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Subject: Conventional Fusion FAQ Section 1/11 (Fusion Physics)

This article was archived around: 14 Nov 1999 10:27:51 GMT

All FAQs in Directory: fusion-faq
All FAQs posted in: sci.physics.fusion
Source: Usenet Version

Archive-name: fusion-faq/section1-physics Last-modified: 7-Aug-1994 Posting-frequency: More-or-less-monthly Disclaimer: While this section is still evolving, it should be useful to many people, and I encourage you to distribute it to anyone who might be interested (and willing to help!!!).
------------------------------------------------------------------ 1. Fusion as a Physical Phenomenon Last Revised August 7, 1994 Written by Robert F. Heeter, rfheeter@pppl.gov, unless otherwise cited. ------------------------------------------------------------------ ### Please let me know if anything here is unclear. ### *** A. What is fusion? "Fusion" means many things when discussed on the newsgroup. Technically, "fusion" is short for "Nuclear Fusion," which describes the class of reactions where two light nuclei fuse together, forming a heavier nucleus. This heavier nucleus is frequently unstable, and sometimes splits (fissions) into two or more fragments. "Fusion" also refers to the type of energy produced, and a "fusion reactor" describes an energy-producing facility which generates power via fusion reactors. Finally, "fusion" can also be used to refer to the scientific program aimed at harnessing fusion for clean, safe, and hopefully inexpensive energy production - a collaborative international program which has been carried on for the past 40-some years. Each of these three uses - the technical, the energy source, and the scientific research program - is discussed in a separate section of this FAQ. The technical aspects of fusion are discussed below in this section. *** B. How does fusion release energy? If you add up the masses of the particles which go into a fusion reaction, and you add up the masses of the particles which come out, there is frequently a difference. According to Einstein's famous law relating energy and mass, E=mc^2, the "mass difference" can take the form of energy. Fusion reactions involving nuclei lighter than iron typically release energy, but fusion reactions involving nuclei heavier than iron typically absorb energy. The amount of energy released depends on the specifics of the reaction; a table of reactions is given further below to give an idea of the variety of fusion reactions. Another way to look at this is to consider the "binding energy" of the elements in question. If the reactants are bound more weakly than the products, then energy is released in the reaction. "Binding energy" is the amount of energy you would have to put into a system in order to pull its components apart; conversely, in a system with high binding energy, a lot of energy is released as the components are allowed to bond together. Suppose you had two balls connected by a long, thin rubber band, so that they are not very tightly connected, and the rubber band can be broken easily. This is a system with low binding energy. Now here's an analogy to what happens in fusion: imagine the long, thin rubber band suddenly being replaced by a short, thick one. The short thick one has to be stretched a lot in order to connect to the two balls, but it wants to bind them more tightly, so it pulls them together, and energy is released as they move towards each other. The low-binding energy, long rubber band system has been replaced by a high-binding energy, short rubber band system, and energy is released. *** C. Where does fusion occur in nature? The conditions needed to induce fusion reactions are extreme; so extreme that virtually all natural fusion occurs in stars, where gravity compresses the gas, until temperature and pressure forces balance the gravitational compression. If there is enough material in the star, pressures and temperatures will grow large enough as the star contracts that fusion will begin to occur (see below for the explanation why); the energy released will then sustain the star's temperature against losses from sunlight being radiated away. The minimum mass needed to induce fusion is roughly one-tenth the sun's mass; this is why the sun is a star, but Jupiter is merely a (large) planet. (Jupiter is about 1/1000th the sun's mass, so if it were roughly 100 times bigger, it too would generate fusion and be a small, dim star.) Stellar fusion reactions gradually convert hydrogen into helium. When a star runs out of hydrogen fuel, it either stops burning (becoming a dwarf star) or, if it is large enough (so that gravity