Funny balloon thought experiment

[Q_Goest]

A vacuum at the top of the container (localized) is hard to visualize for me also because according to Pascal, pressure applied at the top is transmitted undiminished and does not "localize" in little areas. If this were true a hydraulic jack would not work. Unless, this is a different situation... let me know.

Do you think the pressure is the same wherever you go in the fluid? (it’s not) Is the pressure at the top of the container the same as the pressure at the bottom? (it’s not) Are you familiar at all with Bernoulli’s equation? (Pressure in a verticle column under gravity varies by rho*g*h)

So the column of water does not have a constant pressure throughout. The pressure is constant at any height, h, but increases the farther down we go.

I visualize that vacuum pulling on the water from the top, also pulling on the balloon.. and hence, the "shrinking" of the balloon is defeated by balloon expansion - and we are back at zero - canceled out.

If you’re interested in learning about physics, please stop saying the vacuum pulls. A vacuum, regardless of how strong it is, does not pull. Are you familiar with absolute pressure? If the pressure is 1 psia (absolute pressure) it is producing a compressive force on the water equal to one pound of force over every square inch. If the pressure drops to zero (0 psia), there is no longer a compressive force caused by that pressure on the water. Force simply goes to zero when pressure equals zero absolute. At atmospheric pressure however, the force on the water is 14.7 pounds for every square inch of surface.

A layman however, may think of there being some kind of ‘pulling’ force caused by subatmospheric pressure, which is wrong. That’s just an incorrect mental representation of what’s happening, because what we often see is that a fluid such as air or water will flow into a vacuum. But the fluid isn’t being ‘pulled’ into the vacuum at all. It is being PUSHED into that space by atmospheric pressure acting at another location on the fluid and the pressure in the ‘vacuum’ space isn’t strong enough to keep it out!

By "at first" do you mean that even under tiny amounts of vacuum not significant enough (which is what a balloon might cause), water will find a way to vaporize? I think it will not be the case and only super duper thick strong potentially powerful balloons at super depths could work for that.

By “at first” I mean that assuming the balloon is at 14.7 psia at the top of the container, there is some depth to which the balloon can be taken during which the size of the balloon does not change significantly. The balloon pressure will actually increase very, very slightly, assuming no disolved air in the water, and the volume of the container is not extreamly large compared to the volume of the balloon. As the balloon goes down deeper into the water, the pressure of the water at every point also goes down. In so doing, the density of the water will decrease some very minute amount which is accommodated by the increase in the balloon’s air density.

Can anything in the system heat up or cool down by the way - i.e. some pressure or vacuum causes something else to change temperature. like the air in the balloon, or the water itself, etc. That would be interesting if a refridgerator, freezer, or heater was formed out of this.

If a volume stays constant but pressure increases, a temperature change could occur. i.e. if the balloon is locked into position, but the pressure increases inside - the (specially insulated) balloon becomes hot inside, yet the water starts to freeze or cool down a human. Please disprove this, it can't be true.

As the balloon sinks in the container, the air is compressed very, very slighty. This would cause it to warm very, very slightly. The process would follow a line of constant entropy.

In contrast, the water pressure at every point decays significantly, also following a line of constant entropy. Once the balloon reaches a level at which water at the upper surface can boil, a few things start to happen.
- The surface of the water begins to boil as the water passes through the saturation point. Imagine for example, taking the radiator cap off a car that has been running and is hot. The pressure inside the radiator drops suddenly and the water begins to boil.
- The boiling water actually gets cooler. The temperature where the water is boiling goes down and gets colder.

As the balloon goes deeper:
- Water vapor formed above the surface of the liquid water will begin to get colder faster and there would be a flow of heat from the warmer water to the colder water vapor and also to the colder water surface where the water cools from boiling.
- The balloon begins to get compressed as pressure in the water now is being pushed on by the water vapor at the top surface. Remember that although there is a partial vacuum at the top water surface, that vacuum pressure is PUSHING down on the water, not PULLING UP. As the balloon is compressed by the increased pressure, it gets warmer, again following a line of constant entropy. As it warms, heat transfer from the warmer balloon into the water would also start to occur.

 

[physical1]

As the balloon is compressed by the increased pressure

If you want to learn anything about physics, I suggest you stop using sentences that are doubling up on words. As the balloon is compressed already literally states that there is increased pressure - stating "compressed by pressure" is useless, like a double negative but instead a double positive. I forget the term for this error in English. As a human, I usually look over things like this - but since you nitpicked me, I am returning the favor.

I suggest if you want to have a conversation with a human being, you look in the mirror and nitpick your own pimples.

Are you familiar at all with Bernoulli's equation? (Pressure in a verticle column under gravity varies by rho*g*h)

I suggest if you want to learn physics you spell verticle using the "vertical" word from my dictionary.

As it warms, heat transfer from the warmer balloon into the water would also start to occur.

I mentioned insulation.

It is funny though, I wonder what the lurkers of this forum will think after reading these messages. Everything I said in my first few posts has been ridiculed, and yet everything is turning out to be quite close to what my hypothesis was. Sure I make a few statements in english that are not perfect, like every human. But to have some of my hypothesis(es) repeated back in my ear that were originally ridiculed.. is quite non-surprising actually. I am told to learn physics and what in fact has happened is many others on this forum, some with doctorate degrees, have learned more than I have. Where is that quotation about ridicule and self evidence. I have it stored somewhere. And one about self reflection.

 

[Q_Goest]

My post was not intended to ridicule you. In responding it seems clear you're not familiar with some basics, so showing you what misconceptions you have is just part of explaining the answer. Vacuum doesn't pull, and I'm not trying to poke fun of what you don't understanding.

 

[belliott4488]

physical1 - You've got a few misconceptions about this topic, which are causing the confusion that prompted your original question (as well as some follow-up questions you've posted). People are just trying to help you clear up these misconceptions so that you'll understand the answers, so trying to flame them the with insulting responses is not only rude, it defeats the presumed purpose of your post - to get a satisfactory answer to your question.

