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

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mukushla Group TitleBest ResponseYou've already chosen the best response.0
@eliassaab @estudier
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.0
P is not a vector\[\frac{\partial }{\partial t}(\frac{\rho}{2} \vec u^2)=\nabla .(\frac{\rho}{2} \vec u ^2 \vec u)  P\nabla. \vec u\nabla. P \vec u\nabla. [\tau.\vec u]+\tau:\nabla\vec u +\rho \vec u.\vec g\]
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.0
how to got that after dot multiplying by \(\vec u\) ... i mean how to do this???\[\vec u.(\frac{\partial (\rho \vec u)}{\partial t})=\vec u.(\nabla .\rho \vec u ^2\nabla P\nabla .\tau +\rho \vec g)\]
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.0
@mahmit2012
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.0
@TuringTest i need some help here
 2 years ago
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