The maths behind the simulation

This model describes nuclear fusion in a few deliberately simplified equations so that everyone can follow it. The qualitative behaviour stays the same: how temperature, density and magnetic field ignite a plasma, when it stays stable and how much power a plant gains from it. The simulation is built on these equations and lets you try out every effect directly.

Two helper functions are used in many places. The first limits a value to a range, the second forms a smooth transition from 0 to 1.

$$\operatorname{clamp}(x, a, b) = \min(\max(x, a), b)$$

$x$

Input value

$a,\ b$

Lower and upper limit of the range

$$s(x;\ a, b) = u^2\,(3 - 2u), \qquad u = \operatorname{clamp}\left(\frac{x - a}{b - a},\ 0,\ 1\right)$$

$s$

Smooth transition: 0 up to the start, 1 from the end, a smooth rise in between

$x$

Input value

$a,\ b$

Start and end of the transition

$u$

Position between start and end, limited to 0 to 1

Fuels

The model considers four fuel pairs. The values are rounded textbook figures.

$\text{D}$

Deuterium, heavy hydrogen with one neutron

$\text{T}$

Tritium, hydrogen with two neutrons

${}^{3}\text{He},\ {}^{4}\text{He}$

Helium-3 and helium-4

$\text{p}$

Proton

$\text{n}$

Neutron

${}^{11}\text{B}$

Boron-11

With deuterium and tritium, the neutron carries away 14.1 MeV of the energy and the helium nucleus 3.5 MeV. In the equations below, the ignition temperature is called $T_\text{ign}$.

Size of atomic nuclei

The radius of an atomic nucleus grows with the cube root of its mass number. The simulation adopts this proportion and draws deuterium with a radius of 9 pixels.

$$R = r_0\, A^{1/3}, \qquad r_0 = 1.2\ \text{fm}$$

$R$

Nuclear radius; a deuterium nucleus is about 4 fm across

$r_0$

Radius constant, the same for all nuclei

$A$

Mass number, the number of protons and neutrons in the nucleus

$$R_\text{px} = 9\ \text{px} \times \left(\frac{A}{2}\right)^{1/3}$$

$R_\text{px}$

Drawn nuclear radius in pixels

$A$

Mass number, the number of protons and neutrons in the nucleus

A femtometre is a quadrillionth of a metre, so the simulation magnifies the nuclei about a trillion times.

Tokamak: energy balance of the plasma

How readily a fuel fuses depends steeply on temperature. The reactivity rises from a quarter of the ignition temperature and levels off just above it.

$$r(T) = s(T;\ 0.25\, T_\text{ign},\ 1.15\, T_\text{ign})^2$$

$r$

Reactivity between 0 and 1: how readily the fuel fuses

$T$

Plasma temperature in millions of degrees Celsius

$T_\text{ign}$

Ignition temperature of the fuel from the table above

$s$

Smooth transition from the second helper function above

$$R_\text{f} = n^2\, r(T)$$

$R_\text{f}$

Fusion rate, a measure of the number of reactions per unit of time

$n$

Fuel density between 0 and 1; the slider shows it as up to 2 × 10²⁰ particles per cubic metre

$r$

Reactivity between 0 and 1: how readily the fuel fuses

The fusion rate grows with the square of the density, because every reaction needs two nuclei to meet. The temperature changes through heating, through self-heating by helium nuclei and through losses.

$$\frac{\mathrm{d}T}{\mathrm{d}t} = P_\text{ext} + P_\alpha - \frac{T}{\tau_\text{E}}$$

$T$

Plasma temperature in millions of degrees Celsius

$t$

Time in seconds

$P_\text{ext}$

External heating, in millions of degrees per second

$P_\alpha$

Self-heating by the helium nuclei from the fusion reactions

$\tau_\text{E}$

Energy confinement time: a stronger field holds the heat longer

$$P_\text{ext} = 220\, h + 500\, i, \qquad P_\alpha = 180\, R_\text{f}, \qquad \tau_\text{E} = 0.25 + 1.7\,\frac{B}{12}$$

$P_\text{ext}$

External heating, in millions of degrees per second

$h$

Heating power between 0 and 1; the slider shows it as 0 to 50 MW

$i$

Injection pulse: jumps to 1 on injection and decays at 1.4 per second

$P_\alpha$

Self-heating by the helium nuclei from the fusion reactions

$R_\text{f}$

Fusion rate, a measure of the number of reactions per unit of time

$\tau_\text{E}$

Energy confinement time: a stronger field holds the heat longer

$B$

Magnetic field in tesla, between 1 and 12

The simulation solves this equation in time steps of at most 0.05 seconds and keeps the temperature between 2 and 600 million degrees.

