Does Fluid Dynamics Explain Pressure Gradients in a Frictionless Tube?

[Arjan82]

1) The pressure of a fluid exiting the tube will be equal to the atmospheric pressure.

Correct!

• Is the statement correct or does it need some tweaks?

It is correct as long as you are not interested in the pressures within, say, one diameter of the exit.

• Does pressure mean "static pressure"? (I'm pretty sure it means "static".)

Correct!

2) To create a flow in the tube, I've been told I need a difference in pressure.

Correct!

• Is this true for viscous fluids as well? (I think it is; viscous or inviscid, you need a net force to start the flow.)

Exactly as you say, a (static!) pressure difference is the only way you can apply a force on any kind of fluid and a force is the only thing that will make a fluid (or anything for that matter) move (or accelerate, to be more precise).

• Does the term "pressure" here mean "total pressure" or "static pressure"? (I think it's "total pressure"—force is force.)

No! It's static pressure! Total pressure is static pressure plus kinetic energy (dynamic pressure is kinetic energy). So you can use total pressure for energy analysis (Bernoulli is an energy conservation statement). But ONLY static pressure will apply a force!

3) If I have a tube with a uniform cross-section, a steady flow and an incompressible fluid, then I can apply the continuity equation. It tells me that the velocity of the fluid is equal through any cross-section.

You can ALWAYS use the continuity equation (as long as we stay on Earth that is, it fails in near vacuum). Even if the flow is compressible, or time dependent or both. The continuity equation looks a bit different in each case, but it states simply that mass is conserved. Thus the continuity equation is the mathematical equation that enforces mass conservation.

4) Starting from #3, if I also assume the conditions needed for Bernoulli's equation, then the static pressure in the tube is the same at all points.

Yup, that's why applying Bernoulli in a pipe doesn't make much sense. Only when you are analyzing short sections with a contraction or something like that (so a Venturi or just a contraction in the diameter).

This is where the discussion veered into some rather interesting areas. It appears that by having different static pressures at the two ends of the tube and specifying a frictionless flow, I created a theoretical monster. I thought I needed a difference in pressures to get a flow started (#2) and a frictionless flow to apply Bernoulli.

The theoretical monster truly arises if you insist on all of these three points:

• Frictionless flow
• A pressure difference over the tube
• A steady state case, thus the flow looks the same now as in a bit of time

I'm pretty sure that the universe collapses if you try this...

On the piston:

• I'm pretty sure I get a flow.

You do indeed.

• Can I apply the continuity equation? Water is not highly compressible.

As I said, the continuity equation is a mass conservation statement. You can always apply this, even for a compressible flow, although the equation looks a bit different in that case.

• If I can, then if I push 1 L/s in, I should get 1 L/s out and through every cross-section.
• If I know the cross-section, I can calculate the velocity of the flow.
• Going back to #1, the pressure at the end of the tube is whatever the exit pressure is.

Correct, applying the first two points is essentially applying the continuity equation.

• If I incorrectly apply Bernoulli to this, the equation tells me that the static pressure is equal at every point; therefore, it has to be equal to the exit pressure.

Correct.

• I then correct for friction in the tube and from the viscosity of the water. This tells me the static pressure drop over the tube's length. There may be some other corrections that need to be applied.

Correct! Note that also here, if in steady state, the force you apply though the piston balances with the friction in the tube (In the case of a syringe it is a bit more complicated because you are also continuously accelerating flow in the reservoir to the velocity in the needle, this also takes a force to do, so to be more exact: the pressure difference over the tube balances with the friction).

• If I were a real physicist, I would then be able to calculate the force needed by the piston to maintain that 1 L/s flow.

You are already very close. This last step is not the hardest. Pressure is force divided by area, so if you multiply the force by the area of the piston you have your force again.

About the compressor bit:

• I connect a hose to an air compressor's tank. I have now violated the incompressibility requirement of Bernoulli's.

For the bread and butter Bernoulli equation indeed, but there also exists a compressible variant (and even a time dependent variant).

• I have a limitless supply of compressed air.
• This is a guess: the mass flow through any cross-section is equal. It would seem that if x grams of air goes in, x grams has to go out. It can't be more and if it is less, the air would be collecting somewhere and the tube would eventually blow up.

