Open Access Articles- Top Results for Aneutronic fusion

Aneutronic fusion

Aneutronic fusion is any form of fusion power in which neutrons carry no more than 1% of the total released energy.[1] The most-studied fusion reactions release up to 80% of their energy in neutrons. Successful aneutronic fusion would greatly reduce problems associated with neutron radiation such as ionizing damage, neutron activation, and requirements for biological shielding, remote handling, and safety.

Some proponents also see a potential for dramatic cost reductions by converting energy directly to electricity. However, the conditions required to harness aneutronic fusion are much more extreme than those required for the conventional deuteriumtritium (DT) fuel cycle.

Candidate aneutronic reactions

There are a few fusion reactions that have no neutrons as products on any of their branches. Those with the largest cross sections are these:

Deuterium–helium-3 fusion 2D + 3He   4He + 1p + 18.3 MeV
Deuterium–lithium-6 fusion 2D + 6Li 2 4He     + 22.4 MeV
Proton–lithium-6 fusion 1p + 6Li 4He + 3He + 4.0 MeV
Helium-3–lithium fusion 3He + 6Li 2 4He + 1p + 16.9 MeV
Helium-3-helium-3 fusion 3He + 3He   4He + 2 1p + 12.86 MeV
Proton–lithium-7 fusion 1p + 7Li 2 4He     + 17.2 MeV
Proton–boron fusion 1p + 11B 3 4He     + 8.7 MeV
Proton–nitrogen fusion 1p + 15N   12C + 4He + 5.0 MeV[2]

The two of these which use deuterium as a fuel produce some neutrons with D–D side reactions.[citation needed] Although these can be minimized by running hot and deuterium-lean, the fraction of energy released as neutrons will probably be several percent, so that these fuel cycles, although neutron-poor, do not qualify as purely aneutronic according to the 1% threshold. See main article at Helium-3#Fusion reactions. These reactions also suffer from the 3He fuel availability problem, as discussed below.

The next two reactions' rates (involving p, 3He, and 6Li) are not particularly high in a thermal plasma. When treated as a chain, however, they offer the possibility of enhanced reactivity due to a non-thermal distribution. The product 3He from the first reaction could participate in the second reaction before thermalizing, and the product p from the second reaction could participate in the first reaction before thermalizing. Unfortunately, detailed analyses do not show sufficient reactivity enhancement to overcome the inherently low cross section.[citation needed]

The pure 3He reaction suffers from a fuel-availability problem. 3He occurs in only minuscule amounts naturally on Earth, so it would either have to be bred from neutron reactions (counteracting the potential advantage of aneutronic fusion), or mined from extraterrestrial sources. The top several meters of the surface of the Moon is relatively rich in 3He[3] on the order of 0.01 parts per million by weight,[4] but mining this resource and returning it to Earth would be relatively difficult and expensive. 3He could in principle be recovered from the atmospheres of the gas giant planets, Jupiter, Saturn, Neptune and Uranus, but this would be even more challenging. The amount of fuel needed for large-scale applications can also be put in terms of total consumption: according to the US Energy Information Administration, "Electricity consumption by 107 million U.S. households in 2001 totaled 1,140 billion kW·h" (1.14×1015 W·h). Again assuming 100% conversion efficiency, 6.7 tonnes per year of helium-3 would be required for that segment of the energy demand of the United States, 15 to 20 tonnes per year given a more realistic end-to-end conversion efficiency.

The p –7Li reaction has no advantage[citation needed] over p –11B, given its somewhat lower cross section.[citation needed] But this is mitigated by having double the power output.

For the above reasons, most studies of aneutronic fusion concentrate on the reaction, p –11B.[5] [6]

Technical challenges


Despite the suggested advantages of aneutronic fusion, the vast majority of fusion research has gone toward D-T fusion because the technical challenges of proton–boron (p –11B) fusion are so formidable. Proton–boron fusion requires ion energies or temperatures almost ten times higher than those for D-T fusion. For any given densities of the reacting nuclei, the reaction rate for proton-boron achieves its peak rate at around 600 keV (6.6 billion degrees Celsius or 6.6 gigakelvins)[7][better source needed] while D–T has a peak at around 66 keV (765 million degrees Celsius). For pressure-limited confinement concepts, optimum operating temperatures are about 5 times lower, but the ratio is still roughly ten-to-one.

