Fusion energy gain factor

In 1955, John Lawson was the first to explore the energy balance mechanisms in detail, initially in classified works but published openly in a now-famous 1957 paper. In this paper he considered and refined work by earlier researchers, notably Hans Thirring, Peter Thonemann, and a review article by Richard Post. Expanding on all of these, Lawson's paper made detailed predictions for the amount of power that would be lost through various mechanisms, and compared that to the energy needed to sustain the reaction.[1] This balance is today known as the Lawson criterion.

In a successful fusion reactor design, the fusion reactions generate an amount of power designated P_fus.[lower-alpha 3] Some amount of this energy, P_loss, is lost through a variety of mechanisms, mostly convection of the fuel to the walls of the reactor chamber and various forms of radiation that cannot be captured to generate power. In order to keep the reaction going, the system has to provide heating to make up for these losses, where P_loss = P_heat to maintain thermal equilibrium.[2]

The most basic definition of breakeven is when Q = 1,[lower-alpha 4] that is, P_fus = P_heat.

Some works refer to this definition as scientific breakeven, to contrast it with similar terms.[3][4] However, this usage is rare outside certain areas, specifically the inertial confinement fusion field, where the term is much more widely used. Inertial devices, and many similar concepts, do not attempt to maintain equilibrium but simply capture the energy produced. In this case, P_heat considers all of the energy needed to produce the reaction, whether it be direct heating or other systems such as laser or magnetic compression.[5]

Since the 1950s, most commercial fusion reactor designs have been based on a mix of deuterium and tritium as their primary fuel; other fuels have attractive features but are much harder to ignite. As tritium is radioactive, highly bioactive, and highly mobile, it represents a significant safety concern and adds to the cost of designing and operating such a reactor.[6]

In order to lower costs, many experimental machines are designed to run on test fuels of hydrogen or deuterium alone, leaving out the tritium. In this case, the term extrapolated breakeven is used to define the expected performance of the machine running on D-T fuel based on the performance when running on hydrogen or deuterium alone.[7]

The records for extrapolated breakeven are slightly higher than the records for scientific breakeven. Both JET and JT-60 have reached values around 1.25 (see below for details) while running on D-D fuel. When running on D-T, only possible in JET, the maximum performance is about half the extrapolated value.[8]

Another related term, engineering breakeven, considers the need to extract the energy from the reactor, turn that into electrical energy, and feed some of that back into the heating system.[7] This closed loop sending electricity from the fusion back into the heating system is known as recirculation. In this case, the basic definition changes by adding additional terms to the P_fus side to consider the efficiencies of these processes.[9]

D-T reactions release most of their energy as neutrons and a smaller amount as charged particles like alpha particles. Neutrons are electrically neutral and will travel out of any magnetic confinement fusion (MFE) design, and in spite of the very high densities found in inertial confinement fusion (ICF) designs, they tend to easily escape the fuel mass in these designs as well. This means that only the charged particles from the reactions can be captured within the fuel mass and give rise to self-heating. If the fraction of the energy being released in the charged particles is f_ch, then the power in these particles is P_ch = f_chP_fus. If this self-heating process is perfect, that is, all of P_ch is captured in the fuel, that means the power available for generating electricity is the power that is not released in that form, or (1 − f_ch)P_fus.[10]

In the case of neutrons carrying most of the practical energy, as is the case in the D-T fuel, this neutron energy is normally captured in a "blanket" of lithium that produces more tritium that is used to fuel the reactor. Due to various exothermic and endothermic reactions, the blanket may have a power gain factor M_R. M_R is typically on the order of 1.1 to 1.3, meaning it produces a small amount of energy as well. The net result, the total amount of energy released to the environment and thus available for energy production, is referred to as P_R, the net power output of the reactor.[10]

The blanket is then cooled and the cooling fluid used in a heat exchanger driving conventional steam turbines and generators. That electricity is then fed back into the heating system.[10] Each of these steps in the generation chain has an efficiency to consider. In the case of the plasma heating systems, ${\displaystyle \eta _{heat}}$ is on the order of 60 to 70%, while modern generator systems based on the Rankine cycle have ${\displaystyle \eta _{elec}}$ around 35 to 40%. Combining these we get a net efficiency of the power conversion loop as a whole, ${\displaystyle \eta _{NPC}}$, of around 0.20 to 0.25. That is, about 20 to 25% of ${\displaystyle P_{R}}$ can be recirculated.[10]