compresses the helium strongly) it begins burning the helium into heavier elements. Because fusion reactions cease to release energy once elements heavier than iron are involved, the larger stars also eventually run out of fuel, but this time they collapse in a supernova. Gravity, no longer opposed by the internal pressure of fusion-heated gases, crushes the core of the star, forming things like white dwarfs, neutron stars, and black holes (the bigger the star, the more extreme the result). (For more details, try the sci.astro or sci.space.science newsgroups.) *** D. Why doesn't fusion occur anywhere else in nature? Current scientific knowledge indicates that very little fusion occurs anywhere else in nature. The reason is because in order to get two nuclei to fuse, you first have to get them close together. (This is because the nuclear forces involved in fusion only act at short range.) However, because the two nuclei are both positively charged, they repel each other electrically. Nuclei will not fuse unless either (a) they collide with enough energy to overcome the electrical repulsion, or (b) they find a "sneaky" way to circumvent their repulsion (see muon-catalyzed fusion in section 4). The energy required for fusion is so high that fusion only occurs in appreciable amounts once the temperature gets over 10 million degrees Kelvin, so (a) doesn't happen anywhere outside of stars. Current knowledge suggests that the sort of processes that would allow sneaky-fusion as in (b) are very rare, so there just isn't much fusion in the everyday world. *** E. What are the basic fusion reactions? While it is possible to take any two nuclei and get them to fuse, it is easiest to get lighter nuclei to fuse, because they are less highly charged, and therefore easier to squeeze together. There are complicated quantum-mechanics rules which determine which products you will get from a given reaction, and in what amounts ("branching ratios"). The probability that two nuclei fuse is determined by the physics of the collsion, and a property called the "cross section" (see glossary) which (roughly speaking) measures the likelihood of a fusion reaction. (A simple analogy for cross-section is to consider a blindfolded person throwing a dart randomly towards a dartboard on a wall. The likelihood that the dart hits the target depends on the *cross-sectional* area of the target facing the dart-thrower. (Thanks to Rich Schroeppel for this analogy.)) Below is an annotated list of many fusion reactions discussed on the newsgroup. Note: D = deuterium, T = tritium, p = proton, n = neutron; these and the other elements involved are discussed in the glossary/FUT. (FUT = list of Frequently Used Terms; section 10 of the FAQ.) The numbers in parentheses are the energies of the reaction products (in Millions of electron-Volts, see glossary for details). The percentages indicate the branching ratios. More information on each of the elements is given below. Table I: Fusion Reactions Among Various Light Elements D+D -> T (1.01 MeV) + p (3.02 MeV) (50%) -> He3 (0.82 MeV) + n (2.45 MeV) (50%) <- most abundant fuel -> He4 + about 20 MeV of gamma rays (about 0.0001%; depends somewhat on temperature.) (most other low-probability branches are omitted below) D+T -> He4 (3.5 MeV) + n (14.1 MeV) <-easiest to achieve D+He3 -> He4 (3.6 MeV) + p (14.7 MeV) <-easiest aneutronic reaction "aneutronic" is explained below. T+T -> He4 + 2n + 11.3 MeV He3+T -> He4 + p + n + 12.1 MeV (51%) -> He4 (4.8) + D (9.5) (43%) -> He4 (0.5) + n (1.9) + p (11.9) (6%) <- via He5 decay p+Li6 -> He4 (1.7) + He3 (2.3) <- another aneutronic reaction p+Li7 -> 2 He4 + 17.3 MeV (20%) -> Be7 + n -1.6 MeV (80%) <- endothermic, not good. D+Li6 -> 2He4 + 22.4 MeV <- also aneutronic, but you get D-D reactions too. p+B11 -> 3 He4 + 8.7 MeV <- harder to do, but more energy than p+Li6 n+Li6 -> He4 (2.1) + T (2.7) <- this can convert n's to T's n+Li7 -> He4 + T + n - some energy From the list, you can see that some reactions release neutrons, many release helium, and different reactions release different amounts of energy (some even absorb energy, rather than releasing it). He-4 is a common product because the nucleus of He-4 is especially stable, so lots of energy is released in creating it. (A chemical analogy is the burning of gasoline, which is relatively unstable, to form water and carbon dioxide, which are more stable. The energy liberated in this combustion is what powers automobiles.) The reasons for the stability of He4 involve more physics than I want to go into here. Some of the more important fusion reactions will be described below. These reactions are