How does the Lota Bowl water experiment work?

[Xenon02]

Hello!

Can anyone help me understand why this experience works this way?

If we plug the hole on the left side, the water does not affect the cup that has the hole on the bottom. However, if we let go of this hole on the left side, the water will start to pour into the cup again.
Is the law "Connected utensils" isn't implied here? Because normally the water should equal it's water level with the other connected vessels/cups.
I heard something about underpressure creating in the bottom hole but I don't understand it at all.

Could someone please explain me how does it work? I'm very curiouse how it works.

Thanks for reading this post.

 

[jbriggs444]

As I understand the drawing, you have an empty transparent plastic cup seated in a fishbowl so that there is a tight seal between the lip of the cup and the sides of the opening into the bowl. The cup displaces some of the water in the bowl.

As a result, there is a sealed air space inside the bowl, outside the cup and above the water line.

We suppose that the cup has been inserted into the bowl prior to a hole being bored in its bottom. A hole is then bored in the bowl and pressure is allowed to equilibriate before that hole is sealed with a finger.

Now a hole is carefully bored in the bottom of the cup without unsealing the cup from the bowl or the finger from the hole in the bowl.

You begin by asking why water does not flow up into the cup through the hole in its bottom.

Answer:

If water did flow (and a small bit will actually do so) then the volume of the air space will increase and the air pressure in that space will decrease as a result.

That would be the "underpressure" explanation you have been given.

At the top of the water in the bowl, the water pressure there is equal to the air pressure in the sealed air space.

As you go deeper into the bowl, the water pressure increases with depth. The amount by which it increases is given by $\rho g h$ where $\rho$ is the density of the water (1 gram per cc) $g$ is the acceleration of gravity and h is the depth below the water's surface.

If the pressure deficit in the air space reaches $\rho g h$ for the $h$ = the depth of the hole in the cup then the pressure difference across the hole in the cup will be reduced to zero and no further water will flow.

 

[kuruman]

Watch the video.

 

[Xenon02]

Watch the video.

I don't understand the video so I wanted to ask here. He said that the air goes through the outer hole and pushes the water up, but I'm more interested in why doesn't the water go into the cup when the whole is sealed.

As I understand the drawing, you have an empty transparent plastic cup seated in a fishbowl so that there is a tight seal between the lip of the cup and the sides of the opening into the bowl. The cup displaces some of the water in the bowl.

As a result, there is a sealed air space inside the bowl, outside the cup and above the water line.

We suppose that the cup has been inserted into the bowl prior to a hole being bored in its bottom. A hole is then bored in the bowl and pressure is allowed to equilibriate before that hole is sealed with a finger.

Now a hole is carefully bored in the bottom of the cup without unsealing the cup from the bowl or the finger from the hole in the bowl.

You begin by asking why water does not flow up into the cup through the hole in its bottom.

Answer:

If water did flow (and a small bit will actually do so) then the volume of the air space will increase and the air pressure in that space will decrease as a result.

That would be the "underpressure" explanation you have been given.

At the top of the water in the bowl, the water pressure there is equal to the air pressure in the sealed air space.

As you go deeper into the bowl, the water pressure increases with depth. The amount by which it increases is given by $\rho g h$ where $\rho$ is the density of the water (1 gram per cc) $g$ is the acceleration of gravity and h is the depth below the water's surface.

If the pressure deficit in the air space reaches $\rho g h$ for the $h$ = the depth of the hole in the cup then the pressure difference across the hole in the cup will be reduced to zero and no further water will flow.

Kuruman send a great video I was talking about. But I didn't understand your answear. I'll send the video again the item is more likely linked together

I know that the author at the end talks about Connected utensils. But I didn't understand most of his dialoge so I want to know how does it work exactly here.

 

[kuruman]

$\dots$ but I'm more interested in why doesn't the water go into the cup when the whole is sealed.

Because the hole is sealed. For water to go into the central cup from the side reservoir, air needs to come in and take its place. With the hole sealed this cannot happen.

 

[Xenon02]

Because the hole is sealed. For water to go into the central cup from the side reservoir, air needs to come in and take its place. With the hole sealed this cannot happen.

But if the we seal the bowl that already has the air. Can't the air spread in the place where water went to the another cup? The air will decrease in pressure but it can spread into the bigger vessels.

Edit:

I've found this video


But I don't get it what's vapor lock. How is it possible that the water won't pour into the outside vessels without the outer hole/vent or when we lock this hole.

 

[jbriggs444]

But if the we seal the bowl that already has the air. Can't the air spread in the place where water went to the another cup? The air will decrease in pressure but it can spread into the bigger vessels.

Pressure is what drives the water to flow. Decrease the pressure enough and it will not flow.

"Enough" turns out not to be very much: $\rho g h$.

Back of the envelope... One atmosphere is 30 feet of water. If the water is one inch deep on the outside of the cup, that amounts to a pressure of 1/360 of an atmosphere. If water in the bowl drops far enough that the air space increases in volume by a factor of 0.3 percent, the water will stop flowing into the cup.