Good guess, this is what the continuity equation states mathematically (if you take the correct version)

• Another guess: the fluid density is a gradient from entrance to exit. I'm totally guessing; for all I know, the density drops completely right near the entrance.

Phew, this depends on the details (and I usually only deal with incompressible flow, i.e. water...). In general the density varies but it could be that at some point the density hardly changes any further.

• Based on the above: the velocity of the air at any point in the tube will not be a constant. The air will accelerate as it gets closer to the exit.

Correct.

• The pressure drops exactly the amount needed to go from the compressed tank's pressure to atmospheric pressure. Bernoulli's equation has nothing to offer us in this case.

Correct!

• The air accelerates until viscosity and tube friction balance the forces accelerating it.

Correct! Viscosity and tube friction are two sides of the same coin.

The real scenario I want to analyze is just blowing through a tube, but I'm not sure if lungs compress air (making it more like an air compressor) or not (making it more like the piston). Also, maintaining a constant airflow is difficult with lung power, so it makes it harder to set up a realistic problem. One of the two scenarios above should apply.

I would guess it is more like a piston (unless your head turns red and blue, then it may be compressible... )

It seems that Bernoulli's equation is a bit more useful with the piston than with the air compressor.

I think it is not useful for either...

 

[Freixas]

Correct!

Thanks, Arjan!

 

[boneh3ad]

Thank you, everyone! I have been trying to absorb everything and I thought I'd review my original post and see if I actually have answers.

Kudos on actually taking the time to think through all of the responses you've received.

1) The pressure of a fluid exiting the tube will be equal to the atmospheric pressure.

I got this from Arjan82, so I'm pretty sure it's roughly correct, although it's my paraphrasing.

Follow up points:

• Is the statement correct or does it need some tweaks?

I'd argue it needs some tweaks. This is true only for an incompressible flow (which is, admittedly, what we are talking about here).

• Does pressure mean "static pressure"? (I'm pretty sure it means "static".)

In almost all cases, the word "pressure" without any modifier refers to static pressure.

2) To create a flow in the tube, I've been told I need a difference in pressure.

I received an answer from boneh3ad: "... a frictionless fluid ... will [not] spontaneously start to flow without some kind of pressure gradient applied."

Follow-up points:

• Is this true for viscous fluids as well? (I think it is; viscous or inviscid, you need a net force to start the flow.)

Yes. No matter what, and object at rest tends to stay at rest unless acted on by an outside force. That force just needs to be higher for a viscous fluid and needs to be maintained in order to sustain the flow, unlike in the inviscid case.

• Does the term "pressure" here mean "total pressure" or "static pressure"? (I think it's "total pressure"—force is force.)

Static. Static is always the relevant pressure when talking about the force on a fluid element.

3) If I have a tube with a uniform cross-section, a steady flow and an incompressible fluid, then I can apply the continuity equation. It tells me that the velocity of the fluid is equal through any cross-section.

I don't think I saw any dissenting opinions.

That's correct whether the fluid is viscous or inviscid. In the viscous, case, the statement needs the additional qualifiers that the velocity in question is the average velocity.

4) Starting from #3, if I also assume the conditions needed for Bernoulli's equation, then the static pressure in the tube is the same at all points.

This is where the discussion veered into some rather interesting areas. It appears that by having different static pressures at the two ends of the tube and specifying a frictionless flow, I created a theoretical monster. I thought I needed a difference in pressures to get a flow started (#2) and a frictionless flow to apply Bernoulli.

You have to understand that many introductory texts start with the Bernoulli equation, give the conditions for its application and then make claims along the line of "it's not exact, but it's close". I thought I was setting up something along those lines.

All of this is effectively true, but you can apply the Bernoulli equation to viscous flows provided you have some kind of modification to account for the effects of viscosity. Usually this is done by incorporating a "head loss" term that serves as an energy sink on the right-hand side of the equation.

Let's test my understanding with something simpler. My answers depend on some of the points above, so if I got those wrong, I may get off-base pretty quick.

• I have a large piston filled with water connected to a normal tube (with friction).
• I move the piston so as to move a constant 1 L/s into the tube.

