In this space, visitors are invited to post any comments, questions, or skeptical observations about Philo T. Farnsworth's contributions to the field of Nuclear Fusion research.Subject: Nuke Fuels

Date: Jan 18, 04:29 am

Poster: Scott StephensOn Jan 18, 04:29 am, Scott Stephens wrote:

The following is from a web site on advanced fusion fuels, which address I forgot, sorry...

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Alternate Fusion Fuels

Last revised: July, 1995. Status: not to bad.

This is not an exhaustive discussion and tends to elaborate on some technical points. A better introduction to alternate fuels is in the hot

fusion FAQ. Thanks to Robert Nachtrieb, Philip Benjamin Snyder, and John W. Cobb for thoughtful comments and useful numbers. I can't offer

much in the way of references, but you might want to look at

Vol. A271 of Nucl.Instr.Meth.Phys.Research (1988--special issue)

Bosch&Hale,Nucl.Fusion 32(1992)611)

Glasstone and Lovberg, Controlled Thermonuclear Reactions (Van Nostrand, New York, 1960, Chap. 2

McGowan, et al., Nucl. Data Tables A6, 353 (1969), and A8, 199 (1970)

Miley, Towner and Ivich, Fusion Cross Section and Reactivities, Rept. COO-2218-17 (Univ. Illinois, Urbana, 1974)

Cox, Larry T., _Thermonuclear Reaction Bibliography with Cross Section Data for Four Advanced Reactions._, AF-TR-90-053, Edwards

Air Force Base: Phillips Laboratory Technical Services Office, 1991

If you want to use fusion as a source of energy, the reaction chosen 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.

This makes the list of candidates pretty short. The most interesting reactions are the following.

(1) D +T -> He4 (3.5 MeV) + n (14.1 MeV)

(2) D +D -> T (1.01 MeV) + p ( 3.02 MeV) (50%)

(3) -> He3 (0.82 MeV) + n ( 2.45 MeV) (50%)

(4) D +He3 -> He4 (3.6 MeV) + p (14.7 MeV)

(5) T +T -> He4 + 2 n + 11.3 MeV

(6) He3+He3 -> He4 + 2 p

(7) He3+T -> He4 + p + n + 12.1 MeV (51%)

(8) -> He4 (4.8 MeV) + D ( 9.5 MeV) (43%)

(9) -> He4 (0.5 MeV) + n ( 1.9 MeV)

+ p (11.9 MeV) (6%)

(10) D +Li6 -> 2 He4 + 22.4 MeV

(11) p +Li6 -> He4 (1.7 MeV) + He3 ( 2.3 MeV)

(12) He3+Li6 -> 2 He4 + p + 16.9 MeV

(13) p +B11 -> 3 He4 + 8.7 MeV

The cross section, sigma, generally rises dramatically with temperature up to a maximum. A fusion reactor (not only a tokamak) will be limited in

the pressure attainable. The fusion power density, which can be considered a rough surrogate for the cost of power, is proportional to/T^2, where < > denotes an average over a Maxwellian distribution. For any given fuel, you will usually want to operate at the

temperature where this function is maximum.

As a side note, one way to try to get around the limitations I will discuss below is to consider a non-Maxwellian plasma. The migma is one

example of a concept which claims to work better with fuels other than D-T because the plasma is not Maxwellian. Another elegant idea is to

combine the reactions (11) and (12). The He3 from reaction (11) can react with Li6 in reaction (12) before completely thermalizing. This produces

an energetic proton which in turn undergoes reaction (11) before thermalizing. The last I heard is that this won't really work, but it is a good

example of a case where the assumption of a Maxwellian plasma in not appropriate.

It is also important to remember 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-B11, 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, e.g.,

p + B11 -> C12 + gamma

p + B11 -> n +C11

He4 + B11 -> n + N14

He4 + B11 -> p + C14

He4 + B11 -> T + C12

B11 + B11 -> junk

as well as reactions with a possible B10 impurity fraction. Furthermore, some of the energetic product alphas will impinge on the wall and knock

out neutrons.

