What's new
Van's Air Force

Don't miss anything! Register now for full access to the definitive RV support community.

Fluid Dynamics Question

David_Nelson

Well Known Member
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,
 
Quick rough numbers, about 4.8 MSCFD or 3.3 CFM, with a 25psig drop across that 6 feet, assuming no sharp bends.
 
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....:D 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_flow/flow_of_air_in_pipes_14029.htm
 
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.

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

I used Darcy's equation in my original answer, which does take into account compressibility - so therefor I shall stand on my soapbox and taunt you, and proclaim my answer vastly more knowledgable and superior to yours. :D

https://www.youtube.com/watch?v=QSo0duY7-9s

At least we are both in the same order of magnitude!

*sits in the chair with a proper beverage and popcorn, prepares to watch the show*
 
Last edited:
Here is a tip. Test it. Empirical data will set you free. There are unknows like the source of the air (volume/CFM available) and tube surface roughness. You can estimate CFM by velocity and area. Easy.
 
I shall stand on my soapbox and taunt you, and proclaim my answer vastly more knowledgable and superior to yours. :D

I concede defeat ;) David asked for a “back of the napkin” estimate, and that’s what I delivered (including a link to some turbulent flow equations in case anyone felt like being any less lazy than me).

At least we are both in the same order of magnitude!

And that’s what confirmed to me that I was in the ballpark. Your answer is about 2/3 of mine, and you took viscosity (i.e. more losses) into account and I neglected it, so, you getting a lower (but not massively lower) answer was to be expected. Nice work!

Given that the last time that I solved a problem like this was ~20 years ago, I’m feeling pretty good about how I did ;)
 
Given that the last time that I solved a problem like this was ~20 years ago, I’m feeling pretty good about how I did ;)

I cheated - I used an Excel spreadsheet that I wrote about 20 years ago for this exact sort of thing, back when I started in the oilfield engineering world.
 
In the time you slide-rulers take to do the thought experiment, a hands-on guy could've cobbled up a test jig and given us an actual real-world answer based on measurement rather than theory :D
 
Yeah---I'm watching-----you guys make my head hurt! If we put all of you in a room, we could solve 90% +- of the worlds problems before lunch!
This is fun to watch---and NO, I'm going to mix it up with a bunch of guys that are ALOT smarter than I am! :D

Tom
 
I cheated - I used an Excel spreadsheet that I wrote about 20 years ago for this exact sort of thing, back when I started in the oilfield engineering world.

Ha! I was wondering why you didn’t show your work ;)

In the time you slide-rulers take to do the thought experiment, a hands-on guy could've cobbled up a test jig and given us an actual real-world answer based on measurement rather than theory :D

Yeah but could he do it from the comfort of his couch while wearing his pajamas?

If we put all of you in a room, we could solve 90% +- of the worlds problems before lunch!

Ha! That might be true… if everyone agreed to respect expertise. One of the most baffling things to me about the modern world is the fact that if I tell someone “Don’t pull more than 4g on this airplane or you’ll bend metal or pop some rivets” or “With this load and at this elevation, you’re gonna need three times as much runway as usual to get off the ground, and half a mile before you climb past those 50-foot trees”, people trust me on it because I’ve spent my whole life doing tests and analysis about airplane design… but someone who spent even more time doing tests and research about climate science or epidemiology does not get the same trust or respect. It boggles the mind. Pretty sure the world’s problems will remain unsolved, at this rate…
 
In the time you slide-rulers take to do the thought experiment, a hands-on guy could've cobbled up a test jig and given us an actual real-world answer based on measurement rather than theory :D
Oh? How would you measure the flow rate at the exit in practice?
 
Ha! I was wondering why you didn’t show your work ;)



Yeah but could he do it from the comfort of his couch while wearing his pajamas?

Depends how his wife feels about experimentation outside of the four walls of the man-cave, I guess.



Ha! That might be true… if everyone agreed to respect expertise. One of the most baffling things to me about the modern world is the fact that if I tell someone “Don’t pull more than 4g on this airplane or you’ll bend metal or pop some rivets” or “With this load and at this elevation, you’re gonna need three times as much runway as usual to get off the ground, and half a mile before you climb past those 50-foot trees”, people trust me on it because I’ve spent my whole life doing tests and analysis about airplane design… but someone who spent even more time doing tests and research about climate science or epidemiology does not get the same trust or respect. It boggles the mind. Pretty sure the world’s problems will remain unsolved, at this rate…

This right here is going to get a post deleted or a thread locked. Trust me - but don't ask how I know.
 
Here is a tip. Test it. Empirical data will set you free. There are unknows like the source of the air (volume/CFM available) and tube surface roughness. You can estimate CFM by velocity and area. Easy.

So easy you did not provide a quantitative answer?
 
Thanks!

This is perfect. This has been very entertaining and educational.