Let's see if I can get this correct:

• I'm pretty sure I get a flow.

Of course.

• Can I apply the continuity equation? Water is not highly compressible.

The continuity equation applies regardless of compressibility. It's a statement that mass is conserved. The form of the equation simply changes for compressible and/or unsteady flows.

• If I can, then if I push 1 L/s in, I should get 1 L/s out and through every cross-section.
• If I know the cross-section, I can calculate the velocity of the flow.

Correct as long as you note that you get average velocity this way (which is a meaningful distinction for viscous flows, not really for inviscid).

• Going back to #1, the pressure at the end of the tube is whatever the exit pressure is.
• If I incorrectly apply Bernoulli to this, the equation tells me that the static pressure is equal at every point; therefore, it has to be equal to the exit pressure.

That isn't an incorrect application. It's just a trivial one. The result you get is correct under these conditions.

• I then correct for friction in the tube and from the viscosity of the water. This tells me the static pressure drop over the tube's length. There may be some other corrections that need to be applied.
• If I were a real physicist, I would then be able to calculate the force needed by the piston to maintain that 1 L/s flow.

No issues here.

Let's move on to something more complicated:

• I connect a hose to an air compressor's tank. I have now violated the incompressibility requirement of Bernoulli's.
• I have a limitless supply of compressed air.
• This is a guess: the mass flow through any cross-section is equal. It would seem that if x grams of air goes in, x grams has to go out. It can't be more and if it is less, the air would be collecting somewhere and the tube would eventually blow up.

This is all correct.

• Another guess: the fluid density is a gradient from entrance to exit. I'm totally guessing; for all I know, the density drops completely right near the entrance.
• Based on the above: the velocity of the air at any point in the tube will not be a constant. The air will accelerate as it gets closer to the exit.
• The pressure drops exactly the amount needed to go from the compressed tank's pressure to atmospheric pressure. Bernoulli's equation has nothing to offer us in this case.
• The air accelerates until viscosity and tube friction balance the forces accelerating it.

Compressible flows get... messy. There are a lot of effects that can come into play. In general, yes, the flow will have density changes that go along with pressure, temperature, and velocity changes. How those variables couple depends on the nature of the flow, but the usual assumption is that it is a perfect gas (so you can assume $p=\rho R T$). Then you've got to worry about whether the flow is adiabatic or, better, isentropic, which has implications on how those variables couple as well.

In general, though, if you have no area change in your tube, then the pressure is overcoming friction, which acts along the whole tube, so all of those variables ($v$, $\rho$, $p$, $T$) will be changing as a function of distance along the tube. You are correct that Bernoulli is effectively useless here. The "simple" way to analyze these flows would be what is called Fanno flow, which models friction similar to how it's done in incompressible flows with a friction factor.

The real scenario I want to analyze is just blowing through a tube, but I'm not sure if lungs compress air (making it more like an air compressor) or not (making it more like the piston). Also, maintaining a constant airflow is difficult with lung power, so it makes it harder to set up a realistic problem. One of the two scenarios above should apply.

It seems that Bernoulli's equation is a bit more useful with the piston than with the air compressor.

Lungs do compress air, but my understanding is that it's maximally on the order of 1 psi, which means the flow is still going to behave as an incompressible one.

 

[Freixas]

There is one puzzling thing that came up in several answers in one form or another. It is not really relevant to my initial question, but I'm curious.

This shouldn't be entirely surprising. If no outside force acts on an object, it will not change velocity, so a frictionless fluid that is already flowing does not need a pressure gradient to make it flow, nor will will spontaneously start to flow without some kind of pressure gradient applied.

Who said it would begin to flow? All you have been told is that would continue to flow.

If there is a pressure difference the flow will accelerate because there is a force imbalance. If the flow is to be steady state, there simply cannot be a pressure difference over the tube.

The impression I'm left with is that if I have a frictionless fluid:

• To start the flow, I need a (temporary) pressure difference between the entrance and exit.
• If static pressure is then equal at entrance and exit, it will continue to flow.
• If the static pressures remain unequal, the flow will accelerate until horrible things happen.

