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Not so simple buoyant forces in bubbling water

  1. Oct 7, 2006 #1

    OlderDan

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    Many of us have probably seen the TV documentary offering the theory that ships at sea in places like the Bermuda triangle might be sinking because methane gas released at the sea floor creates bubbles that lower the density of the water, resulting in reduced buoyancy. The theoretical foundation for this is Archimedes principle; the buoyant force is equal to the weight of the supporting fluid that is displaced. If the water is filled with bubbles, which weigh essentially nothing compared to water, the vessel can no longer displace its own weight in the bubbly (less dense) fluid.

    The piece documents a couple of experiments performed to prove the theory. One involves a small, open, pleasure-craft loaded with bricks floating in open water above a large grid of pipes pumping air into the water to lower the density. A second experiment involves a model ship in a large pool with a similar grid of bubble generators. The model used has open hatch covers that permit the vessel to fill with water if it turns on its side, or water flows over it.

    I found both of these experiments rather unsatisfying, although the second was more appealing than the first. The small pleasure-craft would not sink when it was surrounded by bubbling water, but it eventually sank when the bow was in more stable water and the open stern was in bubbling water, gradually being filled as water splashed into the boat. There was a noticeable lowering of the back of the boat relative to the waterline that the authors attributed to the lower water density, an effect that became more pronounced as the boat filled with water.

    The model ship in the pool seemed quite buoyant in spite of the bubbling water until the churning water found its way into the hold. An effective method of sinking the model was the release of a giant bubble directly beneath the ship, effectively creating a "hole" in the water directly beneath the boat for it to fall into and then be overflowed by the mounds of water created around the edges of the bubble.

    I have no doubt that a giant bubble could engulf a ship, or that a shock wave from an explosion used to create such a bubble could blow a ship apart, or that a ship might suffer structural damage if supported only at both ends if a huge bubble opened beneath its midsection. And I don’t doubt that a very high concentration of bubbles in a fluid will result in reduced buoyancy. But I question the applicability of Archimedes principle to any of these situations. I’d like to hear what other people think about this.

    What causes an object to float?

    Fundamentally, it is not the displaced liquid that creates the buoyant force. The displaced liquid is not in contact with the object, and there are no long-range forces like gravity involved that enable the displaced liquid to exert a force on the object. The force that supports the weight of the floating object comes from the collisions of water molecules with the submerged portion of the floating object. On the average, these forces are described as the pressure exerted on the object's surfaces by the fluid. Fundamentally, it is because the pressure at depth in a fluid is greater than the pressure above that a buoyant force exists. If these pressure differences did not exist, there would be no buoyancy. It happens that in a fluid of uniform density the pressure variation with depth depends only on the weight of the fluid in a column between any two depths. The magic of Archimedes principle is that even for complex shapes, the net upward force from the fluid pressure is equal to the weight of the fluid displaced by the object.

    But what if the fluid is not homogeneous?

    A thought experiment: Consider a large rectangular tank like the one used in the second experiment in the TV show. Before filling the tank with water, blow up some balloons- lots of balloons- enough balloons to nearly fill the tank. These are special balloons that have no elasticity, so the pressure inside the balloon is the same as the pressure outside. They are not very large balloons, but there are a lot of them. All these balloons are tied with strings to the bottom of the tank in such a way that when the tank is filled with water the balloons will be rather uniformly distributed throughout the liquid. The ones near the bottom of the tank will be smaller than the ones above because the water pressure will compress them. Assume for the sake of discussion that when all is said and done, the volume of water is ½ the volume of the tank, so the average density of the water/balloon combination is only half that of water.

    Two questions:

    How does the water pressure vary with depth?

    Will a block of ice float in this pool?

    I’ll continue this later- Responses to these questions are welcome.
     
