Flow of air through an open tube with a balloon

  • #1
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I hope you guys don't mind a bizarre question from a novice. I've learned just enough about fluid dynamics to be dangerous.

Assume that we have a straight, rigid tube with a constant inner diameter. It's not long, let's say it's around a foot (in case that matters). We cut a chunk out of the middle section of the tube, but add some supports to keep the end pieces in the same spatial position relative to each other. We then use a balloon to connect the ends of the tube. The balloon material is such that at rest it has a constant inner diameter equal to that of the rest of the tube. Essentially, the geometry of the inside of the tube is exactly the same as when we started, but we changed a chunk of the wall material from rigid to stretchable.

We are going to use lung power to blow into the tube. The way I picture this is that the lungs are not capable of any significant compression of air (that assumption could be wrong--it's a guess). As lung space shrinks, what happens is that we impart kinetic energy to the molecules of air and we direct the kinetic energy in the direction of the tube entrance.

Now, somewhere I heard that to get a fluid flowing through a tube, you needed to increase the pressure at the entrance. Fluids flow toward regions of lower pressure.

When we look at the Bernoulli equation, it differentiates between pressure and kinetic energy. As I picture it, pressure comes from the vector of motion of particles that is directed toward the walls of the tube and velocity is the vector of motion left over; i.e. down the tube.

So when we look at the air entering the tube, it doesn't seem that I can use Bernoulli's concept of pressure. The air headed into the tube certainly applies a force to the air sitting in the tube, even though the air stream's velocity is mostly aimed into the tube. Bernoulli's use of pressure and the statement that a fluid flows to regions of lower pressure seem slightly off. If I just analyze it by looking at the kinetic motion of particles, then it's much clearer.

Let's examine the situation when the air blown in is still in the rigid part of the tube and now let's look at it from Bernoulli's point of view. We know that there is a lot of kinetic energy in the air moving down the tube. But what is the pressure on the tube walls relative to the air outside the tube? Do I even have enough information to know?

If we treat air as incompressible, then a change in pressure is equivalent to a change in temperature. So let's say the air temperature is exactly the same as body temperature. My best guess is that the air blown into the tube has a lot of kinetic energy in the direction of the tube's main axis, but the same pressure as the outside air. In other words, if the temperature of the air didn't change, neither did its pressure.

The air reaches the balloon wall. What happens? One theory is: nothing. If the outside pressure = the inside pressure, the balloon will neither stretch nor shrink. Another theory is that the lungs stole some pressure energy in creating the kinetic energy directed toward the tube, so the balloon will shrink. This is interesting in that, as the tube walls shrink, the air should move faster, which will lower the pressure and cause more shrinkage. Eventually, the forward moving particles will strike the shrinking passage walls, increasing the pressure until the system stabilizes. The tube will never completely close off.

Then there is the dark horse of the compression wave created when the blown air first strikes the air in the tube. I'm not sure if that changes anything. The pressure differences are still aimed down the tube and not at the tube walls.

There's a lot I don't know. I should just get a balloon, build the tube and see what happens. It's not something that's quick and easy to build, but it's doable. But even if I found out what happens, there's no guarantee my explanation for it would be correct so I'd appreciate any corrections from someone who actually knows how this all works.
 

Answers and Replies

  • #2
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I hope you guys don't mind a bizarre question from a novice. I've learned just enough about fluid dynamics to be dangerous.

Ok, I'll put my helmet on :).

Now, somewhere I heard that to get a fluid flowing through a tube, you needed to increase the pressure at the entrance. Fluids flow toward regions of lower pressure.

This is indeed correct. There are three common ways you can get a fluid parcel moving. The first is a pressure gradient, the second is friction / viscous effect, and the third is some body force (like gravity, but you also need a density gradient for that, or more fancy, if your fluid is electrically charged a magnetic field will also do)

When we look at the Bernoulli equation, it differentiates between pressure and kinetic energy. As I picture it, pressure comes from the vector of motion of particles that is directed toward the walls of the tube and velocity is the vector of motion left over; i.e. down the tube.