The puzzle is that that mass I accelerated eventually leaves the tube. With the pressures now equal at both ends, there would appear to be no force left to accelerate any further flow. Wouldn't the flow stop?

I would think that to maintain a flow through the tube, I would need to keep applying a force. After all, I am moving an ever increasing amount of mass the length of the tube. But it has been made clear that if I tried this, the world would end.

Of course, another answer is that my three bullet point assumptions are incorrect.

While it's not essential that I understand what's going on here, it'll probably prove useful at some point. I might find some frictionless fluid and inadvertently cause a catastrophe!

 

[boneh3ad]

The puzzle is that that mass I accelerated eventually leaves the tube. With the pressures now equal at both ends, there would appear to be no force left to accelerate any further flow. Wouldn't the flow stop?

I would think that to maintain a flow through the tube, I would need to keep applying a force. After all, I am moving an ever increasing amount of mass the length of the tube. But it has been made clear that if I tried this, the world would end.

Of course, another answer is that my three bullet point assumptions are incorrect.

While it's not essential that I understand what's going on here, it'll probably prove useful at some point. I might find some frictionless fluid and inadvertently cause a catastrophe!

Think of it like pushing a block on a frictionless surface. You need some temporary force to get it going but once you let go and no longer provide a force, it just keeps going. If you keep pushing, it keeps going faster. That's the same principle as what would happen with a frictionless fluid. If you accelerate said fluid and then stop "pushing" with pressure, there is no force to flow it back down so it won't stop. Because fluids are continuous media, not a collection of discrete objects, once you get it going, the remaining fluid will in some sense be dragged along with the fluid in front of it (that's an inexact analogy, but hopefully helps with visualization).

 

[Freixas]

Ok, I've had two smart people tell me this when I asked whether the pressure difference at entrance and exit meant total or static. My guess was total; the correct answer is static:

Static. Static is always the relevant pressure when talking about the force on a fluid element.

Here's my thought experiment:

• I take a tube, sealed at both ends with air at 1 atmosphere. Clearly, there is no flow.
• I take it outside, where the static pressure is also 1 atmosphere, but where there is a steady breeze--remarkably steady, really.
• I line up the tube so it points into the wind.
• I remove the end caps.

Do I get a flow? Since the static pressure everywhere is equal, the answer should be no. That answer doesn't seem correct.

 

[Freixas]

Because fluids are continuous media, not a collection of discrete objects, once you get it going, the remaining fluid will in some sense be dragged along with the fluid in front of it (that's an inexact analogy, but hopefully helps with visualization).

Everything up to this sentence makes total sense. If there is no longer a pressure difference, where is the force coming from that "drags" the remaining fluid?

I'm looking at the case where the pressure difference is temporary. The behavior looks to me like a perpetual motion device. I can find no source of energy, yet mass continues to travel across the length of the tube.

Could it be that some static pressure is being converted to velocity? Hmm...that would create a pressure drop at the inlet and reverse the flow. Eventually, it would stop.

In case it helps, it my model the tube is a bridge between two regions that otherwise have no connection. To be more precise, the tube is filled with fluid at a pressure equal to the outlet region and capped at the inlet. The experiment begins with the inlet region at higher pressure than the outlet and the cap removed. After a brief interval where (presumably) we get a flow, the pressure in the inlet region is set equal to the outlet.

 

[boneh3ad]

Ok, I've had two smart people tell me this when I asked whether the pressure difference at entrance and exit meant total or static. My guess was total; the correct answer is static:
Here's my thought experiment:

• I take a tube, sealed at both ends with air at 1 atmosphere. Clearly, there is no flow.
• I take it outside, where the static pressure is also 1 atmosphere, but where there is a steady breeze--remarkably steady, really.
• I line up the tube so it points into the wind.
• I remove the end caps.

Do I get a flow? Since the static pressure everywhere is equal, the answer should be no. That answer doesn't seem correct.

That has a lot of extra variables not considered in your original analysis such as viscosity and entrance geometry. In this case, the total pressure is important but not what determines the flow.

