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Take a look at this drawing of the situation Picture of the bottle

(Ignore that it is in spanish) it is a bottle of water that can be opened on the top of the bottle and closed. Why does the water exit the bottle faster through the little hole if I leave it open? I'm assuming it has to do with different pressures. On the case of leaving the bottle open, we are having atmospheric pressure on the top of the bottle and on the case of closing it we are having a smaller pressure than the atmospheric one? Which I'm assuming that can affect the equation of the Bermoulli's equation that $$P_0+\frac{1}{2}\rho v^2+\rho gh=constant$$ in some way? And then we can reason it through solving the velocity $v$?


EDIT: According to the comments I can solve for $v$ in the Bermoulli's equation in the closed case: $$\frac{P_1 V_1}{V_0}+\rho gh=P_{atm}+\frac{1}{2}\rho v^2\Rightarrow v=\sqrt{\frac{2}{\rho}(\frac{P_1 V_1}{V_0}-P_{atm}+\rho gh)}$$ as $P_0 V_0=P_1 V_1$ (due to $P_0$ being not constant) and $v=0$ on the top of the bottle and $h=0$ on the bottom of the bottle.

While for the open case, on the top of the bottle and on the bottom we have in both timeds the atmospheric pressure so that cancels out within each other: $$\rho gh=\frac{1}{2}\rho v^2\Rightarrow v=\sqrt{2gh}$$

So then it remains unclear if $\frac{P_1 V_1}{V_0}-P_{atm}>0$ which I don't know how to continue

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  • $\begingroup$ You're using $P_0$, but note that the pressure over the water isn't constant if the top is closed; you'll have an expanding volume with no source of air to maintain atmospheric pressure. Try applying the ideal gas law to relate the pressure over the water to the amount of water that's flowed out. $\endgroup$ Jul 23, 2023 at 18:44
  • $\begingroup$ "Ignore that it is in spanish" I couldn't. It added interest. $\endgroup$ Jul 23, 2023 at 21:05
  • $\begingroup$ @Chemomechanics I'm having troubles understanding why the pressure on the inside isn't constant, but on any case, if I were to apply $P_0 V_0=P_1 V_1$ on the closed case, how would I even compare it with the open case mathematically? $\endgroup$
    – Alysid
    Jul 23, 2023 at 21:49
  • $\begingroup$ "I'm having troubles understanding why the pressure on the inside isn't constant" A constant amount of air occupies an increasing volume as the water flows out, no? In contrast, the pressure is a constant 1 atm when the top is open, correct? $\endgroup$ Jul 23, 2023 at 21:55
  • $\begingroup$ @Chemomechanics oh yeah that makes sense, but still don't know how I would be comparing both cases nor how to continue with this logic $\endgroup$
    – Alysid
    Jul 23, 2023 at 22:50

2 Answers 2

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leaving the top open allows air to enter the container as the water flows out the bottom. if you close the top vent, then as the water leaves the container, it draws a vacuum on the inside of the container which opposes the departure of water from the hole in the bottom. Eventually the flow stops and air blurbles into the container through the bottom hole to cancel the vacuum and then the water starts flowing out the hole again, the vacuum builds, the flow stops, etc., etc.

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  • $\begingroup$ I really appreciate this answer but I'm looking for a more "mathematical" proof, I could believe this is true but I would want to have it written on an physics expression, if you know what I mean. (Also how does "the air blurbles into the container through the bottom hole to cancel the vacuum"?) $\endgroup$
    – Alysid
    Jul 23, 2023 at 21:47
  • $\begingroup$ when there is a vacuum inside the container, water stops flowing out and air can compete with water at the opening in the bottom. the air is sucked in to cancel the vacuum, then water can flow out again. the mathematical condition that establishes this is when the vacuum inside the airspace in the container equals the gravity head of the water standing in the container. when this occurs, the flow stops and air gets gulped in through the hole. $\endgroup$ Jul 24, 2023 at 3:54
  • $\begingroup$ what's that term gravity head? I've never heard it before, and do you have any name for this phenomena? It is a really not known one for me, so that I can look more into it $\endgroup$
    – Alysid
    Jul 24, 2023 at 16:19
  • $\begingroup$ gravity head is the water pressure created at the bottom of a container of liquid due to gravity pulling down on the liquid. it's what creates crushing pressure at the bottom of the ocean. $\endgroup$ Jul 24, 2023 at 17:33
  • $\begingroup$ your answer feels great but somewhat still something left that makes it so it I'm not 100% satisfied, so sorry for the insistance. But do we know why this phenomena ocurrs? Like it stills sounds pulled out of thin air that "the vacuum inside the airspace equals the gravity head of the water in the container", the rest does seem reasonable to me $\endgroup$
    – Alysid
    Jul 24, 2023 at 22:16
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It depends on the volume of air you have in the bottle when you start. Assume it is very small say 10cm, than if the water level goes down 10cm the air volume doubles and the pressure is half atmospheric pressure, when the water level goes down 20 cm , the air volume is 3 times so the pressure 1/3 and so on. So depending on the initial condition when closing the bottle, you can calculate the resulting air pressure with $P_1V_1=P_2V_2 $ also the pressure from the water goes down with diminishing high.

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  • $\begingroup$ so the edit I made in my post is fairly correct? And the answer is that "it depends on the volume of air on the bottle"? Because that would precisely meet with my equation I wrote in the edit, which depends on the initial conditions $\endgroup$
    – Alysid
    Jul 24, 2023 at 10:37
  • $\begingroup$ I can' see your initial condition V depends on h and h(0) or V(0), so you have the height of the bottle : H, over the outflow, the initial h of the water and the area A of the bottle, h(t) depends on v(t) and the aerea of the outflow, so you will get a differential equation. $\endgroup$
    – trula
    Jul 24, 2023 at 14:10
  • $\begingroup$ we will get a differential equation as well if we consider only "taking a glass of $50$ mL water from a filled bottle" for example? I didn't understood much of what you said $\endgroup$
    – Alysid
    Jul 24, 2023 at 16:21
  • $\begingroup$ The volume of air in the bottle depends on the amount of water, which left the bottle, and this amount depends on v. so v(t) and h(t) and P(t) depend on each other. Your "taking a glass of 50 mL water from a filled bottle" has nothing to do with differential equations? $\endgroup$
    – trula
    Jul 24, 2023 at 16:26
  • $\begingroup$ I see, so there is no easy way to "prove" this mathematically with the Bermoulli's equation, not even knowing the volume of air and water in both initial and final moments. Appreciate it $\endgroup$
    – Alysid
    Jul 24, 2023 at 22:02

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