Hydrodynamics: Pressure of water coming out of a glass

AI Thread Summary
To determine the pressure required for water to flow at 2 m/sec from a hole in a glass, Bernoulli's equation is applied, which relates pressure, potential energy, and kinetic energy. The equation can be modified to express the difference in height (y1 - y2) and atmospheric pressure (patm). The conversion between potential energy and kinetic energy leads to the relationship 2gh = v^2, where g is the acceleration due to gravity and h is the height difference. The initial velocity (v1) is not needed if the final velocity (v2) is known, as it can be derived from the given parameters. Understanding these principles allows for the calculation of the necessary pressure to achieve the desired flow rate.
qnney
Messages
10
Reaction score
0
"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 +\rho \cdot g \cdot y1 + 0.5 \rho\cdot V12 = p1 +\rho \cdot g \cdot y2 + 0.5 \rho\cdot 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:
Physics news on Phys.org
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.
 
I'm not sure I understand. Is the pho supposed to be rho or p*rho?
 
rho throughout. But as it turns out it is a common factor leaving 2gh=v^2. As I said, does this look familiar?
 
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?
 
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.
 
Yes... thank you!
 
Back
Top