Does gas flow from low to high pressure?

[erobz]

I please, can I get you to consider my thought experiments 1), 2) and 3) before returning to the aerodynamic solution.

It's not aerodynamics, its thermodynamics. There is a difference. It handles any case. Small pressure, large pressure without really caring too much...I could be misinterpreting the problem, the first law ...not so much.

 

[Sailor Al]

It's not aerodynamics, its thermodynamics.

I stand corrected. It's not my subject.
Which textbook would help me understand your working?

 

[Sailor Al]

@erobz So may I please rephrase my request: can I get you to consider my thought experiments 1), 2) and 3) before returning to the thermodynamics solution?

 

[erobz]

1.) If, instead of having two wide chambers separated by a narrowed one, we had 10, 100 or $\infty$ chambers and the appropriate quantity of water to get the level at 4.9 cm in all of them, when we turned on the air tap, would we see the pattern of 0, 9.5, 0, 9.5, 0, 9.5 ....repeated indefinitely? I think so.

We don't see that happening with a single narrow chamber between two wide chambers...

2) If, in this thought experiment, we then turned on the air tap, how long would it take for the the system to settle down? We'd probably have to decide the length of each section and the length of each nozzle/diffuser. I think the answer would be very similar to the answer to the piston in the infinite tube.

Transient analysis. An entirely more complex problem. The startup is irrelevant. The system can be throttled and brought up to this state gradually without major shocks so long as the compressed air tank has sufficient capacity.

3) And what difference would it make if the system terminated with a wide chamber or a narrow chamber

It means nothing for this experiment. That is a different experiment. I can't use this systems values to predict every systems response when hooked up to the same hose.

Of those three questions, only the first has any relevance to the task at hand, which is determining the mass flowrate in the system, given assumed temp of chamber A $273 \rm{K}$, and these three manometer readings.

 

[erobz]

b) If it was opened slowly, to keep the water being pushed out of the top of tube B, then eventually, when fully opened, the water levels would return to the 0, 9.5, 5.3 of the video.
So yes and no. The pressures would return to near atmospheric pressure like I indicated in #111:
PA = 1033 + 4.9 = 1037.9
PB = 1033 - 4.6 = 1028.4
PC = 1033 - 0.4 = 1033.0
There is a pressure gradient across the system: All around atmospheric pressure but all slightly different.

This (the manometer setup in the experiment):

The reference pressure for the experimental setup could be literally anything. The manometers are coupled.

It is not the same as having three independent manometers as you are treating the problem (like show below):

Where the reference pressures for the setup is atmospheric (at the open top of each manometer).

 

[Sailor Al]

The reference pressure for the experimental setup could be literally anything. The manometers are coupled.

Not quite!
As long as the experiment was run when the environment was in the region of the phase diagram :

where water was liquid and air was a gas, the numbers would be the same. 0, 9.5, 5.5 cm.
Even if the environment was other than air: Argon, Nitrogen., Hydrogen, Helium...
I am sticking with the problem, I think there are more insights on the way!

 

[erobz]

Not quite!
As long as the experiment was run when the environment was in the region of the phase diagram :
View attachment 327256
where water was liquid and air was a gas, the numbers would be the same. 0, 9.5, 5.5 cm.
I am sticking with the problem, I think there are more insights on the way!

As far as I can tell you're completely missing the point...I'm done with this. What is the heck does teh phase diagram have to do with this simple issue!

 

[Sailor Al]

As far as I can tell you're completely missing the point...

Oh, please don't be dismissive. I was just responding to your interesting post.

 

[erobz]

Oh, please don't be dismissive. I was just responding to your interesting post.

What does the phase diagram have to do with this simple issue I'm trying to point out that your "analysis" is completely bungled. The manometers are coupled! They are not independent of each other as I clearly illustrated in #145! Your response was "gotcha" it cant be "anything" because of phase changes, when it should have been..."Oh yeah, I see what you mean"... My patience is now worn completely through the paper and is beginning to dig my own grave... Good by!

 

[Sailor Al]

The point is that the experiment is what it is. The manometers ARE linked, and we only have those numbers to work with.

 

[Sailor Al]

What does the phase diagram have to do with this simple issue I'm trying to point out that your "analysis" is completely bungled. The manometers are coupled! They are not independent of each other as I clearly illustrated in #145! Your response was "gotcha" it cant be "anything" because of phase changes, when it should have been..."Oh yeah, I see what you mean"... My patience is now worn completely through the paper and is beginning to dig my own grave... Good by!

I'm sorry you took it that way. It was not meant to be a "gotcha!" at all, but a considered response to your post.
I said to myself when reading it ,"Oh, yeah, I see what you mean"... but didn't post as it didn't seem appropriate. I did take it very seriously and subjected to the question s- why wouldn't it work in the vacuum of space. And then very cold... or hot...? That's when I thought about the phase diagrams.
I'm sorry if you read it as a "gotcha" - not intended at all. I'm sorry.
Before you go, can I ask you to tell me the name of your thermodynamics textbook please? This is a sincere request and I will make the effort to get into it.

 

[erobz]

I'm sorry you took it that way. It was not meant to be a "gotcha!" at all, but a considered response to your post.
I said to myself when reading it ,"Oh, yeah, I see what you mean"... but didn't post as it didn't seem appropriate. I did take it very seriously and subjected to the question s- why wouldn't it work in the vacuum of space. And then very cold... or hot...? That's when I thought about the phase diagrams.
I'm sorry if you read it as a "gotcha" - not intended at all. I'm sorry.
Before you go, can I ask you to tell me the name of your thermodynamics textbook please? This is a sincere request and I will make the effort to get into it.

