Incompressibility and pressure on a container

  • #1
DaveC426913
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I want to understand the factors and terminology associated with pressure in the following scenario.

I have a balloon filled with water. Inside this balloon is an irregular vessel, say a corked glass test tube, also filled with water. I can poke and squeeze the balloon with impunity, knowing that, due to the nature of incompressibility of the water both outside and inside the test tube, it is immune to breakage. Furthermore, unless the balloon is transparent, I might not even know there is anything floating around inside. It is immune to my poking and probing.


Now I open the balloon, and let 90% of the water out. It is now quite easy to conform the balloon to the test tube and even feel the test tube within the balloon. If I now squeeze the balloon, I am in great danger of breaking the test tube because it will now experience forces directly from my hands against its irregular shape.

I wish to use the correct terminology to apply to these two scenarios when it comes to pressure and forces. i.e. in terms of pressures and forces, why would one say it is better for the test tube to have the balloon filled with water?
 

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  • #2
rcgldr
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The main difference is in the second case, you're applying forces to specific points on the test tube, resulting in much greater pressure at the points of contact, than the first case, where the pressure is evenly distributed across all of the surface of the test tube.

I'm not sure why you mentioned incompressibility, since no real object is truly incompressable. There's no reason a similar experiement couldn't be done with a very compressable gas inside the balloon and test tube, perhaps using a cylinder with movable piston to apply pressure to the gas. In the first case, the cylinder applies pressure to the gas which applies pressure to the balloon, and the gas inside the balloon, which in turn applies pressure to the test tube. In the second case, the piston effectively applies a direct pressure to parts of the test tube, resuting in more pressure at the points of contact.
 
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  • #3
DaveC426913
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There's no reason a similar experiement couldn't be done with a very compressable gas inside the balloon and test tube,
Of course there is! The test tube is glass i.e. rigid. And closed.

If the balloon and test tube were both full of a compressible gas, and I squeezed the balloon to half volume, the test tube would implode. That would be the only way the test tube's internal pressure could equalize with its external pressure.


While I grant that this is technically also true if it were water filled, that's because we don't happen to mention the effort required to actually squeeze a water-filled balloon to half volume - which is beyond practical ability. For all practical purposes, water is incompressible. That why this works.
 
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  • #4
DaveC426913
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The main difference is in the second case, you're applying forces to specific points on the test tube, resulting in much greater pressure at the points of contact, than the first case, where the pressure is evenly distributed across all of the surface of the test tube.
Yes. But technically how does one describe these two types of pressure? Is there a principle? Should I invoke the term 'hydraulic pressure'?
 
  • #5
AlephZero
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I have a balloon filled with water. Inside this balloon is an irregular vessel, say a corked glass test tube, also filled with water. I can poke and squeeze the balloon with impunity, knowing that, due to the nature of incompressibility of the water both outside and inside the test tube, it is immune to breakage.
I agree with the conclusion but I think your reasoning is not quite right. I think what you mean is that because the balloon is too big for you to get hold of all the surface at once, when you poke and squeeze it you are deforming it so it maintans (almost) constant volume, and doing that only requires a small amount of force. Therefore the water pressure in the balloon stays close to atmospheric, and the test tube inside doesn't "feel" any increased forces acting on it

Now I open the balloon, and let 90% of the water out. It is now quite easy to conform the balloon to the test tube and even feel the test tube within the balloon. If I now squeeze the balloon, I am in great danger of breaking the test tube because it will now experience forces directly from my hands against its irregular shape.
Here, you have the balloon directly in contact with the test tube. Most of the force you are applying compresses the balloon material through its thickness and then acts on the glass, just the same as if the test tube were wrapped in a sheet of material without any water. The rest of the balloon will still deform so that the water in it has constant volume, and the pressure in the water outside of the test-tube will still be low.

The pressure of the water inside the tube will be high assuming the tube is sealed, and that could result in the tube breaking not where you were squeezing it, but at the weakest point in its structure (for example where seal was joined to the top of the tuibe.

