last day (15 days later) » 

23:41
1
Q: Is the pressure of water in the ocean enough to cause a nuclear bomb to go critical?

Richard CI understand that most nuclear weapons work on the basis of a standard explosive forcing the 2 fissionable parts together causing the nuclear reaction. If a nuclear warhead or bomb fell into the deepest part of the ocean would the pressure eventually be such that it would force the 2 parts of the...

Re, "...forcing the 2...parts together..." The only weapon that ever worked that way was the Little Boy, of which only seven were ever built, and only one ever exploded. All subsequent designs used the implosion principle in which an "explosive lens" creates a spherical shock wave that turns the spherical core into compressible "shocked matter."
the pressure at the bottom of the deepest ocean on earth is insufficient by a huge margin. So is the tremendously greater pressure at the center of the earth's core.
@niels nielsen: I'd tend to agree for the ocean, but the Earth's core? It is several times larger than the the Young's modulus of plutonium (Earth core ~ 360 GPa, Pu Young's modulus ~ 96 GPa). I imagine there would thus be significant compression of that plutonium core - and given it's symmetrical static pressure, that just might do the trick, no? Though the heat at depth may pose problems in that it might melt that core before it got deep enough if the melting point doesn't rise fast enough with compression, which would then disperse it by mixing with mantle and/or outer core material.
See a Serber plot. You need pressures and temperatures equivalent to the center of the sun.
@Jon Custer: Then it would be impossible to ignite a nuclear fission weapon with ordinary chemical explosives at all. We're talking fission, not fusion, here, remember.
23:41
I’m well aware of that. Remember that the Serber plots were applicable to Fat Man and Little Boy. B. Cameron Reed has a nice article on it, with a version there. Google should find it.
@Jon Custer : This is the only article that comes up with a graph showing "pressure" - link.springer.com/chapter/10.1007/978-3-030-61373-0_2 - though the picture appears only in the Google result, else you need a pass. That plot does indeed show a peak around ~$10^{16}$ Pa, which is "like the center of the Sun" alright, but that is also at the maximum of fission, which means the pressure is being generated by fission, i.e. it is explosive, not implosive, pressure. I.e. the core is already very much lit at that point. OP is asking about lighting the core.
(Note though this pressure can, in effect, be converted to implosive pressure with a suitable bomb design - that's how a fission-ignited fusion bomb, better known as the "hydrogen bomb", works. But we're just talking here about making a sphere of plutonium go boom by dropping it into the abyss.)
Use the Serber plot. Particularly the yield vs pressure comparison. If all you want is a tiny fizzle, well, fine.
@Jon Custer: Tell me how that a chemical explosive manages to equal the conditions inside the Sun, then, since that's how you ignite plutonium in a real fission (not fusion) bomb. If we could do that, seems fusion would be a lot easier than it actually is. Detonation pressure of chemical explosive is maybe around ~10 kbar from a quick google search (I looked up RDX, can't imagine there's something many OoM higher that's still chemical), which is just 1 GPa. If that is enough to do it, then definitely it seems deep Earth pressures should be far more than enough.
Fusion is not that hard mind you. Igniting a burn without a primary to drive it is harder. Look, read the Los Alamos Primer by Serber.
@Jon Custer : Yes, of course. That's the point. We're thus talking about igniting what would, in that configuration, constitute a "primary" here. And you're saying to do even that would require Sun-like conditions.
23:41
To keep it together and get reasonable yield, yes. Serber is a good read, or Reed on the Serber plot.
@JonCuster, serber's book is one of my all-time favorites.
@The_Sympathizer, nope nope nope. see Jon Custer's responses here.
@niels nielsen: I don't have access to any of the links beyond what I've said, so that doesn't help. Can you explain though how a chemical explosive can generate conditions like that under this configuration (millions of times the detonation pressure, thousands of times the temperature)? Otherwise, it seems to me like there's a systematic confusion here similar to saying that because paper ignites at about 233 degrees C, then the flame is only 233 degrees C hot, or conversely that because a flame is much hotter, then you need that much to ignite paper.

last day (15 days later) »