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Calculate the Flux through the Pipe Part 2

  1. Dec 31, 2008 #1
    1. The problem statement, all variables and given/known data
    Okay, so this was Part 1:

    ... which I solved as follows:

    [tex]\phi=\int_{Surface} v \cdot da[/tex]

    [tex]=\int_{Surface} v*da\cos\theta[/tex]

    [tex]=\int_{Surface} v*da[/tex]

    [tex]=\int_s\int_l v*(dsdl)=v*s*l[/tex]




    Now this is Part 2: Figure 1.29

    We slice the end of the pipe off at some angle [itex]\theta[/itex]. This does not change [itex]\Phi[/itex]. Express your formula for [itex]\Phi[/itex] in terms of the dimensions s andl' and [itex]\theta[/itex].

    [​IMG]

    So is the main idea of this to use only the normal component of A in the integral? (normal to v, that is).

    Casey
     
  2. jcsd
  3. Jan 1, 2009 #2
    So how about this: if da is the normal to v

    [tex]\phi=\int_{Surface} v \cdot da[/tex]

    [tex]=\int_{Surface} v*da\cos\theta[/tex]

    [tex]=\int_{Surface} v*(ds)(dl'\cos\theta)[/tex]

    [tex]=v\cos\theta\int_s\int_{l'} *(dsdl')[/tex]

    [tex]\Rightarrow\Phi=vsl'\cos\theta[/tex]

    Now the next part of the problem says "Write the formula more compactly, in terms of the vector area a and the velocity vector v in figure 1.30"

    [​IMG]

    I think this is just the integral I did above, but skipping the part where I substitute for da

    That is [itex]\Phi=vacos\theta[/itex] correct?

    And now the last part:

    What if velocity varied (in magnitude and/or direction) from point to point, AND the end of the pipe were cut off in some arbitrary way as in fig 1.31; how would you compute the flux in the general case?"

    I am thinking there is a summation inside of the integrand for this one... something like:

    [itex]\Phi=\int_{Surface}\sum v_o\cos\theta_o\, da_o[/itex] but I dunno... looks ugly to me
     
    Last edited: Jan 1, 2009
  4. Jan 1, 2009 #3
    Stupid question but is this the same as electric flux?? Or can you view it the same way?
     
  5. Jan 1, 2009 #4

    Dick

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    phi=v*a*cos(theta) is not correct. For the first part you got v*s*l and for the second part you got v*s*l'*cos(theta). s*l' is not 'a'. Think about it some more. What's the relation between l and l'?
     
  6. Jan 1, 2009 #5

    Dick

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    You use the same sorts of mathematical tools, if that's what you mean. Things like the divergence theorem apply.
     
  7. Jan 1, 2009 #6
    s*l'=a' ---> the actual area

    So.... I want the component normal to v

    How is s*l'*cos[itex]\theta[/itex] not the component that is normal to v?
     
  8. Jan 1, 2009 #7

    Dick

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    Don't you mean the component that is parallel to v? I think the point they are trying to make here is that v*s*l=v*s*l'*cos(theta). The amount of water coming out of the pipe doesn't depend on the shape of the end. Your scan of the problem is sort of cut off. So I hope that's the point they are trying to make.
     
  9. Jan 1, 2009 #8
    Isn't that the component of v normal to the surface a'?

    Also, I'm wondering if the answer to your second question is just to use the method you used for the first questions, but without being able to factor out the terms involving theta or v since they would not be constant but instead functions of s and l. I think you could get the answers to the first questions without using a surface integral
     
  10. Jan 1, 2009 #9
    Are we looking at the same diagram? Post #1 figure 1.29 ?

    [​IMG]

    If v runs along the longitudinal axis of the pipe and I want the component of a that is perpendicular to it then I want v*s*l where l=l'cos[itex]\theta[/itex]

    So how is [itex]\Phi=v*s*l'*\cos\theta[/itex] not correct? What (big idea) am I missing Dick?

    Casey
     
  11. Jan 1, 2009 #10

    Dick

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    v*s*l'*cos(theta) IS correct. v*s*l is also correct for the first case. And they are equal. ANY surface that cuts across the end of the pipe has the same flux through it. I think that's the 'big idea'. Or did you already know that and am I missing the point of this exercise?
     
  12. Jan 1, 2009 #11
    Oh wait... I forgot what we are talking about :redface:

    It's the writing it more compactly thing I messed up. Wait, I just noticed this...

    and now I am confused and I do not know why....

    I think this is why, the definition is [itex]\int_{surface}\mathfb{v}\, d\mathbf{a}[/itex] where a is the area parallel to v, but the direction of a is given by the normal direction to a..... okay, my mistake.

    Back to the question of how write v*s*l'*cos(theta) more compactly....
     
  13. Jan 1, 2009 #12
    Dick, looking at this diagram

    [​IMG]

    why wouldn't [itex]\Phi=va\cos\theta[/itex] be correct?
     
  14. Jan 1, 2009 #13

    Dick

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    Sometimes questions can be a little obscure, but it says write it in terms of the vectors 'v' and 'a'. In v*s*l'*cos(theta) v is the length of the vector 'v', s*l' is the length of the vector 'a' and cos(theta) is, uh, the cosine of the angle between them. It's dot product, right?
     
  15. Jan 1, 2009 #14

    Dick

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    It depends on which 'a' you mean. I thought you meant a=l*s from the original problem. If you mean the new area l*s' then it's fine. Looks like a dot product, doesn't it?
     
  16. Jan 1, 2009 #15
    Sure didily does. How about this one:
    [​IMG]

    The area is arbitrary and v varies with position.

    I started with this: [itex]\Phi=\int_{Surface}\sum v_o\cdot\cos\theta_o\, da_o[/itex] but it looks kind of lame to me....
     
  17. Jan 1, 2009 #16

    Dick

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    How about just writing the integral of [itex] \vec v \cdot \vec n da[/itex] where n is the normal vector? Or [itex]\vec v \cdot d \vec a[/itex] where [itex] d \vec a=\vec n da[/itex]? Is that what you are trying to express?
     
  18. Jan 1, 2009 #17
    Oh sure. Do it the easy way :smile: This was the original question:

    [​IMG]

    "What if velocity varied (in magnitude and/or direction) from point to point, AND the end of the pipe were cut off in some arbitrary way as in fig 1.31; how would you compute the flux in the general case?"

    So [itex] \int_{s}\vec v \cdot \vec n da[/itex] would take care of it? Is that integral actually do-able? (not that it said it had to be... just curious) Since you would need to know the angle between v and a at every point....
     
  19. Jan 1, 2009 #18
    Like I said, I think it's just the same integral you used before except now v and theta aren't constant
     
  20. Jan 1, 2009 #19

    Dick

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    If you know an actual equation for the surface A and the vector field v then you could certainly write down an integral.
     
  21. Jan 1, 2009 #20
    Cools. I have never see you use itex before Dick..... I am honored. This book is starting to piss me off. But I feel like I have to do all of the exercises anyway.

    These are just the introductory "math chapters" to Griffith's Intro to Electrodynamics.... what a silly way to spend my X-mass break. I must really hate myself this year.

    Thanks
     
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