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Thread: Incompressible Navier-Stokes Problem

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    Incompressible Navier-Stokes Problem

    I am stuck on this problem:

    Incompressible N-S

    $\displaystyle \frac{d\vec{v}}{dt} + (\vec{v} \cdot \nabla)\vec{v} = - \nabla P + \vec{b} + \mu {\nabla }^{2} \vec{v}$

    $\displaystyle \mathrm{div}{\vec{v}} = 0$

    Incompressible,

    $\displaystyle \vec{b} = 0$

    $\displaystyle a < r < b$

    Pressure is uniform with

    $\displaystyle \frac{\mathrm{dP} }{\mathrm{dz} } = -m$

    $\displaystyle \vec{v} = u \vec{{i}_{r}} + v \vec{{i}_{\theta}} + w \vec{{i}_{z}}$

    $\displaystyle u=v=0$

    $\displaystyle w=w(r)$

    Show that

    $\displaystyle \frac{\mathrm{{d}^{2}w}}{\mathrm{d{r}^{2}}} + \frac{1}{r}\frac{\mathrm{dw}}{\mathrm{dr}} = -\frac{m}{\mu}$

    with general solution

    $\displaystyle w = A + B \log{r} - \frac{1}{4}\frac{m}{\mu}{r}^{2}$

    Choose A and B to satisfy the no-slip because at r = a, r = b. Derive

    $\displaystyle w(r) = \frac{1}{4}\frac{m}{\mu}[({a}^{2} - {r}^{2}) + ({b}^{2} - {a}^{2})\frac{\log{\frac{r}{a}}}{\log{\frac{b}{a}} }}]$

    __________________________________________________ _________

    Since posting this I have gotten farther, solving the differential equation as a linear nonhomogenious equation I get

    $\displaystyle {w}_{c}={c}_{1} + \frac{{c}_{2}}{\exp} $

    $\displaystyle {w}_{p}={c}_{3} \cdot r$

    meaning
    $\displaystyle w = {w}_{c} + {w}_{p} $

    plugging w back into the differential equation I get

    $\displaystyle {c}_{3} = -\frac{m \cdot r}{4 \cdot \mu}$

    this equation does satisfy the differential equation but as far as matching the equation for w above I dont understand how they are getting the $\displaystyle B \log{r}$ term.
    Last edited by gamerman315; May 11th 2011 at 03:30 AM.
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