Bernoulli is an energy conservation statement. It states that the total head of a fluid remains equal in absence of any losses, or ... any energy source (see where I'm heading?). The total head consist of the sum of static and dynamic pressure, the first is just the pressure, the second is ##\frac 1 2 \rho v^2## which is the kinetic energy per unit volume (you can add other terms like hydrostatic pressure if gravity is important, but let's ignore that here). One often applied way of using this equation is by stating that on location A the pressure and velocity, and thus the total head is known. On location B only one of the two is known, and you can thus compute the other. But...

What is often forgotten is that there are conditions to be met. Because the total head of location A and B need to be equal to start with. This can be achieved when:
  • A and B are on the same streamline
  • Or the streamlines through A and B have a common origin which have equal total head
  • And between A and B there is no energy source or sink
So when we look at the air entering the tube, it doesn't seem that I can use Bernoulli's concept of pressure. The air headed into the tube certainly applies a force to the air sitting in the tube, even though the air stream's velocity is mostly aimed into the tube. Bernoulli's use of pressure and the statement that a fluid flows to regions of lower pressure seem slightly off. If I just analyze it by looking at the kinetic motion of particles, then it's much clearer.

You should have stopped after the first sentence :) From my previous remark I hope it is clear that Bernoulli does not apply here (at least, you cannot say anything useful about the difference in pressure between the in and outside of the tube, Bernoulli itself is still valid)

Let's examine the situation when the air blown in is still in the rigid part of the tube and now let's look at it from Bernoulli's point of view. We know that there is a lot of kinetic energy in the air moving down the tube. But what is the pressure on the tube walls relative to the air outside the tube? Do I even have enough information to know?

You have indeed enough information to determine this. The tube ends in the surroundings, therefore at the end of the tube the (static) pressure must be equal to this surrounding pressure. But air needs a pressure gradient to flow, this is necessary because the tube has a certain friction which needs to be overcome. This gradient must be in the direction of the flow. Therefore upstream of the entrance the pressure must be higher than at the entrance. So the pressure at the entrance of the tube is highest and decreases steadily towards the end. And at the end the pressure is equal to the surroundings. This means the pressure inside the tube is higher than the surroundings everywhere.

If we treat air as incompressible, then a change in pressure is equivalent to a change in temperature.

Nope, you got it backwards. The change in density (so the compression or expansion) of air changes its temperature. If you treat air as incompressible (which is correct in this case) then temperature has no role in this problem.

So let's say the air temperature is exactly the same as body temperature. My best guess is that the air blown into the tube has a lot of kinetic energy in the direction of the tube's main axis, but the same pressure as the outside air. In other words, if the temperature of the air didn't change, neither did its pressure.

Euhrm... no... I don't even see how this would work for the case when you include compressibility.

The air reaches the balloon wall. What happens?

Since the pressure is higher everywhere in the tube compared to the surroundings, the balloon would expand a little bit.

Another theory is that the lungs stole some pressure energy in creating the kinetic energy directed toward the tube, so the balloon will shrink.

You add energy (total head) to the flow by the muscles in your lungs. This will go both into kinetic energy since the air starts moving and in the static pressure, since that is neccesary to overcome the friction of the tube.

Then there is the dark horse of the compression wave created when the blown air first strikes the air in the tube. I'm not sure if that changes anything. The pressure differences are still aimed down the tube and not at the tube walls.

A compression wave cannot exist in incompressible fluid. Since you've assumed (correctly) that the flow can be regarded as incompressible for this case, this is not an issue. If it was an issue and you would send a compression wave through the tube than the balloon would expand even farther when this wave passes it since you can only get compression by increasing the pressure.

There's a lot I don't know. I should just get a balloon, build the tube and see what happens.

I would be curious!
 
