I Does gas flow from low to high pressure?

[Sailor Al]

The text-editing menu is not available when the BB code mode is off. This control is a pair of brackets [ ] almost all the way to the right in the menu bar at the top of the input pane. This control is a toggle.

Aha! Now I see it!
Maybe a couple pf paras in the INFO section, alerting relative newbies such as me to the fact that this forum implements some pretty fancy features, and a Dummies guide to some of them. Till today, despite 40+ years of S/W development and extensive experience in web app development, I hadn't come across "BB code". Just a thought.

 

[Mark44]

Maybe a couple pf paras in the INFO section, alerting relative newbies such as me to the fact that this forum implements some pretty fancy features, and a Dummies guide to some of them.

Check out the Help menu item under INFO. The Help menu item links to tutorials on LaTeX (near the top) and to the BBCode tags that are supported (down quite a ways in the list).

 

[Sailor Al]

Thats my best crack at it. Hope it helps

I see a huge amount of working, but I still don't see an answer to the stated problem: to determine the mass flow rate through the system.
I still don't have an answer, but am still working on it.
I would expect the answer to me in the order of 10^-2 or 10^-3 grams per sec.

 

[erobz]

I see a huge amount of working, but I still don't see an answer to the stated problem: to determine the mass flow rate through the system.
I still don't have an answer, but am still working on it.
I would expect the answer to me in the order of 10^-2 or 10^-3 grams per sec.

Give me a pressure and temperature of the gas in chamber A and you will have an answer…if you don’t accept the 1st law of thermodynamics, just say so and I’ll stop wasting my time.

 

[Sailor Al]

Give me a pressure and temperature of the gas in chamber A and you will have an answer…if you don’t accept the 1st law of thermodynamics, just say so and I’ll stop wasting my time.

I have specified the temperature in chamber A is 0°C 273 °K.
The pressure can be determined using hydrostatics from the heights of the water columns provided i the problem statement in post #65:

By deducing the scale of the image, I have done some rough estimates of the manometer pipe diameter and thus the heights of the water and wonder how we could calculate the flow rate in Kg/sec of the gas through the system?
Estimate of the diameter of the manometer tubes: 8 mm
Scale: say 1:1,
Measured heights above red line:
A = 0 mm
B = 97 mm - allowance for narrowing at the top, say 95 mm
C = 53mm
95 + 53 = 148 /3 = 47
1 Atm = 10332 mm = 101,325 Pa (x 9.8)
Each mm is 9.8 Pa

It seems obvious that the flow rate in each section has to be the same.
I think we'll need to take into account the temperature of the gas as it passes through the system, and make a guess at the ambient air conditions (20°C, 101.3kPa?) Also something about the diameters of the various chambers through which the gas flows.
My measurements of the water heights in A,B and C are:
S mm
A 0
B 95
C 53
I think the first step is to work out the pressure in A, B and C, but as they are mutually dependent, I'm not sure where to start. I think the heights would be different if the manometer tubes were independent of each other.

and #86
And some assumptions about the system:
a) The air behaves as an Ideal Gas.
b) There is no heat transfer between the gas and the surrounding glass tubes.
c)The temperature of the gas at chamber A is 0°C or 273°K
d)The inside diameter of the cylindrical chambers A and C is 1.6 cm, giving a cross-section area 2.01 cm^2
e)The inside diameter of cylindrical chamber B is 0.5 cm: cross-section area 0.196 cm^2
f) The ambient air has a temperature of 20°C and pressure of 1033 cm water.
g) the heights of the water columns in the manometer tubes A, B and C are 0, 9.5 and 5.3 cm resp.

Absolutely no problem with the first law of thermodynamics: energy cannot be created or destroyed.

 

[erobz]

I have specified the temperature in chamber A is 0°C 273 °K.
The pressure can be determined using hydrostatics from the heights of the water columns provided i the problem statement in post #65

Whatever you say…It’s apparent I’ve already wasted my time. Good day sir.

 

[Sailor Al]

Whatever you say…It’s apparent I’ve already wasted my time. Good day sir.

You asked for the pressure and temperature in chamber A.
The pressure, from hydrostatics is 1017.8 x 10³ dynes/cm²
The temp is 0°C or 273°K

 

[Arjan82]

I see a huge amount of working, but I still don't see an answer to the stated problem: to determine the mass flow rate through the system.