Thus, the fusion energy gain factor required to reach engineering breakeven is defined as:[11]

$${\displaystyle Q_{E}\equiv {\frac {P_{\text{fus}}}{P_{\text{heat}}}}={\frac {1}{\eta _{\text{heat}}\cdot f_{\text{recirc}}\cdot \eta _{\text{elec}}\cdot (1-f_{\text{ch}})}}}$$

To understand how ${\displaystyle Q_{E}}$ is used, consider a reactor operating at 20 MW and Q = 2. Q = 2 at 20 MW implies that P_heat is 10 MW. Of that original 20 MW about 20% is alphas, so assuming complete capture, 4 MW of P_heat is self-supplied. We need a total of 10 MW of heating and get 4 of that through alphas, so we need another 6 MW of power. Of the original 20 MW of output, 4 MW are left in the fuel, so we have 16 MW of net output. Using M_R of 1.15 for the blanket, we get P_R about 18.4 MW. Assuming a good ${\displaystyle \eta _{NPC}}$ of 0.25, that requires 24 MW P_R, so a reactor at Q = 2 cannot reach engineering breakeven. At Q = 4 one needs 5 MW of heating, 4 of which come from the fusion, leaving 1 MW of external power required, which can easily be generated by the 18.4 MW net output. Thus for this theoretical design the Q_E is between 2 and 4.

Considering real-world losses and efficiencies, Q values between 5 and 8 are typically listed for magnetic confinement devices to reach ${\displaystyle Q_{E}=1}$,[10] while inertial devices have dramatically lower values for ${\displaystyle \eta _{\text{heat}}}$ and thus require much higher Q values, on the order of 50 to 100.[12]

As the temperature of the plasma increases, the rate of fusion reactions grows rapidly, and with it, the rate of self-heating. In contrast, non-capturable energy losses like x-rays do not grow at the same rate. Thus, in overall terms, the self-heating process becomes more efficient as the temperature increases, and less energy is needed from external sources to keep it hot.[13]

Eventually P_heat reaches zero, that is, all of the energy needed to keep the plasma at the operational temperature is being supplied by self-heating, and the amount of external energy that needs to be added drops to zero. This point is known as ignition. In the case of D-T fuel, where only 20% of the energy is released as alphas that give rise to self-heating, this cannot occur until the plasma is releasing at least five times the power needed to keep it at its working temperature.[13]

Ignition, by definition, corresponds to an infinite Q, but it does not mean that f_recirc drops to zero as the other power sinks in the system, like the magnets and cooling systems, still need to be powered. Generally, however, these are much smaller than the energy in the heaters, and require a much smaller f_recirc. More importantly, this number is more likely to be near-constant, meaning that further improvements in plasma performance will result in more energy that can be directly used for commercial generation, as opposed to recirculation.[14]

The final definition of breakeven is commercial breakeven, which occurs when the economic value of any net electricity left over after recirculation is enough to pay for the reactor.[7] This value depends both on the reactor's capital cost and any financing costs related to that, its operating costs including fuel and maintenance, and the spot price of electrical power.[7][15]

Commercial breakeven relies on factors outside the technology of the reactor itself, and it is possible that even a reactor with a fully ignited plasma operating well beyond engineering breakeven will not generate enough electricity rapidly enough to pay for itself. Whether any of the mainline concepts like ITER can reach this goal is being debated in the field.[16]

• Entler, Slavomir (June 2015). "Engineering Breakeven". Journal of Fusion Energy. 34 (3): 513–518}. doi:10.1007/s10894-014-9830-2. S2CID 189913715.
• Lawson, John (1957). "Some Criteria for a Power Producing Thermonuclear Reactor". Proceedings of the Physical Society, Section B. 70 (6): 6–10. Bibcode:1957PPSB...70....6L. doi:10.1088/0370-1301/70/1/303.
• Meade, Dale (October 1997). Q, Break-even and the nτE Diagram for Transient Fusion Plasmas. 17th IEEE/NPSS Symposium on Fusion Engineering.
• McCracken, Garry; Stott, Peter (2005). Fusion. Academic Press. ISBN 9780123846563.