I Does gas flow from low to high pressure?

[Sailor Al]

I don't know, because I don't know the pressure in the system now. There's no gas flow, there's no pressure gradient.

 

[erobz]

With the air supply closed. The pressure in the system is Atm - say 1033 cm water. The pressure at $P_{ref}$ is 1033 +$l_{C}$ = 1033 + $l_{C}$

No, the valve is at the end of the glass tube. The entire system is pressurized to whatever lab pressure is. What is $P_{ref}$?

 

[Sailor Al]

No, the valve is at the end of the glass tube. The entire system is pressurized to whatever lab pressure is. What is $P_{ref}$?

Yes, sorry, I corrected my answer after you replied.
I should have said:
I don't know. There is no gas flow. There is no pressure gradient.

 

[Sailor Al]

There is a throttle on the air supply.
Pure speculation, but I suspect, for the video, he adjusted that to get the water level at tube A to zero. (I bet that took a bit of fiddling, and water squirting through the tubes which would explain the bead of water at the top of tube C!)

 

[erobz]

I don't know. There is no gas flow. There is no pressure gradient.

Exactly. It could be 100 psi, could be 5 psi, could be 0.01 psi. Whenever the valve is opened, would the pressure in the system drop to near atmospheric in chamber A regardless of the system pressure?

 

[erobz]

View attachment 327210
There is a throttle on the air supply.
Pure speculation, but I suspect, for the video, he adjusted that to get the water level at tube A to zero. (I bet that took a bit of fiddling, and water squirting through the tubes which would explain the bead of water at the top of tube C!)

Multiple Choice: From the picture, what's the effective area of the "pinched tube"? If the tube has area $A$, is it
a) 80% $A$,
b)50% $A$,
c) 0%$A$

 

[Sailor Al]

@erobz I have responded to your thought experiment in #120.
Would you like to have a look at mine: 1), 2) and 3) in #111?
I think it may throw some light onto the problem.

 

[erobz]

@erobz I have responded to your thought experiment in #120.
Would you like to have a look at mine: 1), 2) and 3) in #111?
I think it may throw some light onto the problem.

No, I think we are about to make some progress here if you answer #126 and #125. Which is exactly why you are desperately attempting to deflect the topic.

 

[Sailor Al]

From the picture, what's the effective area of the "pinched tube"? If the tube has area $A$, is it 80% $A$, 50% $A$, 0%$A$?

I think I specified that in the original question. I suggested the diameter of A was 1.6 cm and B, 0.5 cm, so the cross-section area of A is 2.01 cm² and of B is 0.196 cm². So B is ~10% of A.

 

[erobz]

I think I specified that in the original question. I suggested the diameter of A was 1.6 cm and B, 0.5 cm, so the cross-section area of A is 2.01 cm² and of B is 0.196 cm². So B is ~10% of A.

I'm talking about the throttle you are pointing out. If the feed tube has area $A_{tube}$, what is the area of the "pinched" section.

Also, you have not answered #125.

 

[Sailor Al]

I'm talking about the throttle you are point out.

I'm sorry, there's no way of determining that. And even if we did know that, we still don't know the pressure behind that nozzle/diffuser. It's just adding another set of nozzles and diffusers to the problem.
It's a variation on my question 1) in #111.
I only pointed out the throttle to allay your concerns about the small size of the apparent pressures in the tube that prompted my question in #118: "What's absurd about 4cm pressure difference between A and C?"

 

[erobz]

I'm sorry, there's no way of determining that. And even if we did know that, we still don't know the pressure behind that nozzle/diffuser. It's just adding another set of nozzles and diffusers to the problem.
It's a variation on my question 1) in #111.
I only pointed out the throttle to allay your concerns about the small size of the apparent pressures in the tube that prompted my question in #118: "What's absurd about 4cm pressure difference between A and C?"

The absurdity is a tiny guage pressures we expect almost no flow, and consequently almost no losses ( as a percentage) between A and C. You are trying to tell me that the entirety 100% of the internal energy stored as pressure in A is converted to heat by the time it reaches chamber C under the action of the tiny guage pressure you propose.

 

[Sailor Al]

Exactly. It could be 100 psi, could be 5 psi, could be 0.01 psi. Whenever the valve is opened, would the pressure in the system drop to near atmospheric in chamber A regardless of the system pressure?

This is another thought experiment, and reaches to the heart of the problem.
Here's my thought experiment 4):
a) If the valve was opened abruptly, then, whatever the supply pressure, the build-up of pressurised air would blast the water out of the tubes!
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.