  • #3
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Ok, I'll put my helmet on :).

Can I say I really enjoyed your reply? It's very useful, but with a some nice humor. Let me focus on just one point right now, where I appear to have a big hole in my understanding.

Let's assume we have a straight, horizontal, magically frictionless tube of constant diameter. At the entrance we have ##P_1## and ##v_1##. At the exit, we have ##P_n## and ##v_n##. At any arbitrary point m in between, we have ##P_m## and ##v_m##. Bernoulli's equation, as I understand it, says that $$P_1 + \frac {1}{2} \rho v_1^2 = P_m + \frac {1}{2} \rho v_m^2 = P_n + \frac {1}{2} \rho v_n^2$$If a flow implies a pressure gradient, then $$P_1 > P_m > P_n$$If that's the case, then Bernoulli's equation implies that $$v_1 < v_m < v_n$$The flow seems to increase in speed! But there's also another rule that says $$A_1 v_1 = A_m v_m = A_n v_n$$and since I said that the diameter is constant, we have $$\begin{gather*} A_1 v_1 = A_1 v_m = A_1 v_n\\
v_1 = v_m = v_n\end{gather*}$$ which leads to a contradiction.

Help!

Most of the stuff I find (at an introductory level) talks about Bernoulli's equation. I might find something that mentions the pressure gradient. I haven't seen anything that addresses the contradiction (a problem in trying to learn a topic through web searches, I suppose).

There's actually a practical application of all this. I play the melodica, a wind instrument with a piano keyboard. Some people use hard mouthpieces and some people use flexible tubes, which might expand a tiny bit. People argue about the effects of that tiny bit of stretch in terms of the responsiveness of the instrument. One way to examine the problem is to exaggerate the effect by using a very flexible tube. But even if I built such a tube, I'd still like to understand the physics behind whatever is happening.
 
  • #4
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Can I say I really enjoyed your reply? It's very useful, but with a some nice humor.

Thanks!

Help!

I'll try.

The point here is the assumption of a magically frictionless tube. The only reason the pressure gradient exist is to overcome the friction. If there is no friction then there is no pressure gradient. If there is no pressure gradient then Bernoulli checks out in your example.

So what happens if there ís friction? Note what I said on the application of Bernoulli: it is an energy conservation statement. Meaning it is only valid if between point A and B (or point ##1##, ##m## and ##n## in your case) if and only if there is no energy source or sink. Friction, however, is an energy sink. This means that there is a head loss between point ##1## and ##m##, and between ##m## and ##n##. So the Bernoulli constant differs on each point (that is, the sum of ##p## and ##\frac 1 2 \rho V^2##, which is the total head or total pressure, is lower for point ##m## en lower still for point ##n## compared to point ##1##).
 
  • #5
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Thanks for the response above. It makes sense. Bernoulli's equation always seems to be accompanied by a lot of qualifiers.

The tube ends in the surroundings, therefore at the end of the tube the (static) pressure must be equal to this surrounding pressure.

I saw your qualifier "static" above, so I went searching. I learned about static pressure, dynamic pressure and total pressure. When I said "So when we look at the air entering the tube, it doesn't seem that I can use Bernoulli's concept of pressure," what I was thinking of was clarified when I saw that Bernoulli's P is static pressure. This is different (I think) from someone saying that the pressure must be higher at the entrance than the exit to get a fluid to flow. I think they mean that the total pressure must be different, but I could still be wrong.

I learned the lovely word " inviscid". With an inviscid fluid, the static pressure is constant and the velocity of the fluid is constant all the way through the tube (according to another post by someone with a "mentor" tag). So if I understood that right, the static pressure would not be equal to the surrounding pressure at the end of the tube (for an inviscid fluid). Correct?

