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Need help on pipe-flo pro document

  1. Oct 8, 2012 #1
    hi everyone this is my first thread on this board and sorry for any missing in English because it's not my native language

    i just want to calculate the pressure drop base on pipe-flo pro document which has been share in this thread https://www.physicsforums.com/showthread.php?t=179830.


    my question is about equation1 in this paper.
    dP = fρ(L/D)v2/2g

    Is it correct for gravitational term to be in these d-w equation's form???
    And if it's correct, where is "g" come from???
     
  2. jcsd
  3. Oct 8, 2012 #2

    Q_Goest

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    Hi keng, welcome to the board.
    g is the acceleration due to gravity or 9.8 m/s2 or equivalent. For example:
    http://en.wikipedia.org/wiki/Darcy–Weisbach_equation#Head_loss_form

    See also equation 15 in Pipe-Flo Pro.
     
  4. Oct 8, 2012 #3
    hi Q_Goest thank for reply,
    on pressure loss form(wiki),there is no g on the right hand side but it appear in equation 1 in pipe-flo pro document?

    about equation 15, i'm confuse again because i solve there unit and the first term on the right hand side p/ρ is not unit of hieght ???
     
    Last edited by a moderator: May 6, 2017
  5. Oct 8, 2012 #4

    Q_Goest

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    Ok, I see the dilemma now. The g isn't actually the acceleration due to gravity as they state. I suspect Pipe-Flo Pro put it in there to represent gc which is the conversion factor used for imperial measure. Just ignore it if you're using metric units or use gc if using imperial units.

    The head loss form of the equation has the g in it (acceleration due to gravity:
    8eca882f8e4cf2800ab7d4ddfaae3a97.png

    But you're correct that the pressure loss form of the equation does not have a g in it since they cancel out:
    d507bcd5f5091fb94dabf55d828f59a3.png
     
  6. Oct 9, 2012 #5
    thx a lot, Q_Goest
    i think there is some conflict in unit of pressure in this document.
    in eq 15, if i cancel g out(for my metric units calcuation like you've said)
    the first term on the right hand side still not unit of height and the second term reduce to0.5v2 which is not unit of height like before cancel g out
    i suspect that unit of pressure in this paper maybe Pa/(m/s2) then eq.1 and 15 become D-W pressure loss form for eq.1 and bernoulli's total head form for eq15
     
  7. Oct 9, 2012 #6
    Another question, Is it ok to solve velocity from the set of equation in this document with known of pressure (dP) and pipe dimension (L, D)

    I’ve done on calculation as attach below
    Is it correct to use D-W and Colebrook this way????
     

    Attached Files:

  8. Oct 9, 2012 #7

    Q_Goest

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    Yes, you're right. Equation 15 is just Bernoulli's equation so it should look like this:
    187d6853e6b3183e324fadb92b51735a.png
    (above is from Wikipedia)[/PLAIN] [Broken]
    So equation 15 is missing g in the denominator where it has P/p (pressure / density).
    Note also that g in this equation really IS the acceleration due to gravity, not gc, the constant used in imperial units.
     
    Last edited by a moderator: May 6, 2017
  9. Oct 9, 2012 #8

    Q_Goest

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    I've gone over your work and I agree you've done this correctly. I have a program that does this using imperial measurements and I come up with similar values to what you've calculated with the exception of the friction factor. For friction factor, I have 0.0195. I have a proprietary equation that I believe is more accurate than the Colebrook equation but I wouldn't expect you to accept that. I would suggest however that you try the Moody diagram and see what value of f you get. I would also recommend reviewing the explicit equations provided at eng-tips for other equations for friction factor.
    http://www.eng-tips.com/faqs.cfm?fid=1236
     
  10. Oct 10, 2012 #9
    my problem can't use moody diagram directly because unknow of velocity and also Re.
    but i've try the explicit equation from your link by substitute f and Re with the their velocity relation(f=121.067/v2 and Re=10765.5v from my previous calculation)to the equation and then trial and error to find v. The result is Moody Eq.f=0.0118,Churchill Eq.f=0.01218 and Serghides Eq.f=0.012157

    How did u get f=0.0195 ?????

    If the straight pipe is replace by pipe in attachment below, can i calculate new length like that and then do the same as previous in cal1?
     

    Attached Files:

  11. Oct 10, 2012 #10

    Q_Goest

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    My mistake. I had the surface roughness for pipe as an input instead of drawn tube. When I put in drawn tube I get a friction factor of 0.0121.
    Yes. You just calculate an equivalent length of pipe as you've done and use that in your equation for pressure drop as you did in cal1. Well done! :smile:
     
  12. Oct 10, 2012 #11
    In English units, gc is equal to 32 (lbf/lbm)ft/sec2

    An older equation that gives a pretty good approximation to the friction factor as a function of Re is

    f = 0.0791/Re0.8

    Maybe this can give you a good first approximation to your solution.
     
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