As an illustrating example, you are essentially describing a Pitot probe, which is a tube that is closed on only one end (using a pressure measurement device) and oriented as described so that the flow directly impacts the opening. Because no flow can occur, the fluid gets slowed down at the tip to zero velocity, meaning all of its dynamic pressure gets converted to static pressure. At that point, the static pressure and total pressure (also called stagnation pressure) are identical since velocity is zero and the sensor at the other end measures, in effect, the total pressure in the flow.

Now turn to your example. The tube is open on both ends so flow can occur. The moving air slows down as it approaches the tip but likely not to zero velocity since flow can pass through now. As it slows down, the static pressure rises (though not all the way to the total pressure since it doesn't likely reach $v=0$). That static pressure rise at the tip will be determined by matching the effects of viscosity in the tube so that a steady flow results (i.e. $p_{in} - p_{out}$ will need to be able to exactly cancel out viscosity).

If the flow was inviscid, the fluid would just move through the tube without requiring a pressure increase at the tip. You could even get a pressure decrease if the tube walls are thick since it is going to deflect more flow into the tube, requiring an acceleration (note that this effect can also be present in the viscous case, but I neglected it above). For an inviscid tube with infinitely thin walls, it would be like the tube wasn't even there.

 

[boneh3ad]

Everything up to this sentence makes total sense. If there is no longer a pressure difference, where is the force coming from that "drags" the remaining fluid?

Ultimately, the fundamental assumption here is that this is a continuum, so if one part of the flow abruptly stopped while the part in front of it kept going, you'd have a void form, which can't happen.

I'm looking at the case where the pressure difference is temporary. The behavior looks to me like a perpetual motion device. I can find no source of energy, yet mass continues to travel across the length of the tube.

It is a perpetual motion device. Why? Because we've neglected the only source of dissipation, which is viscosity. Of course comparing this to the real world will cause your brain to start steaming because in the real world, viscosity is not zero.

 

[Twigg]

The wind decelerates when it smacks headlong into the stationary fluid in your pipe. What does Bernoulli tell you when a fluid starts at atmospheric pressure and 10mph then decelerates to 0mph? Higher or lower pressure? (Don't worry about air being compressible. It will behave like an incompressible fluid at typical wind speeds.)

I can find no source of energy, yet mass continues to travel across the length of the tube.

Yep, mass leaves the tube at a constant rate. Mass also enters the tube at a the same rate. So the real question is, does the energy per kilogram change between the inlet and the outlet? If not, then the energy in equals the energy out. If there's no friction, then at steady state all variables are identical from one end of the tube to the other, so you can conclude that energy in = energy out. In the case of flow with friction, the pressure drops from inlet to outlet, and that shows up in the energy balance as an enthalpy drop. In other words, the force of friction is doing (non-conservative) work on the fluid, dissipating energy.

 

[Arjan82]

The impression I'm left with is that if I have a frictionless fluid:

• To start the flow, I need a (temporary) pressure difference between the entrance and exit.
• If static pressure is then equal at entrance and exit, it will continue to flow.
• If the static pressures remain unequal, the flow will accelerate until horrible things happen.

The puzzle is that that mass I accelerated eventually leaves the tube. With the pressures now equal at both ends, there would appear to be no force left to accelerate any further flow. Wouldn't the flow stop?

I would think that to maintain a flow through the tube, I would need to keep applying a force. After all, I am moving an ever increasing amount of mass the length of the tube. But it has been made clear that if I tried this, the world would end.

Ok, let's go back to your earlier problem, one tube, connected to a region at either side which have different pressures.

For this question we need to consider the case that a steady state flow has developed through the tube. Let's look at an individual fluid parcel far away, but not too far away, from the inlet. It has a static pressure of 1000kPa and is pretty much at rest, so its total pressure is almost equal to its static pressure (at the limit of infinity these two are indeed equal). Slowly but surely the parcel starts to move towards the inlet of the tube, this means its dynamic pressure (i.e. kinetic energy) starts to build up, what happens with the static pressure? If we just apply Bernoulli (which we can do here, we can ignore friction in this region) you would see that the static pressure would decrease further and further, i.e. its static pressure is converted to dynamic pressure (i.e. to kinetic energy, i.e. it now has a velocity). This would go on up to the entrance of the tube. In the case of no friction the pressure is now at the pressure of the region at the outlet since the tube cannot maintain a pressure difference. At the outlet the parcel still has a velocity, thus its total pressure is higher than static. A jet forms, and if you assume no friction then this jet will continue to exist until eternity (we've assumed steady state).

