More on Critical Depth

[cantaloup]

A flow of 450 cfs partly fills a concrete pipe with diameter = 8 ft . What is the nearest value of critical velocity in fps?

A/ 14.4

B/ 15.2

C/ 16.0

D/ 16.6

With the correct answer from the above problem, what is the nearest value of critical slope?

A/ 0.0015

B/ 0.0021

C/ 0.0049

D/ 0.0076

 

[Suns Den]

A flow of 450 cfs partly fills a concrete pipe with diameter = 8 ft . What is the nearest value of critical velocity in fps?
A/ 14.4

B/ 15.2

C/ 16.0

D/ 16.6

With the correct answer from the above problem, what is the nearest value of critical slope?

A/ 0.0015

B/ 0.0021

C/ 0.0049

D/ 0.0076

Cantaloup, Let me know where I am messing it up

I got a critical depth of 5.4' from app 19.D CERM

and angle of 285 deg

solving for V gives me 11.5 ft/sec

 

[cantaloup]

Cantaloup, Let me know where I am messing it up
I got a critical depth of 5.4' from app 19.D CERM

and angle of 285 deg

solving for V gives me 11.5 ft/sec

This problem is taken from Schaum's 2500 Hydraulics Solved Problems

Crit D = 5.4 ft is correct.

Now we gotta find V, we already have Q so we want to find A since V=Q/A

d/D = 5.4/8=0.68

Appendix 16A CERM, with d/D= 0.68 give

A/D^2 = 0.569 so A=0.569*8^2 = 36.4 ft^2

V=450/36.4=12.4 fps

So A is correct

Now use App 16A to find hydraulic radius R with d/D=0.68:

R/D = 0.294 So R= 2.35 ft

Now use Appendix 19B nomograph to find S with n=0.013 (concrete pipe) and V=12.4 fps and hydr. radius = 2.35 ft

S=0.0037; therefore C is correct

Instead of using the nomograph I use HP33s to solve for S with the equation solver:

V=1.486/N*R^0.667*S^0.5

and get the result S = 0.0038 , very close

Note: when we encounter Critical depth problem, we have to know if it's rectangular x-section or pipe flow

the formula Vcrit = SQRT(g*Ycrit) is for rectangular x-section