This is a really disheartening thing to say, especially because in his very long post #100 @erobz gives in his equation 10 literally his answer to your question. It literally gives you his equation of the mass flow rate, he's only asking you to do a bit of effort and plug in the numbers yourself... Don't you see that?

 

[erobz]

This is a really disheartening thing to say, especially because in his very long post #100 @erobz gives in his equation 10 literally his answer to your question. It literally gives you his equation of the mass flow rate, he's only asking you to do a bit of effort and plug in the numbers yourself... Don't you see that?

They are actually saying I didn't go far enough with that "huge working". I think it's their presentation of "you're wrong and I'm right, but I can't tell you why" is what strikes a nerve with everyone they seem to interact with...

That being said, In this case (after getting some sleep) It dawned on me that we have another equation that starts the whole process over again. This time it's the cliff notes version.

Assumption:
The gas is undergoing an adiabatic expansion from A to C ( ? I have less confidence that there is no heat transfer with the surroundings between A and C as I do in gong from A to B )

Hydrostatics:
$$ P_C = P_A + \rho_w g ( l_A - l_C ) $$

First Law:
$$ 0 = h_A - h_C + \frac{V^2_A- V^2_C}{2}$$

$$ \implies V_C = \sqrt{2 \left( h_A(T_A) - h_C(T_C) \right) + V^2_A} $$

Adiabatic Expansion:

$$ T_C = T_A \left( \frac{P_C}{P_A} \right)^{ \frac{1-\gamma}{\gamma}} = T_A \left( \frac{ P_A + \rho_w g ( l_A - l_C ) }{P_A} \right)^{ \frac{1-\gamma}{\gamma}} $$

Conservation of Mass (Steady State):

$$\dot m = \rho_C A V_C = \rho_A A V_A $$

$$ \implies \rho_C V_C = \rho_A V_A $$

Sub all that in along with the Ideal Gas Law:

$$ \frac{P_A + \rho_w g ( l_A - l_C )}{R T_A \left( \frac{ P_A + \rho_w g ( l_A - l_C ) }{P_A} \right)^{ \frac{1-\gamma}{\gamma}} } \sqrt{2 \left( h_A(T_A) - h_C(T_C) \right) + V^2_A} = \frac{P_A}{RT_A} V_A $$

Where:

$T_C = T_A \left( \frac{P_C}{P_A} \right)^{ \frac{1-\gamma}{\gamma}} = T_A \left( \frac{ P_A + \rho_w g ( l_A - l_C ) }{P_A} \right)^{ \frac{1-\gamma}{\gamma}}$

$T_B = T_A \left( \frac{P_B}{P_A} \right)^{ \frac{1-\gamma}{\gamma}} = T_A \left( \frac{ P_A + \rho_w g ( l_A - l_B ) }{P_A} \right)^{ \frac{1-\gamma}{\gamma}}$

$V_A = \sqrt{ \frac{ 2 \left( h_B(T_B) - h_A(T_A) \right) }{ 1 - \left( \frac{P_A}{P_B} \right)^{ \frac{2}{\gamma}} \left( \frac{A}{a} \right)^2 } }$

$P_B = P_A + \rho_w g ( l_A - l_B)$

$h_B(T_B) = \int_{0}^{T_B} c_p(T)dT$

$h_C(T_C) = \int_{0}^{T_C} c_p(T)dT$

"In theory" "you might be able to determine $P_A$, and then $\dot m$, but $P_A$ is not only buried in a sea of symbols, but also some of those symbols $h_B(T_B), h_C(T_C)$ it finds itself buried in a non-linear function, which is then buried in a limit of an integral...It's a real pickle of an equation.

At this point I fully expect @Sailor Al to say , "but I still don't see an answer to my question", To which I say

 

[erobz]

You asked for the pressure and temperature in chamber A.
The pressure, from hydrostatics is 1017.8 x 10³ dynes/cm²
The temp is 0°C or 273°K

Just out of curiosity...Are you saying you have solved for the pressure independently and this is what you find from your workings? Or are you saying, "just say the pressure is X"?

After further consideration I believe you are correct that for a particular state of the three manometers taken as a system of measurements (under adiabatic assumption) the initial state in chamber A will be fixed given the temperature of the gas in chamber A. I "imagined" it's the case that if we change the initial state, the state of all three manometers would change accordingly. In other words, I think there is a one-to-one correspondence between the state of the manometers and the initial state of the flow.

