Nuclear fusion: Difference between revisions
m Reverted edits by 211.252.84.247 to last version by Art Carlson |
Art Carlson (talk | contribs) →Figures of merit for terrestrial reactions: new tables |
||
Line 181: | Line 181: | ||
===Figures of merit for terrestrial reactions=== |
===Figures of merit for terrestrial reactions=== |
||
Any of the reactions above can in principle be the basis of [[fusion power]] production. In addition to the temperature and cross section discussed above, we must consider the total energy of the fusion products ''E''<sub>fus</sub>, the energy of the charged fusion products ''E''<sub>ch</sub>, and the atomic number ''Z'' of the non-hydrogenic reactant. |
Any of the reactions above can in principle be the basis of [[fusion power]] production. In addition to the temperature and cross section discussed above, we must consider the total energy of the fusion products ''E''<sub>fus</sub>, the energy of the charged fusion products ''E''<sub>ch</sub>, and the atomic number ''Z'' of the non-hydrogenic reactant. ''E''<sub>n</sub> is the energy of the neutrons, which is naturally the difference of the other two. |
||
Specification of the D-D reaction entails some difficulties, though. To begin with, one must average over the two branches (2) and (3). More difficult is to decide how to treat the T and {{sup|3}}He products. T burns so well in a deuterium plasma that you probably can't get it out even if you want to. The D-{{sup|3}}He reaction is optimized at a much higher temperature, so the burnup at the optimum D-D temperature may be low, so it seems reasonable to assume the T but not the {{sup|3}}He gets burned up and adds its energy to the net reaction. Thus we will count the DD fusion energy as |
Specification of the D-D reaction entails some difficulties, though. To begin with, one must average over the two branches (2) and (3). More difficult is to decide how to treat the T and {{sup|3}}He products. T burns so well in a deuterium plasma that you probably can't get it out even if you want to. The D-{{sup|3}}He reaction is optimized at a much higher temperature, so the burnup at the optimum D-D temperature may be low, so it seems reasonable to assume the T but not the {{sup|3}}He gets burned up and adds its energy to the net reaction. Thus we will count the DD fusion energy as ''E''{{sub|fus}} = (4.03+17.6+3.27)/2 = 12.5 MeV and the energy in charged particles as ''E''{{sub|ch}} = (4.03+3.5+0.82)/2 = 4.2 MeV. |
||
⚫ | |||
:''E''<sub>fus</sub> = (4.03+17.6+3.27)/2 = 12.5 MeV |
|||
⚫ | |||
and the energy in charged particles as |
|||
{| border="1" cellpadding="5" align="center" |
|||
:''E''<sub>ch</sub> = (4.03+3.5+0.82)/2 = 4.2 MeV |
|||
⚫ | |||
|- |
|||
|D-T || 1 || 17.6 || 3.5 || 0.80 |
|||
|- |
|||
|D-D || 1 || 12.5 || 4.2 || 0.66 |
|||
|- |
|||
|D-<sup>3</sup>He || 2 || 18.3 ||18.3 || 0.00 |
|||
|- |
|||
|p-<sup>11</sup>B || 5 || 8.7 || 8.7 || 0.00 |
|||
|} |
|||
The last column is the '''neutronicity''' of the reaction, the fraction of the fusion energy released as neutrons. To the extent that neutrons are "bad", the D-D reaction is a slight improvement over the D-T reaction, particularly since the energy of the neutrons produced is also less, which mitigates some of the problems. The last two reactions are nominally aneutronic, but the D-[[sup|3}}He reaction in particular will always have several percent neutronicity because of D-D side reactions. |
|||
⚫ | |||
Of course the reactants should also be mixed in the optimal proportions. This is the case when each reactant ion plus its associated electrons accounts for half the pressure. Assuming that the total pressure is fixed, this means that density of the non-hydrogenic ion is smaller than that of the hydrogenic ion by a factor 2/(''Z''+1). Therefore the rate for these reactions is reduced by the same factor, on top of any differences in the values of <&sigmy;v>/T². On the other hand, because the D-D reaction has only one reactant, the rate is twice as high as if the fuel were divided between two hydrogenic species. |
|||
⚫ | |||
Thus there is a "penalty" of (2/(Z+1)) for non-hydrogenic fuels arising from the fact that they require more electrons, which take up pressure without participating in the fusion reaction. There is at the same time a "bonus" of a facotr 2 for D-D due to the fact that each ion can react with any of the other ions, not just a fraction of them. |
|||
We can now compare these reactions in the following table. |
|||
{| border="1" cellpadding="5" align="center" |
{| border="1" cellpadding="5" align="center" |
||
!fuel!!<σv>/T²!!penalty/bonus!!reactivity!!Lawson criterion!!power density |
|||
⚫ | |||
|- |
|- |
||
|D-T || 1 || |
|D-T || 1.24e-24 || 1 || 1 || 1 || 1 |
||
|- |
|- |
||
|D-D || |
|D-D || 1.28e-26 || 2 || 48 || 30 || 68 |
||
|- |
|- |
||
|D-<sup>3</sup>He || 2 || |
|D-<sup>3</sup>He || 2.24e-26 || 2/3 || 83 || 16 || 80 |
