Saturday, 27 October 2018

fluid dynamics - Viscous drag proportional to r or r3?


When a spherical object is falling at terminal velocity through a fluid: W=U+F, where W is weight, U is upthrust and F is viscous drag. Rewriting, using Stoke's Law, we get: 43πr3ρobject.g=43πr3ρfluid.g+6πrηvt

6ηvt=43r2ρobject.g43r2ρfluid.g
vt=2r2(ρobjectρfluid).g9η,


where r is the radius of the object, ρ is density, g is the gravitational acceleration, η is the viscosity coefficient, and vt is the terminal velocity of the object.



So vt varies proportionally to r2.


However, clearly upthrust varies proportionally to r3. This implies that F varies proportionally to r so that vt can be proportional to r2. But F=6πrηvt, so as we increase r, vt increases quadratically, meaning that F must increase cubically.


This seems to be a contradiction - vt is proportional to r2 so F is proportional to r, but also F is proportional to r3.


So where is my reasoning wrong, and does F vary proportionally to r or r3?


EDIT:


I found this quote on the Wikipedia page for Stokes' law:



Note that since buoyant force increases as R3 and Stokes drag increases as R, the terminal velocity increases as R2 and thus varies greatly with particle size as shown below.





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