - Discussion

Discussion Forum : GATE Exam Questions - Section 1 (Q.No. 49)
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
3800
475
271
190
Explanation:
No answer description is available. Let's discuss.
Discussion:
5 comments Page 1 of 1.

SAMALA LAXMAN said:   1 decade ago
(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:   7 years ago
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.
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Shivaprasad sajjan said:   5 years ago
(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.

John said:   3 years ago
Why e? Can't we use base 10?

Lt=Lo (1-10^kt).

Ravi said:   5 years ago
What is e here? Please tell me.
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