1. There is nothing wrong with referring to the creation of a vacuum in a thought experiment, since there is nothing physically wrong with a vacuum (speaking classically, at least, and I assume you're trying to invoke Quantum Mechanics to frame your problem). In practical terms, it is difficult to produce extremely low-pressure vacuums, but you can get arbitrarily close with the right equipment. As for answering your OP, however, a vacuum at the top of the container poses no physical problems; in fact it is what solves the problem.

2. Are you familiar with a water barometer? It's like a regular mercury barometer, but since water is so much less dense than mercury, you need a column of water some 10 m tall (at sea level). Nonetheless, in such a device you have a container with water and a vacuum above the surface of the water.

3. As others have mentioned, vacuums don't suck, and neither does your balloon as it contracts due to the increased pressure. Rather, greater pressure pushes from the other side. I used to pose the following problem in Intro. Physics, which I now pose to you: Suppose you are an astronaut on the surface of the Moon (where there is no atmosphere, and thus a vacuum), and you've brought along a soda in cup. Can you drink from this cup by sucking on a straw that goes from your mouth through a well-sealed port in your helmet and has its other end submerged in the soda?

 

[Doc Al]

On intuition, one could be forgiven for thinking that the vacuum at the top would pull the water up, causing the water level to rise and fill the container, and the balloon to keep its original volume.

I think this "common sense" idea of a vacuum "pulling up" is at the root of the confusion in this thread. Q_Goest and belliott4488 have already expounded on that one.

But because there can’t be any air getting into nor out of the container, and because water is essentially incompressible, then as the balloon is moved down toward the bottom of the container, the pressure inside the balloon can’t change.

I'd love to know why you think this. (It's not clear to me at all.)

If pressure in the balloon increased, the volume would decrease, but since water is incompressible and there is no way for air to get inside the container, the balloon can’t (at first) change volume.

Again: Why? Why does air need to get inside the container in order for the balloon to change volume?

The pressure of the water therefore, is equal to the pressure of the balloon at the depth of the balloon. You can think of the balloon as a pressure gage at this point. It is measuring the pressure of the water at the given depth. Whatever the location of the balloon, the pressure of the water at that location must equal the pressure of the balloon,

Right.

and the pressure in the balloon is equal to ambient pressure because the water is assumed to be incompressible.

Not sure what you mean by "ambient pressure".

(in reality, there's some bulk modulus for water which will tend to decrease the water density slightly so that the air pressure increases very, very slightly as the balloon sinks in the container.)

The sealed, rigid container is 1000 ft deep (say). Still think the air pressure within the balloon increases "very, very slightly" as the balloon is pulled to the bottom?

Perhaps you mean as the balloon just begins its descent, the first thing that happens is that the pressure between the water and the top of the container (presumed to equal 1 atm, but depends on how the tank was filled) quickly reduces as the water microscopically decompresses.

Then as the balloon continues to be pulled down, the pressure within it quickly increases as the balloon is compressed, which drops the water level. (As you describe, things are a bit more complicated due to the formation of water vapor--but I still don't see that as essential to the main point of this thought experiment. But well-described, by the way.)

Everything I said in my first few posts has been ridiculed, and yet everything is turning out to be quite close to what my hypothesis was.

Not quite, but I can see where you might have gotten that impression.

 

[LURCH]

A vacuum at the top of the container (localized) is hard to visualize for me also because according to Pascal...

This is where I found it helpfull to think about a barometer. It is a real-life example of a (very tall) tube of water with a vacuum at the top. This of course does not violate Pascal, as pressures remain the same across any cross-section of the container at a particular depth. I think that you would agree that Pascal is not saying pressure must be the same from top to bottom, right?

So, going back to the barometer, pressure is greatest at the bottom, and decreases by degrees as one travels farther up the pressure decreases untill, at the very top, there is vacuum. This is exactly the situation that exists within the imaginary container at the end of the thought experiment.

 

[Q_Goest]

I'd love to know why you think this. (It's not clear to me at all.)

Again: Why? Why does air need to get inside the container in order for the balloon to change volume?

In order for the balloon’s air pressure to change, there has to be some change of volume of the balloon. No volume change means no pressure change.

Imagine a tall, verticle, water filled pipe, 12” in diameter with a balloon just under the surface. Now imagine putting a cap on this so the water is up to the underside of the cap but instead of sealing the pipe, we put a hole in it. Now, as the balloon is pulled down the pipe, it shrinks due to the increased pressure and air is pulled in through the hole in the cap. The volume of air coming in through the hole is equal to the volume that the balloon is decreased by, and the balloon volume is decreased according to an isentropic compression of the air inside. The air warms up, the pressure increases and the volume decreases. For every dV of volume decrease in the balloon, there has to be an equal volume of air admitted through the hole in the cap.

Now if you seal the hole and do the same thing, no air can get in. So as the balloon sinks, if it were to be compressed, there would need to be air or something getting into the pipe to take up the decrease in volume of the balloon. But since nothing can get in, the balloon's volume can’t decrease. If the balloon's volume can’t decrease, its pressure can’t increase. If the pressure can’t increase, the water pressure on the outside of the balloon must be in equilibrium with the pressure of the air inside. It’s this last sentence that’s hard to digest. It’s counterintuitive to think of a balloon going down in some water filled column and the pressure of the water filled column changing equally all along the length as it goes down. But we could do the same thing by removing the balloon, adding a piston to the top of the pipe and simply pulling up on the piston. In this case, the pressure at the top of the pipe drops according to how hard we pull on the piston. As the pressure at the top of the pipe changes, the pressure all along the length of the verticle pipe also changes the same amount.

In the case of the piston, there is a small amount of motion needed to change the pressure. It’s going to be extremely small since water is nearly incompressible, but it will be there. And the more water volume you have, the more this piston is going to move. The piston movement is dependant on the bulk modulus of the water.

Similarly, the balloon is decreasing in volume very slighty as it sinks in the pipe because all along the length of the verticle pipe, the pressure is dropping just as it would if you had a piston on top. The small increase in volume we saw with the piston movement will have to be the same as the small decrease in volume that occurs in the balloon. The water column pressure is dropping along the entire length in both cases, and because of the bulk modulus of the water, there is a very small but finite volume increase in the water which is compensated for by either the movement of the piston or the compression of the balloon.

