Bernoulli's equation and (mostly) sealed containers.

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SUMMARY

This discussion centers on the application of Bernoulli's Equation in a sealed soda bottle scenario. When a hole is drilled and sealed, the pressure at the top of the bottle is initially one atmosphere. However, when the bottle is capped tightly, the pressure at the hole decreases, preventing water from flowing out. The analysis reveals that the pressure at the hole (P2) becomes less than one atmosphere due to the sealed environment, demonstrating the principles of fluid dynamics and pressure differentials.

PREREQUISITES
  • Understanding of Bernoulli's Equation
  • Basic concepts of fluid dynamics
  • Knowledge of pressure measurements (atmospheric pressure)
  • Familiarity with hydrostatic pressure principles
NEXT STEPS
  • Study the implications of pressure differentials in fluid systems
  • Explore applications of Bernoulli's Equation in real-world scenarios
  • Learn about hydrostatic pressure and its calculations
  • Investigate the effects of sealing containers on fluid behavior
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Students in physics or engineering, educators teaching fluid dynamics, and anyone interested in the practical applications of Bernoulli's Equation in sealed systems.

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?

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??)
 
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Assume the pressure at the top of the bottle was iniitally 1 atm when it was sealed, not zero.

Then again, instead of a hole, just assume that the pressure at the bottom of the bottle is 1 atm, and if so, then what is the pressure at the top of the bottle? You can assume that density doesn't change significantly within the height of the bottle.
 
Once the hole is opened, at standard sea level conditions the backpressure at P2 can be assumed to never be less than 14.7 psia.
 

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