Power balance

In addition, the peak reaction rate of p–11B is only one third that for D–T, requiring better plasma confinement. Confinement is usually characterized by the time τ the energy must be retained so that the fusion power released exceeds the power required to heat the plasma. Various requirements can be derived, most commonly the product with the density, nτ, and the product with the pressure nTτ, both of which are called the Lawson criterion. The nτ required for p–11B is 45 times higher than that for DT. The nTτ required is 500 times higher.[8] (See also neutronicity, confinement requirement, and power density.) Since the confinement properties of conventional fusion approaches, such as the tokamak and laser pellet fusion are marginal, most aneutronic proposals use radically different confinement concepts.

In most fusion plasmas, bremsstrahlung radiation is a major energy loss channel. (See also bremsstrahlung losses in quasineutral, isotropic plasmas.) For the p–11B reaction, some calculations indicate that the bremsstrahlung power will be at least 1.74 times larger than the fusion power. The corresponding ratio for the 3He-3He reaction is only slightly more favorable at 1.39. This is not applicable to non-neutral plasmas, and different in anisotropic plasmas.

In conventional reactor designs, whether based on magnetic confinement or inertial confinement, the bremsstrahlung can easily escape the plasma and is considered a pure energy loss term. The outlook would be more favorable if the plasma could reabsorb the radiation. Absorption occurs primarily via Thomson scattering on the electrons,[9] which has a total cross section of σT = 6.65×10−29 m². In a 50–50 D–T mixture this corresponds to a range of 6.3 g/cm².[10] This is considerably higher than the Lawson criterion of ρR > 1 g/cm², which is already difficult to attain, but might not be out of reach in future inertial confinement systems.[11]

In very high magnetic fields, on the order of a megatesla, a quantum mechanical effect may suppress energy transfer from the ions to the electrons.[12] According to one calculation,[13] the bremsstrahlung losses could be reduced to half the fusion power or less. In a strong magnetic field the cyclotron radiation is even larger than the bremsstrahlung. In a megatesla field, an electron would lose its energy to cyclotron radiation in a few picoseconds if the radiation could escape. However, in a sufficiently dense plasma (ne > 2.5×1030 m−3, a density greater than that of a solid[14]), the cyclotron frequency is less than twice the plasma frequency. In this well-known case, the cyclotron radiation is trapped inside the plasmoid and cannot escape, except from a very thin surface layer.

While megatesla fields have not yet been achieved in the laboratory, fields of 0.3 megatesla have been produced with high intensity lasers,[15] and fields of 0.02–0.04 megatesla have been observed with the dense plasma focus device.[16][17]

At much higher densities (ne > 6.7×1034 m−3), the electrons will be Fermi degenerate, which suppresses bremsstrahlung losses, both directly and by reducing energy transfer from the ions to the electrons.[18] If necessary conditions can be attained, net energy production from p–11B or D–3He fuel may be possible. The probability of a feasible reactor based solely on this effect remains low, however, because the gain is predicted to be less than 20, while more than 200 is usually considered to be necessary. (There are, however, effects that might improve the gain substantially.)[citation needed]

Power density

In every published fusion power plant design, the part of the plant that produces the fusion reactions is much more expensive than the part that converts the nuclear power to electricity. In that case, as indeed in most power systems, power density is a very important characteristic.[19] Doubling power density at least halves the cost of electricity. In addition, the confinement time required depends on the power density.

It is, however, not trivial to compare the power density produced by different fusion fuel cycles. The case most favorable to p–11B relative to D–T fuel is a (hypothetical) confinement device that only works well at ion temperatures above about 400 keV, in which the reaction rate parameter <σv> is equal for the two fuels, and that runs with low electron temperature. p–11B does not require as long a confinement time because the energy of its charged products is two and a half times higher than that for D–T. However, relaxing these assumptions, for example by considering hot electrons, by allowing the D–T reaction to run at a lower temperature or by including the energy of the neutrons in the calculation shifts the power density advantage to D–T.