If you put the balloon inside a bigger pressure vessel so you were appling a force to the whole surface of the balloon, then the inside and outside of the balloon would both be at the same pressure, Some part of the pressure would be used in deforming the shape of the test-tube, and creating stress in the tube, and the rest would increase the pressure of the water inside the tube.
 
  • #6
DaveC426913
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I agree with the conclusion but I think your reasoning is not quite right. I think what you mean is that because the balloon is too big for you to get hold of all the surface at once, when you poke and squeeze it you are deforming it so it maintans (almost) constant volume, and doing that only requires a small amount of force. Therefore the water pressure in the balloon stays close to atmospheric, and the test tube inside doesn't "feel" any increased forces acting on it
No. If I put the balloon in a pressure chamber under 2 atmospheres, the test tube would still not break. That water acts as a buffer that, while transmitting pressure, does not compress the test tube due to it owns water-filled interior.


The pressure of the water inside the tube will be high assuming the tube is sealed, and that could result in the tube breaking not where you were squeezing it, but at the weakest point in its structure (for example where seal was joined to the top of the tuibe.
Even moderately high pressures will not break the test tube because moderately high pressures do not significantly affect the volume of the water in the test tube. thus it is at low risk for breakage.

If you put the balloon inside a bigger pressure vessel so you were appling a force to the whole surface of the balloon, then the inside and outside of the balloon would both be at the same pressure, Some part of the pressure would be used in deforming the shape of the test-tube, and creating stress in the tube, and the rest would increase the pressure of the water inside the tube.
Yes but the pressure is not proportional to the compression in water. You need a lot of pressure to get a little compression.
 
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  • #7
AlephZero
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Yes. But technically how does one describe these two types of pressure? Is there a principle? Should I invoke the term 'hydraulic pressure'?
Actually, the best term is probably just "pressure", unless you want to get into detailed modelling.

The pressure in the water in the balloon is not actually constant, it varies with the depth of the fluid. The pressure at any point is given by [itex]P = P_0 + \rho g h[/itex] where [itex]P_0[/itex] is often called the "hydrostatic pressure", and the [itex]\rho g h[/itex] term comes from the weight of the fluid itself.

The term "hydraulic" is usually used in a situations where the fluid is doing work, e.g. a hydraulic ram, or a pump or turbine.
 
  • #8
AlephZero
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No. If I put the balloon in a pressure chamber under 2 atmospheres, the test tube would still not break. That water acts as a buffer that, while transmitting pressure, does not compress the test tube due to it owns water-filled interior.
See the next comment - also, remember that 2 atmospheres (about 0.2 MPa) is tiny compared with the ultimate strength of "ordinary" glass (33 MPa) or perspex (70 MPa).

Even moderately high pressures will not break the test tube because moderately high pressures do not significantly affect the volume of the water in the test tube. thus it is at low risk for breakage.
In both cases, the pressure will change the shape of the test tube even though its internal volume stays the same. The sides will tend to "bow inwards" making the middle thinner, and the ends will bulge out to equalize the volume. The stress in the glass when it deforms is what breaks the tube, and that stress can be much higher than the static pressure in the fluid where the glass has to bend.

Yes but the pressure is not proportional to the compression in water. You need a lot of pressure to get a little compression.
Yes you need a lot of pressure to get a little compression, but that doesn't mean it is "not proportional" :confused:
 
  • #9
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No. If I put the balloon in a pressure chamber under 2 atmospheres, the test tube would still not break. That water acts as a buffer that, while transmitting pressure, does not compress the test tube due to it owns water-filled interior..
I think for a typical test tube, the glass will be stiffer than the water, but it depends on the ratio of the wall thickness to the tube diameter. I downloaded a few pictures of test tubes and blew them up in Photoshop. The R/t I got was approximately 6, corresponding to a 1 mm wall thickness for a 12 mm OD. I found a value for the Young's modulus for http://www.makeitfrom.com/data/?material=Borosilicate_Glass&type=Mechanical", I got an estimate for the bulk modulus of an empty tube

[itex]K_{glass}=\frac{2Et}{(5-4\nu)R}=5.2 GPa[/itex]

Compare this to the http://en.wikipedia.org/wiki/Properties_of_water#Compressibility",

