What are the unknown variables in the flowrate equation?

[Scott Vidicik]

Hello friends!

I got this flowrate equation I wish to translate, into metric system, but somehow I find it hard to do.

It is as follows:

M (lb/s) = A x C x P / √(R)

A: Area (inches)
C: Flow coefficient
P: Pressure (psi)
R: Temperature (Rankine)

So, to get this into SCFM I modify the equation with gas density of air
0.7494 lb/ft³ and 60s/min so I get

V (SCFM) = A x P x C x 60 / [√(R) x 0.7494]
this gives me SCFM. I could ofcourse just go from SCFM to m³/min, but that wouldn't be as awesome as having a formula where you could put in your data in metric.

Any help would be really really helpful.
Feel free to ask if there's anything unclear. Thanks a lot!

 

[Scott Vidicik]

I find it hard to believe that neither the American nor European members can help me out here :)

 

[Scott Vidicik]

Bumping it for the last time.

If no one are able to help me, do you have any suggestions for where I could find some?

Thanks in advance! :)

 

[Scott Vidicik]

I found the answer. Feel free to ask if interested.

Now I got another probem!

I found this equation, but I don't understand it.

P/0.1 x Flow x Hours x 0.35 x 0.007 x cost.

P is in bar, so by dividing by 0.1 we get MPa.
Flow is in L/sec
Hours means operating hours for a compressor.
0.35 is some unknown variable, so is 0.007.
Cost is cost per kwh.

So I get

MPa x L/sec x Hours x 0.35 x 0.007 x € = total cost of a pressure drop.

What the heck is 0.35 and 0.007 values for? I haven't given any other input than what I've said.

Thanks a lot in advance!

EDIT:
There also seems to be a rule-of-thumb which claims that for every 1 psig reduction in compressor output pressure, compressor power consumption will be reduced by approximately 0.5%. How did they find this number?