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Navier-stokes simplifications

  1. Nov 30, 2013 #1


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    hey pf!

    so i have a question concerning navier-stokes equations in a boundary layer, which, as a refresher, is [tex] \frac {D \vec{V}}{Dt} = - \nabla P + \nu \nabla^2 \vec{V}[/tex] where we know the x-component of [itex] \nabla^2 \vec{V}[/itex] may be re-wrote as [itex] \frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}[/itex] (i dismiss the [itex]z[/itex] component due to two dimensions). define the vertical displacement as [itex] \delta[/itex] and the horizontal length scale as [itex] L [/itex]

    okay, now for the question: we know [itex]\frac{\partial^2 u}{\partial x^2}[/itex] disappears, as it is not of much importance via order of magnitude. my question is, in boundary layer analysis, is this true from the fact that [itex]\delta < L \implies \delta^2 << L^2[/itex] and thus the [itex]\frac{\partial^2 u}{\partial x^2}[/itex] component can be thought of as insignificant (from the large denominator) compared to the [itex]\frac{\partial^2 u}{\partial y^2}[/itex] component?

    if so, when we leave the boundary layer are we going to assume that the double partial over x is still insignificant, or are we allowed to assume this (assuming same flow and geometry, just outside the BL)

  2. jcsd
  3. Aug 21, 2015 #2
    Uhhh, it would be nice to have a figure of what your system is or some kind of information describing what's going on in your system. You can't always say that (d^2u/dx^2) is 0. If du/dx is constant or weakly dependent on x, then you may make that assumption.

    Can I assume y is the direction to and from a plate, and x is the direction along the plate? So, that would seem the most logical to me. I'm not sure if the u velocity varies linearly inside the boundary layer or not, I honestly don't remember, it's been a while. However, outside of the boundary layer you can definitely assume the second derivative in the x direction is 0. That is, assuming flow on a plate as I described.
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