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Trouble with fluid dynamics

  1. Mar 18, 2013 #1
    1. The problem statement, all variables and given/known data

    5. A rectangular opening on the side of a water tanks has a width of L. The upper part of the opening it's in a height of h1 underneath the water y and the bottom part it's in h2 . prove that the volume of water that leaves the tank is given by :
    2/3(L)√(2g(h1³-h2³))
    2. Relevant equations
    Bernoulli's law
    e2dd757ae8209ce6fed45947ad9ad43b.png
    Continuity equation

    d8ceecc84d9efdc2ac536c90630b71c2.png
    Volumetric flux
    ce0b80a445b4435285d84193ddf63b32.png
    both V and v are Speed.

    3. The attempt at a solution
    No idea so far.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 18, 2013 #2

    tiny-tim

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    hi juanitotruan77! welcome to pf! :smile:

    imagine the opening is divided into tiny horizontal openings of height dh, and integrate from h1 to h2

    what do you get? :wink:
     
  4. Mar 18, 2013 #3
    I think i get it, i'll try it. But i'm not very familiar with using integrals. Can you explain me how?

    thanks, btw.
     
  5. Mar 18, 2013 #4

    tiny-tim

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    try first

    see how far you get :wink:
     
  6. Mar 18, 2013 #5
    i'm trying, but i don't know how to get to torricelli's law from bernoulli's in a rectangular area.
     
  7. Mar 18, 2013 #6

    tiny-tim

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    if dh is small enough, you can regard the pressure as constant

    so find the speed v from Bernoulli's equation
     
  8. Mar 18, 2013 #7
    i can't, i don't understand where i have to put the differential of h. I think i should ask a teacher tomorrow.
     
  9. Mar 18, 2013 #8

    tiny-tim

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    the rectangle is length L, and starts at height h, and finishes at height h + dh

    you can regard the pressure as being constant, = ρgh

    find v, then multiply by the area (L*dh) to get the flow :smile:
     
  10. Mar 18, 2013 #9
    oh, that sounds reasonable, thanks :D
     
  11. Mar 18, 2013 #10
    well, i tried, but i couldn't make it work.
     
  12. Mar 19, 2013 #11

    tiny-tim

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    (just got up :zzz:)

    show us what you did :smile:

    hint: 2/3(L)√(2g(h1³-h2³)) = (L)√(2g) times [(2/3)h13/2 - (2/3)h23/2] :wink:
     
  13. Mar 19, 2013 #12
    Got it, bro, i already finished, i think i was cofused with the h terms. I used y terms instead. it's quite easy now that i did it. lol. Thanks.
     
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