 

[Sailor Al]

The absurdity is a tiny guage pressures we expect almost no flow, and consequently almost no losses ( as a percentage) between A and C. You are trying to tell me that the entirety 100% of the internal energy stored as pressure in A is converted to heat by the time it reaches chamber C under the action of the tiny guage pressure you propose.

Can I postpone responding to this till you have had a chance to review my #133 and perhaps #97 where I talk about trading the components of the gas energy - pressure, density and temperature:

I think the following statements are a bit wooly and can be tidied up, but I think they are important.
In a venturi, a low pressure occurs in the narrow section when the fluid is either liquid or gas.
With liquid, the pressure change is due to the "trading" of potential energy and kinetic energy by a change in momentum of the fluid. Since liquid is incompressible, i.e its density is constant, and so the mass per unit volume is constant, the momentum change arises from a change in the speed of the fluid.
With gas, the low pressure is due to the "trading" of the components of the gas energy, i.e. PV and nRT, and occurs through a change in density, pressure and temperature . This "trading" can be handled mathematically using the density version of the Ideal Gas Law : PM = dRT .
That's the theory I'm using to work up the answer.

 

[Sailor Al]

Here's what's being traded.

• Compared to A, In B the pressure falls and the density and temperature rise.
• Compared to B, In C the pressure rises and the density and temperature fall.

In each of the three chambers the energy in the gas: PV and/or nRT is constant.
The challenge is to use that in the density version of the Ideal Gas equation P = cdT where c: for air is 2.88 10⁶cm²sec^-2
The equation has two variables: density and temperature. We know the pressures.
The three cross-section areas are defined (2.01, 0.196, 2.01 cm²).
We can make reasonable assumptions about the ambient temperature and pressure (NTP?) and the temp of the air in A: 0°C
It should be soluble for the mass flow rate $\phi$ gm/sec, I haven't got here yet.

 

[erobz]

Here's what's being traded.

• Compared to A, In B the pressure falls and the density and temperature rise.
• Compared to B, In C the pressure rises and the density and temperature fall.

In each of the three chambers the energy in the gas: PV and/or nRT is constant.
The challenge is to use that in the density version of the Ideal Gas equation P = cdT where c: for air is 2.88 10⁶cm²sec^-2
The equation has two variables: density and temperature. We know the pressures.
The three cross-section areas are defined (2.01, 0.196, 2.01 cm².
We can make reasonable assumptions about the ambient temperature and pressure (NTP?) and the temp of the air in A: 0°C
It should be soluble for the mass flow rate $\phi$, I haven't got here yet.

For near ambient gauge pressures in chamber A I expect this:

Not This:

 

[Sailor Al]

For near ambient gauge pressures in chamber A we expect this:

View attachment 327211

Not This:

View attachment 327212

But that IS what we're getting!
l_A < l_B
l_B >l_C
l_C >l_A
It's just you are surprised just how large are the height differences, caused by such small pressure differences.
Remember Air pressure is 1033 cm
We're looking at 5 cm differences in the water level in the tubes . 5 cm is only 0.5% of atmospheric pressure.
It only takes a tiny amount of pressure to lift quite a lot of water.

 

[erobz]

But that IS what we're getting!
$l\{A}$ < $l\{B}$
$l\{B}$ > $l\{C}$
$l\{C}$ > $l\{A}$

It's just you are surprised just how large are the height differences, caused by such small pressure differences.
Remember Air pressure is 1033 cm
We're looking at 5 cm differences in the water level in the tubes . 5 cm is only 0.5% of atmospheric pressure.

No, my concern is not for absolute differences. Its for the relative loss between chamber A and C. Its an indication of high thermal generation. Thats not going to be the case in a near ambient chamber A.

Anyhow, tomorrow I'll solve my equation...somehow.

 

[Sailor Al]

No, my concern is not for absolute differences. Its for the relative loss between chamber A and C. Its an indication of high thermal generation. Thats not going to be the case in a near ambient chamber A.

Anyhow, tomorrow I'll solve my equation...somehow.

Sorry we're 17-odd hours apart (I'm UTC+10), it makes the conversation a bit fragmented.
I edited my last post which makes it more readable.
I please, can I get you to consider my thought experiments 1), 2) and 3) before returning to the aerodynamic solution. It's not the relative differences in play here, it's the absolute differences.
Very small pressure changes - very small temperature and density changes.

 

[erobz]

Yeah, I have considered it.

Let me get things scaled write in the context of this experiment:

Small pressure chamber A implies:

However, In the experiment we have this:

This is indicative of significant relative thermal losses. You aren't going to see this without significant pressure gradient.

 

[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.