also described in Section 2 in the context of their usefulness for energy-producing fusion reactors. *** F. Could you tell me more about these different elements? (Note: there's more information in the glossary too.) Hydrogen (p): Ordinary hydrogen is everywhere, especially in water. Deuterium (D): A heavy isotope of hydrogen (has a neutron in addition to the proton). Occurs naturally at 1 part in 6000; i.e. for every 6000 ordinary hydrogen atoms in water, etc., there's one D. Tritium (T): Tritium is another isotope of hydrogen, with two neutrons and a proton. T is unstable (radioactive), and decays into Helium-3 with a half-life of 12.3 years. (Half the T decays every 12.3 years.) Because of its short half-life, tritium is almost never found in nature (natural T is mostly a consequence of cosmic-ray bombardment). Supplies have been manufactured using fission reactors; world tritium reserves are estimated at a few kilograms, I believe. Tritium can be made by exposing deuterium or lithium to neutrons. Helium-3 (He3): Rare light isotope of helium; two protons and a neutron. Stable. There's roughly 13 He-3 atoms per 10 million He-4 atoms. He-3 is relatively abundant on the surface of the moon; this is believed to be due to particles streaming onto the moon from the solar wind. He3 can also be made from decaying tritium. Helium-4 (He4): Common isotope of helium. Trace component of the atmosphere (about 1 part per million?); also found as a component of "natural gas" in gas wells. Lithium-6 (Li6): Less common isotope of lithium. 3 protons, 3 neutrons. There are 8 Li-6 atoms for every 100 Li-7 atoms. Widely distributed in minerals and seawater. Very active chemically. Lithium-7 (Li7): Common isotope of lithium. 3 protons, 4 neutrons. See above info on abundance. Boron (B): Common form is B-11 (80%). B-10 20%. 5 protons, 6 neutrons. Also abundant on earth. Note: Separating isotopes of light elements by mass is not particularly difficult. *** G. Why is the deuterium-tritium (D-T) reaction the easiest? Basically speaking, the extra neutrons on the D and T nuclei make them "larger" and less tightly bound, and the result is that the cross-section for the D-T reaction is the largest. Also, because they are only singly-charged hydrogen isotopes, the electrical repulsion between them is relatively small. So it is relatively easy to throw them at each other, and it is relatively easy to get them to collide and stick. Furthermore, the D-T reaction has a relatively high energy yield. However, the D-T reaction has the disadvantage that it releases an energetic neutron. Neutrons can be difficult to handle, because they will "stick" to other nuclei, causing them to (frequently) become radioactive, or causing new reactions. Neutron-management is therefore a big problem with the D-T fuel cycle. (While there is disagreement, most fusion scientists will take the neutron problem and the D-T fuel, because it is very difficult just to get D-T reactions to go.) Another difficulty with the D-T reaction is that the tritium is (weakly) radioactive, with a half-life of 12.3 years, so that tritium does not occur naturally. Getting the tritium for the D-T reaction is therefore another problem. Fortunately you can kill two birds with one stone, and solve both the neutron problem and the tritium-supply problem at the same time, by using the neutron generated in the D-T fusion in a reaction like n + Li6 -> He4 + T + 4.8 MeV. This absorbs the neutron, and generates another tritium, so that you can have basically a D-Li6 fuel cycle, with the T and n as intermediates. Fusing D and T, and then using the n to split the Li6, is easier than simply trying to fuse the D and the Li6, but releases the same amount of energy. And unlike tritium, there is a lot of lithium available, particularly dissolved in ocean water. Unfortunately you can't get every single neutron to stick to a lithium nucleus, because some neutrons stick to other things in your reactor. You can still generate as much T as you use, by using "neutron multipliers" such as Beryllium, or by getting reactions like n + Li7 -> He4 + T + n (which propagates the neutron) to occur. The neutrons that are lost are still a problem, because they can induce radioactivity in materials that absorb them. This topic is discussed more in Section 2. *** H. What is aneutronic fusion? Some researchers feel the advantages of neutron-free fusion reactions offset the added difficulties involved in getting these reactions to occur, and have coined the term "aneutronic fusion" to describe these reactions. The best simple answer I've seen so far is this one: (I've done some proofreading and