Not sure what you mean by "ambient pressure".

The sealed, rigid container is 1000 ft deep (say). Still think the air pressure within the balloon increases "very, very slightly" as the balloon is pulled to the bottom?

Assuming the bulk modulus of water is always positive (pretty sure that’s a solid assumption), then regardless of what pressure the verticle pipe starts out at, if you increase the volume of the pipe, either using a piston or a balloon, the pressure along the length of the pipe will drop. Similarly, decreasing the volume of the pipe increases pressure, though the change in volume is small for such an incompressible media as water.

Note also that as soon as the pressure at the top of this verticle column drops to the boiling point of the water, it will be forced to boil. Regerdless of whether a piston or balloon is used to change the volume of the container, the water is going through an isentropic process as long as there is no heat transferred from the environment (ie: adiabatic container).

 

[Doc Al]

In order for the balloon’s air pressure to change, there has to be some change of volume of the balloon. No volume change means no pressure change.

Sounds good.

Imagine a tall, verticle, water filled pipe, 12” in diameter with a balloon just under the surface. Now imagine putting a cap on this so the water is up to the underside of the cap but instead of sealing the pipe, we put a hole in it. Now, as the balloon is pulled down the pipe, it shrinks due to the increased pressure and air is pulled in through the hole in the cap.

Sure. (Better to say that the air is pushed in through the hole by the atmosphere, not pulled in, of course.)

Now if you seal the hole and do the same thing, no air can get in.

This is true.

So as the balloon sinks, if it were to be compressed, there would need to be air or something getting into the pipe to take up the decrease in volume of the balloon.

Why is that? (This is the point that I'd like to see physically justified.)

But since nothing can get in, the balloon's volume can’t decrease.

Again, I'm not seeing this.

If the balloon's volume can’t decrease, its pressure can’t increase.

I agree with this.

If the pressure can’t increase, the water pressure on the outside of the balloon must be in equilibrium with the pressure of the air inside. It’s this last sentence that’s hard to digest. It’s counterintuitive to think of a balloon going down in some water filled column and the pressure of the water filled column changing equally all along the length as it goes down.

Quite counterintuitive, if I understand what you're saying. Are you saying that if I pull the balloon down 1000 meters (a difference in hydrostatic pressure of about 100 atmospheres) in that closed container, that somehow that huge water pressure at the bottom is negated and that the balloon is not compressed?

 

[Q_Goest]

Sure. (Better to say that the air is pushed in through the hole by the atmosphere, not pulled in, of course.)

<laughing> Touche, you got me there!

q_goest said: So as the balloon sinks, if it were to be compressed, there would need to be air or something getting into the pipe to take up the decrease in volume of the balloon.

Why is that? (This is the point that I'd like to see physically justified.)

Ok, good question. Note the assumptions we’ve made:
- The container is sealed so that nothing can get in or out.
- Container is infinitely rigid, therefore internal volume can’t change.
- Water is essentially incompressible. Let’s neglect water's bulk modulus for now and add that in later.
- The balloon’s pressure is in equilibrium with the water pressure. If the balloon’s pressure was more or less than the water, it wouldn’t be in static equilibrium and the balloon would have to expand or shrink respectively until it was in equilibrium. By equilibrium, I mean that the pressure on the inside of the balloon is essentially the same as the pressure on the outside.*

Now, let’s assume that as the balloon goes down, it shrinks in volume.
1. Since the container is sealed, neither air nor water can get into the container to make up for the volume left by the shrunken balloon.
2. Since the container is rigid, the container’s volume can’t decrease to make up for the volume.
3. Since the water is incompressible, the water can’t expand to make up for this decrease in volume.
4. All we’re left with is the knowledge that the water outside the balloon must be equal in pressure to the gas inside the balloon. So if nothing in the system can make up for the shrinking balloon volume, and if the pressure inside is the same outside the balloon, then we’re left with the assumption that the balloon pressure hasn’t changed but the water pressure did!

#4 leads us to the realization that the pressure all along the column must change as the balloon sinks. And we also know that when the local water pressure drops below the saturation point, it must boil, and the lowest pressure is at the upper most surface. So when the upper most surface reaches saturation, it must boil which creates water vapor above the column.

Quite counterintuitive, if I understand what you're saying. Are you saying that if I pull the balloon down 1000 meters (a difference in hydrostatic pressure of about 100 atmospheres) in that closed container, that somehow that huge water pressure at the bottom is negated and that the balloon is not compressed?

You’ve left out some assumptions. What should the pressure be at the top of the column when the balloon is at the top of the column?

For example, if the pressure at the top of the column was 15 psia, then the balloon would have to have 15 psia in it to be in equilibrium. Assuming roughly 2 feet per psi for water, then 30 feet down the pressure would be 30 psia. Now if the balloon sinks 30 feet, the pressure at that level has gone to 15 psia and the top of the column has dropped to 0 psia and begins to boil. Until it did that, the balloon couldn’t shrink. But once it passes that point, the water surface begins to boil producing water vapor at a much lower density than the liquid water. Now you have something that can expand and make up for the difference in the incompressibility of the water and rigidity of the container. Now as the balloon sinks, the decrease in balloon volume can be taken up by the increase in vapor space above the water. So in this case, when the balloon is 1000 feet down, it must have shrunken to the size of a pea.

The above is simplified by assuming the bulk modulus of water is zero. But since there is some finite bulk modulus, the water actually decreases in density slightly as pressure drops. Checking a database for water shows that the density decrease between 0.4 psia (aprox. boiling at 70 F) and 15 psia is only about 0.003%. If this is taken into account, we find that the balloon shrinks very slightly with a corresponding increase in pressure and the water expands very slightly with a corresponding decrease in pressure.

* Note that in real life, there is some minor difficulty with this assumption. Rho*g*h of air inside the balloon is not equal to rho*g*h of water outside. The difference has to be made up by internal stresses of the balloon, but that’s inconsequential. One could for example, instead of using a balloon, use a small cylinder containing air with a piston. In fact, it may be easier to visuallize using a cylinder and piston arrangement instead of a balloon.

