Bottle with a hole, with straw through the lid

bernoulli-equationflowfluid dynamicsfluid-staticspressure

Consider a bottle
filled with water,
with an air tight lid,
with a hole at height h.
It's well known that the water doesn't flow out when the lid is closed.
Now, if a straw is inserted through the lid, while maintaining air tightness,
water flows out if and only if the bottom end of the straw is higher than the hole.

An illustration: https://twitter.com/Rainmaker1973/status/1773432241089032434

enter image description here

Can some one explain with equations? (please also describe symbols used.)

Note:

  • I see that this can be partly simplified to "bottle with two holes" problem (related paper )

Best Answer

It is clear that the water tap works by regulating the air supply through bubbles in the red straw, but how it does that is interesting.

The bubble formation only happens at a threshold pressure. Since water pressure has a linear vertical gradient, he can control when this threshold is crossed by moving the straw up and down. Let me explain.

"For any pipe immersed in a fluid, the fluid will flow out only if the pressure inside the pipe exceeds the pressure outside (plus surface tension, which we neglect here)" Let's call this argument $(1)$.

enter image description here

Let $a$ be the distance between the water surface and the blue pipe, h be the total water depth, $\rho$, $g$, etc have the usual meanings.

enter image description here

Water pressure at any depth is given by $\rho g d + P_0$, where $P_0$ is the pressure exerted by air pocket on the water surface ($P_0<P_{air}<P_0+\rho g h$).

enter image description here

So when water flows, the air pocket expands, and $P_0$ reduces, finally reaching an equilibrium $P_0$ where water flow through the blue pipe stops.

This happens when the water pressure at depth $a$ equals air pressure (apply $(1)$ to blue pipe).

$$P(a)=P_0+\rho g a=P_{air}\tag 2$$

Coincidentally, this is also the pressure needed to make air bubbles in the red pipe (apply $(1)$ to the red pipe).

$$P(d)<P_{air}\tag 3$$

Combining $(2)$ and $(3)$, we get

\begin{align*} P(d)&<P(a)\\ P_0+\rho g d&<P_0+\rho g a\\ d&<a \end{align*}

In other words, bubbles can be made on the red pipe only when the tip of the red pipe is above the level of the blue pipe, which is exactly what we observe. These bubbles reduce the pressure deficit in the air pocket, allowing water to flow.

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