modified the notation a bit.) [ Clarifying notes by rfheeter are enclosed in brackets like this.] >From: johncobb@emx.cc.utexas.edu (John W. Cobb) >Risto Kaivola <rkaivola@mits.mdata.fi> wrote: [[ Sorry I don't have the date or full reference for this anymore; this article appeared in sci.physics.fusion a few months ago.]] >>Basically, what is aneutronic fusion? The term aneutronic >>confuses me considerably. Could you give me an example of >>an aneutronic fusion reaction? How could energy be produced >>using such a reaction? Can there be a fusion reaction in which >>a neutron is never emitted? > >Examples: > >D + He3 --> He4 + p + 18.1MeV >(deuteron + helium-3 --> helium-4 + proton + energy) > >p + Li6 --> He4 + He3 + 4.0MeV >(proton + lithium-6 --> helium-4 + helium-3 + energy) > >D + Li6 --> 2 He4 + 22.4MeV >(deuteron + lithium-6 --> 2 helium-4's + energy) > >p + B11 --> 3 He4 + 8.7Mev >(proton + boron-11 --> 3 helium-4's + energy) > >All of these reactions produce no neutrons directly. [[ Hence "aneutronic." ]] >There are also other reactions that have multiple branches possible, >some of which do not produce neutrons and others that do >(e.g., D + D, p + Li7). > >The question is how do you get a "reactor" going and not get >any neutrons. There are 2 hurdles here. The first is getting the >fuel to smack together hard enough and often enough for fusion >to occur. >The easiest fusion reaction is D + T --> He4 + n (the D-T fuel >cycle). A magnetic reactor can initiate fusion in one of these >things at about a temperature of 10keV. [1 keV = 1000 eV = 11,000,000 (degrees) kelvin, more or less]. >The other reactions require much higher temperatures (for example >about 50KeV for the D+He3 reaction). This is a big factor of 5. >The second hurdle is neutron production via "trash" (secondary) >reactions. That is, the main reaction may be neutron-free, >but there will be pollution reactions that may emit neutrons. [ The products of the main reaction, e.g. He3, can be trapped in your reactor temporarily, and fuse with other ions in the system in messy ways. ] >Even if this is only a few percent, it can lead to big neutron >emission. For example, the D+He3 reaction will also have some D+D >reactions occuring. [ Because in your reactor you will have a lot of Ds and He3s, and the Ds will collide with each other as well as with the He3s. ] >At 50Kev temperatures, the reaction >cross-section for D+D reactions is about 1/2 of the D+He3 >cross-section, so there will be some generation of neutrons from >the 50% branch reaction of D + D-->He3 + n. >Also, the other 50% goes to T+p, The triton (T) will then undergo >a D-T reaction and release another neutron. [ Because the cross-section for D-T reactions is much higher.] >If the reactor is optmized (run in a He3 rich mode) the number >of neutrons can be minimized. The neutron power can be as low >as about 5% of the total. However, in a 1000 megawatt reactor, >5% is 50 MW of neutron power. That is [still] a lot of neutron >irradiation. This lower neutron level helps in designing >structural elements to withstand neutron bombardment, but it >still has radiation consequences. > >On the other hand, it is my understanding that the p-B11 reaction >is completely neutron free, but of course it is much harder >to light. *** I. What sort of fusion reactor is the sun? Fortunately for life on earth, the sun is an aneutronic fusion reactor, and we are not continually bombarded by fusion neutrons. Unfortunately, the aneutronic process which the sun uses is extremely slow and harder to do on earth than any of the reactions mentioned above. The sun long ago burned up the "easy" deuterium fuel, and is now mostly ordinary hydrogen. Now hydrogen has a mass of one (it's a single proton) and helium has a mass of four (two protons and two neutrons), so it's not hard to imagine sticking four hydrogens together to make a helium. There are two major problems here: the first is getting four hydrogens to collide simultaneously, and the second is converting two of the four protons into neutrons. The sun evades the first problem, and solves the second, by using a catalyzed cycle: rather than fuse 4 protons directly, it fuses a proton to an atom of carbon-12, creating nitrogen-13; the N-13 emits a neutrino and a positron (an antielectron, that is an electon with positive instead of negative charge) and becomes carbon-13. (Effectively, the Carbon-12 converted the proton to a neutron + positron + neutrino, kept the neutron, and became C-13). The C-13 eventually fuses with another proton to become N-14. N-14 then fuses with a proton to become oxygen-15. Oxygen-15 decays to N-15 (emitting