Is it possible to find out the speed of air, given pressure and size of exit

[Tugberk]

For an experiment, I sent specific pressures of air strams through a tube and I was wondering if there was a way to calculate the speed of the air since I know the pressure and the radius of the tube where it left from.

 

[Simon Bridge]

Look up "free expansion".
Though you could try a related problem - firing a slug from an air-gun.
Knowing the pressure each side of the slug, and the dimensions of the barrel, you know the muzzle-speed of the slug right?

 

[Chestermiller]

For an experiment, I sent specific pressures of air strams through a tube and I was wondering if there was a way to calculate the speed of the air since I know the pressure and the radius of the tube where it left from.

If the pressure drop is not too large (say less than about 10%), you could approximate the air as being incompressible, and use the basics for the pressure drop/flow rate relationship: The pressure gradient is equal to the friction factor f times ρv²/2 times 4/D, where v is the cross section average velocity and D is the diameter, and where the friction factor is a known correlated function of the Reynolds number (unless the flow is laminar (Re<2100), in which case the friction factor is exactly 16/Re). If the flow is compressible (i.e., the pressure drop is too large), the calculation is a little more difficult, but no big problem. To find out how to do it for a compressible gas, see the oil and gas handbooks.

 

[Tugberk]

If the pressure drop is not too large (say less than about 10%), you could approximate the air as being incompressible, and use the basics for the pressure drop/flow rate relationship: The pressure gradient is equal to the friction factor f times ρv²/2 times 4/D, where v is the cross section average velocity and D is the diameter, and where the friction factor is a known correlated function of the Reynolds number (unless the flow is laminar (Re<2100), in which case the friction factor is exactly 16/Re). If the flow is compressible (i.e., the pressure drop is too large), the calculation is a little more difficult, but no big problem. To find out how to do it for a compressible gas, see the oil and gas handbooks.

Is the p for density, or pressure? Also, how can I multiply by 16/Re, where Re is not a number? Thank you though ^^ :)

 

[Chestermiller]

Is the p for density, or pressure? Also, how can I multiply by 16/Re, where Re is not a number? Thank you though ^^ :)

The ρ is density. The Reynolds number Re is equal to the density times the velocity times the diameter divided by the viscosity. You look up the viscosity of air in any book on fluid mechanics or in a handbook, or on line. The density can be calculated from the ideal gas law.
Did you not get the connection that Re is the abbreviation for the Reynolds number?

 

[Tugberk]

The ρ is density. The Reynolds number Re is equal to the density times the velocity times the diameter divided by the viscosity. You look up the viscosity of air in any book on fluid mechanics or in a handbook, or on line. The density can be calculated from the ideal gas law.
Did you not get the connection that Re is the abbreviation for the Reynolds number?

I got that Re is Reynolds number, but I'm in high school, this is the first time I've heard of Re :/, also, I don't get how this will help me because the pressure isn't used anywhere, I'm using low pressures, 1.5 Bars, 2 Bars etc., so the change in density is negligible

 

[Chestermiller]

You know the pressure at the inlet of the tube and at the outlet of the tube (1 bar), correct? If the flow is laminar, then the pressure drop is related to the cross section average velocity by:

ΔP = η(32v/D²)L

where L is the tube length and η is the viscosity of air at the gas temperature.

Use this equation to solve for v (in consistent units). Then check the Reynolds number to see if the flow is really laminar.

Re = ρvD/η

where ρ is the air density at the average of the inlet and outlet pressures, evaluated using the ideal gas law.

If Re < 2100, you're done. If Re > 2100, then you have to recalculate because the flow is going to be turbulent.

For turbulent flow, the pressure drop is

ΔP = (ρv²/2) (0.079/Re)^0.8(4L/D)

where Re is again given by Re = ρvD/η, but with the new velocity you are now calculating for turbulent flow being used in the equation (not the one from the laminar flow calculation)

You now need to use your algebra to solve the above turbulent ΔP equation for the velocity v in terms of ΔP. Then substitute the data into the equation to get v.

 

[russ_watters]

Why not just use a simplified form of Bernoulli's equation? I think what you are suggesting may reduce to it.

 

[boneh3ad]

Depending on the pressure involved, it is entirely possible that none of this will work. Pretty much all instances of compressed air leaving through a hole are, in fact, compressible. That means that Bernoulli's principle does not apply and it means that things like Hagen-Poiseuille flow do not apply either. The best way to start thinking about this is to have the OP answer a few simple questions.

1) What is the pressure in your reservoir and is it just exhausting to the standard atmosphere?

2) How long is this tube through which you are sending your flow? Is it of constant cross-sectional area?

 

[Chestermiller]

Why not just use a simplified form of Bernoulli's equation? I think what you are suggesting may reduce to it.

If doesn't. This describes the viscous pressure drop, which Bernoulli's equation does not handle. Bernoulli's equation assumes that the system is conservative.