 

[Doc Al]

Ok, good question. Note the assumptions we’ve made:
- The container is sealed so that nothing can get in or out.
- Container is infinitely rigid, therefore internal volume can’t change.
- Water is essentially incompressible. Let’s neglect water's bulk modulus for now and add that in later.
- The balloon’s pressure is in equilibrium with the water pressure. If the balloon’s pressure was more or less than the water, it wouldn’t be in static equilibrium and the balloon would have to expand or shrink respectively until it was in equilibrium. By equilibrium, I mean that the pressure on the inside of the balloon is essentially the same as the pressure on the outside.*

All good.

Now, let’s assume that as the balloon goes down, it shrinks in volume.
1. Since the container is sealed, neither air nor water can get into the container to make up for the volume left by the shrunken balloon.
2. Since the container is rigid, the container’s volume can’t decrease to make up for the volume.
3. Since the water is incompressible, the water can’t expand to make up for this decrease in volume.

All good.

4. All we’re left with is the knowledge that the water outside the balloon must be equal in pressure to the gas inside the balloon. So if nothing in the system can make up for the shrinking balloon volume, and if the pressure inside is the same outside the balloon, then we’re left with the assumption that the balloon pressure hasn’t changed but the water pressure did!

(1) You assume that something must "make up" for the shrinking balloon volume. Why is that? This is the key assumption that requires justification--at least to me.
(2) As the balloon is pulled down, both water pressure and balloon pressure increase as the balloon is compressed. No mystery there.

(I understand your point about water vapor forming at the surface, but to me that is a side issue.)

 

[Q_Goest]

(1) You assume that something must "make up" for the shrinking balloon volume. Why is that? This is the key assumption that requires justification--at least to me.

V_a + V_w = V_c
where
V_a = Volume of the air (balloon)
V_w = Volume of the water
V_c = Volume of the container

V_c doesn't change.
Water is incompressible, so density remains constant and V_w is a constant
V_a can't change unless there's a change in either the container or water.
Since the container can't change volume, V_a can only change if V_w changes.

 

[Doc Al]

V_a can't change unless there's a change in either the container or water.

I don't see a justification for that statement.

The only volumes that are fixed are V_w & V_c; V_a can certainly change. It happens to be true that V_a + V_w = V_c at the beginning, when the balloon is at the top of the container, but it's not true in general. Generally, V_a + V_w ≤ V_c.

 

[Q_Goest]

V_a + V_w ≤ V_c.

Why "less than" or equal to? The less than part makes no sense to me. How can V_a decrease without a change to either V_c or V_w? The volume of the container is the simple summation of the volume of the parts.

 

[Doc Al]

Why "less than" or equal to? The less than part makes no sense to me. How can V_a decrease without a change to either V_c or V_w? The volume of the container is the simple summation of the volume of the parts.

Simple: V_a + V_w + Remaining "Empty" Space = V_c.

 

[Dadface]

Why "less than" or equal to? The less than part makes no sense to me. How can V_a decrease without a change to either V_c or V_w? The volume of the container is the simple summation of the volume of the parts.

Because Va +Vw+Vs=Vc

Here Vs=volume above the water surface caused by the balloon shrinking

 

[Dadface]

Stretching the thought experiment further if the water was replaced by a perfectly non volatile liquid the space above the water surface would be vacuum.

 

[belliott4488]

V_a + V_w = V_c
where
V_a = Volume of the air (balloon)
V_w = Volume of the water
V_c = Volume of the container

V_c doesn't change.
Water is incompressible, so density remains constant and V_w is a constant
V_a can't change unless there's a change in either the container or water.
Since the container can't change volume, V_a can only change if V_w changes.

Why "less than" or equal to? The less than part makes no sense to me. How can V_a decrease without a change to either V_c or V_w? The volume of the container is the simple summation of the volume of the parts.

I'm not sure if Doc Al is really not understanding your objection, of if he's just trying to draw you out. I think I see what you're objecting to, and since I think it's the same thing that was bothering the OP, I'm going to go ahead and address it.

My impression is that you believe that since V_a + V_w = V_c at the start that this must still be true after the balloon has been pulled to the bottom of the container. Moreover, I get the impression you object to the notion that there is a vacuum at the top of the container, of volume equal to change in V_a as the balloon is compressed. Is this correct?

The point I and others have been trying to make is that there is nothing wrong with a vacuum being created as the balloon is compressed by the greater hydrostatic pressure at the bottom of the container - in fact, it must be created in order for the balloon to compress, as you and physical1 have pointed out. This means that when the balloon is at the bottom of the container you have,

V'_a + V_v + V_w = V_c

where V'_a is the (new) smaller volume of the compressed balloon
and V_v is the volume of the vacuum at the top of the container.

I think there's an intuitive resistance to the creation of a vacuum since our experience in life is that it takes a lot of work to lower the air pressure in a container - by pulling on a piston, for example. That's only true if there is atmospheric pressure on the other side of the piston, however. The force you feel is not due to the low pressure "pulling", but rather on the higher atmospheric pressure on the other side "pushing". Air pressure only ever pushes; it cannot pull.

[Note that I am ignoring the presence of some water vapor in the new volume at the top of the container that has been mentioned in this thread. There will undoubtedly be some vapor, so of course it will not be a perfect volume. This has no important effect on the explanation of the outcome of the thought experiment, other than to replace the word "vacuum" with "very low pressure volume of water vapor." Whatever the pressure of this volume - approximately zero, or simply very low - that pressure is added to the pressure of the column of water above the balloon's new depth when calculating the pressure in the balloon.]

 

[Doc Al]

I'm not sure if Doc Al is really not understanding your objection, of if he's just trying to draw you out. I think I see what you're objecting to, and since I think it's the same thing that was bothering the OP, I'm going to go ahead and address it.

I admit that I was trying to draw out the physical reasoning (or lack thereof, as I see it), as I have been throughout this thread. The rest of your post expresses my point exactly. Well done.

 

[Q_Goest]

Simple: V_a + V_w + Remaining "Empty" Space = V_c.