 

[kuruman]

Take a largish juice can. Use a nail to punch two diametrically opposed small holes on the top of the can near the edge. Hold the can over the sink using your thumbs to plug the holes. Turn the can upside down, with both holes plugged. Remove one thumb only. What happens? Remove the second thumb. What happens?

 

[jbriggs444]

I like the playing card over the bottom of an inverted glass of water trick.

Edit: new link substituted. Thanks, @kuruman

 

[kuruman]

I like the playing card over the bottom of an inverted glass of water trick.

I cannot open the link. Here is one I found with a twist that I didn't know about.

 

[Xenon02]

Pressure is what drives the water to flow. Decrease the pressure enough and it will not flow.

"Enough" turns out not to be very much: ρgh.

Back of the envelope... One atmosphere is 30 feet of water. If the water is one inch deep on the outside of the cup, that amounts to a pressure of 1/360 of an atmosphere. If water in the bowl drops far enough that the air space increases in volume by a factor of 0.3 percent, the water will stop flowing into the cup.

Because the hole is sealed. For water to go into the central cup from the side reservoir, air needs to come in and take its place. With the hole sealed this cannot happen.

To sum it up.
I don't know if what I'll say is right (so you will correct me if I'm wrong).
For the water to flow it needs enough pressure. The more water it pours the less pressure it has and the air space increases. When I seal the hole that allows to get air from atmosphere the water will not pour because it doesn't have enought pressure to pour it out.

As you go deeper into the bowl, the water pressure increases with depth. The amount by which it increases is given by ρgh where ρ is the density of the water (1 gram per cc) g is the acceleration of gravity and h is the depth below the water's surface.

If the pressure deficit in the air space reaches ρgh for the h = the depth of the hole in the cup then the pressure difference across the hole in the cup will be reduced to zero and no further water will flow.

That's how I insert pgh equation here? When the h = the depth of the cup it's normal that it won't pour anymore inside the cup but why the air is important in this equation? Because water pressure is exactly pgh and for the water to flow it needs it's pressure and why if air losses pressure not water but air then the water will not flow?

Or if the air pressure equals pgh (water pressure) then the water won't flow because the air from outside has higher pressure than the air inside the vessel?

Does Water tower also requires air and water pressure to flow?

 

[jbriggs444]

That's how I insert pgh equation here? When the h = the depth of the cup it's normal that it won't pour anymore inside the cup but why the air is important in this equation? Because water pressure is exactly pgh and for the water to flow it needs it's pressure and why if air losses pressure not water but air then the water will not flow?

It is the pressure difference between two points in the water that is given by $\Delta P=\rho g \Delta h$

If the pressure at the top of the water is less than atmospheric by $\rho g h$ then the pressure at depth $h$ will be exactly equal to atmospheric.

When water pressure at the hole in the cup is equal to atmospheric pressure at the hole in the cup, nothing flows through the hole.

Fundamentally, hydrostatics is nothing but Newton's laws and common sense.

 

[Xenon02]

It is the pressure difference between two points in the water that is given by $\Delta P=\rho g \Delta h$

If the pressure at the top of the water is less than atmospheric by $\rho g h$ then the pressure at depth $h$ will be exactly equal to atmospheric.

When water pressure at the hole in the cup is equal to atmospheric pressure at the hole in the cup, nothing flows through the hole.

Fundamentally, hydrostatics is nothing but Newton's laws and common sense.

I thought that the water surface pressure equals = 0 because h = 0 so pgh = 0 but it seems that the water pressure in the surface of a vessel is equal air pressure.
So when the water goes out of the vessel when it's sealed the air pressure reduces to pgh that is not a lot as I read and blocked the water to flow out side because pgh is equal to air pressure outside of the vessel. Like in Newton first law.

Usually when I was looking at most of the experiments I was looking only to water pressure. I didn't know that
air pressure will be very important

But what is funny is that when the water pour out the vessel its pressure also decreases so for the air to reach pgh can be really fast.

 

[jbriggs444]

I thought that the water surface pressure equals = 0 because h = 0 so pgh = 0 but it seems that the water pressure in the surface of a vessel is equal air pressure.

Correct. Whenever you have a boundary between two fluids and an equilibrium, you know that the two fluids have the same pressure at the boundary.

That is an important principle to remember.

So when the water goes out of the vessel when it's sealed the air pressure reduces to pgh

No. It does not reduce to $\rho g h$. It reduces by $\rho g h$.

That means that its resulting pressure at the water surface becomes ambient minus $\rho g h$.

That means that the pressure at depth $h$ below the surface then is equal to ambient.

At which point no fluid flows through the hole in the cup (at depth h below the surface) because the pressures on both sides are equal.

The pressure in the air space reduces to $\rho g h$ below ambient precisely because that is the pressure that results in no further water flow.