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Homework Help: 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
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