Tokamak: stability

A hot, dense plasma pushes hard against the magnetic field. If the field is too weak for this pressure, the plasma collapses. What matters is how the pressure compares with a minimum field.

$$p = \frac{n\, T}{1000}, \qquad B_\text{min} = 0.6 + 5\sqrt{p}$$

$p$

Plasma pressure in simplified units

$n$

Fuel density between 0 and 1; the slider shows it as up to 2 × 10²⁰ particles per cubic metre

$T$

Plasma temperature in millions of degrees Celsius

$B_\text{min}$

Minimum field that can just about hold this pressure

$$T \leftarrow T \times 0.18^{\Delta t}, \qquad R_\text{f} \leftarrow 0.3\, R_\text{f}$$

$T$

Plasma temperature in millions of degrees Celsius

$R_\text{f}$

Fusion rate, a measure of the number of reactions per unit of time

$\Delta t$

Length of one time step in seconds

If the field is at least as strong as the minimum field, the plasma stays stable. Otherwise a disruption occurs: temperature and fusion rate drop within fractions of a second, as the second line describes.

Tokamak: power and Q value

A burning plasma delivers up to 1500 megawatts of electrical power. The power follows the fusion rate with a time constant of half a second.

$$P_\text{el} \leftarrow P_\text{el} + (1500\ \text{MW} \times R_\text{f} - P_\text{el}) \times \min(1,\ 2\,\Delta t)$$

$P_\text{el}$

Electrical power of the plant

$R_\text{f}$

Fusion rate, a measure of the number of reactions per unit of time

$\Delta t$

Length of one time step in seconds

$$Q = \frac{1500\ \text{MW} \times R_\text{f}}{P_\text{heiz}}, \qquad P_\text{heiz} = 50\ \text{MW} \times h + 1\ \text{MW}$$

$Q$

Q value: ratio of fusion power to heating power

$R_\text{f}$

Fusion rate, a measure of the number of reactions per unit of time

$P_\text{heiz}$

Heating power in megawatts

$h$

Heating power between 0 and 1; the slider shows it as 0 to 50 MW

The Q value is the ratio of fusion power to heating power. The plasma counts as ignited when it is stable, has reached the ignition temperature and the fusion rate is above 0.15.

Tokamak: nuclei in the plasma

Nuclei with the same charge repel each other. The repulsion follows Coulomb's law, smoothed so that it does not become infinite at very small distances.

$$a = \frac{k\, Z_1 Z_2}{d^2 + \varepsilon^2}, \qquad k = 14\,000\ \text{px}^3/\text{s}^2, \qquad \varepsilon^2 = 64\ \text{px}^2$$

$a$

Acceleration with which two nuclei repel each other

$k$

Strength of the repulsion, scaled to the display

$Z_1,\ Z_2$

Charge numbers of the two nuclei; for boron the repulsion is five times as strong as for hydrogen

$d$

Distance between two nuclei; the repulsion acts up to 55 pixels and is capped at 2200 px/s²

$\varepsilon$

Smoothing length that keeps the repulsion finite at small distances

$$v \leftarrow v + 0.07\,(v_\text{th} - v), \qquad v_\text{th} = 18 + 150\, \operatorname{clamp}\left(\frac{T}{T_\text{ign}},\ 0,\ 1.3\right)\ \text{px/s}$$

$v$

Speed of a nucleus in pixels per second

$v_\text{th}$

Thermal speed, towards which a thermostat pulls every nucleus

$T$

Plasma temperature in millions of degrees Celsius

$T_\text{ign}$

Ignition temperature of the fuel from the table above

Two nuclei fuse when they come closer than 1.13 times the sum of their radii, the plasma is stable and hotter than 45 per cent of the ignition temperature. The light reaction product then flies off at 320 px/s in a random direction. For deuterium-tritium and deuterium-deuterium this is a neutron, for deuterium-helium-3 a proton and for proton-boron a helium nucleus. In the simulation, the number of nuclei shown follows the density.

$$N = \operatorname{round}(6 + 20\, n)$$

$N$

Number of nuclei shown

$n$

Fuel density between 0 and 1; the slider shows it as up to 2 × 10²⁰ particles per cubic metre

Stellarator: confinement without plasma current

The stellarator follows the same equations as the tokamak, with two differences. Its twisted coils hold the plasma without any plasma current, so no disruption can occur. And how well the heat stays confined also depends on the shape of the coils.

