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Boundary Conditions for an inviscid fluid at a fixed boundary

  1. Mar 24, 2014 #1
    This is my first post so I hope this in the right place. I am fairly sure this is quite a straight forward question but I having trouble working out the details of it.

    "State the boundary conditions for an inviscid fluid at an impermeable fixed boundary

    [tex] x_3-h(x_1,x_3)=0 [/tex]

    where we do not permit normal velocity but permit non-zero tangential velocity"


    So my attempt is quite limited as I feel if I could get started I'd be able to make some progress. But I believe that since the wall is stationary then the velocity of the fluid at the wall must also be stationary. And since it says that normal velocity is zero i guess this means that in the [itex] x_3 [/itex] direction the velocity is always zero.
    so:
    [tex] \mathbf{u}\cdot \mathbf{\hat x}_3 =0[/tex]
    Then the velocity of the fluid only has components in the other two directions...
    I was then also thinking that the [itex] \hat x_1 [/itex] component of velocity wouldn't change with [itex] x_1 [/itex] but I am not really sure why I think that. I am just getting a bit confused as you might be able to tell, any help would be really appreciated!

    Edit: I am also not sure why my tex commands aren't working so if somebody could point out what I am doing wrong that would also be great...
     
    Last edited: Mar 24, 2014
  2. jcsd
  3. Mar 24, 2014 #2

    pasmith

    User Avatar
    Homework Helper

    That boundary seems an unnecessary complicated way to write [itex]g(x_1,x_3) = 0[/itex]. Does the question actually say
    [tex]
    x_3 - h(x_1, x_2) = 0
    [/tex]
    which would make sense as the upper boundary of a fluid layer of variable depth?

    For an inviscid fluid only the normal component must be stationary relative to the wall.

    If the normal to the boundary were in the [itex]x_3[/itex] direction this would be correct, but that's not necessarily the case.

    Following the question as written, the first thing is to define
    [tex]
    g(x_1,x_3) = x_3 - h(x_1,x_3)
    [/tex]
    so that the boundary is at [itex]g(x_1,x_3) = 0[/itex]. The boundary is then a contour of [itex]g[/itex], so the direction normal to the boundary is given by
    [tex]
    \nabla g = \frac{\partial g}{\partial x_1} \hat{\mathbf{x}}_1 + \frac{\partial g}{\partial x_3} \hat{\mathbf{x}}_3.
    [/tex]

    The closing tag is [/tex] not [\tex].
     
  4. Mar 24, 2014 #3
    Yeah, sorry! It is:
    [tex] x_3-h(x_1,x_2)=0[/tex]

    So then in this case my direction normal would be:
    [tex]\nabla_h=\hat x_3 -\frac{\partial h}{\partial x_1} \hat x_1 -\frac{\partial h}{\partial x_2} \hat x_2 [/tex]
    and I can dot this with the velocity of the fluid to get zero. So that:
    [tex]\mathbf{u}\cdot\nabla_h= \mathbf{u}\cdot\hat x_3 -\mathbf{u}\cdot \frac{\partial h}{\partial x_1} \hat x_1 - \mathbf{u}\cdot\frac{\partial h}{\partial x_2} \hat x_2=0 [/tex]
    This surely isn't the whole whole answer though is it?

    Thanks for the reply.
     
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