So...Short of solving the equation developed in #109 above, we should be able to specify a pressure $P_A$ and use it to determine the manometer C reading iteratively adjusting $P_A$ making sure $P_C + \rho_w g l_C =P_A + \rho_w g l_A$ is satisfied.

Taking the kids fishing! Maybe tomorrow.

 

[Sailor Al]

Just out of curiosity...Are you saying you have solved for the pressure independently and this is what you find from your workings? Or are you saying, "just say the pressure is X"?

I believe the pressure solution is straightforward hydrostatics.
Here's my working:
With the air tap open, the water levels are 0, 9.5 and 5.3 cm above datum. [1]
When the air tap is closed and the system in equilibrium, the chambers are filled with air at, shall we say NTP (293°K, 1033 cm H₂O) and so , since there's a fixed amount of water in the tubes, the tubes will have the same water level of 4.933 cm (average of [1]) . Let's round that it to 4.9
With the tap closed, the pressure at the datum level in the water will be the sum of the Atmospheric pressure and the hydrostatic pressure : P = 1033 + 4.9 = 1037.9 cm of H₂O
With the tap open then, in cm of H₂O, the pressures at the datum level will be:
P_A = 1033 + 4.9 = 1037.9 (no water, so pressure due entirely to air pressure)
P_B = 1033 - 4.6 = 1028.4
P_C = 1033 - 0.4 = 1033.0
And 1037.9 cm of H₂O converts to 1017.8 dynes/cm²
I was deeply worried by the apparent simplicity of the solution, but it seems ok to me.
A couple of thoughts.
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.
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.
[ADD] 3) And what difference would it make if the system terminated with a wide chamber or a narrow chamber? [/ADD]
But let's press on with this one.
I'm deeply mired in attempting to balance mass flow rate - which I now identify as flux $\phi$, flux density B, and linear density $\mu$. I'm working strictly in CGS units as the pressure and Ideal Gas constants get too confusing. I am re-phrasing PM = dRT, to P = cT, where c = R/M or 2.88 10⁶ cm²/sec² for Air with a Molar mass of 28.95

Since this thread has passed the 100 post mark, should I request it be closed and start afresh?

 

[PeterDonis]

Since this thread has passed the 100 post mark, should I request it be closed and start afresh?

No. A new thread would imply a new question/topic, but that is not the case here. We have no hard limit on how many posts a thread can have. The question is (a) whether you will ever be satisfied with any answer anyone else gives you, (b) whether you will find an answer that satisfies you on your own, and (c) whether, if you do find an answer on your own, it will actually be correct.

Based on the thread so far, I'm very pessimistic about (a) since you have already been given valid answers and have not accepted them. I am not even going to try and speculate about (b) and (c).

 

[Sailor Al]

No. A new thread would imply a new question/topic, but that is not the case here. We have no hard limit on how many posts a thread can have. The question is (a) whether you will ever be satisfied with any answer anyone else gives you, (b) whether you will find an answer that satisfies you on your own, and (c) whether, if you do find an answer on your own, it will actually be correct.

Based on the thread so far, I'm very pessimistic about (a) since you have already been given valid answers and have not accepted them. I am not even going to try and speculate about (b) and (c).

OK, I just suspect the length of the thread might discourage the involvement of new members.
And, if my may, with respect point out that the answers so far have shown workings, but no solutions. The problem posed was to derive the mass flow rate in Kg/sec from the presented experiment.
Please be assured that I will be satisfied with (a). I have not yet found (b) and (c) if I do, I will certainly subject it to the forum for scrutiny to determine if it is correct.

 

[PeterDonis]

I just suspect the length of the thread might discourage the involvement of new members.

That might well be the case. You should perhaps consider that as a sign that you have gone astray, and that rather than continuing to post lengthy calculations that no one else will read, you should look at the valid answers you have already been given and see whether they might not be sufficient, so that the thread would not need to be prolonged further.

with respect point out that the answers so far have shown workings, but no solutions

You're quibbling. You have already been pointed at explicit formulas for the quantities of interest. If you are unable or unwilling to do the work yourself of plugging numbers (all of which you already have) into those formulas to get numerical answers, then this thread will indeed be closed forthwith. It is not any other poster's job to hold your hand and do all your work for you. We expect you to be able to take valid equations and work with them yourself to get a final numerical answer.

