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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.

The pressure at the top of the bottle is now zero, right? And the velocities at both points is zero. So Bernoulli's equation goes from

$$P_1 + p g h_1 + (1/2)p v_1^2 = P_2 + p g h_2 + (1/2) p v_2^2$$

to

$$p g h_1 = P_2 + p g h_2$$

where point 1 is at the top of the bottle, and point 2 is at the hole.

It seems to me that because of this, $P_2$ must be less than one atmosphere, right?

But why is the pressure at point 2 reduced because the bottle is capped? Doesn't the atmospheric pressure still act on point 2? Am I doing something wrong conceptually, and/or from the beginning? I just can't wrap my head around it.

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When the cap is off the pressure at the top of the water is indeed atmospheric. When the cap is sealed, the pressure will be depended on the ideal gas law. As the volume is stretched by fluid leaving the pressure drops.

If there was a small volume of atmospheric pressure at the top, then water would leak out until the gas had expanded and dropped in pressure enough to balance the height difference.

However, it's likely you didn't see any water leak out because when you were closing the lid you were applying pressure to the hole which probably deformed the bottle, so when you released the pressure the bottle returned to its natural shape expanding the air inside to the point where the pressure was already balanced.

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There are two cases to consider:

Very tall column of water

The instant after you release the hole, the water pressure at the top drops to zero. (We are ignoring evaporation and surface tension.) The water pressure at the hole is then:

$$P_2 = \rho g (h_1 - h_2)$$

If the column is tall enough, then $P_2$ will be greater than atmospheric and the water will flow out. As the water continues to flow, the height of the top of the liquid, $h_1$, continues to decrease. When $h_1$ has decreased to the point where $P_2$ is atmospheric, then Bernoulli equation predicts that the water flow stops.

For atmospheric pressure at sea level, this case applies to water columns 34 feet high or taller.

Not-so-tall column

The instant after you release the hole, the water pressure at the hole, $P_2$, is atmospheric. Consequently, the pressure at the top of the column is found from:

$$P_1 = P_2 - \rho g (h_1 - h_2)$$

As long as $P_1$ is greater than zero, then the Bernoulli equation says that the water stays where it is.

In practice, due to factors not in the Bernoulli equation, such as turbulence or vibration, air bubbles will likely enter the hole. As they do, an equivalent volume of water will be released out of the hole. This processes continues until the the water level reaches the height of the hole.

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  • $\begingroup$ I doubt turbulence could contribute to air bubbles coming in , but Rayleigh-Taylor instability could, or deformation of the bottle, and vibration could help, but for small holes the surface tension would prevent that, or at least stop the effect before the water level reaches the height of the hole. $\endgroup$ – Rick Oct 8 '15 at 12:30

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