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Rewrite flowrate equation

  1. May 2, 2010 #1
    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!
     
  2. jcsd
  3. May 3, 2010 #2
    I find it hard to believe that neither the American nor European members can help me out here :)
     
  4. May 12, 2010 #3
    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! :)
     
  5. May 13, 2010 #4
    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?
     
    Last edited: May 13, 2010
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