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Oxygen flow

  1. Jul 17, 2008 #1


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    I am on home oxygen and receive the gas through a 50feet long tube about .1" inside diameter.If the Machine delivers a flow rate of two liters per second, what is the pressure drop through the tubing; What is the flow rate at my end of the tubing?
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  3. Jul 17, 2008 #2

    Andy Resnick

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    If Poiseuille flow holds for your system (and it most likely does if the flow is laminar and steady), the relevant formulas are:

    [tex]\Delta P = \frac{8LQ\mu}{\pi R^{4}}[/tex], where

    [tex]\Delta P[/tex] is the pressure drop between inlet and outlet, L the length of tube, Q the volumetric flow rate (liters/min, for example), R the tube radius, and [tex]\mu[/tex] the viscosity (in Poise, or equivalent)

    So, you have mixed units- convert everything into MKS. Air at room temperature has a viscosity of about 17 *10^-6 Pa*s. Now you can calculate the pressure drop.

    The flow rate at the outlet is the same as the inlet- conservation of mass.

    If there are viscous losses (entirely possible given the aspect ratio of tubing and flow rate), then YMMV.
  4. Jul 18, 2008 #3


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    You can use the Bernoulli losses formula to calculate it.

    ([tex]\frac{P1}{\gamma}[/tex] +Z1) -( [tex]\frac{P2}{\gamma}[/tex] +Z2) = [tex]\frac{8fLQ^2}{\ g(Pi)^2D^5}[/tex]

    Q is the flow rate.
    P is the pressure at point X.
    f is the friction factor of the pipe.
    L is the length of the tube.
    gamma is ro*g.
    Z is the elevation at point X.
    Last edited: Jul 18, 2008
  5. Jul 18, 2008 #4


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    Hi wjt, could you please re-check that data as it seems a bit unreasonable. Fluid at 2L/second through 0.1 inch diameter would give a required mean flow velocity of 395m/s which is supersonic.

    In SI units :

    Flow Rate, Q = 2E-3 m^3/s
    Radius, r = .1 * 2.54E-2 / 2 = 1.27E-3 m
    Cross section : A=pi r^2 = 5.07E-6 m^2

    Therefore the mean velocity is, Q/A = 395 m/s

    Are you sure it's not 2L/min instead of 2L/s ?
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