Don, The answer for your student is in the question. The Reynolds equation is determining the point at which flow becomes turbulent in a tube. You can solve that riddle for any diameter tube but once you choose the diameter of the tube you are solving for it is no longer a variable but rather a fixed constant in the formula. I guess if you were questioning the diameter required to keep a certain fluid of known viscosity and volume from becoming turbulent as you deliver it from point a to b than yes your student question would have some application, but you would first have to determine a starting diameter and it's associated Reynolds number for that specific fluid to be able to apply it. I think.

On Oct 31, 2012 12:05 AM, "Don Ridgway" <[log in to unmask]> wrote:

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I got asked an awkward question by a student today, regarding the Reynolds equation:

Re = V p 2r

n

My primitive school email access won't allow me to make that lower-case p into a rho for density, nor that lower-case n into an eta for viscosity, but you get the idea.

The question was this: since we know from another equation—Q = V x CSA—that a reduction of cross-sectional area means an increase of mean velocity, then why don''t those two balance out in the Reynolds equation?

For example, if the radius is doubled, won't that bring about a decrease to 1/2 of the velocity, leaving Re unchanged?

My clever response: life is complicated. And I'll get back to you.

I'm obviously angling for a bit of perspective from Dr. Beach from up there in Seattle (whence I'm flying Friday, as it happens).

Thanks in advance to anyone with help on this.

Don Ridgway

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