neutrons - How large should an optically large fusion reactor be?

This is just theoretical consideration. I expect that size of the reactor would be impractical to build on Earth, but I'm interested how much. EDIT: perhaps it could be good for some huge spacecraft in far future, this thought was my original motivation

Consider that you do not want neutrons to take away energy from your nuclear reactor because you want to use some form of direct energy conversion with high power density and Carnot efficiency. There are in principle 2 sulutions:

1. Use aneutronic nuclear reaction ( like $^1$H+$^{11}$B or $^3$He+$^3$He ). However, there the ratio between Bremsstrahlung power loss $P_{Xray}$ and fusion reaction power $P_{Fusion}$ is just $P_{Fusion}/P_{Xray} \approx 0.5 ... 0.8 < 1 $. In such case you would like to confine bremsstrahlung inside your plasma. So the plasma has to be optically thick for X-ray of energy $0.1-10 MeV$. ( let aside non-equlibrium concepts like Inertial electrostatic confinement which are controversial )


2. Use $^3$H+$^2$H reaction which has the ratio $P_{Fusion}/P_{Xray} \approx 140$. However it produces 80% of energy in form of fast $ 14 MeV $ neutrons. So you can ask how big should be the reactor to be optically thick for neutrons of this energy.

Answer

I think that this is an interesting question and I will try to give a very rough answer (warning: it might include some approximations).

Instead of calculating the optical thickness of the D-T plasma, we can try to simply estimate the mean free path $\lambda$ of a neutron in the plasma. It can be estimated as (see for example here for a nice explanation) $$\lambda = (n \sigma)^{-1},$$ where $n$ is the number density of the target nuclei and $\sigma$ the neutron cross section. The plasma density in ITER will be approximately $n=10^{20}\,\mathrm{m}^{-3}$, so we have that value. What about the cross section?

We can look at some old papers [1-3], or use the Ramsauer model to estimate it. Whatever we decide to do, an important thing to note is that the cross section decreases with increasing energy (most pronounced for low energies, for high energies, it can be approximated as constant). As a rough estimation of any of the above mentioned methods, we get something like $\sigma \approx 3\,\mathrm{barn}$ (maybe 1, maybe 5, so let's say 1). $1\,\mathrm{barn}$ corresponds to $1\cdot10^{-28}\,\mathrm{m}^2$.

Using those values, we can approximate the mean free path to $$\lambda \approx (10^{20}\mathrm{m}^{-3} \cdot 10^{-28}\,\mathrm{m}^2)^{-1}=10^8\,\mathrm{m}.$$

Which is indeed rather large. Similar to the sun, as mentioned in the comments by @CuriousOne (note, however, that the number density in the sun is much larger).

Keep in mind that those are rough estimations, but you get the idea.

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1. [1] doi:10.1103/PhysRev.98.666


2. [2] doi:10.1016/0375-9474(72)90930-X


3. [3] doi:10.1103/PhysRevC.22.384