- #1
frypan99
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(This is more of a conceptual question than a real homework question; thank you all so much for your help though!) :D
1. Homework Statement
Let's say that I have a large soda bottle. I drill a small hole through the side of it, put my finger over it to seal the hole, and fill the bottle up with water. When I let go of the hole, water flows out of the hole. The pressure at the top of the bottle and at the hole is both one atmosphere.
Now, I plug the hole back up and fill it with water. This time, I cap the water bottle tightly (assume the cap perfectly seals the top of the water bottle). When I let go of the hole this time, the water does not flow out of the hole.
What are the new pressures (relative to the old pressures) at the top of the bottle, and at the hole?
Bernoulli's Equation (or so says my problem)
P1 + p * g* h1 + 1/2 * p * (v1)^2 = P2 + p * g* h2 + 1/2 * p * (v2)^2
So the pressure at the top of the water bottle is zero because it experiences no pressure from water or pressure from the atmosphere. So that part's done.
Now, the velocities at both ends of the water bottle is zero because no water is flowing.
Let point 1 be at the top of the bottle, and point 2 be at the hole. So our new Bernoulli's equation is:
p * g* h1 = P2 + p * g* h2
Therefore, the pressure of point 2 is less than one atmosphere?
(I guess something's wrong with either my method or the way I'm thinking? Why is the pressure reduced at the hole when I seal the cap??)
1. Homework Statement
Let's say that I have a large soda bottle. I drill a small hole through the side of it, put my finger over it to seal the hole, and fill the bottle up with water. When I let go of the hole, water flows out of the hole. The pressure at the top of the bottle and at the hole is both one atmosphere.
Now, I plug the hole back up and fill it with water. This time, I cap the water bottle tightly (assume the cap perfectly seals the top of the water bottle). When I let go of the hole this time, the water does not flow out of the hole.
What are the new pressures (relative to the old pressures) at the top of the bottle, and at the hole?
Homework Equations
Bernoulli's Equation (or so says my problem)
P1 + p * g* h1 + 1/2 * p * (v1)^2 = P2 + p * g* h2 + 1/2 * p * (v2)^2
The Attempt at a Solution
So the pressure at the top of the water bottle is zero because it experiences no pressure from water or pressure from the atmosphere. So that part's done.
Now, the velocities at both ends of the water bottle is zero because no water is flowing.
Let point 1 be at the top of the bottle, and point 2 be at the hole. So our new Bernoulli's equation is:
p * g* h1 = P2 + p * g* h2
Therefore, the pressure of point 2 is less than one atmosphere?
(I guess something's wrong with either my method or the way I'm thinking? Why is the pressure reduced at the hole when I seal the cap??)