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Steady state in fluid mechanics

  1. Oct 3, 2013 #1
    Hey! When a stream is steady-state, you can cancel the acceleration term in navier-stokes equation, right?


    [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.
  2. jcsd
  3. Oct 3, 2013 #2
    Your premiss is incorrect. the acceleration won't necessarily be zero for a steady state flow.
  4. Oct 3, 2013 #3
    At 02:03 or so the professor says the flow is steady-state and removes the acceleration term.
    Last edited by a moderator: Sep 25, 2014
  5. Oct 3, 2013 #4


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    Yes the [itex]\frac{\partial \vec{V}}{\partial t}[/itex] term will be zero on account of it is a time dependent quantity and is defined as zero in a steady-state flow. It is an acceleration term and it goes to zero, so in that sense you are right. But the larger "acceleration term" you are trying to zero out is the "convective acceleration" and is not necessarily zero.
  6. Oct 3, 2013 #5
    Even for a steady state flow, an observer moving with the flow velocity along a streamline can be accelerating (Lagrangian frame of reference). His velocity vector is changing with time. The acceleration equation you wrote down (which is the so called Material Derivative of the velocity) determines his acceleration quantitatively.

  7. Oct 5, 2013 #6
    Okay, but why did the professor in the video in post #3 just say "steady state" and zero-out the
    [tex] \frac{d\vec{V}}{dt} [/tex] term then? He said nothing about assuming the convective acceleration is zero..

    I'm really confused about this... :(

    EDIT: AND OOPS, I see I made a typing mistake in the picture in the OP. The velocity differentiated with regards to Z should be multiplied by w.
  8. Oct 5, 2013 #7


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    Because [itex]\frac{d\vec{V}}{dt}[/itex] is not the same as [itex]\frac{D\vec{V}}{Dt}[/itex]. For a steady state flow, all [itex]\frac{d}{dt}[/itex] terms are zero, but not all [itex]\frac{D}{Dt}[/itex] terms are necessarily zero.
  9. Oct 5, 2013 #8
    oh crap. I just noticed that the professor wrote the PARTIAL derivative of u with respect to time, NOT the total derivative, as his acceleration term. How careless of me.

    When I watched it I thought the professor wrote up the complete navier-stokes equation, but apparently he zeroed all the convective acceleration terms.

    Okay, thanks everybody.
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