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Basic fluid mechanics question

  1. Dec 7, 2005 #1
    Let's say we have a horizontal pipe with steady, incompressible, two dimensional flow. There is friction at the top and bottom of the pipe which causes viscous effects.
    The flow is fully developed. The velocity profile is parabolic.

    Obviously there is an x-component (u) of velocity.

    Is there also a y-component of velocity (v), in this fully developed region?

    My intuition says no, but from reading my book it implies that there is, although it is not very well explained.

  2. jcsd
  3. Dec 8, 2005 #2


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    If there was a 'y' component, where would it go?

    EDIT: To expand on that...

    Draw a control volume inside the pipe. You have no velocity crossing the boundary on either the top or the bottom. If you have any sort of 'y' velocity inside the volume, it will cause a curl in the flow field, and you've already stated that the flow is fully developed.

    Which book are you using, and how does it imply that there is?
    Last edited: Dec 8, 2005
  4. Dec 8, 2005 #3


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    Apart of the integral explanation of enigma....

    Take continuity equation:

    [tex]\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0[/tex]

    If the flow is fully developed: [tex]\frac{\partial u}{\partial x}=0[/tex]

    Therefore [tex]\frac{\partial v}{\partial y}=0[/tex]

    and integrating it from 0 to y:

    [tex]v(y)=v(0)[/tex]. Taking into account there is no suction-inflow through walls [tex]v(y)=v(0)=0[/tex]

    As an advice, try to work out a steady solution when there is flow suction through walls [tex] v(0) \neq 0[/tex] and see what happens. It is very easy.
  5. Dec 8, 2005 #4
    thanks for your replies.

    i understand the part of the continuity equation that says
    change in u / change in x =0 when fully developed. This makes sense because the velocity is now constant.
    And i understand why u is slowing down initially (because of friction).
    So according to the continuity equation, if u is slowing down, v has to be speeding up. and since v initially reaches zero, it must be a negative number to start with. correct?

    the part i don't understand is that its supposed to be LAMINAR flow. so if i took any particle in the stream and followed it through the pipe, it would always be running horizontally (and not have any v component).
    So i understand why there is a v-component to satisfy the continuity equation, but not if i try to picture what is going on in the pipe.
  6. Dec 11, 2005 #5
    The continuity equation is actually

    What is going on in the pipe is that the wall slows the fluid down and the fluid near the center needs to speed up to "make up" for the tardines of the fluid near the wall. Note that viscosity and a velocity gradients also generates vorticity. however the assumtion is that the fuid elements translate horizontally while rotating, just like a bunch of ball bearings on top of each other
  7. Dec 11, 2005 #6
    Thanks guys (or gals), i understand it now.
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