Empty space? If it's empty, it's vacuum. But a vacuum can't be in equilibrium with water, it will boil. For the case of a liquid that can't boil, that's a slightly different issue. It doesn't really matter as long as the surface of the water doesn't boil.

Regardless; let's change the liquid for the sake of discussion. I think this should help shed some light on things.

Thus far we’ve discussed water which has some vapor pressure at ambient temperature that’s well above a perfect vacuum. It’s around 0.3 to 0.4 psia at 70 F.

What if we used a liquid that could not convert to a gas? Vacuum pump oils for example are highly refined, long chain hydrocarbons with a vapor pressure in the micron range. For the sake of argument, let’s just consider a hypothetical liquid such as hypothetical vacuum pump oil that does not boil/does not vaporize at any low pressure. We'll aslo hypothesize that this liquid is completely incompressible so it rules out any bulk modulus affects. I think these affects are probably getting in the way of understanding what's going on anyway, so let's just ignore them.

For our experiment, we use the balloon and this non-boiling liquid. Let’s start at atmospheric pressure at the upper surface, say 15 psia for the sake of argument, and the balloon is at the top of the container. Let’s also assume the liquid is roughly the same as water, so for every 2 feet of head we get 1 psi. So 30 feet down in this liquid, the pressure is 30 psia (ie: rho*g*h +P_s where P_s is the pressure at the surface = 15 psia at the surface, plus 15 psi head).

As we pull the balloon down 2 feet, we might think the balloon shrinks due to increased head. But if it did, we’d need to come up with that volume somewhere else in our system (ie: We may think there's an additional volume = V_v (vacuum volume)).

-> If a vacuum space formed above the liquid, and assuming Bernoulli’s still holds (which it does) then the surface pressure would be 0 psia, so the pressure 2 feet down (per rho*g*h + P_s where P_s =0 for vacuum) would be only 1 psia. But the balloon had 15 psia in it to begin with, hence it won’t be at 1 psia if it shrinks. Pressure has to be greater than 15 psia and the conclusion is it hasn’t shrunk.

In other words, if the surface pressure drops instantly from 15 psia to 0 psia, because there's a vacuum there, we have a sudden change in pressure ALL THE WAY DOWN to the bottom of the container.

So we are left with the balloon not shrinking and not expanding. We are left with the surface pressure decreasing by 1 psi, not suddenly going into vacuum. There is no sudden step change in pressure at the upper surface.

When the balloon gets down to 30 feet, the pressure at the top of the column finally reaches 0 psia (rho*g*h = 15 psi). At this point, a vacuum just starts to form, and the pressure 30 feet down is only 15 psia instead of the 30 psia originally. Everything is still in equilibrium, and the balloon hasn't changed volume yet.

As the balloon goes deeper, a vacuum space finally begins to form above the liquid as the depth of the balloon goes down more than 30 feet and pressure in the balloon exceeds 15 psia. Now, as the balloon goes deeper and contracts, the volume that it decreases by can be made up by the volume of the vacuum at the top of the container.

What everyone is missing is the Bernoulli equation for a static column of liquid, P(at some point below the surface) = P_s + rho*g*h

 

[Doc Al]

In other words, if the surface pressure drops instantly from 15 psia to 0 psia, because there's a vacuum there, we have a sudden change in pressure ALL THE WAY DOWN to the bottom of the container.

So we are left with the balloon not shrinking and not expanding. We are left with the surface pressure decreasing by 1 psi, not suddenly going into vacuum. There is no sudden step change in pressure at the upper surface.

When the balloon gets down to 30 feet, the pressure at the top of the column finally reaches 0 psia (rho*g*h = 15 psi). At this point, a vacuum just starts to form, and the pressure 30 feet down is only 15 psia instead of the 30 psia originally. Everything is still in equilibrium, and the balloon hasn't changed volume yet.

As the balloon goes deeper, a vacuum space finally begins to form above the liquid as the depth of the balloon goes down more than 30 feet and pressure in the balloon exceeds 15 psia. Now, as the balloon goes deeper and contracts, the volume that it decreases by can be made up by the volume of the vacuum at the top of the container.

I agree 100%! In fact I started to make the same point--somewhat obliquely--in post #55. (All the talk about vapor formation must have obscured the issue.) The first thing that happens as the balloon lowers is that the overpressure (the 1 atmosphere at the top of the water) must be reduced throughout the fluid. As I was imagining the balloon being lowered 1000 m, that bit was minimized in my mind as it was resolved in the first 10 m.

So I think we agree after all. And thanks for your patience and clarification. (Whew!)

 

[belliott4488]

... For the sake of argument, let’s just consider a hypothetical liquid such as hypothetical vacuum pump oil that does not boil/does not vaporize at any low pressure. ...

Let’s start at atmospheric pressure at the upper surface, say 15 psia for the sake of argument, and the balloon is at the top of the container. ...

As we pull the balloon down 2 feet, we might think the balloon shrinks due to increased head. But if it did, we’d need to come up with that volume somewhere else in our system (ie: We may think there's an additional volume = V_v (vacuum volume)).

-> If a vacuum space formed above the liquid, and assuming Bernoulli’s still holds (which it does) then the surface pressure would be 0 psia, ...

In other words, if the surface pressure drops instantly from 15 psia to 0 psia, because there's a vacuum there, we have a sudden change in pressure ALL THE WAY DOWN to the bottom of the container.

Okay, that was a pretty good set-up - well thought-through - but I see a critical step where I think your reasoning goes astray. Not surprisingly, it's right where you say "the surface pressure drops instantly from 15 psia to 0 psia". Discontinuous changes like that are usually a sign that something needs be looked at carefully, in case you're side-stepping a process that must be considered in more detail (sometimes it's okay, and the change is just due to a simplifying assumption that does not affect the issue being investigated).

First, where does the 15 psi pressure come from at the start? We might consider that there's a tiny amount of gas at this pressure above the fluid (I know you didn't, but just bear with me for a moment). In that case, it's clear what happens when the balloon is drawn down - the level of the fluid drops, causing the volume of the tiny amount of gas to increase, its pressure decreases, Bernoulli applies, and the rest is as you've described. The final state is such that the pressure of the tiny amount of gas has been reduced to whatever value is necessary to balance all the equations.

