Efficiency, acceleration and fusion

[Ford Prefect]

Now that my exams are over and the academic year has concluded, I think I can bring up a topic I have been curious about for some time. Relatively recently I began working on some writing, and inevitably it involved space travel. As is to be expected, I conjured up a kind of mystical fusion drive that passes raw energy through hopes and dreams and turns into whatever I consider the most exciting acceleration for any given moment. However, I became increasingly curious about what practical limits actually exist when it comes to power generation through fusion, particularly the 3He+3He reaction, as well as using it fling a tin can across the cosmos.

Going by the Constants page of the main site, deuterium fusion has an energy density of 6.2E14 J/kg. My, admittedly cursory, research has not turned up a solid figure on how much energy you could expect out of a kilogram of helium-3, but my understanding is that the 3He+3He reaction produces more energy per particle than the 2D+2D reaction, so I believe that the energy density would be higher; however, I have not covered any physics in several years and I don't think that my understanding is correct. I would actually be interested in understanding how to properly calculate this, even if the mathematics is probably beyond me. Setting that aside for a moment, I come to the question of efficiency. I have heard estimates that suggest that a rate of conversion of close to 70% is possible, the source being a 2001 lecture (note, this is a .pdf link) by this man. However, I'm more curious about the potential efficiency of the reaction itself. I assume that if I had a working fusion reactor in my cellar and fed it a kilogram of Helium-3 fuel I would not actually get every erg promised by the mathematics, which is lucky because I don't really know what I'd do with a hundred kilotons; why is this so? What are the unavoidable inefficiencies that are to be expected in a working fusion reactor? Do they differ from design to design? I understand that I'm asking for a lot of practical information on devices that more or less don't exist, but any information is potentially worthwhile.

So assuming that you have an almost magical fusor at the very edge of impossible efficiency, which will not suddenly flash into steam when you turn it on, and happily produce a hundred terawatts on command, how does that translate to space travel? Assuming you are confined to an actual method of propulsion, as opposed to a magic torchdrive, what sort of performance could you squeeze out of some of the fusion based designs listed on the Atomic Rockets drive table? Obviously, a lot of performance is dependant on things like your payload, deltaV and mass fraction, but I am mostly interested in how much influence raw power output has on propulsion.

Anyway, I think that's more than enough questions for the time being. Any and all answers are appreciated.

[Zixinus]

About efficiency: with aneutronic fusion, you can reach a pretty high efficiency because you can convert the fusion product straight to electricity, unlike neutron-producing ones where you have to go trough the steam-dynamo-generator cycle, thus saving space and costs (which is also more ideal for spacecrafts).

What are the unavoidable inefficiencies that are to be expected in a working fusion reactor?

The greatest will be managing heat, especially for less-advanced neutron-producing reactions. You'll be cooling something that is next to something that is hot enough to be plasma.

Do they differ from design to design

Yes, certainly. For example, the Polywell scheme I was rabid about a few years back, had you producing a considerable rate of fusion in a cube made out of copper rings. There you had the problem that you had to cool, power and protect these rings simultaneously. This is less of a problem with aneutronically-focused reactions, but only just.
This is less of a problem for Tokamak designs, but they also have their own problems.

So assuming that you have an almost magical fusor at the very edge of impossible efficiency, which will not suddenly flash into steam when you turn it on, and happily produce a hundred terawatts on command, how does that translate to space travel?

PM me either an e-mail or MSN address and I can send you over some files about the Polywell space travel, written way back by Brussard.

[Ariphaos]

The reason Deuterium has a higher energy density than Tritium and He3 is because the Deuterium fusion process generates those isotopes and produces a pretty significant amount of energy doing so.

[Surlethe]

The reason Deuterium has a higher energy density than Tritium and He3 is because the Deuterium fusion process generates those isotopes and produces a pretty significant amount of energy doing so.

Which isotopes? How does it generate them? How much energy does it produce generating them?

[Starglider]

Here are the relevant reactions;

D+T → He4 ( 3.5 MeV ) + n ( 14.1 MeV )
D+D → T ( 1.01 MeV ) + p ( 3.02 MeV )
D+D → He3 ( 0.82 MeV ) + n ( 2.45 MeV )
D+He3 → He4 ( 3.6 MeV ) + p ( 14.7 MeV )
T+T → He4 + 2n ( 11.3 MeV )
He3 + He3 → He4 + 2p ( 12.9 MeV )
He3 + T → He4 + p + n ( 12.1 MeV )
He3 + T → He4 ( 4.8 MeV ) + D ( 9.5 MeV )
He3 + T → He4 ( 0.5 MeV ) +n ( 1.9 MeV ) + p + ( 11.9 MeV )

On average, fusing 4 deuterium atoms will produce about 19.3 MeV (the exact statistical distribution of reaction paths is complicated), a helium atom and assorted fast protons and neutrons. Fusing two deuterium and two tritium will produce about 35.2 MeV, a helium atom and two neutrons. Thus D/T fuel is about 150% as energetic per unit mass as pure D (the D-D and T-T side reactions normally have a low cross section), as well as being easier to fuse and almost certainly allowing a smaller, lighter reactor.

D + Li3 → 2 He4 + 22.4 MeV
p + B11 → 3 He4 + 8.7 MeV

Deuterium/lithium reactions produce less neutrons but about 60% as much energy per unit mass as pure D; the D will also react with itself, so this isn't terribly useful. Hydrogen/boron is aneutronic but has a horrible energy/mass efficiency of about 15% of D-D. It also requires very high plasma energies, probably implying a larger/heavier reactor (though you may save quite a lot of mass by omitting neutron shielding). Pure He3 fuel is a much better option for aneutronic fusion, with a energy/mass efficiency of about 45% of D-D, but we'd probably have to mine the lunar surface or gas giants to power more than the odd very expensive probe.

One could easily see military ships using D-T for maximum delta-v and minimum reactor size, fast couriers and exploration ships using He3 for decent efficiency without the radiation hazard, and slow cargo transports using low-efficiency but (relatively) cheap and low-maintenance boron reactors.

[Kuroneko]

Going by the Constants page of the main site, deuterium fusion has an energy density of 6.2E14 J/kg. My, admittedly cursory, research has not turned up a solid figure on how much energy you could expect out of a kilogram of helium-3, but my understanding is that the 3He+3He reaction produces more energy per particle than the 2D+2D reaction, so I believe that the energy density would be higher; however, I have not covered any physics in several years and I don't think that my understanding is correct. I would actually be interested in understanding how to properly calculate this, even if the mathematics is probably beyond me.

[edit]Whoops, I misread the exponent.[/edit] As for sources of inefficiency, there's radiation loss, thermal loss, and hard upper bound of Carnot cycle efficiency even under perfect insulation.

Reaction
²D + ³T → ⁴He + n
²D + ²D → ³T + p or ³He + n
²D + ³He → ⁴He + p
³T + ³T → ⁴He + 2n
³He + ³He → ⁴He + 2p

Energy
17.59 MeV
3.65 MeV
18.35 MeV
11.33 MeV
12.86 MeV

Unionized Reactant Mass
4.686 GeV/c²
3.752 GeV/c²
4.686 GeV/c²
5.619 GeV/c²
5.619 GeV/c²

Energy density
3.38E14 J/kg
8.74E13 J/kg
3.52E14 J/kg
1.81E14 J/kg
2.06E14 J/kg

Note that since typically the products can in turn undergo fusion, what actually happens in a reactor depends on the reaction cross-sections and reactor conditions. The masses of the the individual particles are as follows:

Particle
p
n
²D

Mass (GeV/c²)
0.938272
0.939565
1.875613

Particle
³T
³He
⁴He

Mass (GeV/c²)
2.808921
2.808391
3.727378

I calculated them from the WIkipedia values, adjusted for lack of electrons and electron binding energy (e.g., helium has 2 electrons massing 0.5110MeV each, and the total electron binding energy is -79eV, which was in turn calculated from the ionization figures). So, for example,
(²D + ³T) - (⁴He + n) = 17.59 MeV,
(²D + ²D) - (³T + p) = 4.03 MeV,
(²D + ²D) - (³He + n) = 3.27 MeV.
The average of the last two is 3.65 MeV, since there is a 50% split between them (this not an a priori assumption, but an empirical fact).