rannyfash
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fusion of the atmosphere
does anyone know how to calculate what temperature the composition of air would need to be heated up to to initiate fusion and destroy the whole world
 mad science
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Endimion17
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I know. None.
It is an impossible process. The guys working at Manhattan project were joking about it, too.
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neptunium
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stars fusion oxygen and nitrogen at several hundreds of million degree so in theory the atmosphere would too if kept at these temperature for a few
millisecondes .
when a nuclear weapon detonate it reaches about 15 million degrees ...enough for hydrogen (1, H2 and H3) lithium (in H bombs) and helium 3 )to fusion
but far from enough to trigger oxygen and nitrogen fusion.
some scientist at the manatthan project in 1945 thought this would be a pretty big problem!
[Edited on 1-5-2012 by neptunium]
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rannyfash
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yeah in theory but pressure has to be taken into account, anyone know of an equation, i havent yielded any results from utfse,
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neptunium
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as long as the temperature is reached (meaning energy per particles) preassure is irrelevant .this phenomenon could be achieve in a bottle.
the Farnsworth's experiment proves it well.
according to the oxygen cycle in stars
16O+16O=28Si+4He 45%
or
16)+16)=32S +gamma
this takes place at about 2e9 degrees
over 3e9 we enter the silicon burning phase
this is not an easy question with an easy answer. the rae of decay the temperature the composition of the gases and a lot of other factors come to
play.
nothing like this has ever been encounter in the lab,
stars provide the only knonw model for these kind of temperature and conditions!
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Endimion17
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Fusion of the contained gas is not a problem as long as you have extreme temperatures, but this thread is about a runaway fusion of the atmosphere of
our planet. That's impossible. Period.
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neptunium
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impossible practicaly but not theoryticaly. a high enough temperature in one point (say 100 billion degreeas for several months ) could trigger a
runaway fusion of the oxygen (and maybe nitrogen in the air) heating it slowly all arround the earth. slowly but surely the whole atmosphere would
indeed fusion to silicon
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bfesser
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No, theoretically impossible, as well—at least as the original question was stated.
neptunium, you should look into acquiring/using a spell check function.
[Edited on 5/1/12 by bfesser]
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neptunium
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yes i have this old computer and keyboard i will get a new one soon..
in his question if the atmosphere could undergo thermonuclear fusion i guess Bfesser is right .
it would need some confinment the earth's atmosphere cannot provide.
correct
[Edited on 1-5-2012 by neptunium]
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Dr.Bob
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Yes, the yield of a nuclear reaction is dependent on keeping the materials in close proximity. It is very difficult to confine a gas at millions of
degrees, so therefore it is difficult to do without an intense gravitational field. That is why the bombs built by North Korea and such often
don't work well, as they blow apart before the fission process can get going well, since they are small and not symmetric enough, and then they
quickly loss critical mass as they blow apart. They still explode, since they are made with a large amount of high explosives, but the amount of
nuclear boost is low due to not staying confined tight enough.
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rannyfash
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Quote: Originally posted by Dr.Bob  | | Yes, the yield of a nuclear reaction is dependent on keeping the materials in close proximity. It is very difficult to confine a gas at millions of
degrees, so therefore it is difficult to do without an intense gravitational field. That is why the bombs built by North Korea and such often
don't work well, as they blow apart before the fission process can get going well, since they are small and not symmetric enough, and then they
quickly loss critical mass as they blow apart. They still explode, since they are made with a large amount of high explosives, but the amount of
nuclear boost is low due to not staying confined tight enough. |
i see, probably a good thing knowing some mad people out there
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Aperturescience27
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I think even if you heated one region of the atmosphere to a high enough temperature, you couldn't get a continuous reaction, the heat would dissipate
too quickly, and pressure would push the particles too far apart anyway. I don't have the data on that, though, I could be wrong. Even if you heated
the entire planet to a sufficient temperature, the heat would probably radiate into space too quickly. Again, I don't have the data to back that up.
