01-31-02, 06:01 AM #1
Let's do a Thought Experiment !
The following is a continuation of a fascinating SSSF forum debate going on in Australia.
It will be a great opportunity to see if Canadians and Americans can out-think the Aussies.
The thought experiment is this:
A very long sealed cylinder is pressurised to X psi with air.
At horizontal, the pressure is tested at each end of the cylinder at test point A, which stays at sea level, and point B, which is to be elevated.
The pressures, as expected, are identical, X psi.
By miraculous means, (remember it's a thought experiment), the very long cylinder is hoisted into the air until its top is level with say Mt Everest, around 28,000 ft.
We test the pressure at the low end, A, and at the same time, at B, the high end.
Will the pressures be the same, or different.
Reply with your reasoning.
Cheers from Australia !
01-31-02, 06:29 PM #2
Error! Error! Error!A very long sealed cylinder is pressurised to X psi with air.
The pressures, as expected, are identical, X psi.
... its top is level with say Mt Everest, around 28,000 ft.
identical even when horizontal.
01-31-02, 06:30 PM #3
You could look at this piece by piece.
Firstly air-pressure is comprised by molecular weight and gravity around the earths atmosphere. At sea level the pressure is going to be higher than that of higher altitude, and possibly have a differing gases ratio.
To at first take this as mechanical, you would say that a Pressurised cylinder would stay at the same pressure, because the cylinder itself would sheild it's contents from external forces.
It wouldn't be like an air bubble rising up from the depths of an oceans atmospheric pressure. (expanding)
So there should be no difference of course this is dependant upon the pressure. Namely if the cylinder has been filled with a compression ratio where no more gas can fit, it has no room for internal movement (To the point a cylinder is filled with liquid.)
01-31-02, 06:52 PM #4
You have brought into the extremely simple thought experiment variables which merely serve to confuse matters.
The original question is very simple;
A closed Very Long Cylinder was pressurised at Sea Level. allowed to equilibrate, sealed , and then hoisted by miraculous means, ( remember, this is a thought experiment), through 90 degrees.
The top of the Cylinder is then as high as the top of Mt.Everest, 28000 ft.
The pressures at each end are now measured.
Will there be a pressure difference between the pressure test points at each end?
What do you say, and , more importantly,
What is your Reasoning?
01-31-02, 06:59 PM #5
Thanks for your contribution to what I hope is a vigorous exchange of ideas.
I am somewhat surprised at your statement that there will be a pressure difference even at sea level between the ends of the Very Long Cylinder.
Would you like to give your reasoning for this statement?
Would you like to do the same for the situation where the V.L.C. is hoisted through 90 degrees, and its top is as high as Everest?
Do you think there will be a pressure difference at test point A at sea level, and test point B, at 28000 ft?
Best Wishes from The Land Down Under.
01-31-02, 08:58 PM #6
If the cylinder contains a perfect vacuum then the internal pressure at the two ends remains the same, zero.
If the cylinder contains some air mass, then the bottom must support the weight of a 28,000 ft. column of air whose density will vary from bottom to top but whose average density will equal the density of the air in the horizontal cylinder. Hence the bottom will be at higher pressure than will the top.
To calculate the internal cylinder pressure difference one would also need the original gas density.
01-31-02, 09:16 PM #7
Your reply brings variables into an extremely simple scenario that are unnecessary, and confusing.
In essence, the question is:
If a very long sealed pressurised cylinder is lifted from Horizontal to Vertical, will there be a pressure difference between Top and Bottom?
The answer can only be Yes or No; The reasons are the interesting part.
Would you care to try the problem again, within the very simple rules?
01-31-02, 09:46 PM #8
Unless perfectly parallel to the spin axis of the planet, Coriolis force
will result in a pressure differential. Admittedly small, but present.
01-31-02, 09:49 PM #9
Between top and bottom?
(You REALLY want to hear my reasoning? LOL!)
(Okay, brace yerself... )
(Musing to self first) So... if a small bubble of gas rises from the bottom of a lake where the temperature and pressure are 8 degrees Celsius and 6.4 atm, for instance, but at the water's surface, the temp is 25 degrees Celsius and 1.0 atm, then the volume of the gas has changed within the bubble. If the volume changes then the pressure will change. Boyle's Law.
Sooooo....the cylinder is now turned (EDIT: to vertical), and the outer hull of it's top end is exposed to a lower temperature in the higher atmosphere. The temperature within the top end of the cylinder is lowered and the pressure decreases, too, but only in relation to the area of the cylinder exposed to a lower temperature. SO, wouldn't the pressure be less at the top of the cylinder than at the bottom as the temperature is higher at sea level--which would cause the pressure to increase at sea level? Charles's & Gay Lussac's Law.
And somewhere in all of this fuzzy thinking, gravity MUST play a more significant role, I should think.
Anyway, fun question!
Last edited by Counterbalance; 01-31-02 at 09:56 PM.
01-31-02, 10:30 PM #10
The measured pressure would remain the same both at the top and at the bottom of the cylinder if measured by the common variety of pressure gauge.