||
|- |
|- |
||
|p-<sup>11</sup>B || |
|p-<sup>11</sup>B || 3.01e-27 || 1/3 || 1240 || 500 || 2500 |
||
|} |
|} |
||
The maximum value of <σv>/T² is taken from a previous table. The "penalty/bonus" factor is that related to a non-hydrogenic reactant or a single-species reaction. The values in the column "reactivity" are found by dividing (1.24e-24) by the product of the second and third columns. It indicates the factor by which the other reactions occur more slowly than the D-T reaction under comparable conditions. The column "Lawson criterion" weights these results with ''E''{{sub|ch}} and gives an indication of how much more difficult it is to achieve ignition with these reactions, relative to the difficulty for the D-T reaction. The last column is label "power density" and weights the practical reactivity with ''E''{{sub|fus}}. It indicates how much lower the fusion power density of the other reactions is compared to the D-T reaction and can be considered a measure of the economic potential. |
|||
==See also== |
==See also== |
Revision as of 21:55, 20 March 2005
In physics, nuclear fusion is a process in which two nuclei join, forming a larger nucleus and releasing energy. Nuclear fusion is the energy source which causes stars to shine, and hydrogen bombs to explode.
This article deals with the fusion reaction itself. For information on controlling the fusion reaction to produce useful power, see the article on fusion power.
It takes considerable energy to force nuclei to fuse, even those of the least massive element, hydrogen. But the fusion of lighter nuclei, which creates a heavier nucleus and a free neutron, will generally release even more energy than it took to force them together — an exothermic process that can produce self-sustaining reactions. If the energy to initiate the reaction comes from accelerating one of the nuclei, the process is called beam-target fusion, if both nuclei are accelerated, it is beam-beam fusion. If the nuclei are part of a plasma near thermal equilibrium, one speaks of thermonuclear fusion.
The energy released in most nuclear reactions is much larger than that for chemical reactions, because the binding energy that glues a nucleus together is far greater than the energy that holds electrons to a nucleus. For example, the ionization energy gained by adding an electron to hydrogen is 13.6 electron volts -- less than one-millionth of the 17 MeV released in the D-T reaction shown below.
Requirements for fusion
A substantial energy barrier opposes the fusion reaction. The positive electrical charges of the nuclei repel each other via the electrostatic force, attempting to break any nuclei apart. Opposing this is the slightly more powerful strong nuclear force, which tries to hold them together. It is the tension between these two powerful forces that makes nuclear reactions so powerful.
The strong force only operates over short distances. When a nucleon (proton or neutron) is added to a nucleus, it is attracted by the strong force to other nucleons, but only to those in direct contact. The nucleons in the interior have neughbors on all sides, but those on the surface only have neighbors on one side. Since smaller nuclei have a larger surface to volume ratio, the binding energy per nucleon due to the strong force increases with size up to a limit.
The electrostatic force, on the other hand, is an inverse-square force, so a proton added to a nucleus will feel an electrostatic repulsion from all the other protons in the nucleus. The electrostatic energy per nucleon due to the electrostatic force thus increases without limit as the nuclei get larger.
The net result of these opposing forces is that the binding energy per nucleon generally increases with increasing size, namely up to the element iron, and then starts decreasing again. Eventually, at the element uranium, the binding energy becomes negative and the nuclei are no longer stable.
A notable exception to this trend is the element helium, whose nucleus consists of two protons and two neutrons. In a sense, protons and neutrons are two states of a single particle, and each of these states can have spin up or spin down. Consequently, all the nucleons in a helium nucleus can be in the ground state, making it extremely tightly bound. If any nucleons are added, they have to go into a higher energy state due to the Pauli exclusion principle.
The situation is similar if two nuclei are brought together. As they approach each other, all the protons in one nucleus repell all the protons in the other. Not until the two nuclei actually come in contact can the strong nuclear force take over. Consequently, even when the final energy state is lower, there is a large energys barrier that must first be overcome. In chemistry, one would speak of the activation energy. In nuclear physics it is called the Coulomb barrier.