Now, what if there is no gas at all, other than what's inside the balloon, as I believe you had in mind? Well, if there is 15 psi of pressure on the surface of the fluid, the only thing that can be exerting this pressure is the top of the container where it contacts the fluid. It cannot exert such a pressure unless it is less than perfectly rigid, which means that as the balloon is lowered and the level of the surface of the fluid drops, the top of the container will flex downward, exerting less pressure (think of it as an extremely stiff spring). It will do this until it reaches its mechanical equilibrium point, i.e. it is exerting zero pressure, after which time the vacuum appears.

This might seem like an unnecessarily arcane detail - we don't usually worry about the sides of a sealed vessel flexing - but if you insist that there was 15 psi on the surface of the fluid at the start, then I believe you must take this into consideration.

 

[Doc Al]

Now, what if there is no gas at all, other than what's inside the balloon, as I believe you had in mind? Well, if there is 15 psi of pressure on the surface of the fluid, the only thing that can be exerting this pressure is the top of the container where it contacts the fluid. It cannot exert such a pressure unless it is less than perfectly rigid, which means that as the balloon is lowered and the level of the surface of the fluid drops, the top of the container will flex downward, exerting less pressure (think of it as an extremely stiff spring). It will do this until it reaches its mechanical equilibrium point, i.e. it is exerting zero pressure, after which time the vacuum appears.

This might seem like an unnecessarily arcane detail - we don't usually worry about the sides of a sealed vessel flexing - but if you insist that there was 15 psi on the surface of the fluid at the start, then I believe you must take this into consideration.

The way I see it is that the pressure applied by the top (and sides) of the container compresses the water ever so slightly. But the assumptions that the container is rigid and the water is incompressible mean that the change in volume is microscopic and can be ignored. So "for all practical purposes" the water level (and balloon volume) won't change until that top pressure is reduced to zero. (It doesn't happen instantly.)

 

[Q_Goest]

Okay, that was a pretty good set-up - well thought-through - but I see a critical step where I think your reasoning goes astray. Not surprisingly, it's right where you say "the surface pressure drops instantly from 15 psia to 0 psia". Discontinuous changes like that are usually a sign that something needs be looked at carefully, in case you're side-stepping a process that must be considered in more detail (sometimes it's okay, and the change is just due to a simplifying assumption that does not affect the issue being investigated).

First, where does the 15 psi pressure come from at the start? We might consider that there's a tiny amount of gas at this pressure above the fluid (I know you didn't, but just bear with me for a moment). In that case, it's clear what happens when the balloon is drawn down - the level of the fluid drops, causing the volume of the tiny amount of gas to increase, its pressure decreases, Bernoulli applies, and the rest is as you've described. The final state is such that the pressure of the tiny amount of gas has been reduced to whatever value is necessary to balance all the equations.

Now, what if there is no gas at all, other than what's inside the balloon, as I believe you had in mind? Well, if there is 15 psi of pressure on the surface of the fluid, the only thing that can be exerting this pressure is the top of the container where it contacts the fluid. It cannot exert such a pressure unless it is less than perfectly rigid, which means that as the balloon is lowered and the level of the surface of the fluid drops, the top of the container will flex downward, exerting less pressure (think of it as an extremely stiff spring). It will do this until it reaches its mechanical equilibrium point, i.e. it is exerting zero pressure, after which time the vacuum appears.

This might seem like an unnecessarily arcane detail - we don't usually worry about the sides of a sealed vessel flexing - but if you insist that there was 15 psi on the surface of the fluid at the start, then I believe you must take this into consideration.

An air bubble or gasseous volume of any sort is not needed at the top of the container. Consider for example, a perfectly rigid container filled with a perfectly incompressible liquid. Now we put this 15 psia balloon in, just under the surface and cap it so there's no air or gas of any kind inside. What's creating the 15 psia at the surface? Well, the balloon is! If we imagine taking out 1 molecule of liquid, the balloon volume increases by some infintesimal amount along with a corresponding infintesimal decrease in pressure. The entire system is completely stable for minor changes in the container's volume. There's no need to hypothesize about an extremely stiff spring. We can make the container infinitly rigid and the liquid infinitely incompressible. There's no details here that need clarification because of the rigidity of the container nor the incompressibility of the liquid. The balloon's ability to accommodate small changes in volume can account for that. As the balloon is pushed down, the pressure at the upper surface changes gradually because the balloon's volume is able to take up any hypothesized miniscule variations in pressure at the upper surface.

 

[Phrak]

A very interesting phenomena. It would be nice to see a simple experimental arrangement As a balloon in a sealed, constant volume container is pulled downward, the pressure in the balloon remains nominally constant and the pressure of the container drops.

As long as the pressure at the top of the container remains above vapor pressure, is this correct by you, Q Goest?

 

[Phrak]

Nice probelm Physical1, to keep so many (myself included) scratching their heads!

The result is very counterintuitive. How could a little bubble pushed to the bottom of a container cause a large change in pressure?

It might be more intuitive to consider the following. You have a large rigid container 32 feet deep. It's filled with pure water and caped, without any air, at one atmosphere of pressure. Someone kicks the container loosening a bubble loose from a crack in the bottom. It rises, but doesn't significantly expand. The pressure rises to twice atmospheric pressure to keep the bubble at it's original volume. I find this scenario more fathomable.

Very small bubbles shouldn't have the same effect, due to the compressibility of water and expansion of the container.

 

[Q_Goest]

A very interesting phenomena. It would be nice to see a simple experimental arrangement As a balloon in a sealed, constant volume container is pulled downward, the pressure in the balloon remains nominally constant and the pressure of the container drops [as] long as the pressure at the top of the container remains above vapor pressure, is this correct by you, Q Goest?

Yes, that's correct. There are other phenomena that will affect this however, such as disolved gas and the liquid's bulk modulus, but what you're saying here is essentially correct.

