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diff --git a/source/know/concept/lawson-criterion/index.md b/source/know/concept/lawson-criterion/index.md new file mode 100644 index 0000000..c1848dc --- /dev/null +++ b/source/know/concept/lawson-criterion/index.md @@ -0,0 +1,127 @@ +--- +title: "Lawson criterion" +date: 2021-10-06 +categories: +- Physics +- Plasma physics +layout: "concept" +--- + +For sustained nuclear fusion to be possible, +the **Lawson criterion** must be met, +from which some required properties +of the plasma and the reactor chamber can be deduced. + +Suppose that a reactor generates a given power $P_\mathrm{fus}$ by nuclear fusion, +but that it leaks energy at a rate $P_\mathrm{loss}$ in an unusable way. +If an auxiliary input power $P_\mathrm{aux}$ sustains the fusion reaction, +then the following inequality must be satisfied +in order to have harvestable energy: + +$$\begin{aligned} + P_\mathrm{loss} + \le P_\mathrm{fus} + P_\mathrm{aux} +\end{aligned}$$ + +We can rewrite $P_\mathrm{aux}$ using the definition +of the **energy gain factor** $Q$, +which is the ratio of the output and input powers of the fusion reaction: + +$$\begin{aligned} + Q + \equiv \frac{P_\mathrm{fus}}{P_\mathrm{aux}} + \quad \implies \quad + P_\mathrm{aux} + = \frac{P_\mathrm{fus}}{Q} +\end{aligned}$$ + +Returning to the inequality, we can thus rearrange its right-hand side as follows: + +$$\begin{aligned} + P_\mathrm{loss} + \le P_\mathrm{fus} + \frac{P_\mathrm{fus}}{Q} + = P_\mathrm{fus} \Big( 1 + \frac{1}{Q} \Big) + = P_\mathrm{fus} \Big( \frac{Q + 1}{Q} \Big) +\end{aligned}$$ + +We assume that the plasma has equal species densities $n_i = n_e$, +so its total density $n = 2 n_i$. +Then $P_\mathrm{fus}$ is as follows, +where $f_{ii}$ is the frequency +with which a given ion collides with other ions, +and $E_\mathrm{fus}$ is the energy released by a single fusion reaction: + +$$\begin{aligned} + P_\mathrm{fus} + = f_{ii} n_i E_\mathrm{fus} + = \big( n_i \Expval{\sigma v} \big) n_i E_\mathrm{fus} + = \frac{n^2}{4} \Expval{\sigma v} E_\mathrm{fus} +\end{aligned}$$ + +Where $\Expval{\sigma v}$ is the mean product +of the velocity $v$ and the collision cross-section $\sigma$. + +Furthermore, assuming that both species have the same temperature $T_i = T_e = T$, +the total energy density $W$ of the plasma is given by: + +$$\begin{aligned} + W + = \frac{3}{2} k_B T_i n_i + \frac{3}{2} k_B T_e n_e + = 3 k_B T n +\end{aligned}$$ + +Where $k_B$ is Boltzmann's constant. +From this, we can define the **confinement time** $\tau_E$ +as the characteristic lifetime of energy in the reactor, before leakage. +Therefore: + +$$\begin{aligned} + \tau_E + \equiv \frac{W}{P_\mathrm{loss}} + \quad \implies \quad + P_\mathrm{loss} + = \frac{3 n k_B T}{\tau_E} +\end{aligned}$$ + +Inserting these new expressions for $P_\mathrm{fus}$ and $P_\mathrm{loss}$ +into the inequality, we arrive at: + +$$\begin{aligned} + \frac{3 n k_B T}{\tau_E} + \le \frac{n^2}{4} \Expval{\sigma v} E_\mathrm{fus} \Big( \frac{Q + 1}{Q} \Big) +\end{aligned}$$ + +This can be rearranged to the form below, +which is the original Lawson criterion: + +$$\begin{aligned} + n \tau_E + \ge \frac{Q}{Q + 1} \frac{12 k_B T}{\Expval{\sigma v} E_\mathrm{fus}} +\end{aligned}$$ + +However, it turns out that the highest fusion power density +is reached when $T$ is at the minimum of $T^2 / \Expval{\sigma v}$. +Therefore, we multiply by $T$ to get the Lawson triple product: + +$$\begin{aligned} + \boxed{ + n T \tau_E + \ge \frac{Q}{Q + 1} \frac{12 k_B T^2}{\Expval{\sigma v} E_\mathrm{fus}} + } +\end{aligned}$$ + +For some reason, +it is often assumed that the fusion is infinitely profitable $Q \to \infty$, +in which case the criterion reduces to: + +$$\begin{aligned} + n T \tau_E + \ge \frac{12 k_B T^2}{\Expval{\sigma v} E_\mathrm{fus}} +\end{aligned}$$ + + + +## References +1. M. Salewski, A.H. Nielsen, + *Plasma physics: lecture notes*, + 2021, unpublished. |