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How does the addition of a weight change the speed of an air rising in water?

  1. Dec 23, 2011 #1
    Air rises in water at about one foot per second of time. If a container of air has a lifting force of 100 Lbs and rises at one foot per second what happens when a 10 lb weight is attached to it? What is its speed in the water then? What about 50 Lbs? or 90 Lbs? What is the speed reduction to lifting capacity ratio?

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  3. Dec 23, 2011 #2


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    andy1, Welcome to Physics Forums!

    When you say air rises in water are you referring to a bubble of air? If yes, then how could we attach some weight to it? Can you describe your apparatus in more detail? Thank you.
  4. Dec 24, 2011 #3
    I think you need to back up and re-examine your assumptions. When you say, "Air rises in water at about one foot per second of time," that may be a convenient rule of thumb for certain circumstances, but the real picture is much more complicated. If we are talking about bubbles of air in water, then the speed depends strongly on the size of the bubble, as well as many other factors. (And bear in mind that if we're talking about a significant depth of water, the size of the bubble will vary inversely with the depth.)

    Basically, the speed at which a bubble, or any buoyant object, rises depends on the relationship between the buoyant forces and the drag forces.

    It's simpler to first think about a fixed volume. Let's say you have a certain volume of air at a pressure of one atmosphere. If you enclose that in a rigid, closed container, say a metal cylinder, then its buoyancy will be relatively constant with depth. Its rate of rise will then be a function of its volume and shape (i.e., its buoyancy and how streamlined it is). Adding more non-buoyant weight to the cylinder without changing its outer dimensions will simply subtract from the buoyant force without changing the streamline characteristics, and will thus slow its rise. Also remember that drag increases with velocity, which is why a rising buoyant object will have a definite terminal velocity, just like an object falling through the atmosphere does. As the velocity increases the drag increases but the buoyancy remains constant, so at some velocity they are equal and the velocity will not increase further.

    Unconstrained bubbles are much more complicated. First off, as I mentioned above, with unconstrained bubbles the size increases as the depth decreases, and hence the buoyancy also increases. Also, very small bubbles tend to be spherical, but as they get larger their shape changes, which alters their drag characteristics. To make it even more complicated, you can think of a bubble not as a "thing" at all, but a pattern of liquid circulating around a (nearly) void space, and the flow patterns of the liquid at that interface surface can be very complicated, and this affects the rate of propagation of the void.

    So, there isn't some magic impetus that causes air to rise in water at about one foot per second. Their are buoyant forces and viscous drag forces and lots of complicated fluid dynamics. But in a very simple model, adding more weight simply subtracts from the buoyant force and slows the rise correspondingly. If the weight equals the buoyant force, this condition is called "neutral buoyancy", and the object will neither rise nor sink.

    One place to look for more detailed information (and a totally cool domain name) is http://www.bubbleology.com/Hydrodynamics.html [Broken]
    Last edited by a moderator: May 5, 2017
  5. Dec 24, 2011 #4
    Bobbywhy, The air is in a cylinder and does not expand. By itself it will rise in the water at 1 foot per second. It has the lifting power of 100 Lbs. Meaning that 100 Lbs of force/weight will stop its rising. Question is, how fast will this same cylinder rise in the water with a 10 Lb streamlined weight attached to it? 50 Lbs weight? 90 Lb weight?
  6. Dec 24, 2011 #5
    Greetings entropivore; The air is in a hard cylinder and by itself will rise at 1 foot per second. Its buoyancy is 100 Lbs. Question, what happens to it rate of rise if a 10 LB streamlined weight is attached to it? 50Lbs? 90Lbs?
    Last edited by a moderator: May 5, 2017
  7. Dec 24, 2011 #6
    Is the one foot per second the terminal velocity? Because the cylinder will accelerate until the force of gravity and drag equal the buoyancy force. If you increase the weight I would expect the terminal velocity to decrease. But to get actual numbers you have to do some non trivial calculations.
  8. Dec 24, 2011 #7


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    I don't know the math, but adding weight to it would reduce the speed of whatever was rising since it would reduce the buyouncy of the object overall. I expect it would be fairly simply to find out if you already know how to calculate buyouncy. Of course, this is assuming the weight is itself much more dense than water. I could easily add 10 pounds of air, but obviously that wouldn't do anything but help lol.
  9. Dec 25, 2011 #8
    You definitely need to know the density of the weight you're adding, like Drakkith said 10lbs of air won't slow you down much, what if you attached another canister filled with 10lbs of helium?
  10. Dec 27, 2011 #9
    The question as you've posed it is not specific enough to allow more than a vague answer (or a textbook-length tome covering all of the possibilities).

    Consider two cylinders of equal volume and weight, one long and thin and the other short and fat. They will have equal buoyancy. As they travel through water, the thin one will probably have less drag, and hence rise faster. I say "probably" because it also depends what the aspect ratio of the short one is. If you decrease the length and correspondingly increase the diameter, as it more closely approximates a disc at some point it will tend to rise "edge on", rather than along the cylinder axis, and in that configuration could have less drag than the long thin cylinder. Various intermediate shapes will likely be unstable and flutter or tumble as they rise, further slowing them. This, of course, is assuming free travel. If the cylinder were constrained to travel along a wire in a fixed attitude, for example, the result would be different.

