# How to calculate the bremsstrahlung limit in the fusion triple product diagram?

The fusion triple product is a commonly used figure of merit to quantify the progress in fusion research, it is the product of plasma density $n$, temperature $T$ and energy confinement time $\tau_E$.

It can be used in combination with a power balance to formulate a criterion for ignition, which is defined as the point when the energy released by the fusion reactions is large enough to sustain them (by proving enough heat to keep the fusion processes going). The criterion reads $$nT\tau_e > \frac{12\,T^2}{\left<\sigma_{fusion}u\right>\epsilon_\alpha - 4c_{br}Z_{eff}\sqrt{T}},$$ with $\left<\sigma_{fusion}u\right>$ the fusion reactivity (which is also a function of $T$, approximately of $T^2$ for the temperatures of interest here), $\epsilon_\alpha=3.52\,\mathrm{MeV}$ the energy released by the $\alpha$-particles (assuming here that we have a DT-fusion reaction), $c_{br}\approx1.04\cdot 10^{-19}\,\mathrm{m}^3\mathrm{s}^{-1}\sqrt{\mathrm{eV}}$, $Z_{eff}$ the effective charge number which we assume to be 1 here. The left term in the denominator describes the heating by the $\alpha$-particles produced in the DT fusion reaction and the right term the losses due to bremsstrahlung.

The fusion triple product is often plotted as a function of the temperature, as shown in the following plot.

Some more interesting plots are shown for example here or here or here. I have chosen those particular examples because they include a limit for bremsstrahlung losses (the linear function on the top left).

My question is how to calculate this limit such that it can be included in those plots?

(I have the feeling that the answer is quite obvious but I can't get it at the moment...)

• You mean how to graph the bremsstrahlung contribution from the denominator? Besides from substituting the values for the experimental conditions, what should be the problem? – Germán May 7 '17 at 7:29
• @Germán no, I was referring to the plots from the links, there they have a bremsstrahlung limit included (above which the plasma collapses due to too large bremsstrahlung losses). I just can't get my head around how to include that limit in the $nT\tau_E$ diagram... – Alf May 7 '17 at 9:52
• I got that, what I mean is, wouldn't you take the 4c_{br}Z_{eff}\sqrt{T}} factor of the denominator, replacing the values and coefficients accordingly, and you end up with a sqrt{T} relation (times a coefficient) to plot in the diagram? – Germán May 8 '17 at 10:24

The above definition of the Lawson criterion derives from the following inequality $$$$P_f \geq \frac{W}{\tau'_E} + P_{rad}\,,$$$$ where $$P_f= \frac{1}{4} n^2 \langle \sigma v\rangle \epsilon_{\alpha}$$ is the fusion power density, $$W=3 n T$$ is the total energy density of the D-T plasma, $$\tau'_E$$ is an energy confinement time and $$P_{rad}=c_{br} n^2 \sqrt{T}$$ is the power density loss due to Bremsstrahlung. Solving for $$n \tau'_E$$ and multiplying by $$T$$ gives the following lower limit on the triple product $$$$n T \tau'_E\geq \frac{12 T^2}{\langle \sigma v\rangle \epsilon_{\alpha}-4 c_{br}\sqrt{T}}\,.$$$$
I suspect that the plots in the links above use a different definition of the Lawson criterion, $$$$P_f \geq \frac{W}{\tau_E}\,,$$$$ where $$\tau_E$$ is the total energy confinement time (including radiative losses). Re-arranging again gives a different lower limit on the triple product, $$$$n T \tau_E\geq \frac{12 T^2}{\langle \sigma v\rangle \epsilon_{\alpha}}\,.$$$$ Now, by our definition we also require that the power density loss due to Bremsstrahlung is less than the total energy loss rate per unit volume, $$$$P_{rad} \leq \frac{W}{\tau_E}\,,$$$$ Re-arranging gives the following upper limit on the triple product $$$$n T \tau_E \leq \frac{3 T^{3/2}}{c_{br}}\,.$$$$ Using reactivities $$\langle \sigma v\rangle$$ from Table VII in a paper by Bosch and Hale (1992), I have plotted the lower and upper limits on the triple product (see below).
• Thanks a lot for your detailed answer! So you are saying $\tau_E^`$ is the confinement time taking into account only transport losses and $\tau_E$ is the total confinement time. That makes sense, I am just a bit puzzled why you plot them on the same graph here, as the y-axis should, strictly speaking, only be true for one of them...? – Alf Oct 16 '18 at 18:18