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I was pouring soda, when the act made me think of a question. If you have an unopended $2L$ bottle of soda, and you wish to fill up a generic cylindrical glass cup - if you hold the $2L$ bottle at a constant angle, assuming you don't have to tilt the bottle to continue to get the soda to come out, at what angle should you hold the bottle such that you minimize the time it takes to fill up the glass with soda while not overfilling it?

  • There must not be a "layer" of $\mathrm{CO}_2$ bubbles when the pouring is complete. The rate of the pour is such that the cup does not overfill, and is slow enough that the $\mathrm{CO}_2$ bubbles disipate enough for soda to fill the volume the $\mathrm{CO}_2$ layer was occupying.

This question is very generic, and has a lot of variables - some of which include but are not limited to:

  • glass volume;
  • size of the glass's opening (assume the cup is thin);
  • size of the $2L$ soda bottle's opening (assume US standard);
  • multiple variables in fluid dynamics;
  • room pressure and temperature (assume earth surface pressure and room temperature);
  • carbonization of the soda;
  • elevation (assume sea-level);
  • humidity (I do not know if this would affect anything, but I think this can be ignored).

Any simplification would be good.

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  • $\begingroup$ The tricky part is likely finding the transition point between smooth flow and the pulsatile flow when the airflow into the bottle starts a limit cycle plus estimating the average flow rate in this state (especially since it transitions to turbulent flowif the angle is big enough). Another issue that has come up is whether one can apply maximum entropy principles from nonequilibrium thermodynamics here (being a nonexpert on that field I have no clue). In short, we may have to gather empirical data first. $\endgroup$ – Anders Sandberg Mar 24 at 8:32
  • $\begingroup$ Interesting - although it seems that makes the question ever more complicated than I wanted it. I would say "assume pulsatile flow throughout the entire pour", but then it may be so overly simplified as to be inaccurate, making the question pointless to answer. – $\endgroup$ – QHZ Mar 24 at 15:50

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