The Energy Problem and Fusion

The energy problem is the question of “how much energy we obtain, in what way, and at what cost.” Starting from the world’s energy demand and the need for decarbonization, this page considers where fusion might fit in and how it could be useful, drawing on numbers such as fuel resources, energy balance, and electricity cost. The mechanism of fusion itself is covered in What Is Fusion, so here we focus on “why we aim for fusion.”

Start with Intuition (High School)

Our daily lives are supported by enormous amounts of energy. Lighting, heating and cooling, charging a smartphone, trains and factories, and even the production and transport of food are all energy consumption. Across the world, much of this is still supported by fossil fuels such as oil, coal, and natural gas. Fossil fuels are like “a savings account of energy that ancient living things stored underground over long stretches of time.” Use them and they shrink, and burning them releases carbon dioxide (CO2).

Here two troubles arise at the same time. One is that the savings will eventually run low. The other is that CO2 warms the planet and changes the climate. So the world is trying to shift to “energy sources that emit almost no CO2.” This is called decarbonization.

There are several candidates. Solar and wind are very good options, but the sun does not shine at night, and the wind does not always blow at the same strength. Because the amount of power generated goes up and down at the mercy of the weather, this is called intermittency. On the other hand, society also needs power sources that keep supplying stably all day long regardless of the weather, and these are called baseload power. Factories, hospitals, and urban infrastructure are all supported by this “electricity that is always there.”

Fusion is a challenge to make happen on the ground the same reaction by which the sun shines: “light atomic nuclei fusing into heavier nuclei and releasing energy in the process.” Put in one line, its appeal is that the fuel is very familiar and abundant, it emits no CO2, and it could serve as a baseload power source unaffected by the weather. The main fuel of fusion, deuterium, is a special form of hydrogen present in small amounts in ordinary water, and it can be extracted from seawater. Because the ocean is virtually inexhaustible, the worry about fuel is very small. That said, let us also honestly note that “no fusion reactor that operates as a power plant exists yet.” It is still on the way to being established as a technology.

Understand the Physics (Undergraduate)

First, let us pin down the scale of demand with numbers. World primary energy consumption is on the order of about 600 EJ (exajoules, $1\,\mathrm{EJ}=10^{18}\,\mathrm{J}$) per year, with about 80% of it accounted for by fossil fuels. “Primary energy” refers to the energy of the resource itself before it is converted into electricity or the like. Due to population growth and the economic development of emerging countries, demand is expected to keep rising as an underlying trend.

The reason decarbonization is urgent lies in greenhouse gases. The atmospheric CO2 concentration has risen from about 280 ppm before the Industrial Revolution to around 420 ppm today, and the global average temperature has risen by about 1.1 degrees relative to pre-industrial levels. To hold the temperature rise to 1.5 degrees, achieving net zero, in which CO2 emissions and removals balance out, around 2050 is considered necessary. Fusion matters here because the reaction itself emits no CO2.

Next, let us estimate how “extraordinarily blessed fusion is as a fuel resource.” The reaction first targeted for realization in a fusion reactor is that between deuterium ($\mathrm{D}$) and tritium ($\mathrm{T}$).

$$\mathrm{D} + \mathrm{T} \rightarrow {}^4\mathrm{He}\,(3.5\,\mathrm{MeV}) + \mathrm{n}\,(14.1\,\mathrm{MeV})$$

A single reaction releases $17.6\,\mathrm{MeV}$ of energy. Since $1\,\mathrm{MeV}$ is about $1.6\times10^{-13}\,\mathrm{J}$, each reaction yields roughly $2.8\times10^{-12}\,\mathrm{J}$. This is small per reaction compared with fission, but compared on a per-mass-of-fuel basis it is on the order of millions of times that of chemical combustion (the reaction of burning oil or coal). The fact that “far more energy can be extracted from the same weight of fuel” is fusion’s resource advantage.

Deuterium occurs naturally at a ratio of about 1 in every 6700 hydrogen atoms and can be extracted from seawater. Because seawater is vast, the total amount of deuterium can be regarded as effectively inexhaustible relative to humanity’s energy demand. Tritium, on the other hand, barely exists in nature and has a short half-life of about 12 years, so designs are premised on producing (breeding) it by striking lithium with neutrons inside the reactor. Therefore the resource actually consumed in the D-T approach is lithium, and the amount of lithium resources governs long-term supply. It is important to distinguish precisely here that “deuterium is inexhaustible; the actual constraints are lithium and the tritium cycle.”

