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  #1  
Old 08-25-2021, 01:36 PM
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David_Nelson David_Nelson is offline
 
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Default Fluid Dynamics Question

I thought this would be a seemingly simple question but I keep going down complex rabbit holes ala google. I'm looking for "back of the napkin" kinda of solutions and how you got there (I am trying to learn something here). So, given:

- 6 feet of 1/8" ID Tygon tubing ( say https://www.mcmaster.com/6516T14/ )
- 25 PSIG of air at 80 degF is connected to one end of the tubing above
- The other end of the tubing is free hanging in air at atmospheric pressure

How many CFM of air will flow through the tubing?

Thank you,
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  #2  
Old 08-25-2021, 02:18 PM
KayS KayS is offline
 
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is there a prize for solving the equation? :-)
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  #3  
Old 08-25-2021, 02:37 PM
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airguy airguy is offline
 
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Quick rough numbers, about 4.8 MSCFD or 3.3 CFM, with a 25psig drop across that 6 feet, assuming no sharp bends.
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  #4  
Old 08-25-2021, 02:57 PM
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Default First year college

Freshman English finally put to a good use.
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  #5  
Old 08-25-2021, 03:29 PM
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gmcjetpilot gmcjetpilot is offline
 
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Quote:
Originally Posted by David_Nelson View Post
I thought this would be a seemingly simple question but I keep going down complex rabbit holes ala google. I'm looking for "back of the napkin" kinda of solutions and how you got there (I am trying to learn something here). So, given:

- 6 feet of 1/8" ID Tygon tubing ( say https://www.mcmaster.com/6516T14/ )
- 25 PSIG of air at 80 degF is connected to one end of the tubing above
- The other end of the tubing is free hanging in air at atmospheric pressure

How many CFM of air will flow through the tubing?

Thank you,
Why.... I have flashbacks from my undergrad days in the Collage of Engineering LSU... 20 problems and an hour to solve them for a final...

https://www.engineersedge.com/fluid_...ipes_14029.htm
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  #6  
Old 08-25-2021, 03:30 PM
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skylor skylor is offline
 
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Quote:
Originally Posted by airguy View Post
Quick rough numbers, about 4.8 MSCFD or 3.3 CFM, with a 25psig drop across that 6 feet, assuming no sharp bends.
Since the end of the tube is in free-air, you need to account for exit loss too which will reduce the flow rate further. I don't have time too look it up right now, though. Entrance loss may also be a consideration, depending on how the source air is connected.

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  #7  
Old 08-25-2021, 03:37 PM
BobTurner BobTurner is offline
 
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Fluid dynamics remains difficult with paper and pencil. Modern jets’ wings are designed with massive numerical computer simulations. There is a solution if viscosity can be set to zero (sarcastically referred to as using ‘dry water’). There is also a solution in the limit that viscosity dominates everything (think of molasses). But for real fluids, it remains intractable, with much of the difficulty lying in the fact that the basic equations are intrinsically non-linear, and the theory of non-linear diff eq is no where near as advanced as for linear equations. BTW, even the fundamental equations (Navier-Stokes) have never been proven to have the expected properties of ‘uniqueness’ (for any given starting parameters, there is only one solution) and ‘completeness’ (for any given starting parameters there is a solution). Such a proof is worth a cool $1 million dollars, as this is one of the so-called ‘millenium problems’ .
Probably more than you cared to know.
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  #8  
Old 08-25-2021, 03:49 PM
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AlexPeterson AlexPeterson is offline
 
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Quote:
Originally Posted by David_Nelson View Post
I thought this would be a seemingly simple question but I keep going down complex rabbit holes ala google. SNIP
That's because compressible gas flow is complex! Some iteration is often required, to determine the flow regime (laminar vs turbulent), are shock waves present, tube friction, etc. I'm quite a few decades out from knowing this stuff well, and would have to look up the best approach to solve it.
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  #9  
Old 08-25-2021, 04:39 PM
AeroEngineer AeroEngineer is offline
 
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There are two stages to the flow here. During the first stage, there is some pressure loss (due to viscosity a.k.a. friction) and the air comes out the tube at some lower pressure than it went in but at the same speed (neglecting changes in density). During the second stage, the remaining pressure causes the air to come out at some speed, according to Bernoulli. These two things are coupled: How much pressure is lost depends on how fast the air is flowing, but how fast the air is flowing depends on the pressure at the exit which depends on how much pressure was lost. So you get a system of equations: Two unknowns (speed and pressure at the tube exit) and two equations (one for viscous pressure loss, then the Bernoulli equation at the exit, both of which connect pressure and speed).

If you assume incompressible laminar flow (that is the “back of the napkin” part… Turbulent and/or compressible flow would give a more accurate answer but would be more complicated), the head loss (in units of pressure) is 32VLν/D² where V is speed, L is pipe length, D is pipe diameter, and ν is kinematic viscosity.[1]

Now, the speed is determined by the difference between Pexit and Patmosphere. According to Bernoulli, Pexit minus Patmosphere is (1/2)ρV², right? (…which is how pitot tubes give us indicated airspeed).

This means that so far we have:

Pcontainer - Pexit = 32VLν/D²

Pexit - Patmosphere = (1/2)ρV²

The unknowns are Pexit and V.

I’m going to convert this all to metric because imperial units are an abomination, a curse from the devil, so I avoid them at all costs unless work makes me use them.

172369 Pa – Pexit = 32 * V * 1.8288m * .0000148 m²/s / (0.003175m)²

Pexit – 101325 Pa = 0.5 * 1.225 kg/m³ * V²

Rearrange the first one,

172369 Pa – 32 * V * 1.8288m * .0000148 m²/s / (0.003175m)² = Pexit

Substitute,

[172369 Pa – 32* V * 1.8288m * .0000148 m²/s / (0.003175m)²] – 101325 Pa = 0.5 * 1.225 kg/m³ * V²

For Wolfram Alpha (because I'm starting to get lazy): (172369-32*y*1.8288*.0000148/(0.003175*0.003175))-101325=0.5*1.225*y^2

y = 277.582 m/s … which is like 80% the speed of sound, again showing that the laminar incompressible flow equations are probably not appropriate for this much pressure.

In any case, if the ID is 1/8”, then the area is 0.000007917304361m², which times 277.6 m/s gives 0.0021977m³/s which is about 4.66 cubic feet per minute.

But, again, this will not be very accurate at all due to neglecting compressibility and turbulence.

And someone please check my math because I could have easily made a mistake somewhere. But I'm within a factor of 1.5 of Greg Niehues' estimate, which is reassuring.

[1] Where did I get “32LVν/D²”? Googling the equations for head loss in pipe flow will return many sets of lecture notes such as this one. Specifically:

ΔP=2*(16/Re)(L/D)(ρV²)

Re = DVρ/ν

ΔP=2*(16ν/DVρ)(L/D)(ρV²) = 32νLρV²/DVρD = 32νLV/D²

If you want, you can go into those lecture notes and use the turbulent flow equations. They involve taking out the "16/Re" and replacing it by the Colebrook or the Zigrang-Sylvester Equation, which include a roughness factor ε. (The last page of the PDF includes a table of ε values to plug in). I think this is enough fun for me for today

PS: Do you know what "nerd sniping" is?

Last edited by AeroEngineer : 08-25-2021 at 04:51 PM. Reason: Added "nerd sniping"
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  #10  
Old 08-25-2021, 05:15 PM
Dave12 Dave12 is offline
 
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Am I the only one with popcorn? Watching the smart guys battle this out?
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