# Civil Engineering - Water Supply Engineering - Discussion

Discussion Forum : Water Supply Engineering - Section 1 (Q.No. 13)
13.
Distribution mains of any water supply, is normally designed for its average daily requirement
100%
150%
200%
225%.
Explanation:
No answer description is available. Let's discuss.
Discussion:
23 comments Page 1 of 3.

Moyir said:   2 years ago
For daily it's 180%, for hourly it's 150%, for monthly it's 140%.
(1)

Sheikh Ovais Ahmed said:   2 years ago
225% is correct.

Water mains are Designed for a maximum of ( Max hourly demand or Coincident draft).

Maximum hourly demand = 270% of av hourly demand (i.e. 2.7 * Qav).
Coincident Draft includes Maximum daily demand which is 180% of Average daily demand (i.e. 1.8 * Qav).

Thus we take an average of these two (270+180)/2 = 225%.

I hope this is clarified.
(13)

Sheikh Ovais Ahmed said:   2 years ago
225% is correct.

Water mains are Designed for a maximum of ( Max hourly demand or Coincident draft).

Maximum hourly demand = 270% of av hourly demand (i.e. 2.7 * Qav).
Coincident Draft includes Maximum daily demand which is 180% of Average daily demand (i.e. 1.8 * Qav).

Thus we take an average of these two (270+180)/2 = 225%.

I hope this is clarified.
(1)

Abhishek Das said:   3 years ago
If daily then 180% and if hourly then 150%.

Ashwani Kumar said:   4 years ago
The question is wrong.

Generally distribution system should be designed for maximum hourly consumption of maximum day i.e. Peak Demand or Coincident draft, whichever is more.

God-rich formula to find peak demand to an average demand ratio.

P = 180 x t^ (-0. 1)

Where P = percent of the annual average draft for the time 't' in days.

So, I think 180% is the right answer.

Chaman said:   4 years ago
Thanks everyone for explaining it.

Monika said:   4 years ago
I don't understand? Please explain me clearly.

Aravind said:   5 years ago
Design criteria - coincidental draft or max hourly demand.
Max hourly demand = 1.5 x max daily demand.
= 2.7 x avg daily demand.
= 270%.

Jay said:   5 years ago

Vinay Kumar said:   5 years ago
Maximum daily demand = 1.8 x average daily demand.

Maximum hourly demand of maximum day i.e. Peak demand.
= 1.5 x average hourly demand.
= 1.5 x Maximum daily demand/24.
= 1.5 x (1.8 x average daily demand)/24.
= 2.7 x average daily demand/24.
= 2.7 x annual average hourly demand.
(1)