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Momentum behavior

  1. Mar 23, 2014 #1
    Can somobody help me how to get the 2.72 and 2.73 equation??

    http://imageshack.com/a/img89/5594/kiw6.jpg [Broken]
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Mar 23, 2014 #2

    Simon Bridge

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    2-72: have you tried: starting from 2-70 and take the divergence of both sides - note constant density and div(V)=0

    2-73: this equation is the definition of "fluid vorticity" so it is not a derived thingy.
    The equation just before it comes from 2-70 by taking the curl.
     
  4. Mar 23, 2014 #3
    i tried but it not looks simple to do
     
  5. Mar 23, 2014 #4

    Simon Bridge

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    Most of the terms turn out to be zero - where do you get stuck?
     
  6. Mar 23, 2014 #5
    I really confused.... There are a lot of terms in dV/dT and gravity that become 0, but in vicosity how could i do?
     
  7. Mar 23, 2014 #6

    Simon Bridge

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  8. Mar 24, 2014 #7
    [itex]\nabla p=\rho g + \mu \nabla^2 V-\rho\frac{\partial V}{\partial t}[/itex]

    [itex]\nabla^2 p=\nabla .(\rho g) + \mu\nabla . (\nabla^2 V)-\rho\nabla . \left(\frac{\partial V}{\partial t}\right)[/itex]

    [itex]\nabla^2 p=\nabla .(\rho g) + \mu\nabla^2(\nabla V)-\rho\nabla . \left(\frac{\partial V}{\partial t}\right)[/itex]

    Using [itex] \nabla . V=0[/itex] and [itex] \nabla .(\rho g)=0[/itex]

    [itex]\nabla^2 p=-\rho\nabla . \left(\frac{\partial V}{\partial t}\right)[/itex]

    And i got stuck
     
    Last edited: Mar 24, 2014
  9. Mar 24, 2014 #8
    Can i assume if [itex]div(V)=0[/itex] then it's a steady flow problem?
     
  10. Mar 24, 2014 #9
    The other i got [itex]\frac{\partial \omega}{\partial t}=\frac{\mu}{\rho}\nabla \times (\nabla^2 V)[/itex]

    What now?
     
  11. Mar 25, 2014 #10

    Simon Bridge

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    You are told that div.V=0.
    You can take the div inside the time-partial
     
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