The problem posed was to derive the mass flow rate in Kg/sec from the presented experiment.

And you have been given an explicit equation for the mass flow rate. Use it.

 

[PeterDonis]

Please be assured that I will be satisfied with (a).

You can say that all you want, but your actual behavior in this thread has been to ignore valid answers because the people who gave them too you assumed you had enough intelligence and common sense to know how to plug numbers into formulas for yourself. Is that not the case?

 

[Sailor Al]

No. A new thread would imply a new question/topic, but that is not the case here.

I realise that at post #65, I did indeed pose a new question. The original question was how the gas appeared to be flowing from low to high pressure despite the (now clearly incorrect) assumption that, like water flowing uphill, gas would only flow from high to low. That was comprehensively and satisfactorily resolved by pointing out that my assumption was wrong. Water does flow uphill in a syphon or a U-bend. Gas does flow from low to high pressure in a venturi.
The question I then posed in #65 was to work out the actual mass flow rate in the venturi experiment presented in the video. That was a different question and maybe it is appropriate to present it in a new thread.

 

[erobz]

I believe the pressure solution is straightforward hydrostatics.
Here's my working:
With the air tap open, the water levels are 0, 9.5 and 5.3 cm above datum. [1]
When the air tap is closed and the system in equilibrium, the chambers are filled with air at, shall we say NTP (293°K, 1033 cm H₂O) and so , since there's a fixed amount of water in the tubes, the tubes will have the same water level of 4.933 cm (average of [1]) . Let's round that it to 4.9
With the tap closed, the pressure at the datum level in the water will be the sum of the Atmospheric pressure and the hydrostatic pressure. In cm of H₂O
P = 1033 4.9 cm
With the tap open then, in cm of H₂O, the pressures at the datum level will be:
P_A = 1033 + 4.9 = 1037.9
P_B = 1033 - 4.6 = 1028.4
P_C = 1033 - 0.4 = 1033.0
And 1037.9 cm of H₂O converts to 1017.8 dynes/cm²
I was deeply worried by the apparent simplicity of the solution, but it seems ok to me.
A couple of thoughts.

No,No, and No.

You are trying to tell me that lab pressure is 0.06 psi gage... and there is a flow in this apparatus that causes 100% thermal loss of pressure in that short distance! There is about 4 cm H2O difference in pressure between chambers A and C ~0.06 psi. I'm sorry, this is just absurd. The beginning of the outlet would be vacuum...

What do you think happens if lab pressure is 30 psi gauge(which it is far more likely to be), what would the manometers read? Imagine this setup, but with a valve at the outlet initially closed. The initial equilibrium pressure is 30 psig. All three initial columns of water are (for all intents and purposes) unchanged from the last scenario, but the initial state is 30 psig, not 0.05 psig. So how do you reconcile that? It's like you are looking at that system and saying, the columns are equal, must be atmospheric pressure inside those chambers. Once the system is open it doesn't matter where the valve is placed.

Once the system is open, you cannot say anything about the pressure inside the manometers until you've done the analysis I proposed.

 

[Sailor Al]

you are trying to tell me that lab pressure is 0.06 psi gage

0.06 psi is 4136.854 dynes/cm ²
My calculations give the over-pressure in chamber of 4802 dynes/cm which is 0.069 psi, so we're in the right ballpark.
[EDIT] there will be a pressure drop as the air enters chamber A from the supply.[/EDIT]
What's absurd about 4cm pressure difference between A and C?
I think, for the experiment, the gas pressure was seriously throttled to get the water level exactly to the base of tube A.
P.S. did you catch any fish?

 

[PeterDonis]

The question I then posed in #65 was to work out the actual mass flow rate in the venturi experiment

Yes, and you have been given a formula for that. That question is not identical to the question in the OP, but it's closely enough related that there doesn't need to be a new thread on it (particularly since it's already been discussed in this thread).

 

[erobz]

0.06 psi is 4136.854 dynes/cm ²
My calculations give the over-pressure in chamber of 4802 dynes/cm which is 0.069 psi, so we're in the right ballpark.
What's absurd about 4cm pressure difference between A and C?
I think, for the experiment, the gas pressure was seriously throttled to get the water level exactly to the base of tube A.
P.S. did you catch any fish?

We have the same system with a valve on the outlet, what is the pressure $P_{ref}$ at the base of the manometers?

 

[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.