Timeline for Can I make the water boil simply by spinning it in a glass?
Current License: CC BY-SA 3.0
19 events
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| Apr 1, 2017 at 11:09 | comment | added | Nat | Correction: Instead of "Here, the simplification was omitting that the phase compositions vary with temperature"; I'd meant "Here, the simplification was omitting that the phase compositions vary with pressure". Temperature effects are another branch of things that'd invite more possibilities, but I'd meant to discuss only pressure changes since there's enough confusion already. | |
| Apr 1, 2017 at 11:01 | comment | added | Nat | @Robin I won't keep spamming you with examples; the point's just that a lot of physical phenomena aren't considered in the above answer, allowing for the train of logic to incorrectly find that any sort of pressure-driven vaporization is "impossible". Still +1 for your answer since you presented a lot of the classical transport phenomena theories very well; my only point's that real-world effects aren't fully explored. | |
| Apr 1, 2017 at 10:59 | comment | added | Nat | @Robin As another example, consider when the cup's spun so fast that it creates a vortex in the local atmosphere, creating a vacuum at the center (Wikipedia's explanation, since another user didn't believe this one). | |
| Apr 1, 2017 at 10:57 | comment | added | Nat | @Robin As another example, the amount of air that water can absorb increases with pressure. Spin the cup fast enough, the water'll absorb the gas phase (down to the equilibrium point), at which point the water at the boundary can start to experience arbitrarily strong vacuums. Here, the simplification was omitting that the phase compositions vary with temperature. | |
| Apr 1, 2017 at 10:50 | comment | added | Nat | @Robin Just a quick example, in your answer you note that the pressure at the center can drop. What if water splashes into that region of low pressure? | |
| Apr 1, 2017 at 9:04 | comment | added | Robin | @Nat, can you please point where exactly I'm simplifying and missing real-world behaviour? Also, what does the non-classical framework for viewing these problems look like? Finally, I've not made an assumption for the flow to be not turbulent. I don't see where turbulence contradicts my arguments. | |
| Mar 31, 2017 at 22:35 | comment | added | hmakholm left over Monica | +1, I think the last few paragraphs here really go to the core of the OP's misunderstanding. | |
| Mar 31, 2017 at 22:09 | comment | added | Nat | +1. While the physical concepts presented in this answer are too limited to fully model real-world behavior and thus can't be used to make absolute statements about real-world physical systems, the presented analyses are informative and provide insight into the classical framework for viewing these problems. | |
| Mar 31, 2017 at 21:52 | comment | added | Nat | @PeterA.Schneider Good checking the phase diagram, though - you're right, depending on the exact assumptions behind this problem, various other phases can be entered, e.g. the fluids may go supercritical. This is one of those problems where lots of stuff could happen in different iterations of it; there's not one single scenario from the problem statement. | |
| Mar 31, 2017 at 21:49 | comment | added | Nat | @PeterA.Schneider Vortices displace their inner count to the boundaries, causing a stronger inner pressure drop. While we may've generated a similar phenomena through some sort of local updraft if no one was spinning the glass, that's not necessary here. | |
| Mar 31, 2017 at 21:45 | comment | added | Nat | @Robin Water being spun around violently can enter turbulent flow regimes, splash, etc., allowing parts of it to enter the low-pressure area. It's kind of a spherical cows issue; typical simplifying assumptions can't be said to always be accurate. | |
| Mar 31, 2017 at 19:44 | comment | added | Peter - Reinstate Monica | Hm... looking at the phase diagram one can see that stirring fast enough always creates ice, which I think is the opposite of what the OP wanted ;-). | |
| Mar 31, 2017 at 19:35 | comment | added | Peter - Reinstate Monica | Actually, if you have a concave wall and stir really fast, wouldn't the water heat up from compression? I know it's not very compressible, but it's not completely incompressible either. | |
| Mar 31, 2017 at 19:34 | comment | added | Peter - Reinstate Monica | @Nat No, the pressure would not drop. Why would it? In weather, the updraft creates a local pressure drop. There is no updraft because air is spinning circularly in a glass, I think. | |
| Mar 31, 2017 at 18:32 | comment | added | Robin | @Nat, I extended my answer. Yes, the tornado can be generated. No, this can never lead to a pressure decrease at the surface of the water. It can lead to a pressure decrease in the center of the tornado, but there is no water there to boil. | |
| Mar 31, 2017 at 18:30 | history | edited | Robin | CC BY-SA 3.0 |
re-arranged and extended
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| Mar 31, 2017 at 14:13 | comment | added | Nat | I think that people are looking at this question with too many simplifying assumptions. For example, consider an extreme case in which a cup is continuously spun until it creates a vortex in the local environment and generates a literal tornado around it. Then pressure drop, right? Not that someone's likely to do that, just "impossible"'s going too far. That said, I agree that it'd take a lot of contriving to get there. =P Anyway, there're a few more routes toward getting pressure drops; it's definitely possible. | |
| Mar 31, 2017 at 14:09 | history | edited | Robin | CC BY-SA 3.0 |
added 165 characters in body
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| Mar 31, 2017 at 14:00 | history | answered | Robin | CC BY-SA 3.0 |
