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- Path: senator-bedfellow.mit.edu!faqserv
- From: rfheeter@pppl.gov
- Newsgroups: sci.physics.fusion,sci.answers,news.answers
- Subject: Conventional Fusion FAQ Section 1/11 (Fusion Physics)
- Supersedes: <fusion-faq/section1-physics_934723436@rtfm.mit.edu>
- Followup-To: sci.physics.fusion
- Date: 14 Nov 1999 10:27:51 GMT
- Organization: none
- Lines: 573
- Approved: news-answers-request@MIT.EDU
- Expires: 26 Feb 2000 10:24:18 GMT
- Message-ID: <fusion-faq/section1-physics_942575058@rtfm.mit.edu>
- References: <fusion-faq/section0-intro/part2-outline_942575058@rtfm.mit.edu>
- Reply-To: rfheeter@pppl.gov
- NNTP-Posting-Host: penguin-lust.mit.edu
- Summary: Fusion energy represents a promising alternative to
- fossil fuels and nuclear fission for world energy
- production. This FAQ answers Frequently Asked Questions
- (from the sci.physics.fusion newsgroup) about conventional
- areas of fusion energy research. It also provides other
- useful information about the subject. This FAQ does NOT
- discuss unconventional forms of fusion (like Cold Fusion).
- X-Last-Updated: 1995/02/26
- Originator: faqserv@penguin-lust.MIT.EDU
- Xref: senator-bedfellow.mit.edu sci.physics.fusion:44325 sci.answers:10876 news.answers:170948
-
- 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.
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