Expert: Paul Soderman - 7/20/2010

Question
QUESTION: Hello Mr. Soderman,

If it is not an inconvenience, I would like to respectfully ask for you help on a fluid mechanics problem involving water tunnel testing, the Reynolds number and the Strouhal number. Normally I would try to solve it on my own, but we unfortunately got the teacher mad at us, and were consequently assigned one of the harder problems in the chapter, over a subject we were never once lectured about. The question goes as follows:

"Some students want to visualize flow over a spinning baseball. Their fluids laboratory has a nice water tunnel into which they can inject multicolored dye streaklines, so they decide to test a spinning baseball in the water tunnel. Similarity requires that they match both the Reynolds number and the Stouhal number between their model test and the actual baseball that moves through the air at 80 mi/h and spins at 300 rpm. Both the air and the water are at 20 degrees Celsius. At what speed should they run the water in the water tunnel, and at what rpm should they spin their baseball? Answers: 5.30 mi/h, 19.9 rpm." (Essentials of Fluid Mechanics by John M. Cimbala and Yunus A. Cengel, Ch. 7 PR 7-44, ISBN 978-0-07-313835-0)

We have calculated and worked with the Reynolds number before, but only for simple, non-rotating objects. I am not particularly sure what role the rotation actually plays in the problem, nor am I sure how to find the frequency of vortex shedding needed to calculate the Strouhal number (I hear it has something to do with shape and the magnitude of the Reynolds number, but am not sure).

Any type of help or advice would be greatly appreciated by this confused and dumbfounded student.

Very sincerely,
David,
Student

ANSWER: David
It is not good to make the professor mad.  Lesson number 1.  Now, you know the definition of Reynolds number so to match Reynolds number in air and water you have to find the velocity ratio air to water:

v1/v2 = mu1 rho2/(mu2 rho1) where you know the definitions of the parameters. The question remains - what velocities are to be used:  the free stream velocity or the velocity at the surface of the baseball or both.  Since Reynolds number relates to boundary layer growth, I suspect you need to match velocities at the surface of the baseball.  On one side the free stream velocity is added to the surface speed, on the other the free stream is subtracted from the surface speed. (That is why the pressures are established that make the ball curve.)  So, I would pick one side or the other and compute the water speed and rpm necessary to match the above ratio knowing the free stream and rotational speeds in air.  

As to Strouhal number, which you can look up via google, I believe your Reynolds numbers are high enough that the Strouhal number is approximately 0.21 for both air and water. I don't see how spin enters into Strouhal number.  Schlichting's book:  Boundary-Layer Theory explains Strouhal number in more detail.  I assume you can find the library.

Good luck.
Paul

---------- FOLLOW-UP ----------

QUESTION: Hello again,

Thank you for your very swift answer to my question. Your explanation sounds quite reasonable, and I have a feeling it will definitively prove to be so after a little more research in the library; however, another odd quirk has just come up regarding the same problem, and I hoped that I might be able to bug you one more time.

I managed to guess at how the book arrived at their given answer, but I am not so sure that I agree with it. Apparently, after matching the Reynolds numbers and coming up with a model fluid velocity of 5.30 mi/h, the authors assume that the vortex shedding frequency (f) in the Strouhal equation is exactly the same thing as the rotation of the sphere in rpm.

St1 = St2
(f1L1)/v1 = (f2L2)/v2

where f is defined by numerous sources as the vortex shedding frequency in Hz, L is a characteristic length, and v is a characteristic velocity. Since it is the same baseball, the L terms cancel.

f1/v1 = f2/v2
f1 = (f2)(v1/v2)
f1 = (300 rpm)((5.30 mi/h)/(80 mi/h))
f1 = 19.9 rpm

But how on earth do the authors equate vortex shedding frequency with simple rpms of a sphere?

Answer
You raise a good question.  We know that spinning a cylinder in flow has little effect on the vortex shedding rate, though it can eliminate the shedding.  See:  Mittal, S. and Kumar, B.:  Flow Past a Rotating Cylinder.  J. Fluid Mechanics, vol 476, pp 303-334, 2003.  So, I don't think the authors were connecting spin rate to vortex shedding rate.

So, forget vortex shedding.  Assume you had a 1 meter object in 100 m/s airflow and you perturbed the inflow, say  be injecting a small gust at 100 times per second. The disturbances would be spaced one meter apart or one body dimension apart.  Then you put the same object in water at 1 m/s and wanted to perturb it at the same rate; i.e., one perturbation per object dimension per second.  Then you would use the same equation you show and perturb the flow 1 time per second.  So the baseball in air and the baseball in water would see the same flow disturbance rate or they would create the same disturbance rate in the boundary layer in both cases (rotating stitches ?).

I admit this is a stretch, but it is all I can come up with given the information provided.
Paul