3 Law on Reynolds momentum transport equation

  • Thread starter Plott029
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  • #1
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I want to know how Reynolds Equation of Momentum is accord with the 3rd law of Newton, the law of action-reaction forces. For an ideal fluid, we have that

[tex]
\frac {\partial m \vec {v}} {\partial t} = \vec v [ \frac {\partial {\rho }} {\partial t} + \nabla [ \rho \vec v ] + \rho [\frac {\partial {\vec {v}} {\partial t} + ( \vec {v} \dot \nabla ) \vec v
[/itex]

\frac {\partial m \vec {v}} {\partial t} = \vec v [ \frac {\partial {\rho }} {\partial t} + \nabla [ \rho \vec v ] + \rho [\frac {\partial {\vec {v}} {\partial t} + ( \vec {v} \dot \nabla ) \vec v

Takint 2 points into the flow, I want to know if its posible to make that F12 = F21

Is this posible, or is a nonsense?
 
Last edited:

Answers and Replies

  • #3
13
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Thanks arildno.

I think that has not sense talk about force relative to one particle from another, and third law is about all flow, which every mass point with equilibrium of forces. Is this?
 

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