[itex]K_{water}=2.2 GPa[/itex]

So at these dimensions, the tube itself is roughly twice (2.4x) as stiff as the water it contains. When the tube is compressed, the compression of the water will provide a compensating pressure on the inner wall of the tube, helping it to resist breakage, but not preventing it entirely. (I'm assuming here that the seal is comparable to the glass in strength and not just a rubber stopper!) The net pressure compressing the tube will be

[itex]q_{net} = q_{o}-q_{i} = (K_{glass}-K_{water})dV/V[/itex]

Comparing this to an empty tube,

[itex]q_{o} = K_{glass}dV/V[/itex]

we see that the fracture pressure for the filled tube is higher by a factor of

[itex]Q=\frac{q_{o}}{q_{net}}= \frac{K_{glass}}{K_{glass}-K_{water}}=1.7[/itex]

Your mileage may vary: Q depends on the tube dimensions.

Even moderately high pressures will not break the test tube because moderately high pressures do not significantly affect the volume of the water in the test tube. thus it is at low risk for breakage..
Well, the water helps, but the glass can still be fractured by a high enough ambient pressure. However, the crux of your question, I think, is the fact that glass has a much higher compressive strength than tensile strength. Referring to http://www.makeitfrom.com/data/?material=Borosilicate_Glass&type=Mechanical", the compressive strength of borosilicate glass is about σC=295 MPa, more than 10 times higher in magnitude than the ultimate tensile strength of σU=29 MPa. The fracture pressure under compression is therefore

[itex]q_{fracture}=\frac{σ_{C}}{R/t}=49 MPa[/itex]

or roughly 500 atmospheres. That's the pressure in the ocean at a depth of roughly 5 km. One way to visualize this is that if you dropped an empty sealed test tube (appropriately weighted with some nuts and bolts) into the ocean, it would break at roughly 5 km depth. If you completely filled it with water before sealing it, it would break at a depth of 8.5 km.

The difference when squeezing the tube between your fingers is that squeezing induces a bending stress in the tube wall, placing the inner surface of the glass right below your fingers in tension. The glass is only about 1/10 as strong in tension, so it is easier to break the glass with a finger squeeze than a uniform pressure field. The amount of force is takes to break the glass with a finger squeeze depends on the tube dimensions and the fatness of one's fingers, so I'll leave that as an exercise.

Yes but the pressure is not proportional to the compression in water. You need a lot of pressure to get a little compression.
Au contraire -- you're confusing stiffness with proportionality. The compression of water is fairly linear, with a bulk modulus of K=2.2 GPa as explained above.

This was actually a pretty interesting observation -- thanks!

BBB
 
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  • #10
DaveC426913
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Yeah, you both picked up on the sloppiness I wrote. I knew it as sloppy even as I wrote it, but couldn't find a better word. Compression is proportional to pressure. What I was trying to say was that pressure can go up a lot while compression only goes up a little i.e. there's a large multiplier.
 
  • #11
DaveC426913
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OK, now that I have inadvertently led you guys on a wild goose chase:

The balloon and test tube demo is a model of a fetus in the womb. I'm exploring the physics of it, not the biology of it .which is why I introduced a generalized scenario. Sorry 'bout that...

The uterus contracts rhythmically, squeezing the sac. If the sac is unbroken and filled with amniotic fluid, the fetus will experience pressure with no untoward effects (since it is, itself a fluid-filled sac), and the pressure of the uterus will go directly toward opening the cervix.

If the bag of waters is broken, the fluid drains and all the compressive forces are applied directly to the fetus, and every protruding elbow, foot and cranial dome.

(Granted the sac is 80% fetus and 20% fluid, so it's pretty much filling the sac but the principle remains.)

I'm trying to come to grips with how to explain the forces involved using correct terminology.
 
