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Bernouilli's law help

  1. Nov 29, 2004 #1
    A tank is filled with water to a height H. A hole is punched in one of the walls at a depth h below the water surface. (a) Show that the distance x from the base of the tank to the point at which the resulting stream strikes the floor (b) Could a hole be punched at another depth to produce a second stream that would have the same range? If so, at what depth? (c) At what depth should the hole be placed to make the emerging stream strike the ground at the maximum distance from the base of the tank?

    i know the ans of (a) is x = 2[h(H-h)] ^1/2, i wanna ask.....how to prove
    (b) & (c) also
     
  2. jcsd
  3. Nov 29, 2004 #2
    Don't you need the velocity of the water escaping through the hole before you can solve any of the problems?
     
  4. Nov 29, 2004 #3
    what about Bernouilli's law or the continuity law...

    marlon
     
  5. Nov 29, 2004 #4
    Still, don't you need the velocity at which the height of the water in the tank is decreasing? Unless the tank is big enough so that you can effectively set this to zero, I don't see how to go about solving part (a) by just knowing heights.
     
  6. Nov 29, 2004 #5

    Doc Al

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    Bingo! That's the typical assumption. :smile:
     
  7. Nov 29, 2004 #6
    OK. So given this assumption we have:

    [tex]\rho g H = \rho v^2 + \rho g (H-h)[/tex]

    right? Then I get for x:

    [tex]x = \sqrt{2(H-h)(H+h)}[/tex]

    This is different from what mousesgr got. Is this right?
     
  8. Nov 29, 2004 #7

    Doc Al

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    You forgot a factor of 1/2. The efflux speed is [itex]v = \sqrt{2 g h}[/itex].
     
  9. Nov 29, 2004 #8
    Oh yeah. You'll have to forgive me. I'm a little rusty (actually I'm rusted out but anyways). So the answer to part (b) would be no and for (c) we would just maximize x using calculus.
     
  10. Nov 29, 2004 #9

    arildno

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    Note also that if you couldn't do this approximation, you couldn't use the approximation of STATIONARY FLOW.
    (the domain would change in time, and hence, the velocity would change locally in time as well..)
     
  11. Nov 29, 2004 #10
    i know how to find the velocity, but how to find x?
     
  12. Nov 29, 2004 #11
    i know how to do part(A)
    thx everybody
     
  13. Nov 30, 2004 #12
    part (b) is no???
    why?
     
  14. Nov 30, 2004 #13
    Look at your expression for x. If (b) is yes, then there are two values of h, say h1 and h2 such that x(h1) = x(h2) and h1 != h2. Just from looking at the expression I can tell there are no such h1 and h2 (but I'm no math god so maybe I'm totally wrong). If you're unsure, graph x and find out.
     
  15. Nov 30, 2004 #14

    Doc Al

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    For part b: Use the answer for part a and solve it backwards. Looks like a quadratic equation to me.
     
  16. Dec 1, 2004 #15
    by observation...the ans is h
     
  17. Dec 1, 2004 #16

    Doc Al

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    What does that mean? A little voice spoke to you? :rofl:

    Just teasing... Why don't you try actually solving it now. Let [itex]h_1[/itex] be your original height h, so [itex]x = 2\sqrt{h_1(H-h_1)}[/itex]. Now solve the quadratic equation [itex]x^2 = 4h(H-h)[/itex] for h. The quadratic will have two solutions. (One solution is [itex]h = h_1[/itex], of course. But what's the other?)
     
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