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## Homework Statement

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

## Homework 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]

## 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]

[tex]u_r(\infty)=u_{\theta}(\infty)=0[/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?**

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