I suppose the opposite term is "viscous". I can understand that a viscous fluid would lose total pressure along the way (total pressure = dynamic pressure + static pressure = total energy in the system, some of which is converted to heat and lost due to friction). But why would the loss of energy (total pressure) in the tube from friction be exactly the amount needed to make the static pressure at the exit equal to the surrounding pressure?
 
  • #6
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I saw your qualifier "static" above, so I went searching. I learned about static pressure, dynamic pressure and total pressure. When I said "So when we look at the air entering the tube, it doesn't seem that I can use Bernoulli's concept of pressure," what I was thinking of was clarified when I saw that Bernoulli's P is static pressure. This is different (I think) from someone saying that the pressure must be higher at the entrance than the exit to get a fluid to flow. I think they mean that the total pressure must be different, but I could still be wrong.

Both the static and total pressure need to be higher at the entrance of the tube. You need a gradient in the static pressure to get an air parcel to move.

I learned the lovely word " inviscid". With an inviscid fluid, the static pressure is constant and the velocity of the fluid is constant all the way through the tube (according to another post by someone with a "mentor" tag). So if I understood that right, the static pressure would not be equal to the surrounding pressure at the end of the tube (for an inviscid fluid). Correct?

The static pressure is indeed constant, but why would it not be equal to the surrounding pressure? At the exit of the tube the pressure of the tube and surrounding must be equal (there cannot be a jump in pressure, let's leave shocks aside for now)

I suppose the opposite term is "viscous". I can understand that a viscous fluid would lose total pressure along the way (total pressure = dynamic pressure + static pressure = total energy in the system, some of which is converted to heat and lost due to friction). But why would the loss of energy (total pressure) in the tube from friction be exactly the amount needed to make the static pressure at the exit equal to the surrounding pressure?

The pressure at the end of the tube must be equal to the surroundings, otherwise there would be a pressure jump. If that jump would be there, this would immediately equalize since a jump in pressure means a huge pressure gradient which then means a huge force on the air parcels which would then shoot away equalizing the pressure. So you need to reason from the end of the tube upstream, the pressure at the end is fixed, and the gradient upstream is whatever is needed to overcome the friction.
 
  • #7
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The pressure at the end of the tube must be equal to the surroundings, otherwise there would be a pressure jump. If that jump would be there, this would immediately equalize since a jump in pressure means a huge pressure gradient which then means a huge force on the air parcels which would then shoot away equalizing the pressure.

Thanks. A lot of people stop at the declaration and skip the explanation.

The reason I and others get confused is that if I blow into a short tube and hold a strip of paper a short distance from the exit, the paper will bend. If I don't blow, the paper does not bend. It looks like there is a force being applied to surface of the paper and force over area is the definition of pressure.

If I were mapping the pressure gradient, the curve would have a slight drop (just enough to over come friction) and then a steep drop some distance from the opening. How short a distance would seem to depend on the initial pressure. With the right pressure and tube, it would seem you could bowl over an elephant from 50 feet.

So when you say "end of tube", how close to the end of the tube are we talking about? When you say "immediate", how immediate are you thinking?

I did want to clarify one point:

A compression wave cannot exist in incompressible fluid.

At the macro level, you are correct. At the micro level, you are wrong. Since compression and decompression are defined as changes to the fluid density and fluid density is mass over volume, if we pick a small enough volume (let's say the width of a few molecules) in any incompressible fluid, there will be large changes to the fluid density as molecules zip in and out of the volume.

If I blow into a tube, air comes out the other end almost immediately. The force applied at the opening is transferred through the medium as a compression wave that travels at the speed of sound. At least, that's how I understand it.

Thinking about what you've told me, though, I don't think this pressure wave is significant. It just tells us how long it will take from the time we blow air in before the higher pressure starts to affect the balloon.

If I were using my balloon tube to excite a brass reed and create a tone, I would guess (based on the info you provided about a pressure gradient) that the balloon would expand and the volume would be weak (the air is filling the expanding balloon rather than flowing out) until the balloon reached a state of equilibrium. Then the volume would increase. In musical terms, this would mean a slow attack. Similarly, once I stopped blowing, I would expect a slow decay.

