## How much nuclear fuel is in a nuclear sub?

 Quote by Thermodave I think you're being a little silly. I've worked within the NNPP and I've never heard of such a thing. When I was in grad school my professor used to give us projects like what you describe. It's not illegal to do a paper design for a sub reactor.
Not illegal, but I wouldn't want to accidentally hit close to the real design... I think you probably would meet some people who want to have a long time. Then... you'd probably get a job.

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 Quote by nismaratwork Not illegal, but I wouldn't want to accidentally hit close to the real design... I think you probably would meet some people who want to have a long time. Then... you'd probably get a job.
So the only thing to find out what the real design looks like, is to watch this forum closely, and when some design suddenly disappears, this must be close

 Quote by vanesch So the only thing to find out what the real design looks like, is to watch this forum closely, and when some design suddenly disappears, this must be close
Hmmm... now there's a logical flaw I didn't see coming! I suppose it would be best just to ignore all designs. The only way I can see to settle this is simple... everyone start pitching reactor designs except me of course...

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 Admin Once one of these disappear, we will know order of magnitude.

 Quote by Borek Once one of these disappear, we will know order of magnitude.

I love this site.

 Blog Entries: 2 Recognitions: Gold Member Science Advisor "close to 1000kg" never showed up, so via Borek's logic we have our answer. Unfortunately Borek had to die to achieve it...
 If the power plant has an average output of 100 MW, is meant to last 20 years, and has a burnup limit of 50 GWD/MTU, than the initial fuel loading would be: (20 y * 365 d/y * 100 MW * 1e-3 GW/MW) / 50 GWD/MTU = 15 MTU This seems too high to me so I guess that it is not designed to run at 100% power for 20 years continuously. For reference, a typical commercial nuclear power plant as an MTU loading on the order of 50-100 MTU.

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 Quote by QuantumPion If the power plant has an average output of 100 MW, is meant to last 20 years, and has a burnup limit of 50 GWD/MTU, than the initial fuel loading would be: (20 y * 365 d/y * 100 MW * 1e-3 GW/MW) / 50 GWD/MTU = 15 MTU This seems too high to me so I guess that it is not designed to run at 100% power for 20 years continuously. For reference, a typical commercial nuclear power plant as an MTU loading on the order of 50-100 MTU.
With a usage factor of ~5% (sitting dockside, underway mostly at low power cruise) I'm guessing that's about right at ~.7 MTU / 20 years.

BTW, if I recall correctly US subs use HEU in their reactors. Wouldn't that improve the GWD/MTU burnup with little U238 in the way?

 Quote by mheslep With a usage factor of ~5% (sitting dockside, underway mostly at low power cruise) I'm guessing that's about right at ~.7 MTU / 20 years. BTW, if I recall correctly US subs use HEU in their reactors. Wouldn't that improve the GWD/MTU burnup with little U238 in the way?
Yes, the general rule I use is 0.1 w/o U235 is worth about 1 GWD/MTU (at low enrichments for commercial reactors anyway, not sure if that can be extrapolated up to 100%).

 A bit unrelated, but I remember reading an article how a grad student figured out the layout of the U.S. power grid which is classified, via unclassified sources. I think it was for his Ph.D, but the professor didn't like it. However when the government found out about it (I forget how specifically), they gave him a job.

 Quote by CAC1001 A bit unrelated, but I remember reading an article how a grad student figured out the layout of the U.S. power grid which is classified, via unclassified sources. I think it was for his Ph.D, but the professor didn't like it. However when the government found out about it (I forget how specifically), they gave him a job.
I bet that was an offer that was REALLY hard to refuse!

 880 pounds for 25 years of service.

 Quote by QuantumPion If the power plant has an average output of 100 MW, is meant to last 20 years, and has a burnup limit of 50 GWD/MTU, than the initial fuel loading would be: (20 y * 365 d/y * 100 MW * 1e-3 GW/MW) / 50 GWD/MTU = 15 MTU This seems too high to me so I guess that it is not designed to run at 100% power for 20 years continuously. For reference, a typical commercial nuclear power plant as an MTU loading on the order of 50-100 MTU.
No, the burndown you assume is much too low. Naval power plants use very high enrichment, and get a very high burndown compared to commercial power plants.

 Quote by pcvrx560 How much fissile material, in kilograms, would, say, an Ohio-class submarine carry? If it's classified, what would be about a good estimate?
Well geez, every one seems to be answering for the reactor that is used to propel the ship on its merry way. Were you asking about the fissile material in the armament (like 24 missiles w/8 warheads)?

My specialty was Los Angeles class, but both are classified so a good estimate - enough to get the job done! Borek was close for one of the answers above with one of his estimates and being close counts in nuclear bombs!

 Tags fuel, submarine