  • #12
AlephZero
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I think for a typical test tube, the glass will be stiffer than the water, but it depends on the ratio of the wall thickness to the tube diameter. I downloaded a few pictures of test tubes and blew them up in Photoshop. The R/t I got was approximately 6, corresponding to a 1 mm wall thickness for a 12 mm OD. I found a value for the Young's modulus for http://www.makeitfrom.com/data/?material=Borosilicate_Glass&type=Mechanical", I got an estimate for the bulk modulus of an empty tube

[itex]K_{glass}=\frac{2Et}{(5-4\nu)R}=5.2 GPa[/itex]

Compare this to the http://en.wikipedia.org/wiki/Properties_of_water#Compressibility",

[itex]K_{water}=2.2 GPa[/itex]

So at these dimensions, the tube itself is roughly twice (2.4x) as stiff as the water it contains.
That's interesting, and not something I was thinking about, but I'm not sure it is the whole story. If you take the test-tube to be a cylinder with a hemispherical end, then it is well-known that the effective stiffness of the hemisphere is twice that of the cylinder. Under pressure loading, this means the hemisphere "wants" to expand only half as much as the cylinder, and the result is some local bending at the interface, and a stress concentration there. The same applies but more so, if a cylinder is joined to a flat plate at the end.

The OP didn't way how the "open" end of the test tube was sealed up - though from the latest post we know that was going off-piste from the real question, so I'm not planning to think it through to the bitter end!

Actually, thinking about it some more, I can see the logic that for a perfectly incompressible fluid, there is an equilibrium solution with the same water pressure inside and outside the test-tube and effectively zero stress in the tube (except for the pressure crushing it through its thickness). But my gut instinct from working on engineering dynamcs says that is an unstable equilibrium. And in any case water is not perfectly incompressible.
 
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  • #13
DaveC426913
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If you take the test-tube to be a cylinder with a hemispherical end, then it is well-known that the effective stiffness of the hemisphere is twice that of the cylinder. Under pressure loading, this means the hemisphere "wants" to expand only half as much as the cylinder, and the result is some local bending at the interface, and a stress concentration there. The same applies but more so, if a cylinder is joined to a flat plate at the end.
Let's say the test tube is generally spherical. Let's assume ideal circumstances.


- though from the latest post we know that was going off-post from the real question, so I'm not planning to think it through to the bitter end!
Aaaaaaaaannnnd I lost em... :cry:
 
  • #14
AlephZero
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Now we know what you are asking talking about, all the stuff about test-tubes is irrelevant.

First, some order-of-magnitude numbers, which may not be "obvious" unless you have thought about this before. Probably the most common experience of "force" is through the weight of objects. However atmospheric pressure is a much bigger force than weight, though it is self-cancelling over the entire surface of a body. But atmospheric pressure is also a much smaller force than materials can withstand without damage.

For an inch cube of any structural material (including things like bone), the weight is of the order of about 0.2 lbf, the atmospheric pressure on each face of the cube is about 15 lbf, and the compressive force required to damage the material is of the order of 1000 lbf to 100,000 lbf depending on the material.

Before the sac breaks, the force distribution is almost entirely a function of the geometry of the situation, not of anything else. All the fluid inside the sac is at the same pressure, the internal volume remains (almost) constant, and the sac itself is very flexible.

There are two types of force that balance the fluid pressure. One is the tension (stretching) of the material of the sac itself. The amount of tension required depends on the local curvature of the sac, To support a constant internal pressure, the tension required is larger when the curvature of the sac is less. (It should be obvious that if you are trying to contain a pressure with a "flat" membrane, you would have to stretch it very tight to stop it bulging outwards, compared with a spherical shape.) The other force is the pressure applied to the outside of the sac.

So, you have a sac with a fixed volume, enclosed in a cavity which can apply pressure to it and change its shape via the muscles. Opening the cervix works the same way as a hydraulic jack, or like splitting a piece of wood by driving a wedge into it. When the muscles contract, the only place for the sac to go is into the cervix, and the force applied over a large surface area by the muscles has to be balanced by a bigger force over the small area of the cervix.

The fetus inside is not a "sealed container" like the test tube analogy, it is more like a piece of pipe open at both ends. The fluid pressure inside and outside the fetus is therefore equalized, and going back to the order-of-magnitude numbers, the pressure required to cause any "structural damage" to the fetus would be very high.