And that is the answer I really wanted to find out. It would still be a cool experiment. For instance, if the pressure gradient is just enough to overcome friction in the tube and the tube is short and has little friction, I would expect the balloon to expand only a little, no matter how hard I blew.
 
  • #8
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I find it interesting how you are working on your intuition of a fluid. The questions you ask are actually quite advanced. But I hope I can clarify a few things again.

Thanks. A lot of people stop at the declaration and skip the explanation.

I like to understand fluids myself in this way as well. I like to build an intuitive understanding of what really happens, otherwise I cannot sleep at night :). It is tricky since fluids behave sometimes rather counter intuitively.

The reason I and others get confused is that if I blow into a short tube and hold a strip of paper a short distance from the exit, the paper will bend. If I don't blow, the paper does not bend. It looks like there is a force being applied to surface of the paper and force over area is the definition of pressure.

If I were mapping the pressure gradient, the curve would have a slight drop (just enough to over come friction) and then a steep drop some distance from the opening. How short a distance would seem to depend on the initial pressure. With the right pressure and tube, it would seem you could bowl over an elephant from 50 feet.

So when you say "end of tube", how close to the end of the tube are we talking about? When you say "immediate", how immediate are you thinking?

I think something in the order of the radius of the tube. What you are missing is that, although the static pressure is equal to the surroundings, the dynamic pressure (i.e. the kinetic energy) is not. So what happens if you put a paper in front of the tube is that the flow hits this paper, gets slowed down by it, but then Bernoulli says that the pressure increases again, and there you have your increased pressure again needed to apply a force (in this case a pressure difference over the two sides of the paper).

At the macro level, you are correct. At the micro level, you are wrong. Since compression and decompression are defined as changes to the fluid density and fluid density is mass over volume, if we pick a small enough volume (let's say the width of a few molecules) in any incompressible fluid, there will be large changes to the fluid density as molecules zip in and out of the volume.

Of course I'm not wrong ;). We need to talk about definitions. What do we mean by density? This is actually more complex than you might think.

Fluid is considered a continuum, this is an assumption which eases the development of a mathematical model of the fluid. It means that all the flow variables like mass, velocity, density, temperature, etc. are considered to be smeared out over space. This is a very good assumption and is pretty much the only real assumption in the Navier-Stokes equation. Almost every equation you know about fluid dynamics can be derived from Navier-Stokes, and thus also has the continuum assumption in it.

The continuum assumption is characterized by the Knudsen number which is the mean free path length of the molecules divided by the characteristic length of the smallest flow feature in the fluid. The mean free path length of an air molecule at standard pressure is in the order of ##7\times 10^{-8}m##, so that's 0.07 micro meter. The smallest features in a typical flow are in the order of tenths of millimetres (the smallest scales in a turbulence flow). So there are several orders of magnitude in between which means that you really do not have to be concerned about individual molecules.

So concretely this means that if you consider a very small volume of air in which every variable (speed, density, etc) can for all intents and purposes be considered constant, still contains a massive amount of molecules. The random fluctuations in density as you considered are absolutely tiny at this scale. (and they are also, well, random. A compression wave is not a random change in density but a very well defined flow feature)

Fun fact: at the edge of space the air is so thin that the continuum assumption breaks down. This means Navier-Stokes cannot be used for the first part of the reentry of a space vehicle. Also the thickness of shock waves are in the order of the mean free path length of air molecules.

Further more: incompressible fluid is a fluid model. No real fluid is actually entirely incompressible. Not even water. Water can carry sound waves and sound waves compress and expand the water a tiny bit. Mathematically, incompressible fluid can simply not carry a wave. The speed of sound in a truly, mathematically incompressible fluid is infinite. This means that the influence of a pertubation at point A is instantly felt at point B. Therefore the incompressible fluid model it cannot describe waves of any kind.