It’s funny…posting “I see what you mean now”is exactly what was appropriate given the many hours I ( and others) have put into trying to solve this under constrained problem you posed hoping you would say. “Oh, I see what you are saying”.

Fundamentals of EngineeringThermodynamics

Michael J. Moran
Howard N. Shapiro

Perhaps being an engineering text you will find it a bore…

 

[Sailor Al]

Perhaps being an engineering text you will find it a bore…

Not at all. Thermodynamics has always been a mystery to me. It was one of the subjects I failed at university (along with crystallography). Over the past few weeks I have started to get my head around it and think it will have a profound bearing on arriving at some sort of quantitative analysis of my hypothesis about the source of the pressure differences around a yacht's sail.
I have located and downloaded a digital copy of the 7th edition from [Probable illegal download website URL redacted by the Mentors]
Thanks.

 

[erobz]

Not at all. Thermodynamics has always been a mystery to me. It was one of the subjects I failed at university (along with crystallography). Over the past few weeks I have started to get my head around it and think it will have a profound bearing on arriving at some sort of quantitative analysis of my hypothesis about the source of the pressure differences around a yacht's sail.
[Probable illegal download website URL redacted by the Mentors]
Thanks.

When you get to this type of problem in the textbook you are going to find that irreversibility's are present in the control volume for the manometer problem we've been discussing. Given we were trying to make theory line up with the real measurement at $C$, this has unavoidable consequences i.e. for my efforts... the First law alone is not sufficient to accurately describe what has happened, we need to use the Second Law as well. I realized this after trying to fiddle with the equations I derived. It should have been more apparent irreversibility's were present since the pressure does not return to near $P_A$ in chamber C. I don't think I'm going to try to go any deeper on this though. The many assumptions being made will likely make it very difficult to untangle.

Until next time,
Good luck in your studies.

 

[Sailor Al]

Until next time,
Good luck in your studies.

Copied, thanks.
I'm sorry that my stubbornness for accuracy (pedantry?) has been interpreted as grumpiness or stupidity. I'm just not very good at diplomacy. As my fellow sailing crew-members observe: "with Al, you're never in doubt about what he thinks"! I just was born with a lack of sensitivity. Not my fault, and after 78 years, with much effort, I'm getting better but still short of ideal!
I'll post again when I am closer to the solution.

 

[erobz]

This character trait is not uncommon here. There is a joke one of my prof. told us: Arguing with an engineer is like wrestling a pig in the mud…after a few hours you realize the pig likes it… I think this applies to many of us here…I’ve pushed/push buttons too. I’m not claiming to be a saint. We are all a WIP.

 

[Sailor Al]

With only a slight fudge, I have calculated a value for the mass flow rate of the air supply: 2.0 +/- 0.05 grams/second, the fudge being that the pressure in the third chamber is at the pressure of the ambient air, and not, as in the demo, very slightly higher.
I have assumed that the air is an ideal gas and that the system is thermally insulated and have added three more valves, all of which are considered as adiabatic walls:

• V1 between chamber A and the first narrowed tube isolating the contents of chamber A,
• V2 at the top of manometer B, isolating the contents of manometer B
• V3 at the top of manometer C, isolating the contents of manometer C

I used :

• The ideal gas law PV = nRT ,
• Boyles' Law PV = C,
• Dalton's law of partial pressures and
• classical mechanics in which W = ΔE

And constants:

• Universal Gas Constant R = 8.313 Joules/mole/°K
• Molecular weight of air = 28.96 grams/mole

Before writing up my method, I would welcome a sanity check on the result. 2 grams of air is around 1 ½ litres at STP ( 22.4 * 2/ 28.9 = 1.55). I am not familiar with lab gases so don't know if this is a feasible result. It's my first attempt for many decades to do any such work, so please, be gentle.

 

[Peter034]

TL;DR Summary: Classical Physics states that a gas will only flow from a region of high to one of low pressure and, like water, won't run uphill. This demo of Bernoulli's principle appears to show gas flowing into in the opposite sense, from low to high from the centre constriction to the second wide tube section. What's happening?

Here's a grab at 1:50 from the Bernoulli's Principle Demo: Venturi Tube video
View attachment 326875
It appears to show the air flowing from the viewer's right to left where the pressure in the third manometer appears to be higher than in the middle one.
Is there some complexity from the manometers being interconnected?
Water won't flow uphill. A gas will only flow from high to low pressure.
How can this be a correct interpretation?

This is all news to me. I would have thought that the pressures would be highest on our right, lower in the middle, and lowest our left. But the video seems to be backed up by several authorities. It must be that, at least in the sense of the manometers, the air is flowing from lower to higher pressure when it passes from the middle part to the part on our left. I don't think that they are all paid up on this yet, since, if I was going to guess it, then it is very probably possible to get some manometers to line up the other way, from higher to lower to lowest, left to right, somehow, notwithstanding some much smaller constriction in the set up?

 

[A.T.]

It must be that, at least in the sense of the manometers, the air is flowing from lower to higher pressure when it passes from the middle part to the part on our left.

Which is perfectly fine, as long as it slows down, as explained in post #5 already:

Gas has inertia like any mass. Pressure gradient provides a force, so moving from low to high pressure will slow it down and moving from high to low will speed it up.

This is just Newtons 2nd Law: The net force gives you the direction of acceleration, not the direction of velocity.

 

[Peter034]

I was referring not to the drop of water at the top of the third manometer, but the level of the top of the water column which is indeed lower than in the second column. Isn't the pressure indicated in the second manometer therefore lower than in the third one? Isn't the air moving from a lower pressure in the second one to a higher one in the third one. I don't understand how that can happen.

Right.

 

[PeterDonis]

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