 

[Chestermiller]

Depending on the pressure involved, it is entirely possible that none of this will work. Pretty much all instances of compressed air leaving through a hole are, in fact, compressible. That means that Bernoulli's principle does not apply and it means that things like Hagen-Poiseuille flow do not apply either. The best way to start thinking about this is to have the OP answer a few simple questions.

1) What is the pressure in your reservoir and is it just exhausting to the standard atmosphere?

2) How long is this tube through which you are sending your flow? Is it of constant cross-sectional area?

I guess I made the implicit assumption that the tube is long enough (so that entrance and exit effects are negligible) and that the cross sectional area of the tube is constant. There are, of course, corrections that can be applied to Hagen_Poiseuille (or the turbulent version of Hagen Poiseuille that I gave) to account for entrance and exit effects.

The OP's comment that the pressure is nowhere higher than 1.5 of 2 bars seems to suggest the exit is a atmospheric pressure, and the inlet is at a little higher pressure, and that compressibility can be neglected as a first approximation (although a factor of 2 variation on the pressure is higher than I would like to see if compressibility is being neglected).

 

[russ_watters]

If doesn't. This describes the viscous pressure drop, which Bernoulli's equation does not handle. Bernoulli's equation assumes that the system is conservative.

Since we know virtually nothing about the system, how can we even begin to examine losses?

 

[boneh3ad]

Assuming his reservoir pressure is 2 bar and his exit pressure is around 1 bar, that is more than enough for compressibility to be important. Depending on geometry that could even be supersonic!

 

[Tugberk]

The tube which the air was sent through is approximately 25 cms long. One end is closed and is where the air constantly flows in, at a constant pressure, this is unknown. A valve allows me to control how much air is put into the 25 cm long tube. By covering the exit, I can build up pressure and read it off of a barometer.

 

[boneh3ad]

So you cover one end and fill it, then release it?

At any rate, at a pressure ratio (p₂/p₀₁) of 0.5, your flow will be choked through the tube, meaning that you can calculate the mass flow pretty quickly and easily if you know your reservoir pressure and temperature and the area of your tube (neglecting any viscous effects). It also means the flow through the tube will be exactly Mach 1. The actual exit velocity will depend on whether the tube exhausts straight to atmosphere or has some kind expansion on the end.

When your reservoir pressure is less than about 1.89 atm, you won't reach sonic conditions in your tube and you will need a tiny bit more effort to figure out mass flow and velocity, but it is still pretty easy as long as you ignore viscosity, which may or may not be a reasonable assumption in your case. If you do make that assumption, it is just a matter of determining Mach number based on the isentropic flow assumption and working from there to get temperature, density and velocity and thus mass flow rate.

 

[Chestermiller]

The tube which the air was sent through is approximately 25 cms long. One end is closed and is where the air constantly flows in, at a constant pressure, this is unknown. A valve allows me to control how much air is put into the 25 cm long tube. By covering the exit, I can build up pressure and read it off of a barometer.

Russ_watters is correct... the nature of the geometry and the nature of the experiment is still very unclear.

What does this mean?: "One end is closed and is where the air constantly flows in." If the end is closed, how can air flow in?

Please provide more details of the experiment?

What is the diameter of the tube. Where are the valves located? Do you allow the air to escape slowly from the exit value (if there is an exit valve)?

So far, as best I can understand, you pressurize a cylindrical tube with air through one end while the other end is covered. The pressure reaches a certain value, and then you stop adding air. Then you suddenly open the capped end to the atmosphere and let the air escape. You want to find out what? How much air escapes when you remove the cap? How long it takes to return to atmospheric pressure? What the velocity of the air exiting the tube is as a function of time?

 

[Tugberk]

Russ_watters is correct... the nature of the geometry and the nature of the experiment is still very unclear.

What does this mean?: "One end is closed and is where the air constantly flows in." If the end is closed, how can air flow in?

Please provide more details of the experiment?

What is the diameter of the tube. Where are the valves located? Do you allow the air to escape slowly from the exit value (if there is an exit valve)?

So far, as best I can understand, you pressurize a cylindrical tube with air through one end while the other end is covered. The pressure reaches a certain value, and then you stop adding air. Then you suddenly open the capped end to the atmosphere and let the air escape. You want to find out what? How much air escapes when you remove the cap? How long it takes to return to atmospheric pressure? What the velocity of the air exiting the tube is as a function of time?

I don't stop adding air, the air is still flowing into the tube at the same rate. I attached a picture of the system, The little blue pipe is constantly blowing air into the tube at the same velocity/pressure. You can see the barometer and the valve right next to it, The valve allows me to control how much air I let into the main Tube. Than I cover the exit, read the barometer to see how much pressure builds up in the tube, than I let the air flow again, and I want to find out how fast the air moves when it's leaving the tube. The main blue pipe is still supplying air at this point.