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  3. Oct 7, 2006 #2

    turbo

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    I used to white-water kayak in my younger days, and was one of the first in this area to own a Perception Sabre (a very low-volume boat for it's day). I can vouch for the fact that very bubbly foamy whitewater is less bouyant than plain water, and there were times when I was negotiating heavy rapids with only my torso above water - the boat was entirely submerged. I had previously owned a Taurus, and the low-volume boat was a whole lot more challenging and FUN.
     
  4. Oct 7, 2006 #3

    arildno

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    For a fluid at (macroscopic) rest, any portion of the fluid is at rest as well, hence the force acting upon it from the rest of the fluid must balance the portion's weight. Otherwise, the portion would sink or float, contrary to the assumption that the fluid is at rest (and there are no other forces at work than pressure&gravity).

    A fluid doesn't disinguish between a fluid portion enclosed in itself, or something of exactly the same shape as the fluid portion substituted at the fluid portion's position. Thus, the net force acting upon that object from the fluid would be the same as if we substituted back the "displaced" fluid portion.

    Nowhere here is any use made of uniform density, so this argument is equally valid for a fluid at rest with varying density.

    What IS different between a resting fluid with uniform density and one with non-uniform density is that the pressure distribution will no longer be strictly linear in the vertical in the case of the non-uniform case.
     
  5. Oct 7, 2006 #4

    OlderDan

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    My thought experiment does not require strictly uniform density. I should have allowed for a density gradient, but that is an unnecessary complication. The density of air above the earth's surface is clearly not uniform, but over the distances we are concerned with for floating balloons, the density variation is negligible. The issue at hand is the effect of the regions of drastically different density (bubbles) distributed throughout an otherwise uniform fluid with at most a gradual density gradient.

    In other words, the pressure gradient within the pool is exactly the same with the balloons as it would be without them?
     
  6. Oct 7, 2006 #5

    arildno

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    In the case of stationary balloons, then it would be the same.

    But the problem with highly bubbly water, with bubbles forming and deforming, you actually have a multiple phase situation, and I'm not at all certain that the assumption of a fluid at "rest" is viable here. Certainly the region of the gaseous phase is definitely moving about.

    Thus, the actual pressure gradient in this case might well differ from that one.phase situation.
     
  7. Oct 7, 2006 #6

    arildno

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    Ok, I'll retract a bit, since you said you wanted to nearly fill the pool with balloons (I was thinking of them relatively dispersed).
    I'll think about it some more.
     
  8. Oct 7, 2006 #7

    rcgldr

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    Although the pressure is the same, the density of the air / gas in the water is less than the density of the water. This is why the bubbles are rising, and in the case of the balloons, why they have to be tied down in order to keep from rising to the top of the water.

    > How does the water pressure vary with depth?

    The balloons can be ignored as long as they don't form a boundary completely seperating depths of water. As long as there is any vertical path connecting the water from top to bottom of the pool, the pressure increase versus depth remains the same.

    Note that compressablity of a liquid or gas affects the rate of pressure increase versus depth.

    > Will a block of ice float in this pool?

    Yes, but it could be different, depending on how many balloons the block of ice is resting on.
     
  9. Oct 8, 2006 #8

    OlderDan

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    Yes, I did say that. But I don't want the balloons to be touching. The 50% average density number I offered is approaching the most air volume you could have and still maintain clear separation of the balloons.
    (Hexagonal close packed, or face centered cubic structures result in 74% of space being occupied by hard spheres. For a simple cubic structure, the spheres occupy only about 52%. Random packing experiments with steel balls have yielded about 64% occupancy http://www.tiem.utk.edu/~gross/bioed/webmodules/spherepacking.htm )

    Fewer ballons with wider separation would be fine. I just want enough to lower the average density of the pool to the point where if the water/balloon mixture were replaced by a homogeneous fluid having that reduced density, the ice cube (or some other object that normally floats in water) would surely sink.
     
  10. Oct 8, 2006 #9

    OlderDan

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    I agree with all you say. After reading your hidden post on the "simple buoyancy" thread, I'll let you answer this follow-on question

    How can the pressure at the bottom of the pool be unchanged if half (or some lesser fraction) of the water in the pool has been replaced by air, reducing the weight of the water to half its original value?
     