$$\tau_\text{E} = \left(0.25 + 1.7\,\frac{B}{12}\right) \times (0.55 + 0.5\, o)$$

$\tau_\text{E}$

Energy confinement time: a stronger field holds the heat longer

$B$

Magnetic field in tesla, between 1 and 12

$o$

Coil optimisation between 0.3 and 1

If the pressure exceeds what the field can hold, the confinement time drops instead of the plasma collapsing. It keeps burning, just less well.

$$\tau_\text{E} \leftarrow 0.45\, \tau_\text{E} \qquad (B < B_\text{min})$$

$\tau_\text{E}$

Energy confinement time: a stronger field holds the heat longer

$B$

Magnetic field in tesla, between 1 and 12

$B_\text{min}$

Minimum field that can just about hold this pressure

The simulation calculates the minimum field, the fusion rate, the electrical power and the nuclei in the plasma as for the tokamak.

Laser: gain of a shot

In laser fusion, the symmetry of the beams matters most. The gain factor of a shot is the product of a base value for the fuel and two factors for symmetry and laser energy.

$$G = G_\text{B} \times f_\sigma \times f_\text{E}$$

$G$

Gain factor of the shot: yield divided by laser energy

$G_\text{B}$

Base gain of the fuel: 110 for D-T, 38 for D-D, 72 for D-³He, 30 for p-¹¹B

$f_\sigma$

Factor for the beam symmetry

$f_\text{E}$

Factor for the laser energy

$$f_\sigma = s(\sigma;\ 0.5,\ 0.95)^2, \qquad f_\text{E} = 0.4 + 0.6\, \operatorname{clamp}\left(\frac{E_\text{L} - 0.5\ \text{MJ}}{2.5\ \text{MJ}},\ 0,\ 1\right)$$

$f_\sigma$

Factor for the beam symmetry

$f_\text{E}$

Factor for the laser energy

$s$

Smooth transition from the second helper function above

$\sigma$

Beam symmetry between 0.3 and 1; below 0.7 the implosion fizzles

$E_\text{L}$

Laser energy between 0.5 and 3 MJ

$$Y = E_\text{L} \times G, \qquad P_\text{el} = Y \times 10\ \text{Hz} \times 0.4$$

$Y$

Energy yield of the shot

$E_\text{L}$

Laser energy between 0.5 and 3 MJ

$G$

Gain factor of the shot: yield divided by laser energy

$P_\text{el}$

Electrical power of the plant

A laser power plant fires ten shots per second and converts 40 per cent of the heat into electricity. In the simulation, a shot lasts three seconds. Compression, ignition and burn follow the time axis from 0 to 1 with smooth transitions.

$$e(t) = \begin{cases} 2t^2 & t < 0.5 \\ 1 - \dfrac{(2 - 2t)^2}{2} & t \geq 0.5 \end{cases}$$

$e$

Progress of the animation between 0 and 1

$t$

Timeline of the shot between 0 and 1

Comparison with coal and wind

The comparison extrapolates the plant output to a year. It is based on one megawatt of electrical power at 90 per cent capacity factor and 40 per cent efficiency. That corresponds to about 7.1 × 10¹³ joules of heat per year.

$$m_\text{F} = 0.21\ \text{kg} \times \frac{P}{\text{MW}} \times \frac{3.4 \times 10^{11}\ \text{J/g}}{e}$$

$m_\text{F}$

Fusion fuel per year

$P$

Electrical power of the plant

$e$

Energy density of the selected fuel from the table above

$$m_\text{Kohle} = 2470\ \text{t} \times \frac{P}{\text{MW}}, \qquad m_{\text{CO}_2} = 6160\ \text{t} \times \frac{P}{\text{MW}}$$

$m_\text{Kohle}$

Hard coal per year

$m_{\text{CO}_2}$

CO₂ from burning this coal, per year

$P$

Electrical power of the plant

$$N_\text{Wind} = \frac{P}{4\ \text{MW} \times 0.30}, \qquad N_\text{Kohle} = \frac{P}{400\ \text{MW}}$$

$N_\text{Wind}$

4 MW wind turbines that deliver 30 per cent of their rated power on average

$N_\text{Kohle}$

Coal-fired units of 400 MW each

$P$

Electrical power of the plant

For deuterium-tritium this gives about 210 grams of fuel per megawatt and year. Hard coal with about 29 megajoules per kilogram needs about 2470 tonnes for the same heat, and burning it releases about 6160 tonnes of CO₂.