The most common assumption is to compare power densities at the same pressure, choosing the ion temperature for each reaction to maximize power density, and with the electron temperature equal to the ion temperature. Although confinement schemes can be and sometimes are limited by other factors, most well-investigated schemes have some kind of pressure limit. Under these assumptions, the power density for p–11B is about 2,100 times smaller than that for D–T. Using cold electrons lowers the ratio to about 700. These numbers are another indication that aneutronic fusion power will not be possible with any mainline confinement concept.

Current research

  • The Z-machine at Sandia National Laboratory, a z-pinch device, can produce ion energies of interest to hydrogen–boron reactions, up to 300 keV.[30] Non-equilibrium plasmas usually have an electron temperature higher than their ion temperature, but the plasma in the Z machine has a special, reverted non-equilibrium state, in which the ion temperature is 100 times higher than electron temperature. These data represent a new research field, and indicate that Bremsstrahlung losses could be in fact lower than previously expected in such a design.

None of these efforts has yet tested its device with hydrogen–boron fuel, so the anticipated performance is based on extrapolating from theory, experimental results with other fuels and from simulations.

  • A picosecond laser produced hydrogen–boron aneutronic fusions for a Russian team in 2005.[31] However, the number of the resulting α particles (around 103 per laser pulse) was extremely low.

Residual radiation from a p–11B reactor

Detailed calculations show that at least 0.1% of the reactions in a thermal p–11B plasma would produce neutrons, and the energy of these neutrons would account for less than 0.2% of the total energy released.[32]

These neutrons come primarily from the reaction[33]

11B + α14N + n + 157 keV

The reaction itself produces only 157 keV, but the neutron will carry a large fraction of the alpha energy, which will be close to Efusion/3 = 2.9 MeV. Another significant source of neutrons is the reaction

11B + p → 11C + n − 2.8 MeV

These neutrons will be less energetic, with an energy comparable to the fuel temperature. In addition, 11C itself is radioactive, but will decay to negligible levels within several hours as its half life is only 20 minutes.

Since these reactions involve the reactants and products of the primary fusion reaction, it would be difficult to further lower the neutron production by a significant fraction. A clever magnetic confinement scheme could in principle suppress the first reaction by extracting the alphas as soon as they are created, but then their energy would not be available to keep the plasma hot. The second reaction could in principle be suppressed relative to the desired fusion by removing the high energy tail of the ion distribution, but this would probably be prohibited by the power required to prevent the distribution from thermalizing.

In addition to neutrons, large quantities of hard X-rays will be produced by bremsstrahlung, and 4, 12, and 16 MeV gamma rays will be produced by the fusion reaction

11B + p → 12C + γ + 16.0 MeV

with a branching probability relative to the primary fusion reaction of about 10−4.[34]

Finally, isotopically pure hydrogen fuel will have to be used and the influx of impurities into the plasma will have to be controlled to prevent neutron-producing side reactions like these:

11B + d → 12C + n + 13.7 MeV
d + d → 3He + n + 3.27 MeV

With careful design, it should be possible to reduce the occupational dose of both neutron and gamma radiation to operators to a negligible level.[citation needed] The primary components of the shielding would be water to moderate the fast neutrons, boron to absorb the moderated neutrons, and metal to absorb X-rays. The total thickness needed should be about a meter, most of that being water.[35]

Methods for energy capture

See also: Direct energy conversion

Aneutronic fusion produces energy in the form of charged particles instead of neutrons. This means that energy from aneutronic fusion could be captured using direct conversion instead of the steam cycle which would normally be used for neutrons. Direct conversion techniques can either be inductive, based on changes in magnetic fields, electrostatic, based on making charged particles work against an electric field or, photoelectric, in which light energy is captured. If the fusion reactor worked in a pulsed mode, inductive techniques could be used.[36]

