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Laminarflow in a two-dimensional passage

  1. Oct 31, 2009 #1
    Exact Question from Book
    For laminar flow in a two-dimensional passage, find the relation between the average and maximum velocities.

    Relevant Equations, etc
    I think for circular pipes we have:

    u = u_max - kr^2 (1)

    u: velocity
    u_max: maximum velocity at center of the pipe

    Work Done So Far
    I played around with multiplying (1) by dA = 2 pi r dr and integrating but didn't really achieve much.

    I'm not sure what they mean by "two-dimensional passage" - do they mean a square or rectangle?
     
  2. jcsd
  3. Nov 1, 2009 #2
    If its two dimensional its cross sectional shape doesn't matter right?

    Set up a two dimensional flow with uniform flow at the entrance and and a fully developed profile at the exit of your control volume.

    Use conservation of momentum to find a relation between the velocities.
     
  4. Nov 1, 2009 #3
    like this?
    http://img691.imageshack.us/img691/3605/21166978.jpg [Broken]

    how is this so different than the parabola flow we're used to seeing? because its not a 3-d tube?

    so u_max is at the center and the median of a triangle is 2/3 the height;

    u = 2/3 u_max.
     
    Last edited by a moderator: May 4, 2017
  5. Nov 1, 2009 #4
    No a fully developed profile in a 2d passage will be parabolic.
     
  6. Nov 1, 2009 #5
    Hint: the answer is [tex]u_{max} = 2u_{avg}[/tex]
     
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