Back to/T^2, the maximum values for some of these reactions is given here, with T in keV and in (m^3/sec/keV^2).

fuel T(keV)/T^2

---- ------ -------------

DT 13.6 1.24e-24

DD 15 1.28e-26

DHe3 58 2.24e-26

pLi6 66 1.46e-27

pB11 123 3.01e-27

Li6 is primarily interesting for the potential chain reaction mentioned above. Since the calculations below assume a Maxwellian distribution, and

the/T^2 for p-Li6 is somewhat worse than for p-B11, we will only consider the latter from here on out. That leaves us with four

reactions with varying degrees of neutron problems and varying performance. Let us use the these cross sections to calculate two figures of

merit for a reaction, the minimum n*T*tau which allow ignition and the fusion power density.

Consider any fusion reaction (except D-D!) between a hydrogen species (subscript 1) and another species with atomic number Z (subscript Z),

where Z may equal to 1. The electron density is given by quasineutrality:

n_e = n_1 + Z*n_Z

Assuming all species have the same temperature T, the total pressure is

p = (n_e+n_1+n_Z)*T

Therefore, for a given pressure the maximum value of n_1*n_Z is

(n_1*n_Z)_max = (1/16)*(p/T)^2*(2/(Z+1))

Thus there is a "penalty" of (2/(Z+1)) for non-hydrogenic fuels, which I included in my table. The D-D reaction(s) must be calculated

differently. First, there is only one species of ion, so

n_e = n_D

p = (n_e+n_D)*T

The reaction rate is n_D^2/2, not n_D^2 .

(n_D^2/2)_max = (1/16)*(p/T)^2*2

Thus D-D gets a "bonus" of 2 because each ion can react with every other ion, not just half of the others. The D-D reaction has of course two

pathways which are equally probable, so we have to take the average of the two. It is not clear what should be assumed about the product T

and He3. T burns so well in a deuterium plasma that you probably can't get it out even if you want to. The D-He3 reaction is optimized at a

much higher temperature, so the burnup at the optimum D-D temperature may be low, so let's assume the T but not the He3 gets burned up and

adds its energy to the net reaction. Thus I count the DD fusion energy as

E_fusion = (4.03+17.6+3.27)/2 = 12.5 MeV

and the energy in charged particles as

E_charged = (4.03+3.5+0.82)/2 = 4.2 MeV

With these assumptions and the/T^2 maximum values, we find a relative reactivity of

fuel penalty/bonus/T^2 ratio

---- ------------- -------- -----

DT 1 1.24e-24 1

DDn+DDp 2 1.28e-26 48

DHe3 2/3 2.24e-26 83

pB11 1/3 3.01e-27 1240

For the relative Lawson values, we need to weight with the energy of the charged particles:

fuel E_charged Lawson/Lawson_DT

---- --------- ----------------

DT 3.5 1

DDn+DDp 4.2 30

DHe3 18.3 16

pB11 8.7 500

For the relative power densities, we need to weight with the total fusion energy:

fuel E_fusion P_DT/P

---- --------- ------

DT 17.6 1

DDn+DDp 12.5 68

DHe3 18.3 80

pB11 8.7 2500

The comparison between DD and DHe3 is complicated by several factors. One is the assumption of zero product He3 burnup for the DD

reaction as mentioned above. Another is that a DHe3 plasma will produce a lot of DD side reactions. Fuel availability is an issue for DHe3 but

not for DD. Since the results are within a figure of two of each other and one or two orders of magnitude worse than DT, we don't need to look

at them any closer for present purposes.

The dependence of confinement time on ion mass might improve the pB11 numbers by a factor of two.

The Lawson ratios tell you how much better tau_E has to be to get a given reaction going. Using the empirically determined scaling of tau_E in

tokamaks, it can be shown, if magnetic field and safety factor are held constant, that the volume scales very nearly linearly with the inverse

of/T^3. (See ``Reactor size scaling'') If the ordering of /T^3 is similar to that of /T^2, and the cost of a device

scales somewhat more slowly than the plasma volume, say with the volume to the 2/3 power, then the relative cost for a machine which burns

an alternate fuel is the Lawson ratio to the 2/3 power. For example, if a DT breakeven experiment costs 10 billion dollars, then a breakeven DHe3

experiment might cost 60 billion dollars. At the same time, the fusion output of the DHe3 device will be about five times lower than the other (16

times the volume but 81 times worse reactivity).