another positron), and N-15 plus another proton yields carbon-12 plus a helium-4 nucleus, (aka an alpha particle). Thus 4 protons are tacked one by one onto heavier elements, two of the protons are converted to neutrons, and the result is production of helium and two positrons. (The positrons will undergo matter-antimatter annihilation with two electrons, and the result of the whole process is formation of a helium, two neutrinos, and a bunch of gamma rays. The gamma rays get absorbed in the solar interior and heat it up, and eventually the energy from all this fusion gets emitted as sunlight from the surface of the sun.) The whole process is known as the carbon cycle; it's catalyzed because you start with carbon and still have carbon at the end. The presence of the carbon merely makes it possible to convert protons to helium. The process is slow because it's difficult to fuse protons with carbon and nitrogen, and the positron-emitting nuclear decays are also slow processes, because they're moderated by the weak nuclear force. *** J. Why is it so hard to create controlled man-made fusion reactions? In order to get two nuclei to fuse, you basically have to get them to collide energetically. It turns out that colliding two beams of particles yields mostly scattering collisions, and few fusion reactions. Similarly, blasting a stationary target with a beam of energetic ions also yields too little fusion. The upshot is that one must find some way to confine hot, energetic particles so that they can collide many many times, and finally collide in just the right way, so that fusion occurs. The temperatures required are upwards of 100 million degrees (Kelvin - it would be about 200 million Fahrenheit!). At these temperatures, your fusion fuel will melt/evaporate any material wall. So the big difficulties in fusion are (a) getting the particles hot enough to fuse, and (b) confining them long enough so that they do fuse. *** K. What is plasma physics, and how is it related to fusion? Plasma physics is the area of physics which studies ionized gases and their properties. In most conventional types of fusion (muon-catalyzed fusion being the major exception), one must heat the fusion fuel to extremely high temperatures. At these temperatures, the fuel atoms collide so much and so hard that many electrons are knocked loose from their atoms. The result is a soup of ionized atoms and free electrons: a plasma. In order to achieve the conditions required for controlled fusion, an understanding of how plasmas behave (and particularly how to confine and heat them) is often essential. *** L. Just how hot and confined do these plasmas need to be? (Or, what conditions are needed for controlled fusion?) Basically, the hotter your plasma, the more fusion you will have, because the more ions will be flying around fast enough to stick together. (Although actually you can go *too* fast, and the atoms then start to whiz by too quickly, and don't stick together long enough to fuse properly. This limit is not usually achieved in practice.) The more dense your plasma is, the more ions there are in a small space, and the more collisions you are likely to have. Finally, the longer you can keep your plasma hot, the more likely it is that something will fuse, so duration is important too. More importantly, the slower your plasma loses energy, the more likely it is that it will be able to sustain its temperature from internal fusion reactions, and "ignite." The ratio of fusion energy production to plasma energy loss is what really counts here. Hotness is measured by temperature, and as explained above, the D-T fuel cycle (the easiest) requires temperatures of about 10 keV, or 100,000,000 degrees kelvin. Density is typically measured in particles-per-cubic centimeter or particles-per-cubic meter. The required density depends on the confinement duration. The Lawson product, defined as (density)*(confinement time) is a key measure of plasma confinement, and determines what combinations of density and energy confinement will give you fusion at a given temperature. It is important to note that what you must confine is the *energy* (thermal energy) stored in the plasma, and not necessarily the plasma particles. There's a lot of subtlety here; for instance, you want to confine your fuel ions as well as their energy, so that they stick around and fuse, but you *don't* want to confine the "ash" from the reactions, because the ash needs to get out of the reactor... But you'd like to get the *energy* out of the ash to keep your fuel hot so it will fuse better! (And it gets even more complicated than that!) Regardless, it's true that