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How much faster does the fissile uranium in a power plant 'break down' than an equivalent amount of naturally decaying U-235?

How much faster is the chain reaction in a bomb compared to a power plant? (Again, assuming equivalent amounts of fissile material. U-235 or Pu-239, e.g.)

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  • $\begingroup$ I'm not sure this is neatly answerable, because it depends on the concentration of the U-235 or Pu-239 and if they can exchange neutrons or not. Some power plants, like Chernobyl, were designed to slow down the fission process, not speed it up. That's why it blew up like it did, when the water used to slow down the fission was drained, the plant went critical. I'd be curious if this can be answered by one of the experts but I suspect there are too many variables. $\endgroup$ – userLTK May 13 '18 at 4:36
  • $\begingroup$ This question is like asking: how much faster does a molotov bomb burn gas , compared to a car engine. $\endgroup$ – anna v May 13 '18 at 6:51
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There's no comparison between these two events.

A power plant consumes it's fuel very slowly, over a period of months, whereas every nuclear weapon explosion takes place (for the core material) in a period of less a millisecond. Also note that nuclear weapons are now mostly thermonuclear weapons, not strictly speaking simple fission weapons. It's very difficult to compare these two events, but roughly speaking :

One of the first A-bombs used had about 61 kg of U-235 and let's say it took one microsecond to detonate. So that's of the order of 51,000,000 kg per second.

All the nuclear power plants on Earth use about 500,000 kg ( 500 metric tons ) of U-235 per year. That info is from this Scientific American article and using figures for fuel enrichment from Wikipedia. There are something like 450 nuclear power plants active. So a single plant uses a bit over one metric ton ( 1,100 kg ) of U-235 in a year.

So the 51 kg of U-235 used by a single weapon might power a single power plant for about 17 days compared to a detonation taking something like a microsecond (probably less).

All back of the envelop calculations, of course. :-)

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