 

[Q_Goest]

It might be more intuitive to consider the following. You have a large rigid container 32 feet deep. It's filled with pure water and caped, without any air, at one atmosphere of pressure. Someone kicks the container loosening a bubble loose from a crack in the bottom. It rises, but doesn't significantly expand. The pressure rises to twice atmospheric pressure to keep the bubble at it's original volume. I find this scenario more fathomable.

Very nice! You're quite right, the bubble at the bottom of an incompressible liquid inside an infinitely rigid container rises in that container at a constant volume. Regardless of how high the container is (it can be infinitely tall) then assuming there's no vacuum space at the top of the container, the bubble will not increase in volume as it rises.

 

[Phrak]

Very nice! You're quite right, the bubble at the bottom of an incompressible liquid inside an infinitely rigid container rises in that container at a constant volume. Regardless of how high the container is (it can be infinitely tall) then assuming there's no vacuum space at the top of the container, the bubble will not increase in volume as it rises.

Somebody oughta tell Cyrus. He left on an early flight.

 

[Dadface]

Yes, that's correct. There are other phenomena that will affect this however, such as disolved gas and the liquid's bulk modulus, but what you're saying here is essentially correct.

So let us now consider the dissolved gases, the bulk modulus, the hysteresis properties of the rubber,the speed of descent, etc etc.Just kidding.

 

[pallidin]

So, a bowling ball dropped into a swimming pool will cause the water level to rise.

However, is the water-level rise slightly less when the bowling-ball is just 1-ft underneath the surface versus the ball being at, say, 12-ft underneath the surface?

 

[physical1]

I would start out with a system under 34 feet so I can build something on my stairways in my first home I just got (mortgage free, baby!). Well I may be able to make a system over 34 feet by going out windows with some tubes or pipes.

It is interesting to consider systems 500, 1000, or even 10,000 metres high too, even if the experiment can not be done right away - it would be nice to predict what would happen in those cases.

If water started boiling then it will also take up lots of space - causing... the balloon to collapse even more, and maybe even causing a sterling engine to spin since water needs to extract something in order to boil and gain that molecular movement...

Ultimately what would be interesting is if someone, maybe even myself, could invent a device based on this experiment.

I suspect I was having troubles getting water to boil in a syringe due to lacking a trigger air bubble or a mineral/impurity. I have now read that to get water boiling in a syringe, one needs to snap back the syringe and let things move around a bit, or let a dissolved air bubble pop out of place. Maybe water can therefore be "super steam" before it boils, similar how water can be super cooled and not turn into ice!

I got reading about how water surface actually has a sort of vapor existing above it (water liquid is therefore not just a liquid, to be anal retentive about it - it is a bit both a gas and a liquid in liquid state - therefore it is not a binary 0 or 1). Then I started looking into mercury barometers and they refreshed my memory that in fact even mercury has vapor sitting above it. I can see why some of the quacks are worried about mercury fillings in teeth so much.

I am going to try the balloon experiment with some simple materials. From the day I thought this up, I had planned to take a trip to the dollar store and purchase the materials required.

Some other simpler experiments I have done in the mean time:

1. fill up a 2 litre plastic p.e.t. pop bottle with water
2. squeeze pop bottle and let some water go out the top. Hold the squeeze still.
3. put hand palm tightly on the bottle opening, sealing it off
4. release the "squeeze"
5. bottle stays deformed in the squeezed state

Why wouldn't there be a vacuum space at the top of the container that is formed. Because the bottle "gives way" first before the vacuum space even has a chance to form. The vacuum is transmitted throughout the bottle, and takes the least path of resistance - which is holding the most flexible part of the bottle in deformed imploded state. If there was a balloon in the bottle, I suspect it would latch on to the balloon and pull it. Actually, this is all incorrect technically - since vacuum does not "suck". But you understand what I mean - suction is just an abstraction anyway. (Actually I could try and educate housewives that vacuums do not suck up dirt... forget it ;-) first I need to find one to prey on)

That is why I had trouble seeing how a balloon would just collapse under hydrostatic pressure in a sealed system - because the balloon can expand and contract according to how much vacuum is in the entire system. I think the balloon would more easily expand back to its original size than say a vacuum space being created of vapor/nothing. Why? For the same reason that a plastic pop bottle deforms under vacuum. So the balloon, to me just couldn't change size due to battling itself. I could be wrong and I am not happy until the experiments are completed!

What if we considered the balloon as "part of the container". I.e. consider it an "easily deformable portion of the container", similar to how a p.e.t. plastic bottle is deformable under vacuum. Different way of thinking about it. In the p.e.t pop bottle the container gives way fairly easily to any sort of vacuum. A balloon is even more sensitive and gives way easily.

Is it a matter of a fight for the path of least resistance. In a non rigid container, pop bottle "gives way" easier than water vaporizes/boils into a vacuum. In a rigid container a balloon trapped inside does what? Similar? Is the intra molecular bond strength quite strong, more strong than another path of lesser resistance (deforming something sitting in the container.. and if we can transmit throughout the container according to Pascal, what gives way first?)

Honestly, I am about to stop making any more hypothesis(es). I have had enough! I must experiment now. Experiments to me are a generally a waste of time. I would much prefer to have it done in the mind. But in this experiment the mind only seems to lie, cheat, and mislead. Begin the experiments ASAP.

I try to add some humor to my posts so I do not come off as a total jerk. Sorry for any flames/heat that this post has caused. Cheers to your p.e.t. pop and sparkling water bottles. Bonk!

 

[physical1]

A very interesting phenomena. It would be nice to see a simple experimental arrangement As a balloon in a sealed, constant volume container is pulled downward, the pressure in the balloon remains nominally constant and the pressure of the container drops.

If one places a turbine in the system, then he could extract power from the change of pressure? How does a conservation of energy apply? Something cools down? Temperature? Stealing energy from the atmosphere with the help of gravity, giving Tesla a big spin in his grave?

Disprove what I say - I encourage it.

