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Fancy a cuppa?

  1. Jul 24, 2008 #1
    I hope someone can help with this - its been driving me mad for a while now!

    You will have noticed the shape coffee (or whatever liquid) takes when you stir it. Clearly a vortex getting deeper and steeper the faster it is stirred.

    So - here's my problem. What's the equation that defines this shape? Or, more simply, if you take a vertical slice through the centre of the vortex, what is the equation for the curve defined by the surface of the liquid? In particular, how does density, stirring speed/angular velocity and depth of liquid affect the shape? To make things simple, take a cylindrical mug.

    I have thought about this problem and an approach to solving it by thinking that the liquid will take a shape that minimises its total energy (kinetic and potential). Then work out the total energy of the liquid, differentiate the resulting equation to find a minimum, and see where that gets me. Don't know if that's a good way to think about it.

    Any suggestions and/or solutions very welcome.
     
  2. jcsd
  3. Jul 24, 2008 #2

    Ben Niehoff

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    This is a good way to think about it. You will have to use calculus of variations. If you make some simplifying assumptions, you should be able to work out the solution without too much difficulty. You might try assuming that the fluid rotates as in uniform circular motion; in reality, this is not true, because the parts closer to the center rotate faster. But it should give you a good approximation for low angular speeds. To do better, you would have to take viscosity effects into account and find the variation in angular speed with distance from the center.

    For low speeds and high viscosity (corresponding to the simplifying assumption noted above), I think you should get a parabola. This is how parabolic mirrors are made for large telescopes: by spinning a cylinder of molten glass.
     
  4. Jul 24, 2008 #3

    Andy Resnick

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    The velocity field is known as a vortex, and the velocity as a function of 'r' goes as 1/r, leading to an infinite velocity as r goes to zero (the center of the vortex). The interface goes as 1/r^2.

    http://en.wikipedia.org/wiki/Vortex

    Detailed solutions of fluid flow problems involving a free surface are notoriously difficult, I don't know if a parametric study of your type of problem has been done.

    Note that this is a different flow field than occurs if the cup is set into rotation, dragging fluid with it- the free surface is then a parabola.
     
  5. Jul 25, 2008 #4
    Andy, not sure what you mean by "interface goes as 1/r^2". Which interface are you referring to? But thanks for the response. I did not think in terms of a sophisticated fluid flow model for this - this is just an interesting problem I have thought about and to which I would want to apply some basic mechanics and calculus.
     
  6. Jul 25, 2008 #5
    Ben - agree about the simplifying assumptions. Will need to brush up on the calculus of variations topic though! Do you have any good references to the technique? I understand the basics but have never applied the tool in anger. Also, not sure about the parabola. If you go and stir some coffee when next you have some, the free surface looks as though it goes something like sqrt(mod(x)).
     
  7. Jul 25, 2008 #6

    Andy Resnick

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    The shape of the air-coffee interface.
     
  8. Jul 25, 2008 #7
    Surely the interface cannot be 1/r^2 - as we move towards the centre of the vortex (where r=0) we have an increasing function describing the height of the liquid. ?????
     
  9. Jul 25, 2008 #8

    Borek

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    Make it -1/r^2 then :wink:
     
  10. Jul 25, 2008 #9
    mmmmmm! Me being a bit thick!:rolleyes:
     
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