Greg - Kudos to you for not only remembering that you had done a spreadsheet 20 years ago but that you were able to find it, too. What the heck is 'MSCFD'?? I had to look that one up ... "Abbreviation for a thousand standard cubic feet per day, a common measure for volume of gas. Standard conditions are normally set at 60oF and 14.7 psia." I tried plugging that into my handy 'units' conversion program and it balked. Thanks for converting that to CFM for me. BTW, great Monty Python reference.

Bill and George - This started out as a "that should be simple to figure out" idea that should only take a few minutes using an online calculator. That assumption obviously did not pan out. Now that I know that things are within the realm of possibility, I will go through the trouble to cobble something together and verify.

Conrad - Thanks for the time to spell things out. 'Nerd Sniping' ... hahaha - I hadn't seen that one before. I was expecting something related to 'snipe' from the movie 'Up' but this was much more entertaining and a great reference to XKCD.

When I was trying to work through this on my own, the idea of sonic and subsonic flows came up. The idea that air is flowing at 80% of the speed of sound in this small tubing with only 25 psid still amazes me.

Again, thanks guys! You've been a great help.

We now return you to your normally scheduled program.
 
Flow it into a collapsed bag of known volume, measure the time it takes.

The bag will be above atmospheric pressure while expanding (otherwise it can't do the work required to expand) , so the required exit condition is not met.
 
I also was wondering why Greg did not show his work:)
I was spending my moment thinking how I would measure the flow.
Russ's idea is a good one but I am with DMN that it would not work great.
Things I measure are larger than that so would normally use a flow grid or hot wire anemometer.
 
The bag will be above atmospheric pressure while expanding (otherwise it can't do the work required to expand) , so the required exit condition is not met.

This delta p will be trivial. Only the weight/area that the poly film the bag is constructed from.
 
The internet isn't what it used to be - - what is . . .

Wouldn't a 5 gal water bottle upside down (with water) be easier to measure? fill, turn over, stick the tube inside to displace water. Just record the time-to-empty, then determine flow rate.

450 ft/sec was seeming pretty high for a 1/8" diameter and that long so I looked for some URL's I had to do this instead of the table look up method for all the variables and parameters. Not active anymore.

Online calculators - - there used to be some good calculators what used Re, friction factors etc and did this type of calculation, but I guess they have been googled into the netherworld of demonetized sites.
 
The bag will be above atmospheric pressure while expanding (otherwise it can't do the work required to expand) , so the required exit condition is not met.

Use a soap solution over the end of the tube, record time to inflate bubble to X diameter using time lapse photography, and assume negligible weight/resistance of soap film. Great excuse to purchase a slo-mo digital camera, which can later be used for monetizable YT videos.
 
The bag will be above atmospheric pressure while expanding (otherwise it can't do the work required to expand) , so the required exit condition is not met.

Collapse the bag to atmospheric pressure. Fill to just short of capacity. Tie off and submerge the bag to find the volume. I've used this method for similar experiments before.
 
Yeah---I'm watching-----you guys make my head hurt! If we put all of you in a room, we could solve 90% +- of the worlds problems before lunch!
This is fun to watch---and NO, I'm going to mix it up with a bunch of guys that are ALOT smarter than I am! :D

Tom

Ok Tom, who’s winning this? I’m really digging the bubble blowing idea.
 
Collapse the bag to atmospheric pressure. Fill to just short of capacity. Tie off and submerge the bag to find the volume. I've used this method for similar experiments before.

OK, I thought you were talking about some sort of balloon arrangement. If you use a largish, very light plastic film bag it will be roughly 0.01 kg/m^2, so additional pressure roughly 0.1 Pa if you don't fill to capacity. Negligible compared with standard atmosphere of 101.3 kPa.

Looks like yours is the best experimental solution proposed so far ...
 
Dave---we all are winning! There are some unbelievably smart guys on VAF, some of which unlike me can type and spell==And we all can ask them questions!!!
I'd say that makes us all winners.
Tom
 
Funny off-topic, sorry . . . .

OK, I finally looked up nerd sniping. Kinda sad and funny. Sad because some is so true. Funny for the same reason.

I had some PhD's working for me and one (EE) could not make a 45 min presentation in an hour, not even close. Did you know our internal clock stops when we think? Someone would ask him a question, then the clock stopped counting, he could talk for 15 min and not know time passed. I had a clicker and a laser pointer as a training aid. As soon as the clock stopped I would click and point the laser pointer at the ceiling where others could not see but he could.

Sometimes it worked. Others, people got up and left when the next group wanted the conference room.
 
Half a second - really?

Some interesting ideas around how to measure and confirm this.

In keeping w/ the "back of the napkin" approach, my plan was to take a 5 gal bucket of water and a 1 gal milk jug, also filled with water. Place in the jug upside down in the bucket. Get the hose and air going, keep an eye on a nearby timer and at a convenient time put the hose into gallon jug. Let the jug do its buoyancy thing while keeping it properly oriented and note the elapsed time when the jug fills.

