Fusion Plasma Lawson Triple Product Calculator
Thermonuclear Fusion Power: Compute the Lawson triple product ($n_e T_i \tau_E$), empirical confinement scaling (IPB98(y,2)), fusion gain ($Q = P_{fus}/P_{aux}$), and ignition threshold.
Plasma Confinement & Heating Parameters
Fusion Power, Triple Product & Q Factor
The Lawson Criterion & Thermonuclear Confinement Scaling
Achieving net power from magnetic confinement fusion requires simultaneously mastering three independent parameters: plasma density ($n_e$), core ion temperature ($T_i$), and thermal energy confinement time ($\tau_E$).
1. Core Equations
Triple Product = n_e · T_i · τ_E [10²⁰ keV·s/m³] Q = P_fus / P_aux P_α = 0.20 · P_fus [Self-heating by trapped 3.5 MeV alpha particles] P_n = 0.80 · P_fus [14.1 MeV neutrons captured in blanket] Ignition Criterion: P_α ≥ P_loss (Q → ∞)
2. Confinement Milestones
- Q = 0.67: Record achieved on the JET tokamak (UK) in 1997 during D-T campaign (16.1 MW fusion power from 24 MW heating).
- Q = 10: Primary operational baseline for ITER (500 MW fusion power from 50 MW auxiliary input).
- Ignition (Q = ∞): Plasma self-sustained solely by its own internal alpha particle heating, requiring zero auxiliary sustainment power.
Frequently Asked Questions
What is the Lawson criterion and the fusion triple product?
Formulated by British engineer J.D. Lawson in 1955 and generalized in modern plasma physics, the Lawson criterion establishes the conditions required for a thermonuclear plasma to release more fusion energy than the thermal energy lost to conduction, convection, and radiation. It is quantified by the fusion triple product: $n_e \cdot T_i \cdot \tau_E \ge 3\times 10^{21}\,\text{keV}\cdot\text{s/m}^3 = 30\times 10^{20}\,\text{keV}\cdot\text{s/m}^3$ for Deuterium-Tritium ignition.
What is the difference between scientific breakeven (Q=1) and engineering breakeven?
Scientific breakeven ($Q = P_{fus} / P_{aux} = 1$) occurs when the thermal fusion power released equals the auxiliary heating power injected into the plasma core. Engineering breakeven accounts for plant-level efficiencies: thermal-to-electric conversion ($\sim 35\%$) and heating system wall-plug electrical efficiency ($\sim 40\%$). For an electrical power plant to generate net exportable electricity, the plasma requires $Q \ge 25\sim 40$.
Why does the D-T fusion cross-section peak around 60 keV?
At low energies, the repulsive electrostatic Coulomb barrier between positively charged nuclei prevents them from approaching within reach of the attractive strong nuclear force. As temperature rises, quantum mechanical tunneling through the Coulomb barrier increases exponentially (Gamow factor). Above $\sim 60\,\text{keV}$, the geometric collision time decreases faster than tunneling probability grows, causing the cross-section to level off and decline.