I agree Terry and Joe. It seems to me that the way you state the problem Don, it becomes almost a circular argument. Another large issue in looking at these issues are the assumptions that we make in dealing with Bernoulli and Poiseuille for that matter, that are not consistent with what happens in a patient. Such as: 1. The conduit is a rigid tube with a smooth surface and no irregularities. 2. The fluid is Newtonian and incompressible. 3. No energy loss due to heat. 4. there is steady flow. While Reynolds > 2000 usually results in turbulence, we can find many examples of turbulence with much lower numbers and vice versa. My go to book has become Frank Miele for good models of how this stuff works. Bill William B Schroedter, BS, RVT, RPhS, FSVU Technical Director, Quality Vascular Imaging, Inc 4120 Woodmere Park Blvd Suite 8B Venice, Florida 34293 (941) 408-8855 office <mailto:[log in to unmask]> [log in to unmask] _____ From: UVM Flownet [mailto:[log in to unmask]] On Behalf Of Schneider, Joseph MD Sent: Wednesday, October 31, 2012 7:26 AM To: [log in to unmask] Subject: Re: Reynolds Cross sectional area is pi * r2 . If the radius is doubled then the area quadruples and so velocity would decrease by about 75% to accommodate the same volume flow. More apropos of what we see in disease if the radius decreases by 50% then the velocity roughly quadruples to accommodate the same volume flow. We are used to thinking of turbulence occurring in stenoses so if the radius decreases velocity must increase much more to accommodate the same volume flow and you can see that this will increase the Reynolds number. All the best Joe From: UVM Flownet [mailto:[log in to unmask]] On Behalf Of Terry Zwakenberg Sent: Wednesday, October 31, 2012 5:36 AM To: [log in to unmask] Subject: Re: Reynolds 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 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