# Inverted Pascal's barrel

Pascal's barrel is an experiment that shows the counterintuitive nature of how pressure increases linearly in depth according to $$p=p_0+\rho gz$$, where $$z$$ is the depth. The experiment consists of a barrel and a long hosed attached to it. The hose is raised far above the barrel. By pouring a relatively small amount of fluid inside the hose the depth $$z$$ inside the barrel increases quickly because the depth is measured from the highest point exposed to air. If setup correctly this can cause the barrel to burst using only a small amount of fluid.

Now I imagined what would happend for the "opposite" of this. Imagine the hose goes down instead of up. The hose starts closed off but at a certain moment the seal is broken and the bottom is exposed to atmospheric pressure. I have the following questions

1. What is the pressure distribution before the seal is broken? (left picture)
2. What is the pressure after the seal is broken?
3. Can the pressure go negative given a long enough hose?
4. Will the barrel implode violently? • This is an interesting variation on the original experiment. Dec 27, 2022 at 22:51
• Also, I found a YouTube video demonstrating this here. The commentator is speaking a non-English language, so you will have to turn on subtitles (assuming you cannot speak the language). Dec 28, 2022 at 0:04
• Not an answer to this question, but helpful - How Trees Bend the Laws of Physics. Dec 28, 2022 at 3:13

The culmination of this experiment will be similar to the original Pascal's barrel experiment but the barrel will implode (as opposed to explode as in the original experiment).

What is the pressure distribution before the seal is broken?

It will be as expected, where at any point $$z$$ pressure is $$P=P_0+\rho gz$$

What is the pressure after the seal is broken?

Nothing too spectacular at this instant. The fluid is in motion, so the pressure will be governed by Bernoulli's equation $$P_1+\rho gz_1 +\frac 12 \rho v_1^2=P_2+\rho gz_2 +\frac 12 \rho v_2^2$$ where I guess you assume the fluid to be incompressible and the flow is continuous so $$A_1v_1=A_2v_2$$ ($$A_1,v_1$$ and $$A_2,v_2$$ are the areas and speeds in the upper and lower regions respectively).

Can the pressure go negative given a long enough hose?

The differential pressure ($$P_{\text{inside}}-P_0)$$ will be less than zero since there is a low-pressure region forming in the upper part of the larger container as the fluid is drained from the bottom hose.

Will the barrel implode violently?

This is where things will get interesting. There will be a point where the pressure in the upper part of the container becomes small enough that the atmospheric pressure is enough to cause the barrel to collapse on itself. I would be extremely surprised if the barrel did not suffer this fate (of course you assume the barrel is made from usual materials and not made of some indestructible substance, and the hose is long enough).

• Also, I found a YouTube video demonstrating this here. The commentator is speaking a non-English language, so you will have to turn on subtitles (assuming you cannot speak the language). Dec 28, 2022 at 0:03
• Very nice. I'm still a bit confused about 1) though. When a fluid is exposed to air you know that plane is at atmospheric pressure so at depth $z$ the pressure is $\text{atmpospheric pressure} + \rho g z$. When the container is fully closed there is no reference pressure so that's what confuses me. Dec 28, 2022 at 13:26
• You're right. You're not confused at all, and excuse my shorthand for additional pressure $\rho gz$. Since the fluid was put into the barrel at atmospheric pressure, the top of the barrel and fluid remain at atmospheric pressure. But the pressure inside the barrel at depth $z$ is $P=P_0+\rho gz$ I've made an edit to the answer. Cheers. Dec 28, 2022 at 22:45

In the static case (left image) this is in essence the same as a mercury barometer, only with water. The column would be about 10 meters high to reach close to 0 Pa at the very top. There, a gas would form: water vapour in equilibrium with the liquids phase below, i.e. the gas at vapour pressure of water (approx. 30 mbar=3000Pa at 25C).

The dynamic case reminds me of a Sprengel pump, where falling mercury creates a vacuum above. https://en.wikipedia.org/wiki/Sprengel_pump

Q1: What is the pressure distribution before the seal is broken? (left picture)

Answer 1) If the vessel was filled at ambient pressure completely and you have a stiff container, the pressure inside will be one bar for the liquid. Above a vapour phase will form at the vapour pressure of water at the given temperature for 1bar. Going down the column the pressure rises with depth linearly.

Q2: What is the pressure after the seal is broken?

Answer 2: After the seal is broken, water, initially at rest, will be accelerated, by a large force due to the high hydrostatic pressure. This will quickly increase the gas volume at the top, hence decrease the pressure in this part of the vessel. As a consequence the water at the top could start to boil even at room temperature, filling the vacuum with water vapour.

Q3: Can the pressure go negative given a long enough hose?

No. The pressure might go below vapour pressure while the vessel empties. This will depend on whether the water moves out alone or bubbles can enter to travel up, causing friction and letting water out only slowly.

In a really long column though, the opening of the valve causes a huge pressure drop, which might travel upwards the column (with speed of sound) and hit the top very hard. This effect is known as water hammer in pipes when valves are opened and closed: https://en.wikipedia.org/wiki/Water_hammer.

Q4: Will the barrel implode violently?

Answer 4: Unlikely from static pressure. The load on the vessel at the top is maximal 1 bar from the outside. If it withstands this in the beginning, it should be fine later also.

Dynamic effects like to one above can become important for very long tubes. But they would not cause an implosion, rather an explosion.