  11. Oct 8, 2006 #10
    Static pressure at any point is simply equal to the weigh of water above it. The weight of water is not reduced because there is at least one column of water through the spaces between the balloons that goes to the free surface.
     
  12. Oct 8, 2006 #11

    OlderDan

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    That's not quite the way I would have worded it, but I think all of us are agreed that the pressure at any depth in the pool depends only on the vertical distance to the free surface, regardless of how many balloons are scattered below the surface. The Pascal's Vases demonstration illustrates this phenomenon. http://www.pha.jhu.edu/dept/lecdemo/F-b2a.html Different shaped vessels have different volumes (and hence different weights) of water, but the bottom of the vessel only has to support the weight of water that is not being supported by the non-vertical walls of the vessel.

    In my imaginary pool, the tethered balloons are all experiencing an upward force equal to the weight of water they displace. This buoyant force is counteracted by the tension in the tether, keeping the balloons in place. By Newton III, these balloons are pushing the water downward with a force equal in magnitude to the buoyant force, i.e., the weight of the water they have displaced. At any depth, the water below has to support the weight of the water above, plus the added force from the tethered balloons, which is equal to the weight of the missing water. So there is no effect on static pressure from all of these balloons. A block of ice at the surface will float just as well above a pool full of water and tethered balloons, as it will above a pool filled with water.

    Now, if the balloons are all suddenly released, we know they will rise to the surface. In reality, things get turbulent and complicated, but suppose the process is not chaotic and the balloons rise like the tiny bubbles in a glass of brew. Each balloon is pushed to the surface by the buoyant force, but as long as a balloon is submerged it will still be applying the downward force on the water and the pressure gradient will be maintained. If we could constantly replace the rising balloons with more at the bottom of the tank, the water level would not sink, and neither would the block of ice.

    The balloons are only in the thought experiment to maintain bubble separation. If the balloons are replaced by a steady stream of bubbles that form at the bottom of the tank and rise smoothly and vertically to the surface, the only bubbles the block of ice will ever sense are the ones that hit it on the bottom. Any bubbles temporarily stuck beneath the ice effectively increase the volume and reduce the density of the floating "ice + attached bubbles" system. If anything, these attached bubbles will increase the buoyant force and raise the ice higher in the water.

    Bottom line here is I see no way that you can sink a floating object merely by creating a steady state of bubbly water with an average density that is a bit less than the minimum density of a homogeneous static fluid needed to float the object. For a block of ice, a 10% reduction in density should sink it. I don't doubt that if you turn the whole pool into froth the ice could sink, but I don't think just enough uniformly distributed bubbles to reduce the average density by 10% would have any noticeable effect on the buoyancy of an ice block.

    In open water, a localized source of bubbles will reduce the average density in some region. If you could somehow define a boundary for this region, the region should float above the surrounding water. You should see mountains of "light water" floating on the normal water. But of course being a fluid the light water would always be flowing off the top and somehow finding its way back to the bulk water below, absent the air that was making it light on the way up. Surely there could be strong currents formed, and maybe some vertical currents could be strong enough to tug down one end of a boat, and fill it with water if it was left open, but this is a much more complicated process then the notion of reducing buoyancy by merely poking a bunch of holes in the supporting fluid to reduce its density. If you could poke those holes without creating all the turmoil, things at the surface should float just like they do on homogeneous water.
     
  13. Oct 8, 2006 #12

    rcgldr

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    Because the pressure increase versus depth is related to density, not weight. No matter how small the quantiy of water there is, the density is basically constant (water doesn't compress easily), so the pressure versus depth remains the same. Even if the fluid or gas is compressable, the density and pressure will increase with depth consistently, regardless of the shape of the vessel holding the fluid or gas (this was mentioned already).
     
    Last edited: Oct 8, 2006
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