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Finding the Flux Problem

  1. Dec 10, 2013 #1
    1. The problem statement, all variables and given/known data
    Let v = -j be the velocity field of a fluid in 3 dimensional space. Computer the flow rate through the T of a triangular region with vertices (1,0,0) (0,1,0) (0,0,1) oriented with upward pointing normal vector


    2. Relevant equations



    3. The attempt at a solution
    So can I use stokes theorem here to find the flux?
    ∫∫∇xF dS where F is the velocity field?

    and then since this is a vector suface integral dS will equal n du dv?
    What are my limits of integration? How do I parameterize this triangle in terms of u and v?
     
  2. jcsd
  3. Dec 10, 2013 #2

    tiny-tim

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    Hi PsychonautQQ! :smile:
    You won't have to bother with parametrisation if you choose your surface(s) so that the integral is obvious …

    so can you choose them perpendicular to the three axes? :wink:
     
  4. Dec 10, 2013 #3
    I'm a little confused on what you mean by choose them perpendicular to the three axes. Perpendicular to the triangle you mean?

    On the other hand I was thinking, do I even need calculus for this problem? Can't I just see that the vector only has a k component, and that if I project the triangle down to the xy plane it has an area of (1*1)/2 = .5.
    So could the flux from the velocity vector of -j just be -1*.5 = -.5?
     
  5. Dec 10, 2013 #4

    tiny-tim

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    Hi PsychonautQQ! :wink:
    I meant the x y and z axes.
    Yes! (but don't you need to mention the divergence theorem?) :smile:
     
  6. Dec 10, 2013 #5
    Mention divergence theorem? does divergence theorem give the flux?

    ∫∫∫(∇ dot F) dV?

    that would mean
    ∫∫∫(-1) dxdydz where x y and z all go from 0 to 1? which would give the answer of -1 rather than .5 ;-(

    My final answer is supposed to be a "flow rate" through the triangular area with an upward pointing normal vector. I don't know the units for this, and can a flow rate be negative like the -.5 answer that you approved of?
     
  7. Dec 10, 2013 #6

    LCKurtz

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    There are shortcuts as you and TinyTim are discussing, but at this stage of your work I would suggest you don't use them. Remember your parameters ##u,v## can always be chosen among the original ##x,y,z## variables if convenient. In this problem your flux is in the negative ##y## direction (not ##z##) so is perpendicular to the ##xz## plane. So why not try letting ##x=x,~z=z,~y =1-x-z## be your parameterization:$$
    \vec R(x,z) = \langle x,1-x-z,z\rangle$$and work out the flux integral. You may still see some shortcuts arise.
     
  8. Dec 10, 2013 #7

    tiny-tim

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    yes … that's basically why it's called divergence! :wink:

    no, you're missing the point …

    you don't need to do a ∫∫∫, because divF = … ? :smile:
    yup! :biggrin:

    that's why they specified "upward pointing normal" …

    if it was downward pointing normal, the flux would be minus that, ie +.5

    (and the flux is presumably volume per time: if v was given in metres per second, then that would be cubic metres per second, but v is unrealistically given without units, so the flux will have to be written without units)
     
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