Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

mukushla

  • 3 years ago

momentum transfer equation\[\frac{\partial (\rho \vec u)}{\partial t}=-\nabla .\rho \vec u ^2-\nabla P-\nabla .\tau +\rho \vec g \]

  • This Question is Closed
  1. experimentX
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    *

  2. mukushla
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @eliassaab @estudier

  3. mukushla
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    P is not a vector\[\frac{\partial }{\partial t}(\frac{\rho}{2} \vec u^2)=-\nabla .(\frac{\rho}{2} \vec u ^2 \vec u) - P\nabla. \vec u-\nabla. P \vec u-\nabla. [\tau.\vec u]+\tau:\nabla\vec u +\rho \vec u.\vec g\]

  4. mukushla
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    how to got that after dot multiplying by \(\vec u\) ... i mean how to do this???\[\vec u.(\frac{\partial (\rho \vec u)}{\partial t})=\vec u.(-\nabla .\rho \vec u ^2-\nabla P-\nabla .\tau +\rho \vec g)\]

  5. mukushla
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @mahmit2012

  6. mukushla
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @TuringTest i need some help here

  7. TuringTest
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    *

  8. mukushla
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @Zarkon

  9. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy