How much nuclear fuel is in a nuclear sub?

[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?

 

[OmCheeto]

How much fissile material, in kilograms, would, say, an Ohio-class submarine carry?

If it's classified, what would be about a good estimate?

e=mc^2

25 years between refueling.

If I said more, I'd be arrested, and shot.



we're not supposed to give out the answers to homework questions. PF is brutal regarding that rule.

 

[statphys]

I went to a public lecture on nuclear marine propulsion, the speaker talked for 45 minutes and told us nothing! It's all too classified.

 

[nismaratwork]

I for one, am pleased that this kind of technology is not openly shared. The power plant of a sub is key to its ability to function as a deterrent, or hunter-killer, and comparable only to its means of acoustic insulation and propulsion in importance. This SHOULD be classified.

 

[Borek]

There is a HUGE difference between "sharing" a technology and giving out parts of the information. Amount of fuel doesn't tell you much about boat capabilities - it lasts much longer than any other type of provision needed. Boat sea time is more limited by amount of potatoes on board than by amount of fuel.

 

[minerva]

I have heard the volume of the reactor core fuel in a US nuclear submarine described, by qualified people, as "fits under a desk".

It's not a particularly large volume.

 

[Morbius]

There is a HUGE difference between "sharing" a technology and giving out parts of the information. Amount of fuel doesn't tell you much about boat capabilities - it lasts much longer than any other type of provision needed. Boat sea time is more limited by amount of potatoes on board than by amount of fuel.

Borek,

The amount of sea time is not the sensitive information the Navy is trying to protect.

The real information the Navy wants to keep secret is how fast the sub can go.
If you knew the reactor power - that would give you a big clue to how fast it can go.

If you know the amount of fuel, and can estimate the heat transfer area, you could
get a pretty good idea of what the power of the reactor is.

That's why they don't tell you how much fuel is in the reactor.

E=mc^2 doesn't help - because you don't know the power and hence don't know
the energy that the reactor produces in the 20 years between refueling.

Dr. Gregory Greenman

 

[Vanadium 50]

If I said more, I'd be arrested, and shot.

Maybe not in that order.

 

[russ_watters]

Borek,

The amount of sea time is not the sensitive information the Navy is trying to protect.

The real information the Navy wants to keep secret is how fast the sub can go.
If you knew the reactor power - that would give you a big clue to how fast it can go.
...

E=mc^2 doesn't help - because you don't know the power and hence don't know
the energy that the reactor produces in the 20 years between refueling.

Dr. Gregory Greenman

Well...except that the peak power of the turbines is published, so it is a good starting point to figure on an average output of, say, 50% (though I don't know how well a reactor throttles...) and a plant efficiency of 30% and calculate from there.

 

[nismaratwork]

Wouldn't top speed have a lot more to do with the design of the drive and screws than the power plant? At some point you get cavitation, and therefore noise and loss of efficiency, and I would guess this occurs long before a nuclear pile "poops out", and its related batteries and fuel cells.

 

[apeiron]

http://www.nato.int/acad/fellow/99-01/maerli.pdf

In 1995, with 158 operating U.S. naval reactors, the annual burn-up of U-235 in the entire fleet was reported to be approximately 1.1 tons.117 Thus, as a crude approximation, on average each U.S. reactor used 7 kg of U-235 during that year of operation.

 

[nismaratwork]

http://www.nato.int/acad/fellow/99-01/maerli.pdf

In 1995, with 158 operating U.S. naval reactors, the annual burn-up of U-235 in the entire fleet was reported to be approximately 1.1 tons.117 Thus, as a crude approximation, on average each U.S. reactor used 7 kg of U-235 during that year of operation.

...But the fleet is comprised of vastly different submarines, from hunter-killers to boomers, and of those there are different classes within each, different missions and therefore different depths and speeds.