So what's the moral of this story? The pressure difference between the inlet and outlet region is entirely converted to dynamic pressure, i.e. kinetic energy. Once a parcel has converted all this pressure (essentially a type of potential energy) it will keep this kinetic energy until the end of time, because there is no energy sink in this system.

So, you are right, but only if you consider the region outside of the tube as well (which we haven't done so far). So the force is applied to the fluid before entering the tube, which you intuitively and correctly deemed necessary, to accelerate the flow. Once inside the tube, everything is settled and the fluid keeps on moving until happily ever after.

Note that this is not equal to the case in which I said the world would end. During the earlier discussions an example arose when there was a frictionless flow, as well as a pressure difference over the tube, and an assumed steady state condition. This is what cannot happen.

(I've often insisted that the flow coming out of the tube has the same pressure as its surroundings, this is still true, but it is not true for the inlet!)

 

[Arjan82]

I've realized that I've made a mistake in my explanation earlier!

In the case with the tube connecting two regions of unequal pressure when you include friction I said the pressure difference over the tube was equal to the pressure difference between the two regions, that is not true...

Some of the pressure is needed to accelerate the flow (static to dynamic pressure, see my previous post) and thus the pressure difference over the tube is lower than the pressure difference between the two regions...

 

[Freixas]

First of all, let me thank everyone for trying to help me understand the physics behind a theoretical event. I understand this is time you could all use to do something more productive. And this response is way too long, so I won't fault anyone who decides to skip it.

I'm going to focus on just one thing (with variations) right now and explain why what you're telling me seems to make no sense. This tidbit from Arjan gave me an "aha!" moment that I hope doesn't turn out to be wrong (because I'm going to use it to simplify things):

This would go on up to the entrance of the tube. In the case of no friction the pressure is now at the pressure of the region at the outlet since the tube cannot maintain a pressure difference.

Back a ways, I proposed this:

1,000, 0, 0, 0, 0: Again, this satisfies Bernoulli's and the exit pressure and it allows for a flow since the static pressure is converted into dynamic pressure.

which seems to be match what Arjan is saying. The "aha!" was that I could shorten the tube to zero length and have exactly the same setup. A zero-length tube is certainly frictionless.

So here's the revised setup:

• There are two large but finite regions called "left" and "right". They have a wall between them.
• Region left has higher pressure than region right.
• There is a small hole in the separating wall.
• Before we get involved, the hole is closed. We start observing as we open the hole.
• The fluid in both regions is a frictionless, incompressible gas.
• For further simplification, the atoms/molecules of the gas have perfect elastic collisions.

There are three cases:

1. The pressure in "left" is higher than "right" just briefly after opening the hole. After that brief time, the pressure in "left" is artificially lowered to match "right".
2. The pressure in "left" is artificially maintained higher than "right" by a constant amount.
3. For completeness, no artificial adjustments are made.

What I've been told:

1. A flow will start. Reducing the pressure in "left" will not stop the flow.
2. A flow will start and accelerate without bounds—or until something drastic happens.
3. We haven't talked about this one.

First explanation:

Right by the hole, static pressure is converted to dynamic pressure. In fact, since Bernoulli's equation is about energy conservation, I suspect we could use it to calculate the velocity of the flow: $$v = \sqrt { \frac {2 (P_l - P_r)}{\rho}}$$
Some set of gas molecules are moved some distance by the application of force. This is "work". Those molecules had zero velocity to start, then were accelerated to v. Once they pass into "right", there is no longer a net force applied to them and they continue moving at v.

The next set of gas molecules also need to be accelerated, requiring the application of force to do "work". They also accelerate to v. So:

1. When the pressure in "left" is equalized, there is no longer a net force anywhere. The molecules already accelerated (and in "right") will continue at velocity v, but there will no longer be a flow through the hole.
2. The flow starts and continues at velocity v. It does not accelerate without bounds. We seem to have a steady state.
3. The flow starts at one velocity and slows as the pressure in "left" falls and "right" rises. This one is not steady state, at least not until the pressure balances, in which case we're back at #1.

