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Hydrodynamics: Pressure of water coming out of a glass

  1. Dec 6, 2009 #1
    "What pressure do you need to get water to flow at 2 m/sec coming out of a hole?"

    Here is the visual of a container sitting on top of a glass of water with a hole poked through the bottom: http://i96.photobucket.com/albums/l168/synovial/fluids.jpg

    Given: A1, A2, V2, (y1-y2)

    Find: patm

    Useful equations:
    Bernoulli's Equation:
    p1 +[tex]\rho[/tex] [tex]\cdot[/tex] g [tex]\cdot[/tex] y1 + 0.5 [tex]\rho[/tex][tex]\cdot[/tex] V12 = p1 +[tex]\rho[/tex] [tex]\cdot[/tex] g [tex]\cdot[/tex] y2 + 0.5 [tex]\rho[/tex][tex]\cdot[/tex] V22

    A1V1 = A2V2


    I'm not sure how to modify the Bernoulli's equation to combine the y1 and y2 into (y1 - y2) and to also combine the pressures into patm. Can someone please help me?
     
    Last edited: Dec 6, 2009
  2. jcsd
  3. Dec 6, 2009 #2
    Bernoullis is just a statement of conservation of energy.

    The potential energy is the static pressure = rho*g*h where rho is density and the
    kinetic energy =1/2pho*v^2.

    Look familiar? So we convert from one form to the other.

    Almost like a kinematics problem.
     
  4. Dec 6, 2009 #3
    I'm not sure I understand. Is the pho supposed to be rho or p*rho?
     
  5. Dec 6, 2009 #4
    rho throughout. But as it turns out it is a common factor leaving 2gh=v^2. As I said, does this look familiar?
     
  6. Dec 6, 2009 #5
    Yes it does... especially the 2gh=v^2.

    One last question: do I need the v^1, or can I find it from what is given?
     
  7. Dec 6, 2009 #6
    Well if you mean v^1 is velocity, yes that is what we are trying to solve for as in equal to 2m/second. So yes you need to take the square root if that is what you are asking.
     
  8. Dec 6, 2009 #7
    Yes... thank you!
     
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