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Torque in circular fluid motion

  1. Aug 4, 2007 #1
    I'm reading a book (intro to, by Davidson) about MHD now, but found I'm a bit rusty on tensors and curvilinear coordinates. It is written that for a circular flow the azimuthal component of the NS equations in the steady state gives (with F some body force)

    [tex]\tau _{r \theta} r^2 =-\int _0 ^r r^2 F_{\theta} dr [/tex]

    Shouldn't this read, for axisymmetric flow, without the square on both r's? I would argue that the remaining terms in the NS equations

    [tex]\sigma _{ij,i}+F_j=0[/tex]

    would yield for the azimuthal ([itex] \theta[/itex]) component (any suggestions welcome if the notation is obscure):

    [tex] F_{\theta} = -\nabla \cdot \overline{\overline{\sigma}}_{\theta}=-\frac{1}{r} \frac{d}{dr}(r\sigma_{r \theta})[/tex]

    Now the author continues,

    [tex]\tau_{r \theta}=\mu r \frac{d}{dr}(\frac{u_{\theta}}{r})[/tex]

    In which he says he used Newtons law of viscosity, which I think one can write

    [tex]\tau_{ij}=\mu u_{i,j}[/tex]

    (Is it by the way ok to write this as [tex]\overline{\overline{\tau}}=\mu \nabla \vec{u}[/tex]?)

    But how does one come from that to the (r, theta) component?
    Last edited: Aug 5, 2007
  2. jcsd
  3. Aug 5, 2007 #2
    I managed to get the r,theta component of the stress tensor as it is written in the book now. But I still can't see why there would be squares on the r's in the first formula I posted; anyone?
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