After the sac bursts, I'm don't really understand what sort of description/explanation you are looking for. There are "design features" to reduce the stress levels, like the fact that the bones in the skull are not yet fused together and can slide over each other to reduce the size and therefore the external pressure. But beyond that sort of thing .... ???
 
  • #15
DaveC426913
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The fetus inside is not a "sealed container" like the test tube analogy, it is more like a piece of pipe open at both ends. The fluid pressure inside and outside the fetus is therefore equalized, and going back to the order-of-magnitude numbers, the pressure required to cause any "structural damage" to the fetus would be very high.
Agreed. From pressure there's no danger of damage. That's the point. But once the forces can be applied asymmetrically directly to the fetus (since there's no longer a buffering sac of fluid) the fetus can undergo stresses it was not designed for.


After the sac bursts, I'm don't really understand what sort of description/explanation you are looking for. There are "design features" to reduce the stress levels, like the fact that the bones in the skull are not yet fused together and can slide over each other to reduce the size and therefore the external pressure. But beyond that sort of thing .... ???
This is the crux. The cervix needs to be encouraged to open. It can take many hours for the cervix to relax and change shape. There are two things that can encourage opening of the cervix:

1] the uterine contractions transmit compressive forces to the sac of fluid. Being a fluid its forces are distributed to all surfaces of the uterus, and since the cervix is the only place with "give", the forces translate into encouraging the cervix to open. Note that all this occurs with no impact on the fetus.

2] the uterine contractions transmit compressive forces directly to the fetus. The fetus (namely the head) is the agent* that pushes the cervix open.

* I resisted the urge to use the term "battering ram" :wink:
 
  • #16
boneh3ad
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Agreed. From pressure there's no danger of damage. That's the point. But once the forces can be applied asymmetrically directly to the fetus (since there's no longer a buffering sac of fluid) the fetus can undergo stresses it was not designed for.
This still isn't relevant to your test tube example though. You could certainly break the test tube in your original example provided enough pressure could be applied to the balloon without it bursting or deforming infinitely. Since inside is a closed vessel, the inner pressure will not equalize with the outer pressure and eventually you will break the glass. The water, being incompressible, will definitely be able to resist the deformation of the tube a lot more than if it were air inside the tube, but it wouldn't be able to resist indefinitely as you raise the pressure on the balloon. If you placed it in a pressure chamber pressurized to deep sea pressures, you could certainly break the tube. You would just have to go to higher pressures if it was filled with water than if it was filled with air.

The fetus, being an open container, can equalize pressure inside and out, so it can resist any pressure that is less than the pressure necessary to do cellular or skeletal damage directly, which is much, much higher than what would be needed to do it based on a pressure differential.



This is the crux. The cervix needs to be encouraged to open. It can take many hours for the cervix to relax and change shape. There are two things that can encourage opening of the cervix:

1] the uterine contractions transmit compressive forces to the sac of fluid. Being a fluid its forces are distributed to all surfaces of the uterus, and since the cervix is the only place with "give", the forces translate into encouraging the cervix to open. Note that all this occurs with no impact on the fetus.
Right, though if the uterus was infinitely strong, I suppose you could conceive of a pressure large enough to do damage.
 
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  • #17
DaveC426913
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This still isn't relevant to your test tube example though.
The fetus, being an open container, can equalize pressure inside and out, so it can resist any pressure that is less than the pressure necessary to do cellular or skeletal damage directly, which is much, much higher than what would be needed to do it based on a pressure differential.
Considering the nature of the question, let's limit it to low pressure differentials.

I'm really interested in the uterus/fetus aspect it, but I don't want to dwell on that specifically because it will tend to muddy the waters with incidental biophysical factors.


Right, though if the uterus was infinitely strong, I suppose you could conceive of a pressure large enough to do damage.
Let's rule that out, shall we? :wink:
 
  • #19
DaveC426913
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Found a better model.

Fill a baggie all the way with water. Drop a raw egg in. Seal the baggie closed.
Lay the baggie on the table.
Squeeze the baggie.
See if the seal pops open.

Now, empty the baggie. Drop a raw egg in. Seal the baggie closed.
Lay the baggie on the table.
Squeeze the baggie.
See what pops.
 

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