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Calculating drop in gas pressure

  1. Feb 5, 2013 #1
    Poles Formula (see page 10 of attachment)

    www.mech.hku.hk/bse/MEBS6000/mebs6000_1011_04_steam.pdf [Broken]

    Where have I gone wrong?

    • q = flow (m3/h)
    • d = diameter of pipe (mm)
    • h = pressure drop (mbar)
    • l = length of pipe (m)
    • s = specific gravity of gas (density of gas / density of air)

    h = ( q^2 * s * l ) / ( 0.0071^2 * d^5 )

    q = 6 m3/h
    d = 20mm
    l = 19M
    s = 0.58

    The result is 2.459 i.e. the pressure drops by 2.2459 mb, which on the face of it looks fine, the problem is that if I reduce the flow rate the loss of pressure over the length of the pipe drops. Which in theory means that if I start with 21mb gas pressure and have a lower flow rate I end up with a higher pressure at the end of the pipe; that can't be right can it?
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Feb 5, 2013 #2


    User Avatar

    Staff: Mentor

    Yes, that is correct. Higher flow causes more friction and more pressure loss.
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