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On fluid mechanics

  1. May 20, 2005 #1
    Dear Friends,

    I'm studying the 2D flows, and I have any questions.

    In a square, there's a grid of flow in horizontal an vertical linear flow.

    The flow that entry the square must be the same that exits form it.

    [tex] \frac {\partial \Psi_e}{\partial x} -\frac {\partial \Psi_O}{\partial x}= \frac {\partial \Psi} {\partial x}= 0[/tex]

    Becouse we have presure, we must apply navier-stokes'

    [tex] \rho [ \frac {\partial {v_x}}{\partial t} + ( \vec{v_x} \frac {\partial}{ \partial x} ) \vec{v_x} ] = \mu \frac {\partial^2} {\partial x^2} \vec {v_x} - \frac {\partial p} {\partial x} [/tex]

    How can I vinculate the presure with the flow in x and y directions?




    .
     
  2. jcsd
  3. May 20, 2005 #2

    learningphysics

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    Homework Helper

    I think Navier Stokes in 2D would be:

    [tex] \rho [ \frac {\partial {v_x}}{\partial t} + ( \vec{v_x} \frac {\partial}{ \partial x} ) \vec{v_x} + ( \vec{v_y} \frac {\partial}{ \partial y} ) \vec{v_x} ] = \mu \frac {\partial^2} {\partial x^2} \vec {v_x} + \mu \frac {\partial^2} {\partial y^2} \vec {v_x} - \frac {\partial p} {\partial x} [/tex]

    and
    [tex]\frac {\partial}{ \partial x} \vec{v_x} + \frac {\partial}{ \partial y} \vec{v_y} = 0[/tex]
     
  4. May 21, 2005 #3
    Aha!

    Ah! Thanks learningphysics. Now I'm trying to extract for this formula the flow. Is it possible?
     
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