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Fluid flow through pipe

  1. Jan 22, 2013 #1
    Hi,

    I'm looking at the solution to a question on fluid flow through a rigid pipe.

    Original equation: [tex]\mu u = 0.25r^{2} dp/dx + Aln(r) + B[/tex]
    After applying boundary conditions: [tex]\mu u = 0.25dp/dx (r^{2} - a^{2})[/tex]

    I don't understand how the constants have been solved for. Below is as far as I get:
    Starting with
    [tex]\mu u = 0.25r^{2} dp/dx + Aln(r) + B[/tex]
    Assume a no-slip boundary condition, so
    [tex]u(r = a) = 0: 0 = 0.25a^{2} dp/dx + Aln(a) + B[/tex]

    The notes somehow end up with [tex]Aln(a) = 0[/tex].

    Thanks for any input.
     
  2. jcsd
  3. Jan 22, 2013 #2

    pasmith

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    Homework Helper

    A must be zero, or else [itex]u(0) = A \ln(0) + B[/itex], which is mathematical and physical nonsense.
     
  4. Jan 22, 2013 #3
    I see. So the edge of the cylinder is defined as r = 0, and the centre as r = a?

    Thanks for clearing it up.
     
  5. Jan 22, 2013 #4

    pasmith

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    No, the center of the cylinder is r = 0.

    You have a 2nd order ODE for u(r); it has two boundary conditions. One is that u(a) = 0, the other is that u(0) is finite. It is this condition that requires you to reject the ln(r) complimentary function.
     
  6. Jan 22, 2013 #5
    I see, thanks.
     
    Last edited: Jan 22, 2013
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