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Drag Force at Liquid-Gas Interface

  1. May 28, 2009 #1

    mrjeffy321

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    I am interested in calculating the drag force acting on simple objects (cube and sphere) as they passes from a liquid (water) into a gas (air). I can calculate the drag force as the objects travel through a uniform medium, but I am wondering how the drag force on an object changes as it passes through the liquid-gas interface.
     
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  3. May 28, 2009 #2
    Your question doesnt make any sense. If a cube falls from atmosphere into a body of water, that impact force isnt what we call drag (It's an impulsive force). Then it will sink (if the density is higher than water) and experience a net drag as it falls in the water. Drag occurs before and after impact, not during.
     
  4. May 28, 2009 #3

    mrjeffy321

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    The situation I am asking about is the other way around, a body going from water into air, not falling from the air and splashing into the water.

    For example: Lets say one has a basket ball which one holds under water. After letting the ball go, it will rise to the surface and eventually cross the water-air boundary. As the ball crosses this boundary, where part of the ball is sticking up out of the water and part of it is still submerged, what drag force is acting on it?
    Perhaps the situation is not as simple as I would like it to be for my purposes, as I would imagine that that there is still a “splashing”, of sorts, as the ball pops out of the water.

    What about if the medium which the object is moving into is not a gas but just another liquid? For example from water into a thick layer of oil?
     
  5. May 28, 2009 #4
    What does it matter, it will last so short a time you can just ignore it.

    You need to stop thinking like a physicist and start thinking like an engineer. When its at that point, what do you think is the dominant resistance force? I'd say its the water. In fact, the drag from the water is going to be orders of magnitude larger than the drag from the air (Just take the ratio of the densities). So, just assume its in water (worst case) until its completely out of the water OR, use the drag coefficient with an average density, and weight the density based on the ratio of the exposed surface to air/water.

    Stop making this too complicated. :wink:

    0901_012502.jpg

    http://www.royalnavy.mod.uk/upload/img_400/tomahawk.jpg [Broken]

    I would say the "splashing" you see is after the missile leaves the water and the water falls into itself to fill the void left behind. I.e, that shouldnt change your drag - I wouldnt suspect it to.
     
    Last edited by a moderator: May 4, 2017
  6. May 28, 2009 #5

    mrjeffy321

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    In the case of small / short objects, the time it spends at the liquid-gas interface is, indeed, small.

    However, I am also interested in the case of “big” objects, such as a really long cylinder (or missile, as in the case of your picture) where the time it spends at the interface is considerably longer, ignoring any rocket thrusters of course.

    I can, as you suggest, simplify the problem by making a high-low estimate of what the drag force will be using the assumption that the object is still fully in, or out of, the water as it passes the interface.
    As a physicists who rarely complicates things by taking into account drag, when I saw a discontinuity in my medium I thought, “hey, this might be an issue those engineers have to deal with sometimes”, but it would seem that it is me who is adding unnecessary complications.
     
  7. May 28, 2009 #6
    I would strongly try using weighted density as a more 'educated' guess, as in:

    [tex]\sigma_{air}C_{D_{air}}\rho_{air} + \sigma_{water}C_{D_{water}}\rho_{water}[/tex]

    Where the sigmas are weighting factors that add up to 1. Each value of sigma is the exposed percentage of the body in each fluid medium. The pair of sigmas will start as (0,1) go to (.5,.5) and end at (1,0). It will make a quarter circle arc in the first quadrant.
     
    Last edited: May 28, 2009
  8. May 28, 2009 #7

    FredGarvin

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    The first thing I would ask myself is "is the time that there is a crossing between media really an important analysis?"

    What do you hope to learn between the two extreme states of fully in the water and fully out of the water? If it were something along the lines of a control issue for a seal launched missile, then I would say it's important. If it is for something along the lines of looking at a power requirement for propulsion, then I'd say it's not worth the time.

    Either way, for someone that has the time, it would be a cool problem to work.

    My first pass would be a weighted average like Cyrus mentioned. I am sure there are other ways to make this much more complicated. I would expect the real experts would do this numerically over very short time spans.
     
  9. May 28, 2009 #8

    mrjeffy321

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    The problem I first started considering was similar to the example I gave above, where a basketball is held underwater and then released. While performing this 'experiment' the other day, it seemed like the basketball was not popping out of the water as far I might have expected from the amount of force required to hold it down , and the depth of release. I attributed this to (what I later calculated to be) a low terminal velocity of the basketball in the water, meaning that when it popped up at the surface it was not going very fast and could thus not travel very high into the air.

    But this got me thinking about other objects and more general situations, so I decided that I wanted to try to describe the position of the object as a function of time after release from a given depth. This is relatively simple when the object is fully in the water or fully in the air, even taking into account drag, but since I know virtually nothing fluid dynamics I was unsure if I was justified in taking a weighted average for the drag force as the objected crossed the liquid-gas interface.
     
  10. May 28, 2009 #9
    Well, not you're going to have to include a bouyancy force that changes as it emerges from the water.

    You really are trying to make your life miserable for no reason, lol. This solution is going to require some numerical integration methods. I don't see why you want to do all this work for little to no gain in knowing the answer.
     
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