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hinz
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It's the same problem as the Russians had with their layer cake design. The the explosive pressure would push away and the heat would decrease the
density of the fusionable materials (in the atmosphere and the bomb) before it could be heated to a sufficient temperature for fusion
So the Teller-Ulam design was invented, in which the radiation of the primary fission warhead ablates the case and pushes it outwards while
compression and heating the fission stage (simply Newton third laws of motion). If a object has enough mass, its gravitation can prevent it from
exploding like a H-bomb and thereby turning it into a star.
Does anyone know if H-bombs with 238U-mantle have a higher fusion yield because of its "reactive" mantle. As far as I know, the 238U mantle was later
replaced by a lead mantle to reduce fallout.
The 238U mantle probably does not only apply pressure by becoming an extremely hot gas but also by fissioning by fast neutrons. This would probably
increase the internal pressure, thereby keeping the fusion stage together for a longer time and increasing the fusion yield.
[Edited on 8-5-2012 by hinz]
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franklyn
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www.sciencemadness.org/talk/viewthread.php?tid=7562#pid86902
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SM2
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Burning Air
| Quote: Originally posted by hinz | It's the same problem as the Russians had with their layer cake design. The the explosive pressure would push away and the heat would decrease the
density of the fusionable materials (in the atmosphere and the bomb) before it could be heated to a sufficient temperature for fusion
So the Teller-Ulam design was invented, in which the radiation of the primary fission warhead ablates the case and pushes it outwards while
compression and heating the fission stage (simply Newton third laws of motion). If a object has enough mass, its gravitation can prevent it from
exploding like a H-bomb and thereby turning it into a star.
Does anyone know if H-bombs with 238U-mantle have a higher fusion yield because of its "reactive" mantle. As far as I know, the 238U mantle was later
replaced by a lead mantle to reduce fallout.
The 238U mantle probably does not only apply pressure by becoming an extremely hot gas but also by fissioning by fast neutrons. This would probably
increase the internal pressure, thereby keeping the fusion stage together for a longer time and increasing the fusion yield.
[Edited on 8-5-2012 by hinz] |
AFAIK, Castle Romeo shot utilized the more common 238 for pusher and taper. Of course, a standard fission device wouldn't bring this towards any form
of criticality, but the fusion of the LiD is energetic enough to fission even U-238. Scientists at the time didn't know this, and the yield was
greater than expected. Also, this was one of the dirtiest bombs, getting most of it's bang from secondary fusion of 238. The Russions have detonated
extremely clean H-Bombs using Pb, as you mentioned. And why not? LiD is cheap, and holds such an amazing concentration of energy.
Regarding early fission devices, you can often see a brown red circle close to the middle of the mushroom cap. These temperatures are capable of
burning air mixture to nitric oxide, which then of course picks up another oxygen. With huge fusion deices, the burning of air probably does occur,
but the extreme temperatures and lack of oxygen near the umba, can also reverse the reaction, turning nitric oxide back into air (value deducted air),
lol There are not many chances for nitric oxide to escape to a cool enough air
area, where it would instantly form NO2.
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phlogiston
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While U-238 tampers are common indeed, and in some designs the fissioning of the tamper actually is responsible for more than half of the yield, the
unexpectedly high yield of the Castle Romeo test (and other shots in the castle series, most notoriously the Bravo test) was due to a different
reason:
The designers failed to take into account the contribution of the lithium-7 isotope. It was assumed that only Li-6 would yield tritium on neutron
capture, and for this reason the lithium used was enriched in Li-6. However, the supply of Li-6 was still limited and the lithium used was actually
still 60% lithium-7 (natural level is around 7.5%). Not only does Li-7 also yield tritium on neutron capture (which fuses with the deuterium), it also
yields a neutron that will induce a bit more fissioning of the tamper.
The most powerfull test ever, the russion 'tsar bomba' had a 50 MT yield, used a Pb tamper. But the original design included a U-238 tamper, giving a
designed yield of 100 MT.
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"If a rocket goes up, who cares where it comes down, that's not my concern said Wernher von Braun" - Tom Lehrer |
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SM2
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Thank you. So in other words, the cheaper Li isotope fissioned.
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