 

[Xnn]

about a 1000 lbs

 

[rod_worth]

I'm probably the most qualified to answer this question because I was in the nuclear navy, albeit on an aircraft carrier. My schooling was set up to teach us about submarines however. Here is what I feel I can reasonably tell you:

Potatoes 'do' limit how long a ship is out to sea; a deployment is limited by how much food you can store on the boat.

Most of this material is 'confidential' or 'secret'. They can tell you the general science behind things, but the specific engineering aspects including fuel loading, enrichment, and materials is off limits.

The power output of the reactor will far surpass the boats 'speed limit' which is classified on its own (I don't know it, and if I did I certainly wouldn't tell you). Speed on a submarine is not limited by the power output of the reactor/steam-plants but by materials on the exterior of the hull that will begin to rip off because you're going too fast through the water.

Most of the highly secret aspects lie not on reactors and plant design but harmonics and propulsion plant design to keep the thing from being detected.

The reactor itself is very small, say the size of a bunch of refrigerators stacked on top of each other and side by side (and don't ask me how many, I'm not telling).

I've probably given you enough info to have myself shot and arrested too (and in that order) ;) However, the fact that you're asking a very specific question like that is border-line suspicious, and is probably enough to at least have this entire conversation flagged by the FBI.

 

[pcvrx560]

Wow, thanks fellas. Now I feel like I accidently tried to commit treason by asking this question, haha. But seriously, I think the biggest thing I got from this thread is a visualization of how much energy can be contained in such a small volume. I've always heard "E = mc^2" and that "c^2 is a very large number" but now I can sort of visualize the size of energy/mass equivalency.

Another question: are nuclear reactors 100% efficient, in terms of mass/heat-energy produced? If the laws of thermodynamics took a day off, would nuclear power plants be able to convert ALL fissile fuel into electricity?

 

[rod_worth]

Nothing I know of is 100% efficient, except for my Mom at remembering to mail out birthday cards on time; no, reactors are not 100% efficent. I 'think' in a real plant something like 15-20% of the energy/heat produced by the reactor actually is used to do work in the form of propelling a ship and/or producing electricity.

If the laws of thermodynamics "took a day off" I would create my own castle with 70 virgins and a built in Taco Bell from empty space because energy would no longer be conserved, cold would 'flow' to hot, I would break every piece of glass I could find to watch it spontaneously put itself back together, and then I would fly to the moon on a pink elephant that I gave birth to from the same nothingness as my awesome new bachelor pad!

My point is this, if thermodynamics "took a day off", I believe anything would become possible, which would probably include your 100% efficient nuclear plant! ;) You can consider c^2 as nothing more than a 'conversion factor' which equates a little bit of mass (m) to a lot of energy (E) capable of doing work. I too think it amazing at how much energy can be extracted from a well designed 'hot rock' (aka 'reactor'). Thank you Dr. Einstein for realizing that (1/2)*m*v^2 = (1/2)*(E/c^2)*v^2!

 

[nismaratwork]

Nothing I know of is 100% efficient, except for my Mom at remembering to mail out birthday cards on time; no, reactors are not 100% efficent. I 'think' in a real plant something like 15-20% of the energy/heat produced by the reactor actually is used to do work in the form of propelling a ship and/or producing electricity.

If the laws of thermodynamics "took a day off" I would create my own castle with 70 virgins and a built in Taco Bell from empty space because energy would no longer be conserved, cold would 'flow' to hot, I would break every piece of glass I could find to watch it spontaneously put itself back together, and then I would fly to the moon on a pink elephant that I gave birth to from the same nothingness as my awesome new bachelor pad!

My point is this, if thermodynamics "took a day off", I believe anything would become possible, which would probably include your 100% efficient nuclear plant! ;) You can consider c^2 as nothing more than a 'conversion factor' which equates a little bit of mass (m) to a lot of energy (E) capable of doing work. I too think it amazing at how much energy can be extracted from a well designed 'hot rock' (aka 'reactor'). Thank you Dr. Einstein for realizing that (1/2)*m*v^2 = (1/2)*(E/c^2)*v^2!

You get thermodynamics to take the day off, and you build a taco bell? I'd probably start with something with a little more umph. ;)

 

[rod_worth]

You get thermodynamics to take the day off, and you build a taco bell? I'd probably start with something with a little more umph. ;)

I did say that my castle would come with 70 virgins; is that the "umph" you're looking for!?

 

[nismaratwork]

I did say that my castle would come with 70 virgins; is that the "umph" you're looking for!?

How about 70 really experienced succubi? Virgins are kind of a bore when you get right down to it. Anyway, I was thinking of free energy and world domination, but yeah, taco bell and virgins works I guess.

 

[mgb_phys]

Virgins are kind of a bore when you get right down to it.

You've obviosuly never been to the Pasadena Star Trek convention !

 

[Dav333]

talk of FBI?? get you hand off it.

 

[nismaratwork]

You've obviosuly never been to the Pasadena Star Trek convention !

This is true, but I meant in bed, not as an expression of frustrated desires. :winK: Now, if you have anecdotal evidence with a vulcan virgin, I demand that you tell us all of the details in iambic pentameter, right now.

 

[rod_worth]

This is true, but I meant in bed, not as an expression of frustrated desires. :winK: Now, if you have anecdotal evidence with a vulcan virgin, I demand that you tell us all of the details in iambic pentameter, right now.

lmao!

 

[QuantumPion]

Wow, thanks fellas. Now I feel like I accidently tried to commit treason by asking this question, haha. But seriously, I think the biggest thing I got from this thread is a visualization of how much energy can be contained in such a small volume. I've always heard "E = mc^2" and that "c^2 is a very large number" but now I can sort of visualize the size of energy/mass equivalency.

Another question: are nuclear reactors 100% efficient, in terms of mass/heat-energy produced? If the laws of thermodynamics took a day off, would nuclear power plants be able to convert ALL fissile fuel into electricity?

Commercial nuclear power plants are 30-40% thermodynamic efficiency in converting heat to electricity.

No nuclear reactor or bomb of any kind can be greater than ~97.4% efficient at converting fission energy to heat because ~2.6% of the fission energy is lost to neutrinos.

 

[jae1227]

I know that subs use highly enriched uranium opposed to what power plants use.

 

[Morbius]

http://www.nato.int/acad/fellow/99-01/maerli.pdf

In 1995, with 158 operating U.S. naval reactors, the annual burn-up of U-235 in the entire fleet was reported to be approximately 1.1 tons.117 Thus, as a crude approximation, on average each U.S. reactor used 7 kg of U-235 during that year of operation.

aperion,

A big component of that burnup is due to the reactors in our 11 nuclear powered carriers.

[ 10 Nimitz class + USS Enterprise ]

Dr. Gregory Greenman

 

[mgb_phys]

Commercial nuclear power plants are 30-40% thermodynamic efficiency in converting heat to electricity.

Although that's only reactor heat to electricity, it doesn't include any 'efficency' of the nuclear reaction.

No nuclear reactor or bomb of any kind can be greater than ~97.4% efficient at converting fission energy to heat because ~2.6% of the fission energy is lost to neutrinos.

In practice of course any fission reaction only uses a very tiny proportion of the mass.
A U235 nucleus fissioning 'only' gives around 200Mev, so 235*950Mev/200Mev = 0.1% of the mass energy.

 

[nismaratwork]

My submarines run on the pent-up libido of the Pasadena Convention... 99.998% yield, and a staggering 200 PVV (Pimpled Virgin Volts) per gram!

Kidding aside, as I understand it the draw of practical fission is not super-high efficiencies, but rather the enormously high energy density of the material compared to other fuel sources.

 

[BrianConlee]

but everyone knows all that heat generated from the nuclear power plant leaves a huge wake of heated water that glows on an infrared scan right?

I have some detailed schematics of the power plant and turbine system.

Hold on... there's some men in suits with sunglasses at the door...

I'll be right back to finish the post.

 

[Dr_Zinj]

It's not treason to ask the question. In fact, it's technically not treason to merely answer the question; unless you're providing it to enemies of the United States. It would be a violation of a bunch of regulations, particularly Deparment of Defense regs as applied to DOD operated reactors, and the Uniformed Code of Military Justice.

If some of you bright people decided to do an analysis, study and design on your own from purely unclassified sources, and posted it here; someone in the military might find it and decide that your result was too accurate, slap a classifiication on it, remove it from the boards and purge all references to it. Then they'd show up at your door, demand all originals and copies, and force you to sign a non-disclosure agreement.

And that would really ruin your prospects for foreign employment or vacations for the rest of your life.

 

[NobodySpecial]

If some of you bright people decided to do an analysis, study and design on your own from purely unclassified sources, and posted it here; someone in the military might find it and decide that your result was too accurate, slap a classifiication on it, remove it from the boards and purge all references to it.

Like they did with the alien autopsy stuff?

Then they'd show up at your door, demand all originals and copies, and force you to sign a non-disclosure agreement.

They can try - but they would find that my death star was quite operational <manic laughter>

 

[nismaratwork]

It's not treason to ask the question. In fact, it's technically not treason to merely answer the question; unless you're providing it to enemies of the United States. It would be a violation of a bunch of regulations, particularly Deparment of Defense regs as applied to DOD operated reactors, and the Uniformed Code of Military Justice.

If some of you bright people decided to do an analysis, study and design on your own from purely unclassified sources, and posted it here; someone in the military might find it and decide that your result was too accurate, slap a classifiication on it, remove it from the boards and purge all references to it. Then they'd show up at your door, demand all originals and copies, and force you to sign a non-disclosure agreement.

And that would really ruin your prospects for foreign employment or vacations for the rest of your life.

Anyone here bright enough to do that is bright enough to keep their mouths shut, as evidenced by the fact that such people are here, and they have.

 

[OmCheeto]

Anyone here bright enough to do that is bright enough to keep their mouths shut, as evidenced by the fact that such people are here, and they have.

Did I miss something? What's with this necro? Are we on Wikileaks? ...

hmm... hmm... hmmm...

Woo Hoo!

That was a rap/media frenzy/pf inspired thought/laugh/thought/laugh/thought moment.

 

[Thermodave]

It's not treason to ask the question. In fact, it's technically not treason to merely answer the question; unless you're providing it to enemies of the United States. It would be a violation of a bunch of regulations, particularly Deparment of Defense regs as applied to DOD operated reactors, and the Uniformed Code of Military Justice.

If some of you bright people decided to do an analysis, study and design on your own from purely unclassified sources, and posted it here; someone in the military might find it and decide that your result was too accurate, slap a classifiication on it, remove it from the boards and purge all references to it. Then they'd show up at your door, demand all originals and copies, and force you to sign a non-disclosure agreement.

And that would really ruin your prospects for foreign employment or vacations for the rest of your life.

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.

 

[nismaratwork]

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.

 

[vanesch]

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

 

[nismaratwork]

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...

...
...
...

null result...
...
...
...

 

[Borek]

close to 1 kg

 

[Borek]

close to 10 kg

 

[Borek]

close to 100 kg

 

[Borek]

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

 

[nismaratwork]

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

I love this site.

 

[Mech_Engineer]

"close to 1000kg" never showed up, so via Borek's logic we have our answer. Unfortunately Borek had to die to achieve it...

 

[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.

 

[mheslep]

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?

 

[QuantumPion]

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%).

 

[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.

 

[nismaratwork]

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!

 

[dabble]

880 pounds for 25 years of service.

 

[wizwom]

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.