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Bernoulli's Equation - Efflux question

  1. Dec 3, 2004 #1
    A small circular hole 6.00 mm in diameter is cut in the side of a large water tank, 14.0 m below the water level in the tank. The top of the tank is open to the air.

    A) What is the speed of efflux?

    sqrt(2*g*h) = 16.6 m/s

    B) What is the volume discharged per unit time.

    This is the equation i believe, dV/dt = A*v

    I dont know exactly how to get A..

    any Hints?
     
    Last edited: Dec 3, 2004
  2. jcsd
  3. Dec 3, 2004 #2

    HallsofIvy

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    Are you serious? hole is a circle with radius 3 mm. It's area is pi(r2)=
    9 pi square mm.
     
  4. Jun 13, 2005 #3
    I used (0.009 m)*pi * 16.6 m/s = 0.469 for the volume discharged per unit time. what am I doing wrong?
     
  5. Jun 14, 2005 #4

    HallsofIvy

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    Imagine a "one second" cylinder of water coming out the hole: it will be a cylinder 16.6 m long and with base area the area of the hole.

    But the area of a circle is pi r2, not pi r! The area of the whole is
    (0.009 m)2*pi and so the volume of water discharged in one second is
    (0.009)2*pi*16.6
     
  6. May 6, 2010 #5
    First off

    @hallsofivy: dude that's so retarded. he already squared it it's (3mm)^2 which becomes 9mm squared.

    Final answer: your answer is in fact right you're only off by a degree of 10^-3. such that the final answer should be 4.69*10^-4. I think this happened when you were converting
    mm^2 to m^2. [unit conversion's a ***** isn't it :cool:]
     
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