    If you add mass to these, the result will depend on the size, weight, and shape of the added mass, as well as how and where it is attached. For example, if you add a very dense weight to the end of the long thin cylinder such that the effect is to slightly extend the cylinder, it will have only a small effect on the total drag coefficient. In that case, the added weight has little effect except to subtract from the original buoyancy. On the other hand, if you attach the weight to the centre of one end of the cylinder with a long (say, several times the length of the cylinder) thin wire, it will add its own drag coefficient to the equation.

    If you add the same amount of that material as a lump at the perimeter of one of the ends of the short fat cylinder, it may change the centre of buoyancy in such a way that the attitude of the cylinder is changed, drastically altering its drag (and perhaps its stability).

    Real-world complexities aside, the basic idea you are probably looking for is that the drag tends as the square of the velocity, and the terminal velocity is that at which the buoyancy and drag are equal. Adding weight (disregarding the drag and buoyancy of the weight itself) subtracts from the buoyancy.

    Fundamental principle of science #37: The quality of the answer received will never exceed the quality of the question asked.
  11. Dec 27, 2011 #10
    Entropivore; The question I am asking is just part of a bigger picture. I have a Pat. Pen. on a machine that is over 100 times more efficient than todays Hydroelectric power generating technology. Best part is that the Laws of thermodynamics don't always apply.
    Take a look at it and see.
    Yahoo content "Could this Green Invention Stop Global Warming".
  12. Dec 27, 2011 #11


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    Adrian F. Cerny (Andy 1): When you ask, as you did in your Original Post, “What is the speed reduction to lifting capacity ratio?” it was impossible for any forum reader to know the context or the arrangement of your proposed apparatus. So how could you have expected any reasonable and useful answer?

    It would have been less misleading if, in your Original Post, you had mentioned that you had a patent pending for a “Pneumatic-Electric Power Generating System” and referred us to the site http://voices.yahoo.com/could-green-invention-stop-global-warming-10316831.html?cat=15. There we could have examined your proposed invention, analyzed your drawings, and perhaps contributed some useful suggestions. You were asked for details by at least two members but remained evasive and delayed informing us for three days that you have this proposal for an invention. Enough said about diversionary tactics.

    I visited the website and read your invention description. When I read your statement “Best part is that the Laws of thermodynamics don't always apply” I immediately became skeptical.

    In your description note about the airwheel throttle you give an exact example: “example: If two (2) cubic feet of air were injected into the beginning of the device in page 2, it would start out with 124 lbs. of lifting force and be "full" (1240 lbs) half way up.” Since it’s obvious you have already decided what those lifting forces are in your invention description, it is disingenuous to ask others to verify your work without providing the invention description with which to form reasonable responses.

    You wrote, in post # 4 on 24 December: “Bobbywhy, The air is in a cylinder and does not expand.” But in your invention description you write: “As this compressed air rises it expands due to the now lowering water pressure that surrounds it.” So which is it, Andy 1? Since the volume is fixed at all depths, i.e., rigid cylinder, then how can you expect its buoyant force to increase as depth decreases?

    Finally, if you are correct when you write “It is a slow RPM machine but it has the potential to POWER the WORLD!” then you may become famous and you might even receive the Nobel Prize for your discovery. Otherwise, be ready to be relegated to the dustbin of delusional pseudo-science.
  13. Dec 28, 2011 #12
    Ok, now I feel stupid. I guess I should have seen it coming. It's the Prometheus delusion again.

    Look. That's not the way it works. Thermodynamics is not a Greek god that can be fooled by a few more layers of obfuscation, or a bit of mechanical slight-of-hand. Nor is it a conspiracy by Capital-S-Scientists to keep the little guy down and all the grant money for themselves. It's just a consequence of the way the universe is put together.

    Certainly there are new things to be learned about the structure of matter and energy, and our understanding of the structure the universe keeps evolving. Discoveries at the quantum level do affect what people do at the macroscopic level. There will always be room for more efficient, more sophisticated machines, and machines that exploit hitherto untapped principles. But if you're looking for a way around the current understanding of thermodynamics, it's very unlikely that you're going to find it with a clever arrangement of gears and levers. That's very well trodden ground. The path of science is littered with the broken remains of perpetual motion machines of all sorts.

    In your particular case, if you had a free source of highly compressed gas (say, from a natural reservoir beneath the seabed, or the dissolved carbon dioxide in Lake Nyos), this would be one means of extracting energy from it. Whether it would be more efficient than other means, such as a turbine, is a valid and interesting question, and the answer probably depends on the available pressures and flow rates, among many other things. As you've presented it, you seem to be overlooking the fact that in order to compress the air to inject it at the bottom of the apparatus, you will have to put in at least as much energy as you are going to get out of it. The expansion of the air on the way up gives the superficial illusion of getting something for nothing. This is the common downfall of many such schemes, such as the "overbalanced wheel" class of perpetual motion machines. Our intuitive sense of the geometry of the situation distracts us from a true understanding of the physics involved.

    For much more on this topic, and some entertaining contraptions, see the Museum of Unworkable Devices web site: http://www.lhup.edu/~dsimanek/museum/unwork.htm

    Finally, if you are convinced that it will work, by all means, go ahead and build it. Demonstrate it to the world. But beware. If it works, you may very well find yourself chained to a rock with your liver being eaten out by an eagle.
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