Civil Engineering - Hydraulics - Discussion

Thanks @Nitesh.

A is correct answer.

Reason behind that v^2/2g is a term used in specific energy not for critical depth i.e. E= h+v^2/2g = specific energy.

But critical depth is that depth at which specific energy is minimum so we have to differentiate above equation w.r.t h ... And then term comes dE/dh= 1+ q^2/2g(-2/h^3)=0 ... And simplify this equating we will get h= q^2/g .... i.e. (av)^2/g... i.e. v^2/g.

I hope you will understand what i m saying good luck friends for your future.

V^2/2g is the right answer.

Critical velocity Vc = root of g * hc(Critical depth of flow).

V in the above equation is critical velocity. That is velocity with respect to the critical depth of flow.

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.

No, the right answer is v2/2g.

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.

Please, anyone give me the explanation of the answer.

Froud number equals to 1 for critical so (V/(√gh)):1.