The reason is that a common "gauge reference" pressure gauge measures the pressure of the cylinder against the ambient atmospheric pressure, not against an absolute reference. Due to gravity, both the reference atmospheric pressure and the pressure inside the cylinder are an identical function of the altitude.
If the measurements were instead made with an "absolute reference" pressure gage, the internal pressure of the cylinder would be seen to vary as a function of the altitude.
An example of a "gauge reference" pressure gauge is the device commonly used to check the pressure in tires.
An example of an "absolute reference" pressure gauge is an aneroid baraometer.
Thanks for the fun problem scaevola!
Last edited by orthogonal; 01-31-02 at 11:23 PM.
01-31-02, 11:16 PM #11
Here's what an uneducated (re: physics) programmer thinks.
Presume that the distribution of air molecules within the cylinder is roughly the same. Failing that, presume the distribution of air molecules within the cylinder falls of at a rate slower than that of the outside air. I think these are reasonable presumptions since the cylinder is a confined space, whereas the surrounding air space is not.
Additionally, our riddler has already hinted that distribution within the cylinder is roughly constant.
I presume pressure is a measurement relative to the surroundings.
In which case, the difference in internal pressure to external pressure (these being measured against a standard) is what is relevant.
So, in fact, you can find cases where the pressure (relative to the outside air) is greater either at the bottom or the top (or neither) of the cylinder.
If the internal pressure of the cylinder is near that of the outside air pressure at the bottom, then the relative pressure will be greater at the top. Likewise, if the internal pressure is near that of the outside air pressure at the top, then the relative pressure will be greater at the bottom.
02-01-02, 12:54 AM #12
The major factor affecting pressure in the cylinder in this case is gravity. Assume a sealed cylinder.
The air at the bottom of the cylinder must support the air above it. The weight of the air above pushes down on the air below. This creates a pressure gradient in the cylinder, with lower pressure at the top and higher pressure at the bottom.
02-01-02, 01:14 AM #13
Scaevola, without knowing what part of this hamster’s answer confused you, this hamster’s answer to your original thought experiment must remain the same.
Chagur, this hamster finds it interesting that the answer to this thought experiment would be different if performed on the equator of a fast spinning asteroid.
CounterBalance, this hamster believes that if the cylinder is not insulated and the temperature is allowed to stabilize the temperature gradient does play a role.
Orthogonal, as the cylinder was sealed at sea level the average density of the air in the cylinder is greater than the average density of an equivalent column of atmospheric air. That means more mass to support and hence greater pressure than ambient air pressure.
Methods exist and are used to measure pressure without reference to ambient air pressure. (This hamster agrees that a “trick” answer wouldn’t be surprising. If the problem hadn’t clearly stated that end A remains at sea level, this hamster would have guessed the entire cylinder was lifted to the same level. Likewise one could quibble about external pressure vs. internal pressure.)
James R., this hamster agrees. Perhaps Scaevola will find your answer less confusing than this hamster’s.
The equations to calculate the difference in pressure between top and bottom are fairly complex. If one ignores the variation of gravity with height and ignores centrifugal force and ignores the temperature gradient then the pressure difference would be g times the average air density times 28,000 ft.
What would be the pressure at the bottom instead of the difference in pressure between the top and bottom. This is a gas contained in a sealed container so there is a component of pressure not due to gravity. Is this component just sea level air pressure?
Hmmm…what happens as one considers the same problem for longer and longer cylinders. As gravity is an inverse square law and the mass of air grows linearly it seems that with a sufficiently long cylinder most of the cylinder would be at a pressure slightly lower than sea level. There would be a dense high pressure region on the short end near earth.
Last edited by ImaHamster2; 02-01-02 at 01:48 AM.
02-01-02, 06:03 AM #14
Here's a Gedanken experiment.
First, let me define a "U" tube manometer as a loop of clear tubing in the shape of a "U", half-filled with a brightly colored liquid. A "U" tube manometer allows one to measure the difference in pressure between the ends of the "U" tube.
Imagine the long cylinder is horizontal. Attach one leg of this "U" tube manometer to point "B" of the cylinder. Since the cylinder has previously been charged to a pressure greater than the atmospheric pressure at sea level, the colored liquid in the attached leg of the manometer will be forced downward by some distance.
Now, leaving the manometer attached, lift the end of the long cylinder to the vertical position. The attraction of gravity will result in a decreased average density, and thus decreased pressure of the air molecules in the top of the cylinder as compared with the bottom. But, the open leg of our manometer experiences exactly the same reduction of air density and pressure. The pressure on both legs of the manometer is proportionally reduced by the same amount. The manometer will thus read the same as it did at sea level. An open reference leg manometer is equivalent to the tire pressure gauge that I mentioned in my first post.
Now, return the long cylinder to the horizontal position and attach a flask containing a vacuum to the normally open leg of our manometer. Note the reading at sea level and then again, raise the long cylinder to the vertical position. As before, the inside air pressure in the high side of the cylinder drops with altitude. But in this case the pressure on the other side of the manometer remains fixed at a vacuum level. Thus, as the cylinder is raised, the colored liquid in the manometer will rise in the leg attached to the long cylinder. A manometer with a vacuum on the normally "free" leg is equivalent to the aneroid barometer that I mentioned in my first post.
