Talk:Stokes flow

From Wikipedia, the free encyclopedia
Jump to: navigation, search
WikiProject Physics / Fluid Dynamics  (Rated Start-class, Mid-importance)
WikiProject icon This article is within the scope of WikiProject Physics, a collaborative effort to improve the coverage of Physics on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.
Start-Class article Start  This article has been rated as Start-Class on the project's quality scale.
 Mid  This article has been rated as Mid-importance on the project's importance scale.
This article is supported by Fluid Dynamics Taskforce.
 

This article has comments here.

Are the viscous forces really large in comparison with the advective inertial forces?[edit]

As the headline reads, are the viscous forces really large in comparison with the advective inertial forces? Have this actually been measured or showed somehow? Where does this claim come from? That the Reynolds number is very low is a completely different thing and is much easier to show. Has some Wikipedian made the conclusion that the two are equivalent? I'm not so sure they are, even though they might be. I have put a {{fact}} tag after the statement "Stokes flow ... is a type of fluid flow where advective inertial forces are small compared with viscous forces" until someone can find a source for the claim or show the equivalence I just mentioned. —Kri (talk) 19:34, 14 August 2012 (UTC)

I think you will find the equivalence follows trivially from the definition of Reynolds number as the ratio of inertial forces/viscous forces. Obviously when Re << 1, the numerator (inertial) must be very small relative to the denominator (viscous). Bungeh (talk) 07:20, 20 May 2013 (UTC)
I agree with Bungeh. The wording itself was take from Kirby "Micro- and Nanoscale Fluid Mechanics: Transport in Microfluidic Devices" which says "In this chapter, we discuss Stokes flow (equivalently termed creeping flow), in which case the Reynolds number is so low that viscous forces dominate over inertial forces." Fluellen7 (talk) 03:44, 5 January 2014 (UTC)

Are creeping flow and creeping motion the same as Stokes flow?[edit]

I am confused about "creeping flow" and "creeping motion". Both redirect here, and are mentioned as synonyms to Stokes flow. I actually added "creeping motion" as a synonym, because the article didn't mention "creeping motion" at all.

But I am unsure if that is correct. [1] quotes McGraw-Hill Concise Encyclopedia of Physics. © 2002 by The McGraw-Hill Companies, Inc. for saying that "In creeping flow the Reynolds number is very small (less than 1) such that the inertia effects can be ignored in comparison to the viscous resistance. Creeping flow at zero Reynolds number is called Stokes flow." That means that "Stokes flow" (Re << 1, or maybe even Re = 0) is just a special case of "creeping flow" (Re < 1). Is that true?

And what about "creeping motion"? Is that the same as "Stokes flow", or "creeping flow", or something different from both?

--Jhertel (talk) 01:04, 29 August 2012 (UTC)

http://www.andrew.cmu.edu/course/06-703/Creeping%20Flow.pdf says that "One limiting case is creeping flow which corresponds to the limit in which the Reynolds number is small (i.e. Re → 0)." So, it seems like "creeping flow" is not the same as Stokes flow. My guess is that "creeping motion" is the same as "creeping flow", but I don't know. --Jhertel (talk) 01:16, 29 August 2012 (UTC)

I did a google search for "creeping motion" and found this picture. I guess "creeping motion" is not the same thing as "Stokes flow". —Kri (talk) 11:54, 29 August 2012 (UTC)
creeping motion and stokes flow are the same, they are both derived by neglecting the inertia in the Navier-Stokes equations (i.e. imposing Re = 0, justified by the fact that Re << 1). In Kim and Karrilla's "Microhydrodynamics", they say: "The linearized Navier-Stokes equations for steady (i.e. time-independent) motion are known as the creeping motion or (inhomogeneous) Stokes equations, and these are obtained by neglecting the substantial derivative of v," (the velocity), "in the Navier-Stokes equations" Fluellen7 (talk) 03:16, 5 January 2014 (UTC)