Civil Engineering - Hydraulics - Discussion
Discussion Forum : Hydraulics - Section 4 (Q.No. 20)
20.
Most economical section of a circular channel for maximum velocity, is if,
Discussion:
5 comments Page 1 of 1.
Amar said:
7 years ago
Answer All of the above correct.
Then the discharge through a channel the maximum.
1. The depth of water=0.95d.
2. R=0.286d.
3. Perimeter=2.6 d.
The velocity through a channel circular section is max.
depth of water = 0.810 diameter.
hydraulic mean depth = 0.304 diameter.
wetted perimeter = 2.245 diameters.
Then the discharge through a channel the maximum.
1. The depth of water=0.95d.
2. R=0.286d.
3. Perimeter=2.6 d.
The velocity through a channel circular section is max.
depth of water = 0.810 diameter.
hydraulic mean depth = 0.304 diameter.
wetted perimeter = 2.245 diameters.
Sagar said:
7 years ago
The depth of water =0.95d is for maximum discharge they ask for maximum velocity and its 0.81d.
(1)
Priya said:
7 years ago
As per my knowledge;
1. Depth of water=0.95d.
2. R=0.286d.
3. Perimeter=2.6 d.
1. Depth of water=0.95d.
2. R=0.286d.
3. Perimeter=2.6 d.
Sanju said:
8 years ago
Highest speed at depth 0.81D.
Samantha said:
10 years ago
Should we memorize these.
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