- #1

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- 27

## Main Question or Discussion Point

Hey! When a stream is steady-state, you can cancel the acceleration term in navier-stokes equation, right?

so:

[tex] \rho \vec{a} = 0 = - \nabla P + \rho \vec{g} + \mu \nabla ^2 \vec{V}[/tex]

But there are many terms in the total acceleration which are not dependant on time! The acceleration term in navier stokes can be broken down into https://scontent-b-ams.xx.fbcdn.net/hphotos-prn2/1379985_10201506247874738_1799136895_n.jpg: Why do the acceleration terms with regards to distance (dx, dy, dz) disappear also? This makes no sense to me.

so:

[tex] \rho \vec{a} = 0 = - \nabla P + \rho \vec{g} + \mu \nabla ^2 \vec{V}[/tex]

But there are many terms in the total acceleration which are not dependant on time! The acceleration term in navier stokes can be broken down into https://scontent-b-ams.xx.fbcdn.net/hphotos-prn2/1379985_10201506247874738_1799136895_n.jpg: Why do the acceleration terms with regards to distance (dx, dy, dz) disappear also? This makes no sense to me.