How to determine air velocity in open ended tube?

[rad_controls]

TL;DR Summary

How to determine air velocity in open ended tube given starting pressure and tube length?

Hello,

I am an electrical engineer but was tasked with something a little different this time around and was hoping you guys could help. I would like to determine if my dry air system is sufficient for my purposes and I would like to accomplish that by determining flow rate in CFM. I have been doing my own research but have been having trouble relating Bernoulli's principle to my situation.

If I start with a 50 foot long 0.5in diameter (inner diameter) nylon tube pressurized at 85PSI that is sealed on both ends, and then suddenly opens on one end, how would i determine the velocity of air leaving the tube? I understand as the air begins to leave the tube the velocity would start to decrease so I am interested in the initial, maximum velocity. In actuality, the 50ft long tube is fed by a larger system that is fed by a regulator with a 200 scfm maximum flow rate so I believe that assumption is sufficient.

Follow up questions:
Looking at the problem, I would assume it is a turbulent flow but from what I can tell, without a air velocity I cannot calculate a Reynolds number. do we know what the Reynolds number for nylon tube is?

I believe air should be compressed in this scenario but how does the problem change if we assume it is incompressible?

Thank you in advance!

 

[erobz]

TL;DR Summary: How to determine air velocity in open ended tube given starting pressure and tube length?

Hello,

I am an electrical engineer but was tasked with something a little different this time around and was hoping you guys could help. I would like to determine if my dry air system is sufficient for my purposes and I would like to accomplish that by determining flow rate in CFM. I have been doing my own research but have been having trouble relating Bernoulli's principle to my situation.

If I start with a 50 foot long 0.5in diameter (inner diameter) nylon tube pressurized at 85PSI that is sealed on both ends, and then suddenly opens on one end, how would i determine the velocity of air leaving the tube? I understand as the air begins to leave the tube the velocity would start to decrease so I am interested in the initial, maximum velocity. In actuality, the 50ft long tube is fed by a larger system that is fed by a regulator with a 200 scfm maximum flow rate so I believe that assumption is sufficient.

Follow up questions:
Looking at the problem, I would assume it is a turbulent flow but from what I can tell, without a air velocity I cannot calculate a Reynolds number. do we know what the Reynolds number for nylon tube is?

I believe air should be compressed in this scenario but how does the problem change if we assume it is incompressible?

Thank you in advance!

Just a long pressurized small diameter tube, no supply tank attached? I think it’s a nasty complexity. It’s going to be more involved than “Bernoulli” ( it’s viscous and non steady compressible gas flow you are inquiring about). Even one with a constant pressure supply tank of viscous compressible steady flow makes me cringe.

I’d be willing to say its discharge velocity will be at most the speed of sound, something you expect from a truncated nozzle at over 2 atmospheres.

 

[gmax137]

TL;DR Summary: How to determine air velocity in open ended tube given starting pressure and tube length?

If I start with a 50 foot long 0.5in diameter (inner diameter) nylon tube pressurized at 85PSI that is sealed on both ends, and then suddenly opens on one end, how would i determine the velocity of air leaving the tube? I understand as the air begins to leave the tube the velocity would start to decrease so I am interested in the initial, maximum velocity.

You can estimate the initial velocity by considering the tube to be a vessel with an exit. The head loss thru a straight exit is:

$$h_L=k \frac {v^2} {2g}$$ with k=1. Rearranging,
$$v=\sqrt {\frac {2gh_L}{k}}$$

the head loss here is equal to the pressure difference (85 psig) but in terms of feet at the fluid density. Air at 100 psia and say 70F has a density of about 0.5 lbm/ft^3 so the 85 psi is about 24480 feet.

Then
$$v=\sqrt {\frac {2gh_L}{k}} = \sqrt {\frac {2*32.2*24480} {1}} = 1256 \frac {ft} {sec}$$

Two things:
(1) this exceeds sonic velocity so the calculation should be done for compressible flow, the actual initial velocity will be lower (probably just over one-half of the 1256 ft/sec calculated above)
(2) the volume in your 50 feet of tube is only 0.068 ft^3 so it will be drained in 0.04 seconds (this means that the details of the configuration upstream of the hose will matter. A lot.

