Pressure Drop & Flow Rate for Oxygen Delivery Tube

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Discussion Overview

The discussion revolves around calculating the pressure drop and flow rate in a 50-foot long oxygen delivery tube with a specified inner diameter, focusing on the implications of flow rate and tube dimensions. The context includes theoretical considerations and potential calculations relevant to fluid dynamics in medical applications.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents the problem of determining pressure drop and flow rate through a specified tube length and diameter, assuming a flow rate of two liters per second.
  • Another participant suggests using Poiseuille's law for calculating pressure drop, noting the need to convert units to MKS and providing the relevant formula, while also mentioning the conservation of mass for flow rate.
  • A third participant proposes using Bernoulli's equation to assess pressure losses, introducing additional parameters such as friction factor and elevation.
  • A later reply questions the initial flow rate of 2L/s, indicating that such a flow rate through the given diameter would result in an unreasonably high mean flow velocity, suggesting a possible error in the flow rate specification.

Areas of Agreement / Disagreement

Participants express differing views on the feasibility of the specified flow rate, with some supporting the calculations based on the initial assumption and others challenging the validity of the flow rate as stated. The discussion remains unresolved regarding the correct flow rate and its implications.

Contextual Notes

There are limitations in the assumptions made about flow conditions, such as whether the flow is laminar or turbulent, and the potential impact of viscous losses, which are not fully explored. The discussion also highlights the need for consistent unit usage in calculations.

wjt
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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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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.
 
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:
wjt said:
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 ?
 

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