1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Translating cylinder in viscous fluid

  1. May 1, 2010 #1
    1. The problem statement, all variables and given/known data
    I have a cylinder of radius R translating (in radial direction) with velocity U in a viscous fluid at Re<<1. Find the force exerted on the cylinder per unit length

    2. Relevant equations
    Re<<1 so the Navier-Stokes equation simplifies to the Stokes equation:
    [tex]\nabla P=\mu \nabla^2 v [/tex]
    The cylinder makes use of polar coordinates useful so the stream function for this problem is defined as:
    [tex]\frac{1}{r}\frac{\partial \Psi}{\partial \theta}=u_r[/tex]
    [tex]-\frac{\partial \Psi}{\partial r}=u_{\theta}[/tex]

    3. The attempt at a solution
    Given the problem statement and the choice for polar coordinates, the boundary conditions become:
    [tex]u_r(R)=U \cos\theta[/tex]
    [tex]u_{\theta}(R)=-U \sin\theta[/tex]

    With help of the continuity equation the Stokes equation can be turned into:
    [tex]\nabla^4 \Psi=0[/tex]
    And taking [tex]\Psi=\sin \theta f(r)[/tex] the equation further 'simplifies' to:
    [tex]\left[\frac{\partial^2}{\partial r^2}+\frac{1}{r}\frac{\partial}{\partial r}-\frac{1}{r^2}\right]^2f(r)=0[/tex]

    A general solution to this equation is [tex]f(r)=A r^3+Br \ln r + C r+\frac{D}{r}[/tex]

    Using the definition of the stream function it follows that:
    [tex]u_r=\frac{f(r)}{r}\cos \theta[/tex] and [tex]u_{\theta}=-\sin\theta \frac{\partial f(r)}{\partial r}[/tex].

    The boundary conditions for f(r) then become:
    [tex]f(R)=U R[/tex]
    [tex]\frac{\partial f(R)}{\partial r}=U[/tex]
    [tex]\frac{f(\infty)}{\infty}\cos \theta=0[/tex]
    [tex]-\sin \theta \frac{\partial f(\infty)}{\partial r}=0[/tex]

    The two infinity conditions yield [tex]A=B=C=0[/tex] and for the other two I find:

    [tex]\frac{D}{R}=U R[/tex] and [tex]\frac{-D}{R^2}=U[/tex]

    And that is where I get stuck, because I will not be able to satisfy these conditions, so what did I do wrong?
    Last edited: May 1, 2010
  2. jcsd
  3. May 20, 2010 #2
    mmm, seems like I found the famous Stokes' paradox. For other people reading this and wondering how to solve it: take a look at Oseen's approximation method
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook