Do thermonuclear weapons have infinite fusion energy gain?

Discuss how polywell fusion works; share theoretical questions and answers.

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Do thermonuclear weapons have infinite fusion energy gain?

Post by robertsteinhaus »

The fusion energy gain factor, usually expressed with the symbol Q, is the ratio of fusion power produced in a nuclear fusion reactor to the power required to maintain the plasma in steady state. The condition of Q = 1 is often referred to as breakeven.

Hydrogen bombs, like the Ivy Mike nuclear test back in 1952, are the only known man-made devices to achieve significant fusion energy gain factor above Q=1. When, in a hydrogen bomb (which technically is an instance of inertial confinement fusion) the D-T plasma in the weapon secondary ignites, and the plasma in the secondary heats itself by fusion beyond the energy initially provided by the fission primary. without any other external energy input, does this correspond to infinite fusion energy gain factor Q?

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Post by ladajo »

I think you need to read up on the concept of system analysis boundaries.

In short, no.
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Post by hanelyp »

The fusion energy contribution in an atomic bomb is limited by the available fusion fuel. As I understand it there is no theoretical limit on how much fusion fuel can be packed around and ignited by the fission primary. But since the fusion fuel will always be limited in a real bomb, the gain will not be infinite.

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Post by D Tibbets »

You also need to read up on hydrogen bombs. Ivy-Mike was a fission- fusion- fission bomb. The initial fission bomb provided the heat and density necessary for much of the D_T (or D-Li to T) to burn up. This provided some additional heat, but mostly the neutrons produced by the D-T fusion, converted some of the Ur238 shell to pleutonium239, which then fissioned. I believe most of the explosive force came from this secondary fission reaction. Modern hydrogen bombs probably use more fusion asa percentage of the total explosive power.

In any case the theoretical maximum Q can be determined by comparing the temperature of the reactants, the energy input necessary to heat and maintain the heat of the reactants (confinement efficiency), and the heat released by the fusion products. There are all sorts of complications, but keeping things ridiculously simple, two examples follow:
D-D fusion. input heating= ~80 KeV, output heating = ~3,000 KeV. This implies that under ideal conditions with full fuel burn up and no confinement losses, the net gain that is possible is 3,000 KeV / 80 KeV = ~ Q of ~27. At lower input energy, say 10 KeV the Q could be as high as 300, but the confinement times with no losses would need to be much higher.

For P-B11 , ignoring Bremsstru ung losses, etc, the input heat may be ~ 200 KeV, the heat out may be ~ 9,000 KeV. In this situation the Q may be as much as ~45.

These potential values are maximum limits. Once inefficiencies of confinement are added in, the Q falls.

Some optimistic estimates for Q in P-B11 fusion in the Polywell may reach as high as 5-20. Rider predicted a practical maximum of ~ 0.5 Q.

The practical Q for D-D fusion may be up to ~ 100, but conversion of that raw energy to useful electrical energy would reduce the gain to ~ 30.

Direct conversion, which is possible with P-B11 fuel modifies this final relationship. With direct conversion a P-B11 raw Q of 20 results in a final direct converted Q of perhaps 15. This direct conversion option allows for a less efficient aneutronic fusion Polywell to match a d-d Polywell with only a modest increase in size. Then there are the significant advantages of not needing to deal with the tremendous neutron fluxes, their associated thermal effects and radiation effects.

In short, the reactions, confinement issues, energy extraction issues, engineering issues, etc. all add up to the final maximum efficiency for the system. And this cannot ever exceed the raw temperature (energy) ratio between the reactants and the products. Again the examples above illustrate this.

Note that the ITER Tokamak is hoped to achieve a Q of ~ 5. If this can be achieved, a subsequent larger machine may reach a Q of 10-50. The problem with this is that the size of the machine, which is already huge, becomes unmanageable from an engineering and especially an economic perspective.

These interactions determine the final size and ~ cost of the reactor.
Of course considerations of ignition machines, and non ignition machines (Polywell) changes considerations considerably. This last aspect seems to confuse some.

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Post by WizWom »

Almost right, Dan. Except there is not time in an H-bomb explosion to breed anything; the shell is undergoing fast fission of the uranium. Any absorbed Neutrons, in a weapon, are considered lost.
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Post by kunkmiester »

Nope, the first bomb to use lithium deuteride was MUCH higher yield than they expected. I don't know how much tritium they got out of it, and how that affected yield, but breeding the lithium released extra neutrons in the right energy range to fission more of the tamper.
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Post by MSimon »

I did some BOE calcs on Polywell way back and came up with a maximum Q of around 20. (Maybe 22) I think 45 is excessive even on theoretical grounds.
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Post by D Tibbets »

MSimon wrote:I did some BOE calcs on Polywell way back and came up with a maximum Q of around 20. (Maybe 22) I think 45 is excessive even on theoretical grounds.
The 45 is derived by dividing the fusion yield ~8.8 MeV by the fuel KE of ~ 200 KeV. Though, upon reflection, the fuel KE may need to be multiplied by two, since there are two fuel ions per fusion reaction. That would give a maximum Q of ~ 22.

But, if the cross section resonance peak can be used at 55-75 KeV (depending on the analysis); then the maximum Q, ignoring losses, might be as high as ~ 80. While this target has never been publicly mentioned by anyone within the EMC2 camp, and I take to mean that the "mono energetic" ion characteristics are actually wide enough that taking advantage of this resonance peat is dilluted at best. I wonder though if a target ion energy with an average around this peak value in the P-B11 cross section would bump the performance up enough that in balance once Bremsstrulung and direct conversion effects were factored in, the net benifits would be positive.

PS: Concerning the original question, a very rough estimate of Q would be the fision/ fusion yield divided by the chemical explosive input. This would be perhaps 10 MT of nuclear explosion / 0.1 KT of chemical explosive, or a Q of ~100 million or even a billion.
Last edited by D Tibbets on Sat Aug 18, 2012 8:25 pm, edited 1 time in total.
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Post by D Tibbets »

WizWom wrote:Almost right, Dan. Except there is not time in an H-bomb explosion to breed anything; the shell is undergoing fast fission of the uranium. Any absorbed Neutrons, in a weapon, are considered lost.
Some early non-staged thermonuclear weapon designs

Early thermonuclear weapon designs such as the Joe-4, the Soviet "Layer Cake" ("Sloika", Russian: Слойка), used large amounts of fusion to induce fission in the uranium-238 atoms that make up depleted uranium. These weapons had a fissile core surrounded by a layer of lithium-6 deuteride, in turn surrounded by a layer of depleted uranium. Some designs (including the layer cake) had several alternate layers of these materials. The Soviet Layer Cake was similar to the American Alarm Clock design, which was never built, and the British Green Bamboo design, which was built but never tested.When this type of bomb explodes, the fission of the highly enriched uranium or plutonium core creates neutrons, some of which escape and strike atoms of lithium-6, creating tritium. At the temperature created by fission in the core, tritium and deuterium can undergo thermonuclear fusion without a high level of compression. The fusion of tritium and deuterium produces a neutron with an energy of 14 MeV—a much higher energy than the 1 MeV of the neutron that began the reaction. This creation of high-energy neutrons, rather than energy yield, is the main purpose of fusion in this kind of weapon. This 14 MeV neutron then strikes an atom of uranium-238, causing fission: without this fusion stage, the original 1 MeV neutron hitting an atom of uranium-238 would probably have just been absorbed. This fission then releases energy and also neutrons, which then create more tritium from the remaining lithium-6, and so on, in a continuous cycle. Energy from fission of uranium-238 is useful in weapons: both because depleted uranium is very much cheaper than highly enriched uranium and because it cannot go critical and is therefore less likely to be involved in a catastrophic accident.

This kind of thermonuclear weapon can produce up to 20% of its yield from fusion, with the rest coming from fission and is limited in yield to less than one megaton of TNT (4 PJ) equivalent. Joe-4 yielded 400 kilotons of TNT (1.7 PJ). In comparison, a "true" hydrogen bomb can produce up to 97% of its yield from fusion, and there is no upper limit to its explosive yield.
I may be mistaken that plutonium is an intermediate in fissioning Ur 238 with fast neutrons, otherwise the boosting of fission with fusion produced neutrons applies.

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Post by TallDave »

Well, I'm going to run counter to everyone else in this thread and say: a fairly interesting question!

The answer, imo, is: big, but not infinite. The energy inputs are either the high explosives that drive the fission reaction, or the fission reaction that drives the fusion reaction, depending on how you like to define the inputs. I don't know offhand the Q calculated for either, but I'm sure they've been done, and that the first, at least, is a fairly big number.

Reminds me of a comment from Rick:
Just because Jet hit Q ~ 1 doesn't mean that it is any closer to a practical fusion reactor than the Polywell is. If you want to use that criterion, then Los Alamos beat that a long, long time ago. Furthermore, if you can't make a practical reactor with D-T, then what is the point of the materials development with ITER?
n*kBolt*Te = B**2/(2*mu0) and B^.25 loss scaling? Or not so much? Hopefully we'll know soon...



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