Civil Engineering - Waste Water Engineering - Discussion
Discussion Forum : Waste Water Engineering - Section 1 (Q.No. 34)
34.
The ratio of minimum hourly flow to the average flow of sewage is
Discussion:
5 comments Page 1 of 1.
Arpit shukla ji said:
8 years ago
Min. Hourly flow =1/2 * avg flow.
Xyz said:
8 years ago
Maximum daily flow i= 2 Avg annual daily flow.
max hourly flow = 1.5 max daily flow=3 annual average daily flow.
Minimum daily flow=2/3 annual average daily flow.
Minimum hourly flow=1/2 minimum daily flow=1/3 annual average daily flow.
So the answer is correct.
max hourly flow = 1.5 max daily flow=3 annual average daily flow.
Minimum daily flow=2/3 annual average daily flow.
Minimum hourly flow=1/2 minimum daily flow=1/3 annual average daily flow.
So the answer is correct.
(7)
Har said:
8 years ago
Here average flow of sewage=annual average daily flow.
(2)
Rkv said:
5 years ago
Can anyone explain this clearly?
Liaqat said:
3 years ago
The minimum hourly sewage flow is equal to 50% of average sewage flow =1/2 of average sewage flow.
To calculate maximum or peak sewage flow we take peak factor i,e PF=1+(14/4+ √P).
where P is the population in thousand. And then multiply the peak factor with the average sewage flow to calculate the maximum sewage flow.
To calculate maximum or peak sewage flow we take peak factor i,e PF=1+(14/4+ √P).
where P is the population in thousand. And then multiply the peak factor with the average sewage flow to calculate the maximum sewage flow.
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