Bottle with a hole, with straw through the lid

Question

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

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 )

Answer 1

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 $(1)$.

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.

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 ($P_0<P_{air}<P_0+\rho g h$).

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.

Answer 2

In an otherwise sealed bottle with a single small hole in the side, the pressure due the height of the liquid (h) above the hole and the pressure in the air gap at the top of the bottle ($P_b$) has to be equal to atmospheric pressure ($P_a$) so that there is pressure equilibrium at the side hole.

$P_a = P_b + h \rho g$

where $\rho$ is the density of the water and g is the acceleration of gravity.

If there is a straw that is open to atmosphere with its lower end near the bottom of the bottle, the liquid level inside the straw drops until it reaches a level equal to the height (y) of the hole in the side and the atmospheric pressure acting on the top of the liquid in the straw is also equal to $P_b + h \rho g$.

If the straw is raised until its lower end is above or equal to the height of the hole in the side, air will enter and $P_b$ rises and water starts to flow out the side hole.

This equation is simplified, because it ignores the pressure $P_t$ due surface tension in the surface of water in the side hole and the pressure $P_s$ due to tension in the meniscus of the water in the straw.

Surface tension pressure at the hole is now

$P_t = \frac{2T}{r_h}$

where T is the surface tension of water (0.072 N $m^{-1}$) at the side hole and $r_h$ is the radius of the side hole. Ref: Related paper.

the extended equation for pressure equilibrium at the side hole is:

$P_a + P_h = P_b + h \rho g$

which means at equilibrium the height of the water above the hole is

$h_1 = \frac{P_a - P_b + P_h }{\rho g}$

Surface tension at the meniscus in the straw is similarly:

$ P_s = \frac{2T} {r_s}$

where $r_s$ is the radius of the straw.

the extended equation for pressure equilibrium at the meniscus inside the straw is:

$P_a - P_s = P_b + h_s \rho g$

which means at equilibrium the height of the water above the straw meniscus is

$h_2 = \frac{P_a - P_b - P_s }{\rho g}$

The level where air starts entering the straw when it is raised is therefore $h_1 - h_2$ and this is slightly above the height of the side hole.

For practical purposes, the pressure due to the meniscus in the straw is pretty small because of the larger radius and surface tension of the side hole can be ignored for holes with greater than 2mm radius. However, as the paper points out, the surface tension pressure of a side hole of 1mm radius is significant and water stops flowing when the level of water is about 15mm above the side hole and the pressure in the bottle is atmospheric.

I notice you have fluid dynamics tags on your question. If you are interested in the fluid flow rate versus head height and hole radius this is already adequately covered by the paper linked in your question.

Answer 3

First step: Fill the bottle up to 2/3 with open straw down to near the bottom, hole at the side closed. The level in the straw is equal to the level in the bottle.

Step 2. Seal the upper end of the straw and open the hole at the side. An amount of water flows out until the pressure at the hole inside equals pressure outside. This happens by enlarging the air volume with a reduction of the pressure on the water surface, such that air pressure plus water pressure donw to he hole equals outer pressure.

Step 3. Open the seal on the straw. The level in the straw is now the level of the hole by equal pressure on both openings.

Step 4- Lift the straw. The level inside remains at the hight of the hole. Lift more over the level of the hole, water pressure at the lower end of the straw is lower than the outer pressure, air is pressed into the bottle and water by the difference of hights of wtwer between straw end and hole is pressed through the hole.