Civil Engineering - Waste Water Engineering - Discussion

Discussion Forum : Waste Water Engineering - Section 3 (Q.No. 34)
34.
Afive day B.O.D. at 15°C of the sewage of a town is 100 kg/day. If the 5 day B.O.D. per head at 15°C for standard sewage is 0.1 kg/day, the population equivalent is
100
1000
5000
10000
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
14 comments Page 1 of 2.

MILI said:   3 years ago
I think the Answer is 1000.
(1)

MJM Gcek said:   4 years ago
No, @Jinish.

The answer should be 1000 only.

Soumya sundar said:   4 years ago
No @Jinish.

Simply compare 5 day Bod of a single person to that of the whole city. Ratio them. Get your Population equivalent or for how many persons that city produce the same waste. The answer is 1000.

Er Raj said:   6 years ago
@Jinish. Well explained.

Dheeraj said:   6 years ago
Correct answer is(B) = 1000.
Because,
Population equivalent = Total B.O.D/B.O.D per head.
(100/0.1) = 1000.
(1)

Jinish said:   6 years ago
Bod of 1 day= 0.1 kg but waste per day is 100 so bod = 100*0.1
Waste per day is 100 so 5 days waste = 5*100
And 5 days bod = 5 * 0.1( 0.1= 1 days bod).
Total population = (bod per waste * waste per 5 day)/ 5 days bod.
= (100*0.1 * 5 *100)/( 0.1* 5)
= 10,000.
(5)

Chinmay said:   7 years ago
0.1*1000 = 100 kg/day.

Hence, population is 1000!

Anju said:   8 years ago
100/0.1=1000.
(1)

Dvs said:   8 years ago
Population equivalent is equal to (Total bod)/(per capita Bod).

Gayatri said:   8 years ago
I didn't get it. Please explain anyone.


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