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
Discussion:
14 comments Page 1 of 2.
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.
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)
Anju said:
8 years ago
100/0.1=1000.
(1)
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.
Because,
Population equivalent = Total B.O.D/B.O.D per head.
(100/0.1) = 1000.
(1)
MILI said:
3 years ago
I think the Answer is 1000.
(1)
RANJAN PATRA said:
9 years ago
As per my knowledge 100/.1=1000.
Mukesh maurya said:
9 years ago
Population= .1 * 100 = 1000.
Chintu said:
9 years ago
Population equivalent is a parameter used in the conversion of the contribution of wastes from industrial establishments for accepting into sanitary sewer systems. The strength of industrial sewage is, thus, written as;
Std. BOD5 = (Std. BOD5 of domestic sewage per person per day) x (population equivalent).
Std. BOD5 = (Std. BOD5 of domestic sewage per person per day) x (population equivalent).
Chintu said:
9 years ago
I too think it's 1000.
Gayatri said:
8 years ago
I didn't get it. Please explain anyone.
Dvs said:
8 years ago
Population equivalent is equal to (Total bod)/(per capita Bod).
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