Smallest hole in top of container to allow water to flow out hole size X in bottom of container?

I was dumping some old 1L bottles of soda down the sink by putting the bottle upside down into the drain. The soda was glug-glugging down the drain since air was coming up through the spout while soda was going down. It would drain faster if I punched a hole in the bottom of the bottle since air didn’t have to enter at the spout anymore.

This got me thinking: What was the smallest hole I could make so that the soda still flowed out as fast as possible? So between these ranges:

Max hole: Cut the entire bottom of the soda bottle off. Soda flows out at rate A
Min hole: Use a pin to make a tiny hole in the bottom. Soda flows out at rate B

What is the smallest hole I can make so that the fluid still flows at rate A? And will the answer always be a percentage of the size of the drain hole?

If the shape of a real soda bottle complicates things, just assume it’s a perfect cylinder with a flat top and bottom.

Someone will be along with the math soon (I think it might be as simple as the ratio of mass and viscosity of the two fluids involved) - but I believe it will be a fairly small hole, because if you invert a bottle of water, then insert a bendy drinking straw up into it, it will drain very fast, just from the admission of air through the straw.

Or swirl the contents and the “eye” of the swirl allows air to enter … and looks cool.

It doesn’t answer the actual question (I hope that doesn’t violate GQ rules), but another way to empty bottles quickly is to swirl them around when they are upside down (to make a tornado-like thing in the bottle). Drains much faster than the glug-glugging technique.

Edited to add: Beaten to it by Caught@Work.

I don’t believe you’re going to find that there’s a minimum hole size that “soda flows out at rate A”, unless the hole is exactly what you’ve defined (entire end of bottle cut off) - each hole smaller is likely going to introduce some additional frictional losses, resulting in less head pressure available (and less soda flow). It may be more useful to define the problem as what hole size will result in something like 99% of your max soda flow rate.

On reflection, I think you’ve right - it’s going to plot as a curve - it’ll probably only flatten out to a point where (to human perception) it looks as though an increase in the air hole size makes no difference.

There will probably be a pretty sharp bend in that curve, though, sharp enough that it’s meaningful to ask where the bend is. I suspect that it’ll turn out to be something like the ratio of viscosity of water to air.

As others have said, any air hole will reduce, even if slightly, the rate of water flow. The hole size required so that the influx of air can more or less keeps up, though, can be approximated at first-order by:

r[sup]2[/sup] ~= (2gh)[sup]1/2[/sup]R[sup]2[/sup]/v[sub]sound[/sub]

where r is the radius of the inlet, R is the radius of the outlet, g is the acceleration due to gravity, h is the depth of the liquid in the container, and v[sub]sound[/sub] is the speed of sound in air (about 340 m/s).

For a 1-cm radius outlet with a 20-cm depth of liquid trying to drain, this yields r = 0.7 mm. But that’s for fully choked flow, which will slow your output by about a factor of two from optimum. Make the hole larger to approach optimum. At a factor of ten over the choked flow size (so, say, 7 mm) you’ll be pretty darn close to max flow.

Since the question has been answered (thanks Pasta, how the hell do you know these things? Fluid dynamics? We didn’t get this in EE) I’ll just also say that this will be a really cool thing to show my students.

Thanks for putting that together. But I have to ask… The speed of sound??? Why does that factor into the equation?

Because it’ll affect how quickly the air can flow in to replace the water.

The speed of sound is related to the density and the compressibility of the air, quantities that you might be less surprised to see in the relation. More directly, a fluid moving faster than the speed of sound has thrown off the last vestiges of acting anything like an incompressible fluid.

I should amend my earlier post. For an inlet of radius r the pressure inside will be about half atmospheric, but that won’t yield burp-free, slower flow unless your soda bottle holds many meters (depth) of water. Burping will happen whenever the air pressure inside drops by more than [symbol]r[/symbol]gh, where [symbol]r[/symbol] is the water density. For the 20 cm example, this is only 0.3 psi. So the conclusion that you want a hole a good bit larger (maybe 10x) than r holds not only because of flow restriction (if your soda bottle is enormous) but also – and more relevantly – because of burping (for more traditionally sized soda bottles).

Has that equation a name, that I may Google it?

I’d be surprised if it did. The velocity that fluid can flow out the bottom can be obtained from Bernoulli’s equation, and then the steady-state condition requires that “the rate of water volume out” equals “the rate of air volume in”. Rate of volume out is the velocity out the bottom times the area of the outlet. Thus, for a chosen inlet area, the velocity in is then determined. This needs to stay below the speed of sound, so set it equal to the speed of sound to arrive at a first-order guess for the required inlet area.

A very nifty demonstration (ex-teacher here) of the general concept involved is to blow INTO the straw or tubing that you insert. Providing it doesn’t substantially block the opening, you can force the water out real fast!

If r=R, we have a vertical tube. V=SQRT(2gh), and s varies with the height of the tube. Setting V=Vs would give the height or length of pipe for water to exit at the speed of sound. Where is my error?

Not sure what you mean by s here. Glossing over that: the velocity of water out the bottom is only governed by Bernoulli’s equation if the hole is small compared to the container’s width or, better, if the hole points sideways rather than straight down. If you have a vertical pipe with water in it and you chop off the ends, the speed out the bottom will continually rise since the whole column of water is in freefall. If you have a vertical pipe with water in it and you put an elbow at the bottom to make the water come out sideways, then Bernoulli’s equation is relevant. This works fine for your example, and a pipe about 6 km tall would lead to water spewing out the bottom at the speed of sound in air, ~340 m/s (except that ambient air pressure at the top of the pipe will be lower, slowing the flow.)

ETA: Of course at these extremes, you might hit limits sooner due to turbulent or viscous effects. Depends on assumptions that aren’t critical in the “household” scenario.

Vs=velocity of sound.

The equivalence of resistance, whatever that is.
Can you tell I had problems with flow control in graduate school? :slight_smile:
Thanks, I see now the equation holds true with the difference in relative size that you explained.

You’re either assuming the soda has the same flow characteristics as water or you’re ignoring viscosity.

Since carbonated beverages are mostly water, the hole size is approx .75 in. diameter, and the maximum volume being discussed is 1L, any differences would probably be negligible to the kitchen-sink experimenter.