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Air resistance on rotating cylinder

  1. Sep 1, 2009 #1

    I am interested in the problem of air resistance on a cylinder rotating about its axis (axis of rotation is through the flat parts).
    I used the drag equation and substituted the "cross-sectional" area as the area that's rotating at R (if R is the radius of the cylinder) so it would be the circumference multiplied by the height of the cylinder.
    [tex]\ {F}_d= -{1 \over 2} \rho v^2 A C_d [/tex]
    where [tex]\ A=2 \pi R h[/tex]
    R: Radius of Cylinder
    h: Height of Cylinder
    p: density of air
    v: velocity of that which is moving through the air
    A: cross-sectional area
    Cd: Coefficient of drag

    So I assumed this is the equation that would describe the system, but I am thinking I am wrong here because the cylinder isn't really translating through the air, it seems more like a shear force acting on it. Any ideas?

    Edit: Does anyone know of an equation that describes the system I am talking about?
    Last edited: Sep 1, 2009
  2. jcsd
  3. Sep 1, 2009 #2
    I do not believe the normal Cd applies to a cylinder rotating about its axis of symmetry. Your drag equation applies to turbulent drag, which may not be appropriate here either.
    [Edit] The following paper reviews wind tunnel tests on a 66-cm diameter cylinder rotating at 1600 RPM in 56 m/sec wind (open access)
    http://www.scribd.com/doc/5208081/CFD-Modeling-Of-Wind-Tunnel-Flow-Over-A-Rotating-Cylinder [Broken]
    The Reynolds number for this situation is over 10^6, so it is very turbulent. The drag coefficient (which may be for a non-rotating cylinder) is 0.6 to 0.68. In this case, the area A would be 2R/h, not 2 pi R/h.
    Last edited by a moderator: May 4, 2017
  4. Sep 3, 2009 #3
    Thank you for sending me that article.
    The problem in the article is regarding a rotating cylinder in a wind tunnel.
    I am merely considering a rotating cylinder where there is insignificant air velocity.

    I have discarded the drag equation for the cylinder because I have realized that there is no "pushing" against the air, rather surface boundary interactions. Now I am not really familiar with shear stress calculations but I have tried something which goes like this:

    Wall Shear Stress: [tex] \tau_w = \mu \frac{du}{dy} \right|_{y=R} [/tex]
    where, [tex]\mu[/tex] is the dynamic viscosity, [tex]u[/tex] is the flow velocity parallell to the wall and [tex]y[/tex] is the distance to the wall.
    So I looked at it kind of backwards, rather than the fluid is moving, the wall is moving, so the distance to the wall is from r=0 to r=R (radius of cylinder) and the flow velocity would be the tangential velocity at r=R. So we have [tex]v_t = w*r[/tex] and a derivative with respect to r would give: [tex]\frac{du}{dy} = w[/tex]
    Now we have: [tex] \tau_w = \mu w[/tex]
    The force is related by: [tex]F = \tau_w A[/tex], where A is the cross-sectional area
    So the cross sectional area involved of the cylinder would be the the top, bottom, and the sides. (see attachment picture)
    For top and bottom it would be: [tex]2 \pi R^2[/tex]
    For the side it would be: [tex]2 \pi R h[/tex]
    so the final force equation boils down to: [tex] F = 2 \pi R \mu w(h+R)[/tex]
    This is great because [tex] \mu [/tex] is known.

    Now it seems kind of strange to me that the shear stress is the same for all points of the cylinder even where the tangential velocity of the surface is lower. So far the results of this equation compared to the previous one give me higher terminal velocities (2x, up to 10x) - which is what I would expect. Can anyone tell me if I am wrong/right?

    Attached Files:

  5. Apr 26, 2010 #4

    I don't know if you are still reading this but is you solution worked ?
    I have a similar problem to solve and I pretty agree wit what you are saying.
  6. Apr 26, 2010 #5
    This looks like it only works for t=0 when the cylinder is not moving. A time dependent equation would have a gaussian function as the friction is reduced over time.
  7. Apr 27, 2010 #6
    I don't know if this solution works or not because I would have to experimentally verify it. I still don't know if I derived a correct theoretical model for this case. I agree with LostConjugate that the friction will alter (i.e. decrease) with time as the air currents equilibrate with the rotating cylinder but that is taking the model a step further. I still don't have the basic model that I can trust.
  8. Apr 28, 2010 #7
    You are correct that the drag equation is really derived for the different case of flow going past a body, not for a rotating body. In your next post, you were on the right track I think when you mention the wall shear force of

    [tex] \tau_w = \mu \frac{d u}{dy}. [/tex]

    In polar coordinates, the derivative transforms to

    [tex] \tau_w = \mu \frac{ d u_\theta }{dr}. [/tex]

    The flow field between two rotating cylinders can be solved for exactly in the fluid equations. I've got a book by White, "Viscous fluid flow", page 103, which gives a good reference. This analysis assumes the cylinders are infinitely long and the system is in steady state, i.e., it has already started spinning and now nothing is changing with time. In the limit that the outside cylinder goes to infinite radius and zero angular velocity, it gives the proper velocity that you're looking for. The solution from the book is just

    [tex] u_\theta = \frac{R^2 \omega}{r} [/tex].

    This expression provides

    [tex] \tau_w(R) = \mu R^2 \omega (-\frac{1}{R^2}) = - \mu \omega.[/tex]

    The minus sign just indicates the force is in the clockwise direction, assuming the cylinder rotated counterclockwise, which makes perfect sense of course. Then, integrating around the circumference, for the entire cylinder of length L and circumference 2 pi R, you just get

    [tex] F = \int - \mu \omega R d\theta dL= - 2 \pi \mu \omega R L .[/tex]

    Note -- sorry I made a mistake writing the last night when writing this. Just fixed it -- the answer actually comes out pretty similarly to your answer you already had.

    This analysis doesn't include the drag from the top and bottom of the cylinder, which would have to be calculated differently. You were right before to notice an intuitive issue due to the fact that the top and bottom of the cylinder have a velocity that varies with distance from the center axis. If the cylinder is long, then those sides shouldn't matter too much. Still, this is a good starting point, since you have an exact solution for this case.

    You might be able to account for the top and bottom by considering a different case with a rotating disk, and then adding those contributions to the current case.
    Last edited: Apr 28, 2010
  9. Apr 28, 2010 #8
    Normally in polar coordinates there is an Integral over a Radial function and an integral over the two angles inside this integral.

    Is there a radial function missing here?

    If you increase the radius and keep the period the same, the surface will be going faster which would create more drag, plus there is more surface area and the drag is not a conservative force.
  10. Apr 28, 2010 #9
    Sorry i had an error somewhere else that might have been throwing you off. I just fixed it and now it should be correct. Actually, the integral is correct because you are evaluating at constant R so you don't need to carry out a radial integration.
  11. Apr 28, 2010 #10
    Edit: OK, I looked it over and I'm pretty sure this is all correct now (but I won't make any bets!)
    Last edited: Apr 28, 2010
  12. May 3, 2010 #11
    Thank you very much mordechai9.
  13. May 4, 2010 #12
    I have a question tough,
    If the cylinder have not a perfect shape and have some singularities (like groove) which will lead to some surfaces directly facing the air.
    should I consider a classic drag force for these surfaces ?
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