Recall that the brain trust estimated somewhere between 3.3 - 4.66 CFM. The jug is going to fill in about 2 seconds. Gut feel tells me that that just seems way to fast for 25 psi and a 6' piece of 1/8" ID tubing. Now my curiosity is really getting piqued.
 
Last edited:
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,

I like pie!
 
Some interesting ideas around how to measure and confirm this.

In keeping w/ the "back of the napkin" approach, my plan was to take a 5 gal bucket of water and a 1 gal milk jug, also filled with water. Place in the jug upside down in the bucket. Get the hose and air going, keep an eye on a nearby timer and at a convenient time put the hose into gallon jug. Let the jug do its buoyancy thing while keeping it properly oriented and note the elapsed time when the jug fills.

Recall that the brain trust estimated somewhere between 3.3 - 4.66 CFM. The jug is going to fill in about 0.5 seconds. Gut feel tells me that that just seems way to fast for 25 psi and a 6' piece of 1/8" ID tubing. Now my curiosity is really getting piqued.

Should your calculations be 2 seconds? 4 cfm x 8 gal/cf = 32 gal/min, 60s/min / 32gal/m = 2s.

This still does not pass the reasonableness test.

OK, who's gonna do this excellent test first??
 
OK I think Ross's may be the best idea yet to measure CFM but wouldn't submerging the bag require enough down force to pressurize the content to the point it would shrink.
 
Dave---we all are winning! There are some unbelievably smart guys on VAF, some of which unlike me can type and spell==And we all can ask them questions!!!
I'd say that makes us all winners.
Tom

I must agree. Sorry Mel, all we have is popcorn, or are you talking about the other Pi? We’ve seen some interesting responses, Aero Engineer’s response was quite long, so he must know what he’s talking about. Russ is kind of like the old E.F. Hutton guy, you know, when he talks people listen ( showing my age here). But, since Dan H. has not offered a response that completely blew everyone out of the water, I am going to offer that due to being the earliest respondent that has not been refuted (I think?) and to his added comedic component, the winner is…………..Airguy!

Disclaimer-This is not meant to ruffle any academic feathers.
 
Nope

550955]Am I the only one with popcorn? Watching the smart guys battle this out?[/QUOTE]

[QUOTE=Dave12;1I'm having a bowl of shredded wheat & bran flakes topped with my home grown blueberries...:D:D:D
 
Should your calculations be 2 seconds? 4 cfm x 8 gal/cf = 32 gal/min, 60s/min / 32gal/m = 2s.

This still does not pass the reasonableness test.

OK, who's gonna do this excellent test first??

You are correct. Sigh, this is what I get for doing math in public, in the morning, and not carrying my units all the way through.

Thanks,
 
About 3 seconds ...

I did the "gallon jug in a 5 gal bucket of water" experiment this afternoon. The regulator at the tubing inlet was set to 25 psig while air was flowing. There were many variables (gauge accuracy, timing, barbed fitting into the 1/8" ID tubing, operators, etc, etc). Three runs were performed and the gallon jug filled in about 3 seconds. That comes out to about 2.7 CFM.
 
I did the "gallon jug in a 5 gal bucket of water" experiment this afternoon. The regulator at the tubing inlet was set to 25 psig while air was flowing. There were many variables (gauge accuracy, timing, barbed fitting into the 1/8" ID tubing, operators, etc, etc). Three runs were performed and the gallon jug filled in about 3 seconds. That comes out to about 2.7 CFM.

Nicely done!
 
I did the "gallon jug in a 5 gal bucket of water" experiment this afternoon. The regulator at the tubing inlet was set to 25 psig while air was flowing. There were many variables (gauge accuracy, timing, barbed fitting into the 1/8" ID tubing, operators, etc, etc). Three runs were performed and the gallon jug filled in about 3 seconds. That comes out to about 2.7 CFM.

I think it's a bad test, the exit pressure isn't atmospheric at all, so the delta-p across the tube isn't 25psi.

I'd do it with two pressure transducers, dump the tube into a pressure vessel that starts out at and measure pressure in both versus time. I'd start at say 30 psig, so I could measure the rate of pressure change accurately as it passed thru 25psig. With that I could use the volume of the pressure vessel to calculate the mass flow (ideal gas law) and with that you can easily calculate volume.

Be careful with "CFM" because if you use volume flow you need to specify if that volume flow is at the exit condition or the initial condition, the temperature, etc. Mass flow is more useful.
 
I remember as a soon to be graduate of Texas A&M's aerospace department our highest offers for jobs were coming in from the large oil companies needing folks who could calculate pipe flow and design systems. I thought that inglorious at the time and went for a real airplane company offer for less. What a naive dumb ***.

Joke I overheard at the Garmin campus. "How can you tell an engineer is an extrovert?". A: "When he talks to you instead of looking down at his shoes, he looks at yours".
 
Last edited:
WoW

Popcorn is half empty. Still laughing.
Art
 

Attachments

  • IMG_0919[1].jpg
    IMG_0919[1].jpg
    372.5 KB · Views: 145
Back
Top