# How do simple bird waterers work?

I made a simple bird waterer made from two 1.25L PET bottles. Let's call these two bottles the reservoir and the tray.

Setup

The tray is a bottle with the top cut off to leave only the bottom third. The location of the horizontal slice really only matters to help stabilise the reservoir which will be inverted and placed into the tray. A window is cut into the tray to allow a bird to duck its head in and take a drink.

The reservoir is a whole bottle (compressed in the depicted image for formatting) with the lid removed and a small hole drilled a few cm's up (when inverted) from the neck. The location of this hole is important as it is a negative feedback loop to control the water level in the tray.

Operation

Fill the reservoir with water. Invert it and sit it in the tray. The water will drain from the reservoir through the neck until the tray is full of water and no more. As water is taken from the tray, more water drains into the tank to fill it. Once the tray is full, the water stops draining and remains held in the reservoir.

See a video of it here:

Explanation

The way I thought it would work is the air hole allows air to leak into the reservoir which keeps the air pressure the same as the outside. Water flows out through the neck since the air pressures are matched outside and inside the reservoir. When the water level reaches the air hole, the hole is sealed and the 'vacuum' in the top of the reservoir increases until the air pressure pushing down on the water sitting in the tray prevents the escape of any more water.

This is not what happens though. The water stops rising before it reaches the air hole. You can see this in the video - I've annotated a blue box where air hole is.

So the question is - how does this really work?

I'm thinking it is to do with air pressure but also possibly surface tension. If the bored air hole was too large, intuitively the water would just fall out no matter what. Why though if the air pressure in the reservoir is lower than outside, doesn't the air continue to push through the air hole until it is sealed?

• I'll try to respond later, but in the mean time what you want to do is solve pascal's rule. $P=P_o+\rho g h$. In a connected system such as yours, water at the same level has to have the same pressure. Once this state is achieved you'll see your system stabilise.
– DLV
Dec 9, 2014 at 20:30
• ??? As far as I can tell in the video, the water continues to rise in the tray until the air hole is covered, and does not stop before. (I don't see your blue box, but I think I can make out the air hole on the right side.) So your original explanation is pretty much right.
– pwf
Dec 9, 2014 at 22:14
• I neglected to mention I had to drill the hole twice. The first hole was too low and the water stopped short of the tray window. The hole on the right is the lower one, there is a higher hole more toward the center of the tray window and you can see the bubbles originating from that point. I figure the first lower hole is not part of the problem, it can be approximated away as being the same as the neck hole.
– fiat
Dec 9, 2014 at 22:26
• I see now - the pressure behind the higher hole will be slightly lower than 1 atm when the flow stops, but not enough below to suck in any more air. I guess this is due to surface tension, as you say. Try making the hole bigger to see if there's a size where the surface tension can't overcome it, or adding soap to reduce the surface tension to see if the effect changes.
– pwf
Dec 9, 2014 at 22:36
• I have better footage now. youtu.be/gVNmyBTaJ9s Something I didn't realise until reviewing it is that at equilibrium the bodies of water in reservoir and tray are connected through the air hole trickle
– fiat
Mar 21, 2015 at 12:04

Surface tension is plausible : it implies that, at the level of the hole, the pressure in water will be less than 1 atm. Thus the pressure isolines will look like this:

Quantitatively, curvature $c$ will be of the order $1/(1$mm$)$ because the hole looks about 1 mm size in your video, which with pure water would lead to a pressure difference drop of the order of 70 Pa across the interface. Then the flat interface with atmospheric pressure needs to be $c\gamma/(\rho g)$ lower than the hole, which is about 7 mm with pure water. Because you probably have some impurities, this will be less in practice. As @pwf suggests, you can use surfactants (soap) to reduce it (but that won't be a good thing for the birds!)

• thank you for this, very interesting! My bird waterer is now in production and I can't perform further tests on it but am working my way through some more soft drink. I plan to put the air hole at various heights, increase size until the effect stops (surface tension breaks and it leaks) and will try surfactants. Any other variables I should play with before reporting back?
– fiat
Dec 16, 2014 at 11:15
• If you manage to control accurately the size of the hole, then you can try to make a quantitative check -- and you'll have designed an original measurement apparatus for surface tension. You can also use distilled water so that you can use the known value of water surface tension, but maybe you should check how to clean your apparatus first so that it doesn't get contaminated -- and even then you may have dust or air pollutant contamination in the course of the experiment. (Temperature is also important) Dec 16, 2014 at 12:16
• My Dremel skills didn't let me control hole size very accurately and the water is just tap but I captured better footage of the effect in action youtu.be/gVNmyBTaJ9s
– fiat
Mar 21, 2015 at 12:07
• Nice video. The difference in dynamics may be due to the energetics of bubble break-up: with fixed other conditions, you'll have approximately the same bubble size. But break-up will happen either close to the hole for small bubbles, and away from it for large ones. There is more viscous dissipation through friction with the bottle in the first case, hence one possible reason for slowing down when the pressure force is lower (close to equilibrium). Mar 23, 2015 at 14:59

No surface tension, just equalising pressures.the pressure outside the tray pushing on water surface into the air hole is equal to 1atm + water height above the top edge of air hole gdensity of water and this equals the pressure inside the bottle pushing water on the air hole to go outside which is the height of water * g * water density + air pressure above the surface inside the the bottle which is definitely less than 1atm by an amount equal to the pressure of water column above the edge of air hole outside .To see it better the longest the bottle the higher the column of water above air hole is.