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Two infinite cylinders Couette flow. How to fix the pressure integration constant?

  1. Apr 10, 2012 #1
    1. The problem statement, all variables and given/known data

    Consider the steady flow between two long cylinders of radii [itex]R_1[/itex] and [itex]R_2[/itex], [itex]R_1 > R_1[/itex], rotating about their axes with angular velocities [itex]\Omega_1[/itex], [itex]\Omega_2[/itex]. Look for a solution of the form, where [itex]\hat{\boldsymbol{\phi}}[/itex] is a unit vector along the azimuthal direction:

    [tex]\mathbf{u} = v(r) \hat{\boldsymbol{\phi}}[/tex]
    [tex]p = p(r).[/tex]

    Write out the Navier-Stokes equations and find differential equations for [itex]v(r)[/itex] and [itex]p(r)[/itex].

    Fix the constants [itex]a[/itex] and [itex]b[/itex] from the boundary conditions. Determine the velocity [itex]v(r)[/itex] and pressure [itex]p(r)[/itex].

    2. Relevant equations

    Navier-Stokes equations for steady state incompressible flow are

    [tex](\mathbf{u} \cdot \boldsymbol{\nabla}) \mathbf{u} = -\frac{1}{{\rho}} \boldsymbol{\nabla} p + \nu \boldsymbol{\nabla}^2 \mathbf{u}[/tex]
    [tex]\boldsymbol{\nabla} \cdot \mathbf{u} = 0.[/tex]

    We'll have no-slip boundary value conditions on the surfaces of each cylinder, so

    [tex]\begin{align}v(R_1) &= R_1 \Omega_1 \\ v(R_2) &= R_2 \Omega_2.\end{align}[/tex]

    3. The attempt at a solution

    Working in cylindrical coordinates where the gradient is

    [tex]\boldsymbol{\nabla} = \hat{\mathbf{r}} \partial_r + \frac{\hat{\boldsymbol{\phi}}}{r} \partial_\phi.[/tex]

    I find a pair of differential equations to solve

    [tex]r^2 v'' + r v' - v = 0[/tex]
    [tex]p' = \frac{\rho v^2}{r}[/tex]

    solving these and applying the boundary value conditions I find

    [tex]v(r) = \frac{1}{{R_2^2 - R_1^2}}\left(\left( R_2^2 \Omega_2 - R_1^2 \Omega_1\right) r+\frac{R_1^2 R_2^2}{r} (\Omega_1 -\Omega_2)\right)[/tex]
    [tex]p(r) - p_0 = \frac{\rho }{(R_2^2 - R_1^2)^2} \left( \frac{1}{{2}} \left( R_2^2 \Omega_2 - R_1^2 \Omega_1\right)^2r^2 -\frac{R_1^4 R_2^4}{2 r^2} (\Omega_1 - \Omega_2)^2+ 2 \left( R_2^2 \Omega_2 - R_1^2 \Omega_1\right) R_1^4 R_2^4 (\Omega_1 - \Omega_2)^2 \ln r\right)[/tex]

    That's almost the whole solution, but the part that I am unsure of is what can we use to determine the integration constant for the pressure (I've called it [itex]p_0[/itex] above)? Is there another boundary value constraint that I am missing?
     
    Last edited: Apr 10, 2012
  2. jcsd
  3. Apr 10, 2012 #2
    Re: Two infinite cylinders Couette flow. How to fix the pressure integration constan

    Hello

    I had a homework like this last semester. There were "two long cylinders" like this. However this didn't mean they are infinite. The tops and bottoms of the cylinders were closed and "long" was said to state that there was no radial velocities.

    If you have the same issue, the tops and bottoms are closed, in a sectional area the volume rate is zero (as the volume between cylinders is constant). I used it as the third boundary condition.

    I am not very good at advanced fluid dynamics but just wanted to share something I faced. I hope it helps.
     
  4. Apr 11, 2012 #3
    Re: Two infinite cylinders Couette flow. How to fix the pressure integration constan

    It's not clear to me how one would relate the pressure and the volume flux. We do have the pressure show up in the traction vector. In a later part of the problem, this is used to calculate the torque on the fluid. For example, the torque per unit area on the fluid from the inner cylinder (where the normal is [itex] \hat{\mathbf{r}} [/itex] can be calculated by crossing [itex]r \hat{\mathbf{r}}[/itex] with:

    [tex]\boldsymbol{\sigma} \cdot \hat{\mathbf{r}} = \mathbf{e}_i \sigma_{ij} \hat{\mathbf{r}} \cdot \mathbf{e}_i = -p \hat{\mathbf{r}} + \frac{\mu}{{R_2^2 - R_1^2}}\left( R_2^2 \Omega_2 - R_1^2 \Omega_1+\frac{R_1^2 R_2^2}{r^2} (\Omega_2 -\Omega_1)\right) \hat{\boldsymbol{\phi}}. [/tex]

    but the pressure term drops out doing so, and we don't have to know what that integration constant was.

    ( I'm not going to attempt to derive the above here, but have my attempt at a complete solution to the entire problem here for reference: https://sites.google.com/site/peeterjoot2/math2012/couetteFlow.pdf?revision=1 )

    Perhaps it's possible to utilize this strain relationship to fix the integration constant for the pressure? We've done similar things for two layer flow problems, matching components of the traction vector at the interface (for an interface with an open surface, that ends up incorporating the atmospheric pressure into the mix, but with things closed here we don't have any atmospheric pressure effecting things.)

    However, if I was to match the traction vector with (something?) at the surfaces of the cylinder, it seems to me that there are not enough degrees of freedom, since I've got two surfaces, but only one integration constant.
     
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