Question

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

Answer 1

*

Answer 2

@eliassaab @estudier

Answer 3

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

Answer 4

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

Answer 5

@mahmit2012

Answer 6

@TuringTest i need some help here

Answer 7

*

Answer 8

@Zarkon