Alternate explanation:

I'm going to look at stuff at the molecular level. Pressure is just a bunch of molecules zipping around. Because I don't want to introduce compression (the gas is incompressible), an easy way to get a higher pressure in "left" is to raise the temperature. The molecules move faster in "left" than in "right"; the fluid density is equal.

Let's take a random but average molecule in "left" and call in Joe. And we'll have "Sally" as the average representative for "right". Joe is heading through the hole and runs into Sally—literally. Velocities are swapped, now both are headed back into their respective regions. The pressure is now slightly lower in "left" and slightly higher in "right".

1. I allow the tiny pressure drop to continue for a bit, but then equalize pressures by cooling "left". With equal pressures, there is still plenty of traffic across the hole, but the average velocity is 0.
2. This case is a little harder to analyze this way. I continually warm "left" and cool "right" to maintain the pressure difference. The molecules keep doing what Joe and Sally did. The net average is a flow from left to right in what looks to me like a steady state.
3. Without any artificial adjustments, the molecules mix it up until they the average velocity is 0, the same as what would happen if we removed the entire barrier between "left" and "right". The flow slows and there is no steady state until then.

This is way too long and yet I feel I am really abbreviating my arguments maybe too much. I hope they make sense, even if just to point out where I went wrong. Still, I need to bring this up:

Because fluids are continuous media, not a collection of discrete objects, once you get it going, the remaining fluid will in some sense be dragged along with the fluid in front of it (that's an inexact analogy, but hopefully helps with visualization).

Gases are fluids that are a collection of discreet objects. This long thread I started talks about what happens when a pressure change occurs at the inlet of a tube. While I didn't bring up an inviscid flow, I don't remember anything mentioned that would change what happens if it were.

The idea is that pressure changes travel down the tube at the speed of sound (for that medium). In air, if you blow into a tube, the airflow at the exit begins around 3 ms/m later. If you stop blowing, the airflow stops at the exit 3 ms/m later. So, just because you have a flow in a tube doesn't mean that it creates a vacuum behind it that sucks molecules forward. A tube is always full of fluid; the net velocities at different points in the tube can change (including dropping to 0 or even negative) as pressure waves travel down the tube. While the waves sweep down the tube, Bernoulli's certainly doesn't apply.

 

[Freixas]

So what's the moral of this story? The pressure difference between the inlet and outlet region is entirely converted to dynamic pressure, i.e. kinetic energy. Once a parcel has converted all this pressure (essentially a type of potential energy) it will keep this kinetic energy until the end of time, because there is no energy sink in this system.

The "parcel" converts all the pressure difference into kinetic energy between the inlet and outlet, agreed.

But wouldn't that kinetic energy will be converted back to static pressure at the outlet side after the parcel has collided with a few billion of the molecules on that side?

 

[Arjan82]

But wouldn't that kinetic energy will be converted back to static pressure at the outlet side after the parcel has collided with a few billion of the molecules on that side?

Startup phenomena and inviscid flow are a bit hard to think about. But we are considering steady state. All of the flow that was in the way during startup is already moving when looking at steady state.

 

[Freixas]

Startup phenomena and inviscid flow are a bit hard to think about. But we are considering steady state. All of the flow that was in the way during startup is already moving when looking at steady state.

Ok, I went back and re-read your post #41 in light of this comment.

I don't find anything I disagree with.

• We have two regions with a pressure difference connected by a tube.
• We have no friction.
• We have a steady state.
• We have no end-of-the-world condition.

Sometimes disagreements occur because terms haven't been defined clearly (or, in my case, I misunderstand a term). I should have created a more precise definition of the theoretical experiment and maybe even drawn a picture.

For example, in my mind, the two regions are vary large containers, but containers nevertheless. The jet could flow until it hit a boundary, things would get chaotic and that velocity would translate into static pressure. Thus, my response to your comment.

I also stated that the pressures are artificially maintained. The pressure on one side would increase and the pressure on the other would decrease unless we did this. I didn't talk about how this magic would happen. Different approaches would yield different results, but wouldn't prevent the tube, or even the entire system from reaching a steady state.

