How Do You Shield a Compact Fusion Reactor?

[.Scott]

After Lockheed Martins announcement that they were developing a fusion reactor that might fit on the back of a pickup truck, I got into a conversation with someone about how one would shield such a device.

So I found some tables that described the cross section of protons at different energy levels through different materials. But turning that into a statement such as "you'll loose about half the protons for every 2 inches of Boron-6" involves formulas and conversions that are not presented clearly in any of the sources I ran into.

I'm sure its not as simple as "half the protons every whatever distance", but there should be some way of getting some rough estimate expressed in those terms.

 

[e.bar.goum]

Well, when I want to find the energy loss of nuclei in matter (this happens a lot in my research), I fire up my copy of SRIM/TRIM and do a simulation (don't let the terrible website fool you, it's rather a nice code). You can use the Range Tables feature to find the energy loss in whatever stopping power units you like, for reasonably complex materials and compounds.

Beyond just MeV/(mg/cm^2), where geometry is important, like in a fusion reactor, I'd use Geant4. But I don't know what professional reactor physicists use, but Geant4 is used for an awful lot of things.

Alternatively, if you want a decent estimate without simulations, you can use the Bethe-Bloch equation,

$\frac{dE}{dx} = (\frac{e^2}{4\pi\epsilon_0})^2 \frac{4 \pi z^2 N_0 Z \rho}{mc^2 \beta^2 A}(\ln(2mc^2\beta^2/I) - ln(1-\beta^2) - \beta^2)$

Where $v = \beta c$ is the velocity of the particle, $ze$ the electric charge, $Z$, $A$, $\rho$ the atomic number, weight and density of the stopping material, $N_0$ is Avogadro's number, and m is the electron mass. $I$ the mean excitation energy of the atomic electrons, usually an empirical parameter with a value in eV on the order of $10Z$, in air, $I=86 eV$. You don't do well at low energies though.

You may then calculate the range by integrating this over the energies of the particle.

Krane Introductory Nuclear Physics ch. 7 has a good section on this.

 

[.Scott]

So my neutron (940MeV) kicks off at 14.1MeV or less. So my beta is about 0.12.
ze is zero.
For graphite, Z=6, A=12, p=2.15g/cc.
Avogadro's number is 6.02 x 10^23
m is 9.1x10^-31
I is what?!? 60?

Wait! There's an epsilon 0 in the first denominator! Is that some constant?

 

[.Scott]

Found it
e0= 8.85x10^-12 Furlongs per fort night ... I mean Farads per meter

Okay ... let me work on this.

And I also looked at that SRIM link - which looks a whole lot easier.
Thanks.

 

[e.bar.goum]

Ah, two things: Bethe-Bloch is for charged particles (you were talking about protons in your OP), so you can't use it for neutrons.

$\epsilon_0$ is indeed a constant - it's the vacuum permittivity.

$\frac{e^2}{4 \pi \epsilon_0}$ = 1.44 MeV fm.

 

[Astronuc]

After Lockheed Martins announcement that they were developing a fusion reactor that might fit on the back of a pickup truck, I got into a conversation with someone about how one would shield such a device.

So I found some tables that described the cross section of protons at different energy levels through different materials. But turning that into a statement such as "you'll loose about half the protons for every 2 inches of Boron-6" involves formulas and conversions that are not presented clearly in any of the sources I ran into.

I'm sure its not as simple as "half the protons every whatever distance", but there should be some way of getting some rough estimate expressed in those terms.

Protons are deflected in magnetic fields. It is the fast neutrons that are of most concern, and for that densely hydrogenous material is the best shielding. Then there is the X-ray and gamma shielding for which high-Z metals are best.

Neutrons are not a problem if one uses an aneutronic reaction, however, one pretty has a choice of dd or dt fusion, and dd produces neutrons in 50% or reactions, while dt produces 14.1 MeV neutrons with each reaction.

I believe Lockheed-Martin refers to size, not application. A compact fusion reactor still needs shielding, so a mobile system might be applied in ships.

 

[e.bar.goum]

For neutrons of fixed energy, the probability of interaction per unit path length is constant for a given interaction. You need to take the neutron interaction cross section $\sigma$ (look it up at NNDC), then multiply by the number of nuclei N per unit volume, in your stopping material

$\Sigma = N \sigma$

Then, sum all the types of interaction

$\Sigma_t = \Sigma _{scatter} + \Sigma_{rad. capture} ...$

And then the intensity drops off in the same way as for gammas

$\frac{I}{I_0} = e^{-\Sigma_{t} t}$

For more information, see Knoll, Radiation Detection and Measurement ch. 2 D. (pg 55 in the fourth ed.)

 

[.Scott]

Yes, I missed-typed. Should have said neutrons.

So I should try the H in water and work out the thickness required from that. Funny - cause my friend suggested using engineers for the shielding.

 

[e.bar.goum]

Ah, I see. Well, at least you know how to work it out for charged particles now as well?

Engineers work pretty well, being basically bags of water. ;)

For the record, while water is a pretty good option, a more convenient one is plastic - (CH₂)ⁿ has basically the same number of hydrogens, and you don't have to worry about making things water tight. It is also very common to add some boron (such as in the form of borax) to the mix, as boron has a huge neutron capture cross section.

 

[Astronuc]

A metal hydride might be effective or alternate layers of metal hydride and high-Z metal like W or depleted U. W is quite expensive though, and U is subject to fast fissions.

 

[PAllen]

People who don't believe fusion can ever work could be hired at low cost for shielding.

 

[mfb]

For neutrons of fixed energy, the probability of interaction per unit path length is constant for a given interaction. You need to take the neutron interaction cross section $\sigma$ (look it up at NNDC), then multiply by the number of nuclei N per unit volume, in your stopping material

That works for the first interaction, but most collisions are elastic, so the neutron has a different direction and lower energy afterwards. The actual range is larger than the range to the first interaction.