flow through 3/4 nps sch 80 steel pipe

[JAlton]

If a system has 1500 psi upstream pressure with 400 cu ft of compressed air storage (50,000 cubic feet of air available), will the flow through 45 ft of 3/4 nps sch 80 pipe with four (4) elbows be greater than 10,000 scfm with a pressure drop of 900 or more psi?

MY calculation is 11278 cfm, but I have a colleague who says it is only 2700 cfm.

Any help would be appreciated.

J. Alton Cox

http://www.delucatest.com

 

[Latexman]

With an exit pressure of 600 psig (dP = 900 psi), I got 256 lb/min. At 32 F and 1 atm, this is about 3170 scfm.

With an exit to atmosphere, I got 269 lb/min. This is about 3330 scfm.

Good luck,
Latexman

 

[JAlton]

Thank you for your response. I must have something wrong with my calculation. My colleague wants to use size 2 nps pipe. Do you think we could get 10,000 cfm with size 2 nps piping?

J. Alton Cox

http://www.delucatest.com

 

[katmar]

My program also gives close to 3300 cfm for the 3/4" pipe.

With a 2" pipe I get 37000 cfm.

To get 10000 cfm the program predicts a required diameter of 1.16".

It is interesting to note that with a small diameter like 3/4" the calculation is rather sensitive to the pipe roughness you assume. I got 4500 cfm with drawn tubing (e=0.00006") through 3300 cfm for new pipe (e=0.002") to 2500 cfm with "slightly rusty" pipe (e=0.012").

regards
Katmar

 

[JAlton]

This is for a new system with Sch 80 Carbon Steel Pipe.
I was using 0.051 for roughness factor.
Thanks for the input.

J. Alton Cox

http://www.delucatest.com

 

[katmar]

If you are talking millimetres, then 0.051 sounds about right. I prefer to talk Metric, but because you used cfm and feet I switched to Imperial.

 

[Latexman]

For 10,000 scfm with 900 psi dP, I got 1.178". Essentially the same as katmar.

Good luck,
Latexman

 

[JAlton]

Latexman, Thanks for your confirmation. We have decided to go with 2" pipe instead of 3/4". I appreciate the advice.

J. Alton Cox

http://www.delucatest.com

 

[sailoday28]

For a perfect gas, steady state, adiabatic flow, constant area and specific heats

G=mass flux flow per unit area
V=specific volume
P= pressure
f= moody friction factor (mean value over the flow length)
L= equivalent length of pipe
D= pipe inside diameter
gamma= ratio of Cp/Cv
Subscripts
0 source stagnation conditions
1 pipe entrance conditions
Conservation of mass, energy and momentum yield after integration of differential equations
G^2(gamma+1)/gamma *ln (V/V1) + PoVo(1/V^2-1/V1^2) +G^2fL/D =0

For a known flow rate and fL/D calcualte V
Pressure is obtained from energy equation
P=PoVo/V - (gamma-1)/2/gamma *(GV)^2

As an approximation use V1=V0

NOTE flow is choked if
(GV)= sqrt[2PoVogamma/(gamma+1)]

 

[sailoday28]

CORRECTION TO MY POST
For a known flow rate and fL/D calcualte V
Pressure is obtained from energy equation
P=PoVo/V - (gamma-1)/2/gamma *G^2*V