Note that this is not equal to the case in which I said the world would end. During the earlier discussions an example arose when there was a frictionless flow, as well as a pressure difference over the tube, and an assumed steady state condition. This is what cannot happen.

Ok. I get it. You're talking about one thing and I thought you were talking about something else. But this led to a key point I hadn't understood...

The winner, back from post #15, appears to be the third bullet. I separated the tube into five regions (from inlet to outlet) and tried to guess the pressures at each point:

"1,000, 0, 0, 0, 0: Again, this satisfies Bernoulli's and the exit pressure and it allows for a flow since the static pressure is converted into dynamic pressure. The only problem is that 1,000 kPa of static pressure needs to be converted to dynamic pressure right at the entrance. This seems unlikely to be instantaneous. We could allow for a transition zone, but then we're back to the first option above, just with a shorter tube."

Actually, that's not right. The final piece of the puzzle is that the right pressures are 0, 0, 0, 0, 0—so, as you said, there is no pressure difference over the tube, from inlet to outlet. Bernoulli's is applied outside the inlet, where we don't need to worry about velocity and pressures having to be constant since we don't have the constant of a constant cross-section.

Puzzle solved when there is no friction!

This little correction also makes it easier to analyze the case with friction. Rather than the inlet pressure being 0, it will be whatever is needed to overcome friction. I noted this in response #30 and you said this was correct.

In #30, I didn't provide the breakdown of pressures for my initial set-up. If the total loss from friction is the equivalent of 10kPa, would the numbers be 10, 7.5, 5, 2.5, 0? I assume there is a Bernoulli's equation variant that allows for friction? Quick Google search...yes, there it is. And if I read the equation right, I think my numbers are right assuming that friction is evenly distributed along the length of the tube.

So, really, the question I asked in #1, the thing I was really interested in understanding, is now clear here. Despite the fact that I sounded like I knew what I was talking about in #30, the piece I missed was the application of Bernoulli's outside the inlet. Without that piece, I didn't really have my answer.

 

[Arjan82]

So, just for fun, over the weekend I ran an actual Computational Fluid Dynamics (CFD) simulation using a Reynolds averaged Navier-Stokes solver (since that's what I do for a living anyway). It is more or less the case of @Freixas, two regions of different pressures connected with a tube including friction (i.e. finite viscosity and a turbulence model; k-omega SST if you're interested). I did this 2D to keep it manageable, so it is actually two flat plates at a distance of 2cm from each other. But conceptually that will be equal to a tube. The plates are 1m long. The pressure difference is 1bar (10^5 Pa), and I used water as a fluid (incompressible flow). Since it is CFD this is the truth .

The last picture is a graph with the pressure and velocity over the tube at the centerline, so these are not averaged over the '2D tube's' surface area. Notice that the pressure is reduced for the largest part before it enters the '2D tube' (flat plates) and then drops of linearly once the inlet effects died out. The velocity at the centerline is not yet constant even at the exit, meaning the flow between the plates is not yet fully developed.

Overview picture. The domains attached to the inlet and outlet are larger than shown here. They are 20x20m each. But if I zoom out entirely the '2D tube' in the center is hardly visible.

Grid near the inlet. Each cell (square) contains one value of the pressure and velocity

Grid near the outlet. I coarsen the grid soon after the plates end to keep things manageable.

Detailed flow at the inlet, split over the center. Above the velocity magnitude is plotted, below the pressure. note the streamlines draw air from all directions, as you would expect:

Detailed flow at the outlet, after the flow in the 2D tube exits, you see a jet forming and the velocities mix with the surrounding fluid. Note also that within the tube the velocities go down near the wall, this is because of the no-slip condition (i.e. zero velocity at the wall). So a velocity profile is developing:

These are the pressure and velocity magnitude at the centerline (the dash-dot line above):

 

[Freixas]

So, just for fun,

I like your idea of "fun". What tool(s) did you use?

 

[Arjan82]

Grid made with Hexpress
Solved with code from our own dear developers on 20 cores (220k grid cells)
Postprocessed with Tecplot and Matlab