How much nuclear fuel is in a nuclear sub?

In summary: But if you knew how much fuel was in the reactor, and how long it would last between refueling, you could make a pretty good estimate of boat capabilities. All in all, I think the Navy is trying to protect the technology and the boat's capabilities more than they are protecting the amount of fissile material on the boat.
  • #36
nismaratwork said:
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 :smile:
 
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  • #37
vanesch said:
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 :smile:

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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null result...
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...
...
 
  • #38
close to 1 kg
 
  • #39
close to 10 kg
 
  • #40
close to 100 kg
 
  • #41
Once one of these disappear, we will know order of magnitude.
 
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  • #42
Borek said:
Once one of these disappear, we will know order of magnitude.

:smile:

I love this site.
 
  • #43
"close to 1000kg" never showed up, so via Borek's logic we have our answer. Unfortunately Borek had to die to achieve it...
 
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  • #44
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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  • #45
QuantumPion said:
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?
 
  • #46
mheslep said:
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%).
 
  • #47
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.
 
  • #48
CAC1001 said:
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! :smile:
 
  • #49
880 pounds for 25 years of service.
 
  • #50
QuantumPion said:
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.
 
  • #51
pcvrx560 said:
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!
 
  • #52
Tarr said:
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!

As a rule, the nuclear material in a warhead is trivial, tens of kilograms, vs tons in the reactor.
A ton of TNT is 4.184 gigajoules; that is just over a megawatt-hour. A Sea Wolf sub is officially 45000 HP; that is 33 MW; that means it uses about the same amount of energy as 762 tons of TNT every day at full power.
 
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