The metric that measures performance as an energy source is the Q factor as an energy balance ratio (fusion energy gain factor). Its definition is as follows.

$$Q = \frac{P_\mathrm{fus}}{P_\mathrm{heat}}$$

Here $P_\mathrm{fus}$ is the fusion power produced by the plasma, and $P_\mathrm{heat}$ is the power input from outside to heat the plasma. $Q=1$ is the state where “as much came out from fusion as was put in,” and it is called breakeven. $Q \to \infty$ corresponds to ignition, the point where the reaction becomes self-sustaining without external heating. The target ITER sets is $Q \ge 10$. The achievement condition of fusion itself is closely tied to the Lawson criterion, and the strengths and weaknesses of each device are organized in Types and Approaches of Fusion.

Deepen the Theory (Graduate)

When talking about the Q factor, one must strictly distinguish which Q is being referred to. The $Q$ defined above is a plasma-physics value and is called the scientific $Q$. However, what determines whether it works as a power plant is the engineering $Q$, often written $Q_\mathrm{eng}$. It is defined “wall to wall” as follows.

$$Q_\mathrm{eng} = \frac{P_\mathrm{net,elec}}{P_\mathrm{recirc}}$$

$P_\mathrm{net,elec}$ is the net electric power that can be sent out to the grid, and $P_\mathrm{recirc}$ is the power circulated to run the plant itself, such as heating, magnetic field coils, cooling, vacuum, and tritium processing. Two layers of conversion loss come into play here: the thermal efficiency $\eta_\mathrm{th}$ (typically 30% to 40%) when converting the roughly 80% of fusion output that neutrons carry as heat into electricity, and the efficiency $\eta_\mathrm{heat}$ when converting grid electricity into injected heating power. Because of this two-stage loss, obtaining $Q_\mathrm{eng}>1$ requires the scientific $Q$ to be quite large. It is argued that a commercial reactor requires $Q$ on the order of several tens, roughly $Q \approx 30$ or more as a guideline. The recognition that “even achieving $Q=10$ is not the completion of a power plant but a milestone along the way” emerges from this distinction.

The physics that governs the energy balance is the tug-of-war between fusion power density and losses. The fusion power density is $p_\mathrm{fus} \propto n^2 \langle \sigma v \rangle$, determined by the square of the density $n$ and the reaction rate coefficient $\langle \sigma v \rangle$ (which strongly depends on temperature). Meanwhile, the rate at which the plasma loses heat is characterized by the energy confinement time $\tau_E$, and the loss power can be written as $\sim W/\tau_E$ ($W$ is the stored energy). To approach self-sustained ignition, the triple product $n T \tau_E$ of density, temperature, and $\tau_E$ must be made sufficiently large, and this is the substance of the Lawson criterion. In transport theory, the central problem is that this $\tau_E$ becomes shorter than the classical prediction due to turbulent transport, and how to suppress it.

The way of thinking about cost evaluation is also continuous with this physics. The levelized cost of electricity (LCOE) is often used to compare electricity costs. LCOE is an index that divides construction cost (capital cost), operation and maintenance cost, fuel cost, and so on by the total electricity generated over the operating period, adds time discounting, and levels it into “how much per 1 kWh.” For fusion, fuel cost (deuterium and lithium) is extremely small, while capital cost, including superconducting magnets, the vacuum vessel, and the shielding and tritium processing systems, becomes dominant. Therefore LCOE is mainly determined by “how much the initial construction cost can be lowered and how high the capacity factor can be kept.” In addition, when carbon pricing, which puts a price on CO2 emissions, is factored in, the relative competitiveness of CO2-free fusion rises. Conversely, the understanding at this level is that fusion’s economics is determined by both wheels: physics performance ($Q$, $\tau_E$, capacity factor) and engineering (mass production of magnets and materials, maintainability).

Research Frontier (PhD)

In establishing fusion as an energy source, the unsolved challenges are not limited to plasma confinement. At the research frontier, themes such as the following are being pursued in parallel.