for a special value of the Lawson product, the fusion power produced in your plasma will just balance the energy losses as energy in the plasma becomes unconfined, and *ignition* occurs. That is, as long as the plasma fuel stays around, the plasma will keep itself hot enough to keep fusing. A simple analogy here is to an ordinary fire. The fire won't burn unless the fuel is hot enough, and it won't keep burning unless the heat released by burning the fuel is enough to keep the fuel hot enough. The flame continually loses heat, but usually this loss is slow enough that the fire sustains itself. You can accelerate the heat loss, however, by pouring water on the fire to cool it quickly; this puts the fire out. In fusion, the plasma continually loses heat, much as a fire gives off heat, and if the plasma loses heat faster than heat is produced by fusion, it won't stay hot enough to keep burning. In fusion reactors today, the plasmas aren't quite confined well enough to sustain burning on their own (ignition), so we get them to burn by pumping in energy to keep them hot. This is sort of like getting wet wood to burn with a blowtorch (this last analogy is usually credited to Harold Furth of PPPL). For the D-T fuel cycle, the Lawson ignition value for a temperature of about 200,000,000 Kelvin is roughly 5E20 seconds-particles/m^3. Current fusion reactors such as TFTR have achieved about 1/10th of this - but 20 years ago they had only achieved 1/100,000th of this! How can we improve the Lawson value of a plasma further, so we get even closer to fusion ignition? The trick is to keep the heat in the plasma for as long as possible. As an analogy to this problem, suppose we had a thermos of coffee which we want to keep hot. We can keep the thermos hotter longer by (a) using a better type of insulation, so that the heat flows out more slowly, or (b) using thicker insulation, so the heat has farther to go to escape, and therefore takes longer to get out. Going back to the fusion reactor, the insulation can be improved by studying plasmas and improving their insulating properties by reducing heat transport through them. And the other way to boost the Lawson value is simply to make larger plasmas, so the energy takes longer to flow out. Scientists believe it's technically feasible to build a power-producing fusion reactor with high Lawson value *Right Now*, but it would have to be large, so large in fact that it would cost too much to be able to make electricity economically. So we're studying plasmas and trying to figure out how to make them trap energy more efficiently. *** M. What are the basic approaches used to heat and confine the plasma? (Or, what is magnetic confinement? Inertial confinement?) There are three basic ways to confine a plasma. The first is the method the sun uses: gravity. If you have a big enough ball of plasma, it will stick together by gravity, and be self-confining. Unfortunately for fusion researchers, that doesn't work here on earth. The second method is that used in nuclear fusion bombs: you implode a small pellet of fusion fuel. If you do it quickly enough, and compress it hard enough, the temperature will go way up, and so will the density, and you can exceed the Lawson ignition value despite the fact that you are only confining your pellet for nanoseconds. Because the inertia of the imploding pellet keeps it momentarily confined, this method is known as inertial confinement. The third method uses the fact that charged particles placed in a magnetic field will gyrate in circles. If you can arrange the magnetic field carefully, the particles will be trapped by it. If you can trap them well enough, the plasma energy will be confined. Then you can heat the plasma, and achieve fusion with more modest particle densities. This method is known as magnetic confinement. Initial heating is achieved by a combination of microwaves, energetic/accelerated particle beams, and resistive heating from currents driven through the plasma. (Once the Lawson ignition value is achieved, the plasma becomes more-or-less self-heating.) In magnetic confinement, the plasma density is typically about 1E20 particles per cubic meter, and with a temperature of about 1E8 kelvin, we see that ignition could be achieved with a confinement time of about 4 seconds. (All these numbers in reality vary by factors of 2 or 3 from the rough values I've given.) Currently, magnetic-confinement reactors are about a factor of ten short of the ignition value. (TFTR has an energy confinement time of 0.25 seconds during its best shots.) More information on these different approaches is given in the sections that follow.