Though not required to answer the question posed by scaevola, an approximate equation describing atmospheric air pressure as a function of the altitude is:
P(z)/P0 = (1 - 0.02255z)^5.256
"PO" is the atmospheric pressure
"P(z)" is the pressure at a height of "z" kilometers
I swiped the above equation from a web page located at:
Last edited by orthogonal; 02-01-02 at 06:21 AM.
02-01-02, 08:20 AM #15
Thanks to all Forum Members for their input into the very Long Cylinder (VLC) thought experiment.
A little clarification. The pressures are measured non-intrusively and magically and accurately, as X psi, + or - say .0001psi.
The atmosphere is of constant temperature, and insulation is not a consideration.
The essential concept to be explored is whether a sealed, pressurised cylinder, when turned (magically), through a 90 degree angle, from horizontal to vertical, will have the same internal pressure at the top as at the bottom.
The question being asked is, really :
Does Gravity act on the compressed air inside the vertical cylinder to cause a region of high pressure at the bottom, and a region of low pressure at the top?
02-01-02, 10:08 AM #16
Sorry to make you work so hard with the question
To my mind, the very word "measurement" implies a relationship between the object being measured and some standard. We make measurements with respect to a yardstick, a standard Volt, etc. I have a terrible conceptual difficulty imagining the making of a non-referential measurement of any kind. But, with deference to you let me dispense altogether with the measuring device and concentrate on the physics.
Man muss immer generalisieren!
Hopefully, we agree that the dimensions of the cylinder are unimportant. That said, let us three-dimensionally expand your tall cylinder while keeping the lower end resting on the ground. The result is two concentric spheres. The smaller sphere has a diameter equal to that of the earth, while the larger sphere has a diameter equal to the height of your cylinder. Obviously, we can now forget about the smaller sphere wrapped around the earth.
Again, since the dimensions of your cylinder have no bearing on the problem, I'd like to extend the height of your cylinder up to just outside our atmosphere. The space between the two spheres now encompasses our entire atmosphere. The only difference between this case and our world as-it-is, is the large metal sphere resting just above our atmosphere. Actually, I could have extended the height of your cylinder to the known dimensions of the universe with no change in the outcome.
The thought exercise tells me that the conditions inside the cylinder are exactly the same as appear in our atmosphere.
The question being asked is, really : Does Gravity act on the compressed air inside the vertical cylinder to cause a region of high pressure at the bottom, and a region of low pressure at the top?
Last edited by orthogonal; 02-01-02 at 10:19 AM.
02-01-02, 04:05 PM #17
“The attraction of gravity will result in a decreased average density…”
The average air density of the entire cylinder won’t change as the cylinder is sealed. No air is gained or lost. However this hamster agrees that the air density at the top of the cylinder at 28,000 ft. will be less than the air density at the bottom at sea level.
“But, the open leg of our manometer experiences exactly the same reduction of air density and pressure. The pressure on both legs of the manometer is proportionally reduced by the same amount.”
Is it though? This hamster would agree if the cylinder weren’t sealed. (Of course internal equals external pressure for an unsealed cylinder.) However the air mass in the sealed cylinder is greater than the air mass in an un-sealed vertical cylinder. At horizontal your ambient pressure manometer registers zero psi for both the sealed and un-sealed cylinders. Once the cylinder is lifted the air in the bottom of the sealed cylinder must support more air mass than in the un-sealed case. Hence the pressure at the bottom will now be greater than ambient air pressure.
Now what about the pressure at the top? It will also be greater than ambient air pressure. But by the same proportion? That is not clear to this hamster.
Michael, this hamster admires your thought experiments. In your world enclosing cylinders it seems to this hamster that the cylinder at sea level contained air at sea level density. If one expands the cylinder to enclose the earth one should include sufficient air in order to still have the same average air density. (Otherwise the problem has changed.) So now the cylinder containing earth model has far more air than does the real earth. So the pressure at the bottom of the cylinder is greater than the earth sea level pressure.
This hamster agrees with your final conclusion that the air pressure at the bottom is greater.
02-01-02, 06:31 PM #18
The consensus seems to be that a pressure differential will exist between the top and bottom of our VLC.
So, can I take it that you are comfortable with the proposition that a pressurised gas will segregate itself into regions of high and low pressure, WITHIN A SEALED CONTAINER, just so long as it is under the influence of an acceleration, such as gravity?
Would anyone like to argue against this ; and say that in a sealed container there can NEVER be such a segregation of pressure, under ANY circumstances?
Any teachers, lecturers, professors, like to add to the discussion?
02-01-02, 07:18 PM #19
Scaevola, this hamster believes there will be a continuous pressure gradient from a high at the bottom to a low at the top due to gravity.
02-01-02, 08:11 PM #20
Nope. Pressure is a thermodynamics property. In terms of kinetic theory, the pressure of a gas is due to the rate of change of momentum of molecules when they strike the boundaries of the walls of the cylinder. If the system contained a liquid or other dense material, the effect of gravity would need to be included.
Therefore the pressure will be very close the same at the top as the bottom.