This argument is admittedly very primitive, but it shows the magnitude of the disadvantage that has to be overcome by the advantages of a fuel

which produces fewer neutrons. I have not looked at them myself, but apparently the ARIES III and Apollo reactor studies conclude the the

cost of electricity from a DHe3 reactor would be comparable to or lower than that from a DT reactor. They gain some because the blanket does

not have to breed tritium and does not have to absorb as large a fraction of the fusion energy, so it can be smaller. But it still has to be thick

enough to absorb 14 MeV neutrons. They gain further by assuming a higher magnetic field and a higher plasma current. But it is not fair to

compare a low field DT machine with a high field DHe3 machine. It is true that tritium containment and handling systems are no longer needed,

but I don't believe they drive the cost of a power plant. If the plasma energy can be coupled out directly, e.g. through synchrotron radiation,

then DHe3 will have a higher conversion efficiency and less down time to change out neutron damaged parts. This advantage may be more

decisive at high fields and high betas. It cannot be ruled out that a DHe3 reactor will ultimately be more economical than a DT reactor, but it will

require many developments, and even then it may be a toss-up.

Other points of comparison are safety and conversion efficiency. D-He3 does indeed have weighty safety advantages over D-T. (Of course, if it

isn't able to ignite at all it is even safer.) D-He3 has the potential for a more efficient fusion to electricity conversion. Whether this potential can

be utilized and whether the increased efficiency will pay for the increased capital cost are speculative. We should also keep in mind that T and

He3 do not exist on earth in significant quantities; T must be breed from Li and He3 must be accumulated from the decay of T or mined in space.

Everyone is entitled to an opinion on the ultimately optimum fuel cycle, but only future generations will really have the knowledge base to make

a choice.

Fuels other than DT and DD must also run at significantly higher temperatures. This increases the cyclotron radiation losses dramatically and

possibly fatally. For this reason, some arguments for alternate fuels invoke high beta machines, usually other than tokamaks. It is sometimes

also argued that a particular concept, which is believed to have inherently better confinement than the tokamak, cannot be used with DT

because the neutron flux at the first wall would be too large. Because of its special configuration, The FRC may also be able to use the 14 MeV

protons as an intrinsic current drive. A portion of the protons moving in one direction may be trapped in the FRC while those moving

oppositely are not. Thus selective trapping can lead to self-generated currents from the fusion products. Berk, Momota, and Tajima estimated

that about 25% of the total current drive can be obtained this way. The same effect does occur with the 4 MeV alphas in a DT tokamak.

However, the alpha energy is lower, its mass larger, and the number of rho_i in the device is larger so the trapping area in phase space is smaller

and the generated current is smaller. Even though there is serious discussion of whether the advantages of D-He3 might make up for the one to

two orders of magnitude disadvantages in the long run for a test reactor that is certainly not the case.

Appendix:

Robert Nachtrieb contributed coefficients to curve fits of the reaction rate parameter, sigma-v, for DT, DDn, DDp, DHe3, pLi6, and pB11. The fit

is actually to the logarithm (base 10) of the sigma v curves. The coefficients come from: Cox, Larry T., _Thermonuclear Reaction Bibliography

with Cross Section Data for Four Advanced Reactions._, AF-TR-90-053, Edwards Air Force Base: Phillips Laboratory Technical Services Office,

1991

input: t (in keV)

output: sigma-v (in cm^3/s)

range of validity: 1 < t (keV) < 1000

x=alog10(t)

y= a0 + a1*x + a2*x^2 + a3*x^3 + a4*x^4

sv=10^y (in cm^3/s)

DT

a0=-20.15761

a1=6.318869

a2=-2.421732

a3=.2708006

a4=0.0

DDn

a0=-22.08780

a1=5.701349

a2=-2.082958

a3=0.2981119

a4=0.0

DDp

a0=-21.97105

a1=5.643589

a2=-2.404634

a3=0.5716992

a4=-0.5730514e-1

DHe3

a0=-25.67344

a1=10.10471

a2=-3.310354

a3=0.3535589

a4=0.0

pLi6

a0=-28.12494

a1=11.87649

a2=-4.265839

a3=0.5712202

a4=0.0

pB11

a0=-33.51636

a1=17.40498

a2=-5.833501

a3=0.6879438

a4=0.0

awc

- Re: Nuke Fuels -
Richard HullJan 19, 09:25 am