Print

Print


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

To unsubscribe or search other topics on UVM Flownet link to:
http://list.uvm.edu/archives/uvmflownet.html
To unsubscribe or search other topics on UVM Flownet link to: http://list.uvm.edu/archives/uvmflownet.html