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Hydrodynamics: Pressure of water coming out of a glass

 
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Dec6-09, 09:49 PM   #1
 

Hydrodynamics: Pressure of water coming out of a glass


"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/l1...ial/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?
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Dec6-09, 10:05 PM   #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.
Dec6-09, 10:13 PM   #3
 
I'm not sure I understand. Is the pho supposed to be rho or p*rho?
Dec6-09, 10:15 PM   #4
 

Hydrodynamics: Pressure of water coming out of a glass


rho throughout. But as it turns out it is a common factor leaving 2gh=v^2. As I said, does this look familiar?
Dec6-09, 10:18 PM   #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?
Dec6-09, 10:41 PM   #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.
Dec6-09, 10:46 PM   #7
 
Yes... thank you!
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