The Coulomb barrier is smallest for isotopes of hydrogen, since they contain only a single positive charge in the nucleus. Since a bi-proton is not stable, neutrons must also be involved, ideally in such a way that a helium nucleus, with its extremely tight binding, is one of the products.
In the D-T fuel, the resulting energy barrier is about 0.1 MeV. In comparison, the energy needed to remove an electron from hydrogen is 13 eV, about 7,500 times less energy. Once the fusion reaction is complete, the new nucleus drops to a lower-energy configuration and gives up additional energy by ejecting a neutron with 17.59 MeV, considerably more than what was needed to fuse them in the first place. This means that the D-T fusion reaction is very highly exothermic, making it a powerful energy source.
Fusion cross section
The "easiest" way to provide the energy needed to overcome the Coulomb barrier is to heat the nuclei. Temperature is a measure of the average kinetic energy of particles, so by heating the nuclei they will gain energy and eventually have enough to overcome this 0.1 MeV barrier. Converting the units between eV and kelvins shows that the barrier would be overcome at a temperature in excess of 1 GK, obviously a very high temperature.
There are two effects that lower the actual temperature needed. One is the fact that temperature is the average kinetic energy, implying that some nuclei at this temperature would actually have much higher energy than 0.1 MeV, while others would be much lower. It is the nuclei in the high-energy tail of the velocity distribution that account for most of the fusion reactions. The other effect is quantum tunneling. The nuclei do not actually have to have enough energy to overcome the Coulomb barrier completely. If they have nearly enough energy, they can tunnel through the remaining barrier. For this reason fuel at lower temperatures will still undergo fusion events, at a lower rate.
The reaction cross section combines the effects of the Coulomb barrier and thermal velocity distribution of the nuclei into an "effective area" for fusion collisions. The frequency f of fusion reactions per unit volume is given by the equation
where and are the densities of the two participant nuclei, v is the relative velocity, σ is the cross section, which is itself a function of the relative velocity, and the brackets denote an average over the velocity distribution (normally a thermal distribution). If a species of nuclei is reacting with itself, such as the DD reaction, then the product must be replaced by .
increases from virtually zero at room temperatures up to meaningful magnitudes at temperatures of 10 - 100 keV. At these temperatures, well above typical ionization energies (13 eV in the hydrogen case), the fusion reactants exist in a plasma state.
The significance of as a function of temperature in a device with a particular energy confinement time is found by considering the Lawson criterion.
Fuel confinement
Gravitational confinement One force capable of confining the fuel well enough to satisfy the Lawson criterion is gravity. The mass needed, however, is so great that gravitational confinement is only found in stars. Even if the more reactive fuel deuterium were used, a mass about the size of the Moon would be needed.
Magnetic confinement Since plasmas are very good electrical conductors, magnetic fields can also be used to confine the fusion fuel. A variety of magnetic configurations can be used, the most basic distinction being between mirror confinement and toroidal confinement, especially tokamaks and stellarators.
Inertial confinement A third "confinement" principle is to apply no confining force at all, but to let the fuel expand freely under its own pressure. If the fuel is dense enough and hot enough, the reaction rate will be high enough to burn a significant fraction of the fuel before it is dissipated. To achieve these extreme conditions, the cold fuel must be explosively compressed with external photons or other particles. Inertial confinement is used in the hydrogen bomb, where the driver is a fission bomb, but is also attempted in "controlled" nuclear fusion, where the driver is a laser, ion, or electron beam.
Some other confinement principles have been investigated, such as muon-catalyzed fusion, the Farnsworth-Hirsch Fusor, and bubble fusion.
Important fusion reactions
Astrophysical reaction chains
The most important fusion process in nature is that which powers the stars. The net result is the fusion of four protons into one alpha particle, with the release of two electrons, two neutrinos, and energy, but several individual reactions are involved, depending on the mass of the star. For stars the size of the sun or smaller, the proton-proton chain dominates. In heavier stars, the CNO cycle is more important.
Criteria and candidates for terrestrial reactions
In man-made fusion, the primary fuel is not constrained to be protons and higher temperatures can be used, so reactions with larger cross-sections are chosen. This implies a lower Lawson criterion, and therefore less startup effort. Another concern is the production of neutrons, which are activate the reactor structure radiologically, but also have the advantages of allowing volumetric extraction of the fusion energy and tritium breeding. Reactions that release no neutrons are referred to as the aneutronic.