It takes energy to push the balloon down - but sorry folks - I have a valve in mind. A valve can switch the hydrostatic pressure off and on, controlling the "height" of the system immediately. The balloon does not need to be drawn down with energy. It can be placed at a low level, and the water above it can be connected to this lower level container through a valve. The valve let's massive amounts of hydrostatic pressure be applied to the balloon immediately on an instant. To "recharge" the depressurized water, one might open the container to the atmosphere. The valve that connects to the lower portion of the system where the balloon is sitting is closed off before recharging. This isolates the balloon area from the recharging portion of the container: the upper water (hundreds of litres).

When recharging the low pressure water, the atmosphere pressure finds its way into a low pressure zone. Water and atmosphere find equilibrium. The upper portion container is sealed after recharging, and the valve is opened again to let the hydrostatics contact the balloon in the lower portion of the system.

The balloon is tied on a short string to a weight, to not float up, yet exposing its edges to hydrostatic pressure as much as possible. (does a balloon have edges? hehehe)

Above could be missing some important science that causes this system to fail, form icicles and frost, overheat, or simply "not work at all". On the other hand, refridgerators, heaters, pressure driven turbines, weather changing tools, and an endless amount of items for humanity could be created if this has any potential at all (a few puns on potential here).

Pressure is stolen from atmosphere potential? Any temperature changes or environmental effects?

Tesla where are you? Roll 3 times if you heard me, and get a few humans to prove me wrong please.

 

[Cyrus]

Somebody oughta tell Cyrus. He left on an early flight.

I'll read through Q_Goest's reply when I get some time later in the week or two.

 

[physical1]

What makes nature choose the container to become lower pressure, and not the balloon become higher pressure? Does pressure naturally want to decrease rather than increase?

For the balloon to become higher pressure in the constant volume, it would have to become hotter. Does nature not heat balloon because heat would have to be added by some external source, and cannot be just extracted spontaneously from the surrounding water? Does it not have any reason to want to heat up, or it simply can't due to thermodynamics laws?

The high pressure water molecules are ramming against the balloon which is transmitting through rubber, which rams against the air molecules. Are the air molecules affected at all by this ramming of water molecules against the balloon shell.

 

[EEstudentNAU]

If one places a turbine in the system, then he could extract power from the change of pressure? How does a conservation of energy apply? Something cools down? Temperature? Stealing energy from the atmosphere with the help of gravity, giving Tesla a big spin in his grave?

Disprove what I say - I encourage it.

It takes energy to push the balloon down - but sorry folks - I have a valve in mind. A valve can switch the hydrostatic pressure off and on, controlling the "height" of the system immediately. The balloon does not need to be drawn down with energy. It can be placed at a low level, and the water above it can be connected to this lower level container through a valve. The valve let's massive amounts of hydrostatic pressure be applied to the balloon immediately on an instant. To "recharge" the depressurized water, one might open the container to the atmosphere. The valve that connects to the lower portion of the system where the balloon is sitting is closed off before recharging. This isolates the balloon area from the recharging portion of the container: the upper water (hundreds of litres).

When recharging the low pressure water, the atmosphere pressure finds its way into a low pressure zone. Water and atmosphere find equilibrium. The upper portion container is sealed after recharging, and the valve is opened again to let the hydrostatics contact the balloon in the lower portion of the system.

The balloon is tied on a short string to a weight, to not float up, yet exposing its edges to hydrostatic pressure as much as possible. (does a balloon have edges? hehehe)

Above could be missing some important science that causes this system to fail, form icicles and frost, overheat, or simply "not work at all". On the other hand, refridgerators, heaters, pressure driven turbines, weather changing tools, and an endless amount of items for humanity could be created if this has any potential at all (a few puns on potential here).

Pressure is stolen from atmosphere potential? Any temperature changes or environmental effects?

Tesla where are you? Roll 3 times if you heard me, and get a few humans to prove me wrong please.

Could you draw a diagram of this?

 

[physical1]

Could you draw a diagram of this?

I will try make one soon, maybe even an animated drawing showing how a machine could work based on it. First I am performing some experiments to verify what actually happens in nature with the balloon. If I make a machine drawing up and it is all based on assumptions and bad science, it would be just crackpottery.

The experiments I did yesterday used a valve, custom connections, garden hose, and a p.e.t. 2L sparkling water bottle. Not the best setup since garden hoses and p.e.t. bottles are not so rigid. Still it brought back some results. Until I set up the rigid pipes though, I cannot say for certain what happens physically in nature. I will say, that it appears the balloon doesn't change size (significantly). The garden hose appeared to contract slightly and this threw everything off a bit. When I tried it in an open system situation the balloon contracted a massive and noticeable amount. With the closed system situation and a partly rigid container (garden hose and p.e.t. bottle), the balloon contracts a tiny amount and I think it is from the garden hose contracting.

If the balloon is not changing significantly in a partly rigid system (almost good enough) - then the container pressure must be decreased (similar to how water can hold static pressure, maybe in this case it holds negative static pressure below atmosphere), or the balloon temperature must increase in order to maintain equilibrium. Or, somehow the balloon just acts like a steel ball would (not affected by pressure) and I doubt this since the balloon air is still exposed via soft rubber surface. I have not measured temperature or exact pressure changes yet. Need more materials and time.

I would like to know why nature would choose one or the other and whether we can predict which one it usually or most definitely chooses. By one or the other, I mean either temperature of the balloon must increase to maintain constant volume and increase pressure, or the pressure of the container must drop. Is temperature increase in the balloon a no-no? Nature just wouldn't do that? Why? What law? What causes nature to choose one or the other and why do we know it is one and not the other? Some common behavior we know of?

One interesting tid bit to think about is - where would a "loss of container pressure" go to and come from? Was it transferred from the static pressure of water in the container (static pressure is 0, or atmosphere). And if so, where did this pressure disappear to? Would it be converted into molecular movements holding everything together in the container, a glue like strength? As the pressure of the container drops, the water molecules steal some "pressure" and convert it to intra molecular glue/quantum movements? If the pressure is stolen, it is no longer pressure, so pressure was converted again into WHAT exactly? Molecular stretching/binging/ramming/glue?

 

[Stan_Smith]

I do believe that waiting beside the 1000m tall container on Physical1’s workbench, is another can of worms. This particular can may be sprung as intense pressure causes condensation of the gaseous components of the air in the balloon.