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Navier stokes and pressure on objects

  1. Jan 19, 2010 #1
    Hi every one, im having a few problems with some research im doing. I put this in the PDE section as it seams related, but it is for a specific application and im not sure that it wouldn't be better suited to the mechanical engineering section.

    I am wanting to find the pressure distribution in a 2d flow that moves around a corner (ie bend in a pipe)
    I have been using flexPDE to solve naiver stokes with FEM. however, i am unsure how to incorporate pressure.
    the software examples do one thing my supervisor suggests another.

    First my supervisor is saying to use Bernoulli dynamic pressure. But i feel he is confusing something as that is more related to potential pressure rather than actual

    pressure. also due to "no slip" the pressure on the wall would always be 0.
    also not to mention using this alone mean the software can close on a solution.
    I suspect we need to solve for the static pressure in the system but I dont know how.

    secondly the software examples assume slight compressibility of the fluid.
    they give:

    defining a hypothetical equation of state

    1) p(dens) = p0 + L*(dens-dens0)

    " where p0 and dens0 represent a reference density and pressure, and L is a large
    number representing a strong response of pressure to changes of density. L is
    chosen large enough to enforce the near-incompressibility of the fluid, yet not
    so large as to erase the other terms of the equation in the finite precision
    of the computer arithmetic."

    The compressible form of the continuity equation

    2) dt(dens) + div(dens*U) = 0

    (U is a velocty vector field it seams, and dt(dens)= ddens/dt, i am unsure what they mean by p(dens))

    they then do some magic and come out with

    dt(p) = -L*dens0*div(U)


    3) div(grad(p)) = M*div(U)

    what I understand is vaguely how they formed the first and second equation, and how they got from dt(p)= div(grad(p))

    I dont understand how to get from 1 and 2 to 3, or how to get usable real world value predictions from this as L / M is assumed
    I suspect that this is only used to find a vague distribution and to enforce continuity in the equation.

    any help in unraveling this for me would be much appreciated thank you in advance

  2. jcsd
  3. Jan 21, 2010 #2
    For the incompressible case, you can use:
    [tex]\nabla_{i}\cdot [u_{i,t}+u_{j}\nabla_{j}u_{i}]=\nabla_{i}[-p_{,i}/\rho+\nu \Delta u_{i}][/tex]
    Then you have:
    [tex]\Delta p=-\rho \nabla_{i}\nabla_{j} (u_{i}u_{j})[/tex]
    But I'm quite sure conventional CFD software doesn't calulate pressure that way. Instead, it is included among the dependent variables
  4. Jan 21, 2010 #3
    Hi gato thanks for the reply.

    Just to make sure i understand what you mean...
    deltaP = -p div(uj ui) ?

    also how would you include Pressure as a dependent variable? wouldn't you require 3 equations to solve for 3 unknowns? Vx Vy and P?

  5. Jan 22, 2010 #4
    you DO have 3 equations:
    [tex]u_{t}+uu_{x}+vu_{y}=-p_{x}/\rho+\nu\Delta u[/tex]
    [tex]v_{t}+uv_{x}+vv_{y}=-p_{y}/\rho+\nu\Delta v[/tex]
    Eliminating p using third equation leads to the identity just stated. Anyhow, best way is to check how other people are doing that. There are lots of codes available. If you want to use a predefined one, use comsol multiphysics for example. If you want to do your own, you might try looking this
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