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