If I blow into a tube, air comes out the other end almost immediately. The force applied at the opening is transferred through the medium as a compression wave that travels at the speed of sound. At least, that's how I understand it.

That would indeed be correct.

Thinking about what you've told me, though, I don't think this pressure wave is significant. It just tells us how long it will take from the time we blow air in before the higher pressure starts to affect the balloon.

If what you mean is the wave that forms because you start to blow into the tube then you are absolutely right. The wavelength of this wave is much larger than the tube such that the change at the end of the tube is pretty much instant and there is not really a wave travelling through the pipe.

What I had in mind was a real compression wave with a wave length in the order of the diameter of the tube. This will travel through the pipe with speed of sound and when it passes the balloon the latter will expand a bit. You would not be able to generate such a wave of significant strength with your mouth though :).

If I were using my balloon tube to excite a brass reed and create a tone, I would guess (based on the info you provided about a pressure gradient) that the balloon would expand and the volume would be weak (the air is filling the expanding balloon rather than flowing out) until the balloon reached a state of equilibrium. Then the volume would increase. In musical terms, this would mean a slow attack. Similarly, once I stopped blowing, I would expect a slow decay.

It wouldn't happen in such discrete points in time but more or less simultaneously but indeed you are right. The balloon absorbs a bit of the pressure you put in with your mouth delaying the pressure buildup in the instrument. It will probably also act as a bit of a damper, absorbing or smoothing the changes in the pressure your mouth provides.

And that is the answer I really wanted to find out. It would still be a cool experiment. For instance, if the pressure gradient is just enough to overcome friction in the tube and the tube is short and has little friction, I would expect the balloon to expand only a little, no matter how hard I blew.

That would be an interesting experiment!
 
  • #9
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I find it interesting how you are working on your intuition of a fluid. The questions you ask are actually quite advanced.

Thanks. I was really hoping that I could answer most of my questions by applying some simple principles, but that's looking like a false hope.

What you are missing is that, although the static pressure is equal to the surroundings, the dynamic pressure (i.e. the kinetic energy) is not.

I actually thought this might be the case, but the exact differences between static and dynamic pressure outside of a tube is not yet clear. Let's not worry about this too much as I think I get the gist. Plus, I just got started reading an actual book of fluid mechanics (Introduction to Fluid Mechanics by Yasuki Nakayama) rather than working off of random web pages.

Of course I'm not wrong ;).

Of course! :-)

However, one of the other questions that I need to answer is how long it takes for air blown in one end of a tube to begin a flow on the other end—and why. The why part requires, I believe, looking at very tiny regions of the fluid, probably smaller than tenths of millimeters. If I ignore compression because it factors out on the tenths of millimeters scale, then my explanation vanishes.

The wavelength of this wave is much larger than the tube such that the change at the end of the tube is pretty much instant and there is not really a wave traveling through the pipe.

Now this gets really interesting. Yes, the wavelength is much larger than the tube. Yes, it does affect the other end of the tube "pretty much instantly", but I want to know exactly how much it's not instant and why. I believe it travels at the speed of sound, but I don't yet have a complete understanding of the process.

The best reference I've found is at https://courses.lumenlearning.com/suny-osuniversityphysics/chapter/17-2-speed-of-sound/. As far as I can tell, their analysis of a sound wave doesn't change based on the frequency of the sound. It should apply as well for a wave of 0.1 Hz as for one at 440 Hz, but I haven't tried to work through all the math on my own.

In any case, this is a somewhat different question and I am going off topic.

Thank you for all your time! I appreciate it. Hopefully, the textbook will clear up a lot of things.
 
  • #10
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Ok, one short remark because I suggested that the long wave travels faster than the short wave. That is not true. As you said, all waves of all frequencies travel at the same speed, i.e. the speed of sound. The speed of sound is determined by the compressibility of the medium. That is thermodynamics.
 

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