Civil Engineering - Hydraulics - Discussion
Discussion Forum : Hydraulics - Section 1 (Q.No. 15)
15.
Critical depth (h) of a channel, is
Discussion:
24 comments Page 1 of 3.
Laxman said:
3 years ago
Froudes number= Inertia Force/viscous force.
F= V/(gy)^0.5.
For channel flow to be critical (Fr=1).
Now, 1= V/(gy)^0.5.
=>. (gy)^0.5 = V
=>. gy = V^2
=>. y= V^2/g.
Thus, Option A is correct.
F= V/(gy)^0.5.
For channel flow to be critical (Fr=1).
Now, 1= V/(gy)^0.5.
=>. (gy)^0.5 = V
=>. gy = V^2
=>. y= V^2/g.
Thus, Option A is correct.
(9)
Rohit said:
3 years ago
@Nitesh.
Thanks for explaining the answer.
Thanks for explaining the answer.
M. Shaikh said:
3 years ago
As per Froude Number (Fr):
If Fr < 1 then flow is Sub-Critical flow.
If Fr > 1 then flow is Super-Critical flow,
If Fr = 1 then flow is Critical flow,
So, The formula of Fr is:
Fr = v / ( g d) ^ 1/2.
For critical depth put Fr = 1 and solve for d.
1 = v / ( g d) ^ 1/2,
(g d) ^1/2 = v.
Squaring on Both Sides;
(g d) = v ^2.
d (critical) = (v ^2)/g.
If Fr < 1 then flow is Sub-Critical flow.
If Fr > 1 then flow is Super-Critical flow,
If Fr = 1 then flow is Critical flow,
So, The formula of Fr is:
Fr = v / ( g d) ^ 1/2.
For critical depth put Fr = 1 and solve for d.
1 = v / ( g d) ^ 1/2,
(g d) ^1/2 = v.
Squaring on Both Sides;
(g d) = v ^2.
d (critical) = (v ^2)/g.
(6)
Sanjay Panchal said:
5 years ago
Thanks.
(1)
Sudip Timalsena said:
5 years ago
Thank you @Prassy.
Prassy said:
5 years ago
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.
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.
Priyanka Shah said:
6 years ago
Thank you @Nitesh.
Amare yihunie said:
7 years ago
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.
for rectangular channel = Ec = 1.5 yc.
1.5 yc = yc + v2/2g,
0.5 yc = v2/2g,
yc = v2/g.
Anurag said:
7 years ago
I think [q^2/g]^[1/3].
Thundir said:
7 years ago
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.
Here, v^2/2g is KINETIC energy and "y" is POTENTIAL energy.
So, the given answer is correct.
(1)
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