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Forced vortex

  1. Sep 19, 2010 #1
    Earth's atmosphere is a forced vortex? I guess yes, because an airplane moving to the west does not confront different air resistance than one moving to the east. Correct me if wrong, because I have some doubts: Up to what height? Even the exosphere?

    Suppose you create a forced vortex in a closed box-cylinder full of air or full of liquid. Also, there is no extrernal gravity acting on this box (i.e. the box is not pulled towards the ground). Where will the more dense (than the rest of the air or liquid) bodies go? Towards the centre of the vortex? Outwards? Remain at the same distance from the centre of the vortex?
    Last edited: Sep 20, 2010
  2. jcsd
  3. Sep 19, 2010 #2


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    Luckis11, please don't take this the wrong way. I mean no insult, but the fact is that I absolutely cannot make any sense of your post. That might be just a shortcoming on my part. Can you perhaps rephrase your thoughts?
  4. Sep 19, 2010 #3


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    Staff: Mentor

    Earth's winds go in circles due to the coriolis effect.

    In a vortex, heavier objects go to the outside.
  5. Sep 19, 2010 #4


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    About 30 degrees north of the equator, and at sufficient altitude, there are jet streams that flow from west to east that can flow up to 100 mph or more. At lower latitudes there are trade winds that generally flow from east to west, and near the north pole, there is also a east to west flow.

  6. Sep 20, 2010 #5


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    I stand corrected.
  7. Sep 20, 2010 #6
    Wikipedia at "secondary flow" says that "heavier particles gather at the centre"..."because the linear tangent speeds increase as the distance from the centre of the vortex decreases, thus the Bernoulli principle says there is a pressure towards the centre". I doubt whether the reasoning is correct, but surely they gather at the centre for some reason.
    However, my example is not such a vortex. The linear tangent speeds decrease as the distance from the centre decreases, because the angular speed remains the same because it is a forced vortex. It is a CLOSED box of a cylinder shape, FULL of liquid or gas. ALSO, there is no gravity acting on the box (but to test this experimentally one cannot get rid of gravity). Also, don't tell me that "it is impossible for such a vortex to happen to be forced, because by spinning the box the liquid or gas will tend to rotate faster closer to the centre or faster (than equal angular speed everywhere) further from the centre", because by supposition somehow this will happen at the box, thus you can replace the walls of the box with outside pressure or whatever. By supposition, because I want to see where the denser bodies tend to go when this happens.
    Last edited: Sep 21, 2010
  8. Dec 6, 2011 #7
    Hello ,

    I need to know the pressure distribution in a forced vortex. Can you please tell me how to get an exact solution for pressure distribution in a forced vortex.

    Thanks in advance.
  9. Dec 6, 2011 #8
    The linear tangent speeds decrease as the distance from the centre decreases, because the angular speed remains the same because it is a forced vortex.http://www.uklv.info/g.php [Broken]
    Last edited by a moderator: May 5, 2017
  10. Dec 6, 2011 #9
    For the forced vortex the OP is talking about the fluid rotates as a solid body.


    where w = angular velocity

    The pressure gradient in the flow points inwards because the pressure force balances the centrifugal forces.

    dp/dr = rho*v^2/r = rho*w^2*r which can be easily integrated to get the pressure difference between two points.

    With that said I will explain what I think happens to particles placed in this flow.

    If you place an object in the flow, so that it is moving at the same velocity of the fluid at that point, the direction the object travels in will depend on its density relative to the fluid. For a denser particle the centrifugal force will be greater on the particle then on the fluid. Therefore the centrifugal force on the particle (pointing out) will be greater than the pressure force (pointing in). So the denser particle travels out. For the lighter particle the centrifugal force will be less so the pressure will force the object to the center.

    The Wikipedia article mentions that heavier particles travel to the center. However they are talking about a different situation. Their example is the flow in a cup that has been stirred (such as tea) in this case there will be a boundary layer along the bottom. In this boundary layer the velocity is much lower so the centrifugal force is much less but the pressure gradient is the same as outside of the boundary layer where it balances the centrifugal force. So the fluid and particles travel to the center in the boundary layer. In the example the OP describes (the flow in the rotating cylinder) there will be no boundary layer after an initial start up time as all of the fluid will be rotating with the cylinder.
  11. Dec 6, 2011 #10

    Thank you so much for your quick reply. I understand that integrating between two points can give the pressure difference between two points. Now, actually I am thinking of the surface of the forced vortex. On the concave surface of the forced vortex, at every point the pressure is going to be atmospheric pressure. When I integrate as you suggested, I find the following,

    p2 - p1 = (ro)* (w^2)/2 * (r2^2 -r1^2)

    Now from the above equation, how can I explain that on the free surface of the forced vortex, the pressure is atmospheric (Patm).

    Thank you again in advance.
  12. Dec 6, 2011 #11
    Actually now that I think about it, what I said above is not true in your case of zero gravity if your fluid is a liquid. In zero gravity the liquid would take the form of a sphere as opposed to the the shape of the container as it does on earth. So in zero gravity this experiment probably will not work.

    But if there is gravity we can perform the experiment and the free surface, where the pressure is equal to Patm, will take the form of a parabola. So in this case the equation I had for pressure is not correct because you have to take gravity into account.

    dp/dr = rho*w^2*r
    dp/dz = -rho*g (g = gravitational constant)

    Integrating the first equation

    P = 1/2*rho*w^2*r^2 + C(z) (C1(z) = some unknown function of z)

    Integrating the second equation

    P = -rho*g*z + C2(r)

    Comparing these equations we can see

    P = 1/2*rho*w^2*r^2 - rho*g*z+Constant
  13. Dec 6, 2011 #12

    Thank you again so much for your kind reply. I have also arrived to the final form of equation earlier which is a general solution which has a constant. Now, how do I find an exact solution of this equation, I mean, how can I define the constant, which will tell that, at any point on the free surface , we can have atmospheric pressure.

    Thank you for your time and explanation.
  14. Dec 6, 2011 #13
    You get the constant the same you get any constant when solving differential equations.

    Use the boundary conditions!
  15. Dec 6, 2011 #14
    Hi, I am at a loss about getting this BC. if my coordinate is at the bottom of a tank which has some liquid in it and and I know that at free surface the pressure everywhere will be atmospheric, then how do I get the constant. I have trouble putting the pieces together. thanks in advance.
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