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Rate of flow from a leak in a storage tank

  1. Nov 23, 2006 #1
    "A large storage tank, open at the top and filled with water, develops a small hole in its side at a point 16.0m below the water level. If the rate of flow from the leak is 2.50x10^-3 m^3/min, determine; A) The speed at which the water leaves the hole and (B) The diameter of the hole.

    So I used the equation;
    P(air) + 1/2 d(density)v1^2 + pgh(total height of water) = P(air) + 1/2dv2^2 + pgh(from the bottom of the container to the leak)

    The equation then simplifies to:
    gh(total height) = 1/2V2^2 + gh(from bottom to leak)
    V2^2 = 2g(h(total) - h(form bottom to leak))
    V2^2 = 2g(h(from leak to top of water))

    Since this h is known, I can solve for v.

    I Just can't figure out the diameter part.

    Any help is appreciated.
     
  2. jcsd
  3. Nov 23, 2006 #2

    Pyrrhus

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    You're given the volume flow rate (discharge or flux), Q.
     
    Last edited: Nov 23, 2006
  4. Nov 23, 2006 #3
    okay so the equation is;
    flow rate = A(flow area)V1

    And I can find the area, but how do i find just the diameter.

    A= 4/3 pi r^2 (if it were a circular hole)

    I'm unsure because it doesn't tell us what shape the hole is. Does the shape even matter?
     
  5. Nov 23, 2006 #4

    Pyrrhus

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    Why 4/3? and yes simply use [itex] A = \frac{\pi d^2}{4} [/itex]
     
  6. Nov 23, 2006 #5
    Okay, Thanks
     
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