Timeline for Can a river be put under enough pressure to jet straight up for a mile?
Current License: CC BY-SA 3.0
22 events
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Jul 5, 2016 at 19:51 | comment | added | James K | roughly 15,786.86805 ‽ | |
S Dec 28, 2014 at 21:32 | history | edited | HDE 226868♦ | CC BY-SA 3.0 |
Corrected order of magnitude error as mentioned in comments
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S Dec 28, 2014 at 21:32 | history | suggested | March Ho | CC BY-SA 3.0 |
Corrected order of magnitude error as mentioned in comments
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Dec 28, 2014 at 17:44 | review | Suggested edits | |||
S Dec 28, 2014 at 21:32 | |||||
Dec 7, 2014 at 18:42 | history | edited | HDE 226868♦ | CC BY-SA 3.0 |
Neatened up LaTeX.
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S Oct 17, 2014 at 14:14 | history | suggested | overactor | CC BY-SA 3.0 |
More MathJax
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Oct 17, 2014 at 13:48 | review | Suggested edits | |||
S Oct 17, 2014 at 14:14 | |||||
S Oct 17, 2014 at 13:45 | history | suggested | overactor | CC BY-SA 3.0 |
MathJax up in that answer!
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Oct 17, 2014 at 13:44 | review | Suggested edits | |||
S Oct 17, 2014 at 13:45 | |||||
Oct 14, 2014 at 23:16 | comment | added | trichoplax is on Codidact now | Just to let you know, we now have MathJax | |
Oct 7, 2014 at 18:19 | comment | added | Jay Vogler | Ahhhh, I got it. With those units, P=.5*rho*v^2 gives a result in Pa, not kPa. Divide that last pressure result by 1000 and things line up correctly. | |
Oct 7, 2014 at 17:57 | comment | added | Jay Vogler | @steveverrill: Huh, you're right. That is bizarrely counterintuitive at first. | |
Oct 7, 2014 at 16:00 | comment | added | Level River St | @JayVogler Bernoulli's equation is a purely theoretical calculation, based on no frictional losses. Based on that assumption, we can convert all the pressure energy from a mile high reservoir into kinetic energy, then recover the kinetic energy and convert it back into potential energy (the water will have zero kinetic energy once it gets back up to its original height.) I think both the answer poster and I would agree that it is friction that causes this to deviate from reality. The greatest friction loss will be the wind resistance on the jet, then probably the nozzle, and finally the pipe | |
Oct 7, 2014 at 15:16 | comment | added | Jay Vogler | @steveverrill: In order to raise liquid in a pipe to 1600 meters, you need a reservoir of 1600 meters high next to it. Once you lower the opening of the pipe (by a mile!), you're fighting gravity as the water flows up out of the reservoir, that's where the additional pressure requirement comes in. | |
Oct 7, 2014 at 13:17 | vote | accept | Rowanas | ||
Oct 7, 2014 at 13:17 | |||||
Oct 7, 2014 at 10:45 | comment | added | Level River St |
By conservation of energy (of which Bernoulli's equation is an example) in order to generate a fountain 1 mile (1600m) high you'd theoretically need a reservoir 1600m high. that's 160 bar (16000kPa, not 160000000kPa). You're out by a factor of 1000 on your pressure. Your velocity is correct, though. .5 * 1000kg/m3 * (177.69m/s)^2 = 15786868.05 Pascal (no need to add the kilo, remember SI unit of mass is the kg.) 160 Bar is a manageable pressure, but is sufficient to throw gas pipe fittings 1/4 mile in case of rupture. Of course for water, wind resistance means that it's nowhere near enough.
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Oct 6, 2014 at 22:45 | comment | added | Superbest | @ColinPickard 15 Gpa is 150 kbar, which means that at room temperature the water would be cubic Ice VII. You would have to get to around 500C in order to melt it at that pressure. | |
Oct 6, 2014 at 22:34 | comment | added | Colin Pickard | What would actually happened to water at 15 GPa? Would it remain a liquid? or change to some strange phase like Ice VII? | |
Oct 6, 2014 at 21:36 | comment | added | Mark | As a point of comparison, 15 GPa is roughly the pressure used to make synthetic diamonds. | |
Oct 6, 2014 at 21:16 | comment | added | Bobson | I tried doing this, but my physics weren't fresh enough to pull it all together. | |
Oct 6, 2014 at 21:00 | review | First posts | |||
Oct 6, 2014 at 21:02 | |||||
Oct 6, 2014 at 20:55 | history | answered | guildsbounty | CC BY-SA 3.0 |