Discharge Coefficient for a 1/4" hose 3

[IntrepidLearner]

I'm trying to determine the flow rate for compressed air escaping to the atmosphere via an orifice of known diameter.

Through my search of the web, I managed to come across this link...

http://www.air-dispersion.com/msource.html

It's exactly what I was looking for. With one exception. It suggested an approximate value of 0.72 for the coefficient of discharge (C). However, I've come accross other sources which suggest values anywhere b/w 0.92 and 0.98 for well rounded orifices, down to 0.61 for sharp edges.

I was hoping some gifted individual may actually know what value of C I should use for air escaping through a 1/4" air hose.

 

[sailoday28]

Latexman (Chemical)Can you sketch with dashes etc, what the system/manifold looks like?

 

[katmar]

Latexman, thanks for pointing that out. Yes, I had assumed that the branch was immediately after the regulator and I took the flow as only 7 x branch flows all the way from the regulator to the first branch.

Note to self: Rule #1 - Always state your assumptions!

Harvey

 

[Latexman]

R = regulator
H = hose
T = tee

H H H H H H
!-----------------------------------
!
--R-----T
!
!-----------------------------------
H H H H H H

This is how I interpret it.

Good luck,
Latexman

 

[Latexman]

IntrepidLearner,

Being you are an EE, is there a practical way you can determine the electrical savings if no air is flowing to this air user? Does the air compressor motor have a power meter or VFD that can be recorded for some time with air on to this usage and then valve the air off and determine what you need anyway by difference? You'd just have to be sure this air usage was the only difference.

Good luck,
Latexman

 

[sailoday28]

Latexman, Thanks for the diag. Is the 8psi or 2 psi downstream of R and upstream of T?

 

[Latexman]

sailoday28,

Yes, immediately downstream of R.

Good luck,
Latexman

 

[IntrepidLearner]

You guys have been brilliant! I didn't realize how complicated a problem I was facing! I'm going to try to digest all the information which katmar has been so kind enough to type out. I do, however, feel I'm going to be needing to refer to a fluids book (as I doubt my mechanical colleagues have CTP 410).

BTW, the distance from the regulator to the T is approx 3'.

 

[Latexman]

InterpidLearner,

Assumptions:
½” pipe = 0.622” ID
¼” hose = 0.25” ID

Solutions:
8 psig network 213 iterations 294 lb/hr
2 psig network 20 iterations 51 lb/hr


Good luck,
Latexman

 

[IntrepidLearner]

Hi Latexman,

When you say x iterations, it leads me to believe that you created a spreadsheet for this calculation. Did you? If yes, would you mind sending it to me at lmjwalcott@caribsurf.com. I doublechecked the ID of the '1/4' hose. I found it to be approx 4.5mm (=~3/8").

If not, it's OK, you guys have been more than helpful as it is!

 

[Latexman]

Correction:
1/4” hose = 0.[red]375[/red]" ID

Solutions:
8 psig network 97 iterations 336 lb/hr
2 psig network 20 iterations 64 lb/hr

Sorry, it is not a spreadsheet. It is licensed software. It took me less than 15 minutes to build the two models and get the results.


Good luck,
Latexman

 

[IntrepidLearner]

You're a star, thanks again!

 

[katmar]

There is an error in InterpidLearner's dimensions. 4.5 mm is about 3/16" and not 3/8". Which is correct, the 4.5 mm or the 3/8"?

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[IntrepidLearner]

You're right, it's 4.5mm which is ~3/16" not 3/8" as I mistakenly noted.

 

[Latexman]

Correction:
1/4” hose = 0.[red]177[/red]" ID

Solutions:
8 psig network 456 iterations 198 lb/hr
2 psig network 20 iterations 30 lb/hr


Good luck,
Latexman

 

[IntrepidLearner]

thank you, sir!