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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:

> 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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