Electrostatic direct conversion uses a charged particles’ motion to make a voltage. This voltage drives electricity in a wire. This becomes the electrical power. It is normally thought of in reverse. Ordinarily, a voltage puts a particle in motion. Direct energy conversion does the opposite. It uses a particle's motion to produce a voltage. It has been described as a linear accelerator running backwards.[37] An early supporter of this method was Richard F. Post at the Lawrence Livermore National Laboratory. He proposed a way to capture the kinetic energy of charged particles as they were exhausted from a fusion reactor and convert this into voltage, which would drive current in a wire.[38] Dr. Post helped developed the theoretical underpinnings of direct conversion, which was later demonstrated by Dr. William Barr and Raulph Moir at LLNL. They demonstrated a 48 percent energy capture efficiency on the Tandem Mirror Experiment in 1981.[39]

In terms of photoelectric: aneutronic fusion also loses much of its energy as light. This energy results from the acceleration and deceleration of charged particles. These changes in speed can be caused by charge-charge interaction (Bremsstrahlung radiation) or magnetic field interactions (Cyclotron radiation or Synchrotron radiation) or electric field interactions. The radiation can be estimated using the Larmor formula and comes in the X-ray, IR, UV and visible spectrum. Some of the energy radiated as X-rays may be converted directly to electricity. Because of the photoelectric effect, X-rays passing through an array of conducting foils would transfer some of their energy to electrons, which can then be captured electrostatically. Since X-rays can go through far greater thickness of material than electrons can, many hundreds or even thousands of layers would be needed to absorb most of the X-rays.[40]


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  3. ^ Harrison H. Schmitt, Return to the Moon: Exploration, Enterprise, and Energy in the Human Settlement of Space, Springer 2007, chapter 5.
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  9. ^ Lecture 3 : Accelerated charges and bremsstrahlung, lecture notes in astrophysics from Chris Flynn, Tuorla Observatory
  10. ^ miT = 2.5×(1.67×10−24 g)/(6.65×10−25 cm²) = 6.28 g/cm²
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  32. ^ Heindler and Kernbichler, Proc. 5th Intl. Conf. on Emerging Nuclear Energy Systems, 1989, pp. 177–82. Even though 0.1% is a small fraction, the dose is rate still high enough to require very good shielding, as illustrated by the following calculation. Assume we have a very small reactor producing 30 kW of total fusion power (a full-scale power reactor might produce 100,000 times more than this) and 30 W in the form of neutrons. If there is no significant shielding, a worker in the next room, 10 m away, might intercept (0.5 m²)/(4 pi (10 m)2) = 4×10−4 of this power, i.e., 0.012 W. With 70 kg body mass and the definition 1 gray = 1 J/kg, we find a dose rate of 0.00017 Gy/s. Using a quality factor of 20 for fast neutrons, this is equivalent to 3.4 millisieverts. The maximum yearly occupational dose of 50 mSv will be reached in 15 s, the fatal (LD50) dose of 5 Sv will be reached in half an hour. If very effective precautions are not taken, the neutrons would also activate the structure so that remote maintenance and radioactive waste disposal would be necessary.
  33. ^ W. Kernbichler, R. Feldbacher, M. Heindler. "Parametric Analysis of p–11B as Advanced Reactor Fuel" in Plasma Physics and Controlled Nuclear Fusion Research (Proc. 10th Int. Conf., London, 1984) IAEA-CN-44/I-I-6. Vol. 3 (IAEA, Vienna, 1987).
  34. ^ As with the neutron dose, shielding is essential with this level of gamma radiation. The neutron calculation in the previous note would apply if the production rate is decreased a factor of ten and the quality factor is reduced from 20 to 1. Without shielding, the occupational dose from a small (30 kW) reactor would still be reached in about an hour.
  35. ^ El Guebaly, Laial, A., Shielding design options and impact on reactor size and cost for the advanced fuel reactor Aploo, Proceedings- Symposium on Fusion Engineering, v.1, 1989, pp.388–391. This design refers to D–He3, which actually produces more neutrons than p–11B fuel.
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  40. ^ Quimby, D.C., High Thermal Efficiency X-ray energy conversion scheme for advanced fusion reactors, ASTM Special technical Publication, v.2, 1977, pp. 1161–1165

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