Civil Engineering - Hydraulics - Discussion

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

V^2/2g is the right answer.

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.

Thanks @Nitesh.

E = y + v^2/2g is specific energy.
Here, v^2/2g is KINETIC energy and "y" is POTENTIAL energy.
So, the given answer is correct.

I think [q^2/g]^[1/3].

For critical flow type specific energy = yc + v2/2g.
for rectangular channel = Ec = 1.5 yc.
1.5 yc = yc + v2/2g,
0.5 yc = v2/2g,
yc = v2/g.

Thank you @Nitesh.

hc=( q2/g)^1/3.

Taking cube on both sides hc^3= q2/g.

q=Q/b= A*V/b= (b*h*V)/b= (h*v),
hc^3= (hc * V)^2/g,
hc= V2/g.

Thank you @Prassy.