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Hydrodynamics: Pressure of water coming out of a glass |
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| Dec6-09, 09:49 PM | #1 |
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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? |
| Dec6-09, 10:05 PM | #2 |
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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 |
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I'm not sure I understand. Is the pho supposed to be rho or p*rho?
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| Dec6-09, 10:15 PM | #4 |
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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?
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| Dec6-09, 10:18 PM | #5 |
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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 |
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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.
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| Dec6-09, 10:46 PM | #7 |
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Yes... thank you!
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