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Fluid mechanics - Bernoulli Ball

  1. Nov 6, 2009 #1
    sorry for the "bad" title, I couldn't think of anything else.

    We all have seen the experiment where you take a hair dryer and a ping-pong ball and the hair dryer blows the ball into the air and you can move and tilt the ball around...

    now, that is a very simplified version of my final years engineering project.

    the question I have, and what I cant seem to explain is about the forces acting in on the ball to keep it in the air.

    Bernoulli's law keeps the ball from falling out of the air stream (in this case created by a centrifugal fan), and the drag force on the ball is what keeps it in the air vertically.

    I calculated the velocity of the air needed to produce the required drag force, but is there a way to calculate the theoretical height at which the ball will levitate in the air?

    I know the exit velocity of the air from the nozzle, (for my experiment, just a straight PVC pipe with a 110mm diameter), and the calculated velocity for the drag force is 5.7m/s.
  2. jcsd
  3. Nov 6, 2009 #2
    Well, I think that it might have something to do with where the flow is 5.7m/s along the pipe. this point will be approximately where the drag force equals the ball weight. If we take friction factors into account, we could possibly find the difference between the inlet velocity and the outlet velocity and perhaps a velocity of 5.7m/s somewhere in between.

    To find out how friction affects the flow, we can use the Darcy Weisbach equation to find a pressure loss due to the friction in the pipe. This would be assuming an empty pipe, but the equation looks like this:

    [tex]\Delta P _{friction}=f\frac{L}{D}\frac{\rho v ^{2}}{2}\cdot[/tex]

    The f is a friction factor which can be found using the Colebrook-equation or one of its aproximations or using the moody diagrams.

    To use the moody diagram, you need the Reynolds number and the roughess of the pipe. The roughness of PVC pipe is between 1.5micron and 7micron.

    I worked out your Re to be about 42000 and i used a roughness of 5 micron on a moody chart and got a friction factor of around 0.022.

    Using this information, I think you may be able to go a little further in calcualting the heights using bernoulli's with the known pressure loss included.
    Last edited: Nov 6, 2009
  4. Nov 6, 2009 #3
    thanks for the reply redargon

    The problem is, that the ball is not contained in a pipe.

    The air flow through the pipe and out of a nozzle, the ball then 'floats' in the air above the nozzle.

    So I cant use the friction, because its effect is no longer present once the air exits the nozzle.
    The turbulent air that exits the pipe, slows down as it dissipates into the atmosphere.

    I need a way to calculate the height at which the air has slowed down (in the atmosphere) to 5.7 m/s.

    Is that possible?
  5. Nov 6, 2009 #4
    ah, I see.

    The first thing that comes to mind is doing a CFD simulation, but there must be a way to estimate it, even roughly. I can't find anything on it at the moment.

    Once the air leaves the pipe, the problem becomes somewhat more complex :)-

    My only other suggestion is to actually test your system and do some experiments to gain some empirical data and write a model for your exact situation. Something that relates exit air velocity to ball floating height.
  6. Nov 6, 2009 #5
    thanks will look into it!
  7. Nov 6, 2009 #6


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    Oh, a CFD is probably not necessary. You have the velocity which is needed to keep the ball floating. Which gets me thinking. If you really want a good grade, you won't simply make the assumption that a certain delta P is acting across the entire planform area. You'll usethe potential flow velocity distribution of flow past a sphere:
    U = \frac{3}{2}U_0\sin\phi = \frac{3}{2}U_0\sin\left(\frac{x}{a}\right)[/tex]
    Where a is the radius and phi is the angle measured from the "nose".

    You will then integrate that using both pressures and skin frictions to find the exact mean velocity needed.

    OK, now to the hair dryer. I would assume that the hair dryer is an axisymmetric plane jet. The 2D solution (axial radial) is:
    u(x,y) = u_{max}sech^2 a\eta = u_{max}sech^2\left[0.2752\left(\frac{J\rho}{\mu^2x^2}\right)^{1/3}y\right][/tex]
    J = \frac{16}{9}\rho\nu^{1/2}a^3
    and the width of the jet can be defined where u = 0.01umax, or:
    \mbox{Width} = b \approx = 21.8\left(\frac{x^2\mu^2}{J\rho}\right)^{1/3}
    From there you have your jet velocity as a function of axial distance.
  8. Nov 9, 2009 #7
    nice one minger! Where did you fid the function for the jet velocity in terms of axial distance? Does it also apply for a 3D case or are there some other assumptions?
  9. Nov 9, 2009 #8


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    Viscous Fluid Flow by Frank M. White. It technically is 3D since its assuming axisymmetric flow. So:
    \frac{d}{dr} &\neq 0 \\
    \frac{d}{dz} &\neq 0 \\
    \frac{d}{d\theta} &= 0
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