Hope that helps (and if anyone else want to check what I did, that would be great!)

 

[rad_controls]

You can estimate the initial velocity by considering the tube to be a vessel with an exit. The head loss thru a straight exit is:

$$h_L=k \frac {v^2} {2g}$$ with k=1. Rearranging,
$$v=\sqrt {\frac {2gh_L}{k}}$$

the head loss here is equal to the pressure difference (85 psig) but in terms of feet at the fluid density. Air at 100 psia and say 70F has a density of about 0.5 lbm/ft^3 so the 85 psi is about 24480 feet.

Then
$$v=\sqrt {\frac {2gh_L}{k}} = \sqrt {\frac {2*32.2*24480} {1}} = 1256 \frac {ft} {sec}$$

Two things:
(1) this exceeds sonic velocity so the calculation should be done for compressible flow, the actual initial velocity will be lower (probably just over one-half of the 1256 ft/sec calculated above)
(2) the volume in your 50 feet of tube is only 0.068 ft^3 so it will be drained in 0.04 seconds (this means that the details of the configuration upstream of the hose will matter. A lot.

Hope that helps (and if anyone else want to check what I did, that would be great!)

Thank you so much for your help! I hope its okay to respond with some follow up questions.

I am assuming your first equation is derived from the Weisbach formula. Can I ask how we are okay with assuming k=1?

Also, how did you come up with a rough head loss of 24480 feet?

I am calculating a flow rate (Q = v*pi*r^2) of 246.61 cubic feet per second which would drain 0.068ft^3 in 2.76e-4 sec. How did you get 0.04 seconds?

 

[jack action]

In actuality, the 50ft long tube is fed by a larger system that is fed by a regulator with a 200 scfm maximum flow rate so I believe that assumption is sufficient.

If this is the case you are interested in, I would use the Fanno flow.

In this lesson, you have a pretty detailed explanation of the concept and what I think would be a relevant example for your case.

In the following image, your regulator would be between 1 and 3 (isentropic flow), and your pipe would be between 3 and E. If there is a normal shock (4-5), I think you will also need the normal shock equations.

It's not as simple as you would think and certainly a completely different approach from the case of a pressurized pipe that is sealed on both ends and then suddenly one end opens.

 

[gmax137]

This Thank you so much for your help! I hope its okay to respond with some follow up questions.

I am assuming your first equation is derived from the Weisbach formula. Can I ask how we are okay with assuming k=1?

The exit loss (k=1) is found in any of the engineering handbooks (Crane 410, Cameron Hydraulic Data, etc.). In this case I think it means that the pressure energy in the tube upstream of the opening is converted to kinetic energy of the flow downstream without any loss; there is no swirling or eddies in the open atmosphere. I'm not 100% on that being the correct interpretation.

Also, how did you come up with a rough head loss of 24480 feet?

The upstream pressure (85 psig) is converted to feet of head using

$h_L=\frac {144 \Delta P} {\rho}=\frac {144 * 85} {0.5} \frac {in^2}{ft^2} \frac {lb} {in^2} \frac {ft^3} {lb}=24,480 ft$

This is based on the pressure difference across the exit being upstream pressure minus atmospheric pressure.

I am calculating a flow rate (Q = v*pi*r^2) of 246.61 cubic feet per second which would drain 0.068ft^3 in 2.76e-4 sec. How did you get 0.04 seconds?

I think you're missing a factor of 144 (in^2 to ft^2).

the flow area is $\frac {\pi} {4} (\frac {d} {12})^2 = \frac {\pi} {4}(\frac {0.5} {12})^2=0.00136 ft^2$
then

$q \frac {ft^3} {sec}=v (\frac {ft} {sec^2}) A (ft^2) = 1256 (\frac {ft} {sec}) 0.00136 ft^2 = 1.71 \frac {ft^3} {sec}$

then

$time=\frac {0.068 ft^3} {1.71 \frac {ft^3} {sec}}=0.0397 sec$