In order to be useful as a source of energy, a fusion reaction must satisfy several criteria. It must:
- ... be exothermic. This one is obvious, but it limits the reactants to the low Z side of the curve of binding energy. It also makes helium He-4 the most common product because of its extraordinarily tight binding, although He3 and T also show up.
- ... involve low Z nuclei. This is because the electrostatic repulsion must be overcome before the nuclei are close enough to fuse.
- ... have two reactants. At anything less than stellar densities, three body collisions are too improbable.
- ... have two or more products. This allows simultaneous conservation of energy and momentum without relying on the (weak!) electromagnetic force.
- ... and conserve both protons and neutrons. The cross sections for the weak interaction are too small.
Not many reactions meet these criteria. The most interesting reactions are the following.
(1) | D | + | T | → | 4He | (3.5 MeV) | + | n | (14.1 MeV) | |||||||
(2) | D | + | D | → | T | (1.01 MeV) | + | p | (3.02 MeV) | (50%) | ||||||
(3) | → | 3He | (0.82 MeV) | + | n | (2.45 MeV) | (50%) | |||||||||
(4) | D | + | 3He | → | 4He | (3.6 MeV) | + | p | (14.7 MeV) | |||||||
(5) | T | + | T | → | 4He | + | 2 | n | + 11.3 MeV | |||||||
(6) | 3He | + | 3He | → | 4He | + | 2 | p | ||||||||
(7) | 3He | + | T | → | 4He | + | p |   | + | n | + 12.1 MeV | (51%) | ||||
(8) | → | 4He | (4.8 MeV) | + | D | (9.5 MeV) | (43%) | |||||||||
(9) | → | 4He | (0.5 MeV) | + | n | (1.9 MeV) | + | p | (11.9 MeV) | (6%) | ||||||
(10) | D | + | 6Li | → | 2 | 4He | + 22.4 MeV | |||||||||
(11) | p | + | 6Li | → | 4He | (1.7 MeV) | + | 3He | (2.3 MeV) | |||||||
(12) | 3He | + | 6Li | → | 2 | 4He | + | p | + 16.9 MeV | |||||||
(13) | p | + | 11B | → | 3 | 4He | + 8.7 MeV |
p (proton), D (deuterium), and T (tritium) are shorthand notation for the first three isotopes of hydrogen.
In addition to the fusion reactions, the following reactions with neutrons are important in order to "breed" tritium in "dry" fusion bombs and some proposed fusion reactors:
- n + 6Li → T + 4He
- n + 7Li → T + 4He + n
To evaluate the usefulness of these reactions, in addition to the reactants, the products, and the energy released, one needs to know something about the cross section. Any given fusion device will have a maximum plasma pressure that it can sustain, and an economical device will always operate near this maximum. Given this pressure, the largest fusion output is obtained when the temperature is chosen so that <σv>/T² is a maximum. This is also the temperature at which the value of the triple product nTτ required for ignition is a minimum. This optimum temperature and the value of <σv>/T² at that temperature is given for a few of these reactions in the following table.
fuel | T [keV] | <σv>/T² [m³/sec/keV²] |
---|---|---|
D-T | 13.6 | 1.24×10-24 |
D-D | 15 | 1.28×10-26 |
D-3He | 58 | 2.24×10-26 |
p-6Li | 66 | 1.46×10-27 |
p-11B | 123 | 3.01×10-27 |
Note that many of the reactions form chains. For instance, a reactor fueled with T and 3He will create some D, which is then possible to use in the D + 3He reaction if the energies are "right". An elegant idea is to combine the reactions (11) and (12). The 3He from reaction (11) can react with 6Li in reaction (12) before completely thermalizing. This produces an energetic proton which in turn undergoes reaction (11) before thermalizing. A detailed analysis shows that this idea will not really work well, but it is a good example of a case where the usual assumption of a Maxwellian plasma in not appropriate.
Aneutronic reactions
Several of the reactions listed above have no neutrons as products. It is important to remember, however, that the reactions above never occur in their pure form. This is obvious in the case of reactions (2) and (3), which occur equally often in a deuterium plasma. Thus the aneutronic nature of (2) is spoiled by the neutron from (3). Furthermore, the T produced by reaction (2) has a good chance of reacting with a D in the background plasma to produce a neutron, and an energetic one at that. Or take p-11B, which appears to be completely aneutronic. Although the neutron production rate would be perhaps two to four orders of magnitude smaller than for D-T, there are still a number of side reactions that will produce neutrons or other radiation, e.g.,
p | + | 11B | → | 12C + γ |
p | + | 11B | → | n + 11C |
4He | + | 11B | → | n + 14N |
4He | + | 11B | → | p + 14C |
4He | + | 11B | → | T + 12C |
11B | + | 11B | → | junk |
as well as reactions with a possible 10B impurity fraction. Furthermore, some of the energetic product alphas will impinge on the wall and knock out neutrons. A calculation of the dose rate indicates that shielding will be mandatory even for the cleanest reactions.