One is reactor engineering and materials. The $14.1\,\mathrm{MeV}$ fast neutrons produced by the D-T reaction cause irradiation damage and activation in the reactor wall materials. Research continues on developing long-lived low-activation materials and on material irradiation facilities that simulate a neutron irradiation environment. Along with this, a key step is demonstrating the breeding blanket, which produces tritium from lithium while extracting heat inside the reactor, and tritium self-sufficiency (breeding ratio TBR $>1$). The next-stage device that carries out the demonstration of fusion power generation is covered in The DEMO Reactor.

The second is integration into and operation within the power grid. A fusion reactor is expected to run in baseload operation, but in a power grid where renewable energy becomes the main source, the points of discussion are output adjustment according to supply and demand (grid integration, load following) and how the grid absorbs large power fluctuations. In the tokamak type, achieving steady-state operation, that is, a method that continuously drives the plasma current rather than pulsed operation, affects both economics and grid compatibility.

The third is economics itself. “When, and at what price” fusion electricity can be supplied is not yet at a stage that can be stated as an established fact. In recent years, approaches to miniaturize devices and lower capital costs using high-temperature superconductor (HTS) magnets, as well as the development of diverse reactor concepts by private companies, have been actively studied. Keywords that frequently appear when reading papers include engineering breakeven, tritium breeding ratio, availability / capacity factor, cost of electricity, and balance of plant, all of which connect to the common question of “how to translate physics success into power-plant economics.” To put it while avoiding definitive claims, fusion’s current standing is that while its fundamental advantages are clear, the demonstration of engineering and economics is the real work ahead.

Check Your Understanding

Q1. What is baseload power, and why is fusion expected as a candidate for it?

A It is the cheapest power source to generate, and fusion is a candidate because its fuel cost is low B It is a power source that keeps supplying stably all day long regardless of weather or time of day, and fusion is a candidate because it has no intermittency and can be expected to run continuously C It is a power source that operates only at night, and fusion is a candidate because it is strong at night D It is a power source that operates only during peak demand, and fusion is a candidate because it starts up quickly

Q2. Why is the explanation that deuterium is inexhaustible in seawater inaccurate on its own?

A Because deuterium is heavier than ordinary hydrogen and cannot be extracted from seawater B Because deuterium has a short half-life and decays quickly C Because the tritium actually consumed in the D-T approach barely exists in nature and must be bred from lithium inside the reactor, so the long-term resource constraint lies with lithium and the tritium cycle D Because the deuterium in seawater is bound to salt and cannot be separated

Q3. Which correctly states the difference between the scientific Q and the engineering Q (Q_eng) from the standpoint of losses?

A The scientific Q is fusion output divided by external heating power, while the engineering Q is the net electric power that can be sent to the grid divided by the plant circulating power, and the latter includes the double loss of converting heat to electricity and converting electricity to heating B The scientific Q and the engineering Q are the same quantity called by different names, and their values always match C The engineering Q is always larger than the scientific Q because it includes no losses D The scientific Q represents the efficiency of the whole power plant, while the engineering Q represents the efficiency of the plasma alone

Q4. From the perspective of LCOE (levelized cost of electricity), what mainly determines the cost of fusion electricity?

A The fuel cost of deuterium and lithium B The labor cost of operators and insurance premiums C Only the dismantling cost at decommissioning D How much the initial construction cost can be lowered and how high the capacity factor can be kept

Q5. Which is the correct combination of engineering challenges, other than plasma confinement, needed to establish fusion as a power plant?

A Technology to reduce the mass of the electron, and acceleration technology exceeding the speed of light B Development of low-activation materials that withstand irradiation damage from fast neutrons, and demonstration of a breeding blanket that breeds tritium from lithium and satisfies tritium self-sufficiency C A device to capture the CO2 produced by the reaction, and geological disposal of radioactive waste D Large-scale expansion of solar panels, and cost reduction of storage batteries

Related Topics

• What Is Fusion: Learn the basics of the fusion reaction and the Lawson criterion.
• Types and Approaches of Fusion: Compare the strengths and weaknesses of each approach, such as magnetic confinement and inertial confinement.
• The DEMO Reactor: Covers the next-stage device that demonstrates fusion power generation.