Civil Engineering - Hydraulics - Discussion
Discussion Forum : Hydraulics - Section 1 (Q.No. 15)
15.
Critical depth (h) of a channel, is
Discussion:
24 comments Page 2 of 3.
Anurag said:
7 years ago
I think [q^2/g]^[1/3].
Kajal ncp said:
7 years ago
Thanks @Nitesh.
Javed said:
1 decade ago
As it depends on froudes number. For critical flow froudes number is one and the depth corresponding to critical flow is called critical depth.
Froudes number = Velocity/Root of height*velocity.
Froudes number = Velocity/Root of height*velocity.
Rohit said:
8 years ago
V^2/2g is the right answer.
Santhosh C said:
8 years ago
Critical velocity Vc = root of g * hc(Critical depth of flow).
Santhosh C said:
8 years ago
V in the above equation is critical velocity. That is velocity with respect to the critical depth of flow.
Sandip buktare said:
8 years ago
For critical depth = v/root of height * gravity.
1 = v/root of height * gravity,
The root of height * gravity = v,
Taking square of both side.
h * g = v^2
h = v^/g.
1 = v/root of height * gravity,
The root of height * gravity = v,
Taking square of both side.
h * g = v^2
h = v^/g.
Vaibhav Patle said:
8 years ago
No, the right answer is v2/2g.
Abhishek Anand said:
9 years ago
The depth of water in a channel when vel of flow is critical or when the specific energy is min is called critical depth of the channel.
Critical vel of flow is the vel at which sp.energy is minimum.
Specific energy -> E=h+(v^2/2g).
For minimization, dE/dh =0.
On putting v=Q/A.
We get , 1=v^2/gh.
So, V= (gh)^1/2 or sqrt of gh.
Which is critical velocity;
So, critical depth= v^2/g.
Critical vel of flow is the vel at which sp.energy is minimum.
Specific energy -> E=h+(v^2/2g).
For minimization, dE/dh =0.
On putting v=Q/A.
We get , 1=v^2/gh.
So, V= (gh)^1/2 or sqrt of gh.
Which is critical velocity;
So, critical depth= v^2/g.
Bapugouda said:
9 years ago
Please, anyone give me the explanation of the answer.
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