Figures of merit for terrestrial reactions
Any of the reactions above can in principle be the basis of fusion power production. In addition to the temperature and cross section discussed above, we must consider the total energy of the fusion products Efus, the energy of the charged fusion products Ech, and the atomic number Z of the non-hydrogenic reactant. En is the energy of the neutrons, which is naturally the difference of the other two.
Specification of the D-D reaction entails some difficulties, though. To begin with, one must average over the two branches (2) and (3). More difficult is to decide how to treat the T and 3He products. T burns so well in a deuterium plasma that you probably can't get it out even if you want to. The D-3He reaction is optimized at a much higher temperature, so the burnup at the optimum D-D temperature may be low, so it seems reasonable to assume the T but not the 3He gets burned up and adds its energy to the net reaction. Thus we will count the DD fusion energy as Efus = (4.03+17.6+3.27)/2 = 12.5 MeV and the energy in charged particles as Ech = (4.03+3.5+0.82)/2 = 4.2 MeV.
Another unique aspect of the D-D reaction is that there is only one reactant, which must be taken into account when calculating the reaction rate.
With this choice, we tabulate parameters for four of the most important reactions.
fuel | Z | Efus [MeV] | Ech [MeV] | En/Efus |
---|---|---|---|---|
D-T | 1 | 17.6 | 3.5 | 0.80 |
D-D | 1 | 12.5 | 4.2 | 0.66 |
D-3He | 2 | 18.3 | 18.3 | 0.00 |
p-11B | 5 | 8.7 | 8.7 | 0.00 |
The last column is the neutronicity of the reaction, the fraction of the fusion energy released as neutrons. To the extent that neutrons are "bad", the D-D reaction is a slight improvement over the D-T reaction, particularly since the energy of the neutrons produced is also less, which mitigates some of the problems. The last two reactions are nominally aneutronic, but the D-[[sup|3}}He reaction in particular will always have several percent neutronicity because of D-D side reactions.
Of course the reactants should also be mixed in the optimal proportions. This is the case when each reactant ion plus its associated electrons accounts for half the pressure. Assuming that the total pressure is fixed, this means that density of the non-hydrogenic ion is smaller than that of the hydrogenic ion by a factor 2/(Z+1). Therefore the rate for these reactions is reduced by the same factor, on top of any differences in the values of <&sigmy;v>/T². On the other hand, because the D-D reaction has only one reactant, the rate is twice as high as if the fuel were divided between two hydrogenic species.
Thus there is a "penalty" of (2/(Z+1)) for non-hydrogenic fuels arising from the fact that they require more electrons, which take up pressure without participating in the fusion reaction. There is at the same time a "bonus" of a facotr 2 for D-D due to the fact that each ion can react with any of the other ions, not just a fraction of them.
We can now compare these reactions in the following table.
fuel | <σv>/T² | penalty/bonus | reactivity | Lawson criterion | power density |
---|---|---|---|---|---|
D-T | 1.24e-24 | 1 | 1 | 1 | 1 |
D-D | 1.28e-26 | 2 | 48 | 30 | 68 |
D-3He | 2.24e-26 | 2/3 | 83 | 16 | 80 |
p-11B | 3.01e-27 | 1/3 | 1240 | 500 | 2500 |
The maximum value of <σv>/T² is taken from a previous table. The "penalty/bonus" factor is that related to a non-hydrogenic reactant or a single-species reaction. The values in the column "reactivity" are found by dividing (1.24e-24) by the product of the second and third columns. It indicates the factor by which the other reactions occur more slowly than the D-T reaction under comparable conditions. The column "Lawson criterion" weights these results with Ech and gives an indication of how much more difficult it is to achieve ignition with these reactions, relative to the difficulty for the D-T reaction. The last column is label "power density" and weights the practical reactivity with Efus. It indicates how much lower the fusion power density of the other reactions is compared to the D-T reaction and can be considered a measure of the economic potential.
See also
- Cold fusion
- Helium fusion
- Muon-catalyzed fusion
- Antimatter catalyzed nuclear pulse propulsion
- Timeline of nuclear fusion
- Future energy development
External links
- http://www.fusion.org.uk/ - A guide to fusion from the UKAEA
- SCK.CEN Belgian Nuclear Research Centre Mol, Belgium