Pressure Drop & Flow Rate for Oxygen Delivery Tube

[wjt]

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?

 

[Andy Resnick]

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

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

\Delta P 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 \mu 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.

 

[GT1]

You can use the Bernoulli losses formula to calculate it.

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

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.

 

[uart]

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?

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 ?