starship-design: RE: Starship Design

["L. Parker" <lparker@cacaphony.net>]

> -----Original Message-----
> From: larryf@dcsmail.com [mailto:larryf@dcsmail.com]
> Sent: Tuesday, February 08, 2000 2:09 PM
> To: lparker@cacaphony.net
> Subject: RE: Starship Design
>
>
>   Lee,
>
>   No luck there.  Maybe I just didn't know where to look.
>
>   The most complete list of reaction rate data I have found
> was the Oak
> Ridge National Laboratory/ Astrophysics page.  However, the D+Li6
> reaction was ommitted.  I'm beginning to think that it might be
> classified.
>
>   Let me know I you come across it again.
>

It would seem that Robert Heeter is in the process of rewriting the FAQ...

Is this the information you are looking for?:

"*** 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."

More interesting is the following section:

"*** 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."

My original post delt more with aneutronic fusion, which is described in
more detail in:

"*** 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."


More can be found at:

http://www.faqs.org/faqs/fusion-faq/section1-physics/

which is the older version of the FAQ. I hope this helps, and it is now
forever enshrined in our archive of the list <G>.

Lee Parker