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A Z-pinch fusion drive main electrode, with the radiator in the background. The ion injectors drive a plasma over the electrode (blue) where a current is transferred to the plasma, that is pinched by magnetic pressure at the end of the electrode into an intense fusion arc (purple). One of the potential fusion mechanisms for the Icarus Spaceship drive (ML)
Propulsion a stiction axiale en cisaillement. Les injecteurs poussent un plasma au dessus de l'électrode (bleu). Des électrons sont transférés par un courant électrique et créent une compression du plasma (violet) dans laquelle la fusion nucléaire se produit. Un des mécanismes potentiels pour la propulsion par fusion nucléaire.
Introduction
Introduction
Nuclear fusion is easy.
Nuclear fusion that produces more energy than it uses is impossible hard. In fact, all controlled fusion processes that have been tried out in the last forty years have used more power than they were able to produce, with the notable exception of thermo nuclear weapons, and the recent (end of 2022) success at the NIF facility.
To obtain fusion reactions that produce a net energy output, the correct elements must be present and three conditions must be met: density, confinement time, and temperature. The combination of these three conditions is known as the triple product. If the triple product is not met, then we can still get fusion, but without an overall energy gain. In practice, we have two main examples of systems that achieve it: The Sun and by extension all stars, and the uncontrolled fusion in thermonuclear weapons.
The fusion reactions in the Sun are the result of its immense size and corresponding gravity that create the correct conditions in the core of the star.
Density; the result of the internal pressure created by gravity, that compresses the plasma at the center of the sun to a density of up to 150 tonnes per m3;
Confinement time; that for the Sun is measured in billions of years;
Temperature; that can reach fifteen million K.
But the Sun, for all its might, produces energy at the surprisingly low average volumetric power density of 10 Watts per cubic meter(1). This is less than the power density of a fireplace. So a fusion engine ten cubic meter in volume, operating at that power density, would produce 100 Watts, while a classical nuclear fission reactor core with a ten cubic meter core can produce 1500 MegaWatts, over 1.5 million times more power. Starships need a power source more compact than the Sun.
Humanity has achieved positive energy output for fusion in thermonuclear weapons. And this has been studied as a starship power source, notably in some speculative work done during the Orion study(2) from the late fifties and early sixties and in ICF confinement studies(3). However, in a thermonuclear explosion, fusion is achieved using a fission explosion as driver. The fission driver is heavy, making the propulsion system quite inefficient and wasting large quantities of fissile materials. In the words of the Orion project’ top physicist, Freeman Dyson: ’Orion was a filthy creature’(4)
So how to achieve fusion without the mass and without the mess? That is the focus of this chapter. We will look at fusion reactions first, then the means to achieve them, meeting the triple product criteria and some of the ways these systems could be applied to drive starships. We will also explore why fusion is favored compared to other energy sources.
Fusion reactions
Fusion reactions
Not all fusion reactions provide a net energy gain. In the periodic table, elemental iron serves as the usual cutoff point, where fusion on materials lighter than iron produces energy, while fusion of materials heavier than iron absorb energy. However, even for the low mass element fusion reactions some of the reactions do not produce significant amounts of energy.
Usable Fusion Reactions
Usable Fusion Reactions
The list of binary, exothermic fusion reactions involving low-Z nuclei(5) (A.K.A fusible materials and the products of their reactions) is fairly short:
The reaction cross-sections become orders of magnitude smaller as one progresses down the list in Table 1. Small reaction cross-sections require more input power to create fusion conditions than large cross-sections (e.g. it is easier to hit a baseball than a golf ball with a bat). Consequently, reactions toward the top of the list (particularly the first three) are easier to trigger and most commonly considered for both terrestrial fusion power and fusion propulsion. The last reaction in the list is the cascade fusion of the two D+D fusion reactions and of their by products.
In a fusion reaction, the portion of the fuel that actually reacts is called the burnup fraction. As a first order of magnitude evaluation, we can establish that a fusion drive with a burnup fraction of 10%, using 1 kg of deuterium+he3 per second will produce 353 000GJ/kg x 1kg/s x 0,1 = 35 300 GW of power (reaction 3, table 1).
To find the thrust from the fusion reaction, F=√(ṁ*2Pt)*ηn . For an efficiency (ηn) of 80%, the force is 8 400 000 N. See the example further down on this page.
D+T(Deuterium + Tritium) fusion is the easiest to ignite, though it ejects a high-energy (14.1 MeV) neutron that holds most of the energy and poses problems for engine design. Moreover, Tritium has a very short half-life, 13.5 years, so it cannot be stored for use during the deceleration phase of an Interstellar vehicle. One hundred years after departure most of it will have decayed into He3. DT works fine for Earth based power reactors though, which is why it is used for most of the present research projects on fusion, and since the high energy neutron can combine with lithium to produce a new tritium atom, a D+T reactor can ‘breed’ its own fuel.
D+He3(Deuterium+Helium3) fusion is often favored as a starship fuel because it is intrinsically aneutronic. However, any DHe3 mixture produces unavoidable DD(Deuterium+Deuterium) reactions as well, and at a higher rate in some temperature ranges(32). The DD side-reactions produce their own neutrons, plus Tritium, which reacts immediately with available deuterium in the plasma to produce the aforementioned 14.1 MeV high energy neutrons. This all significantly limits the nominal aneutronic benefit. In addition, He3 is almost completely nonexistent on Earth, with most supplied from tritium decay, and is incredibly scarce in the potential resources of remote places like the Moon and the gas giants(3). Just how the extraction and use of Helium3 might be accomplished is the focus of other sections of this report.
Alternatively, He3 might be produced on Earth using D-D breeder reactors. These could feed higher power D+He3 reactors, and in particular D+He3 engines for space vehicles that would require less shielding than equivalent D+D or D+Tritium engines.
D+D fusion is an interesting option since Deuterium is readily available via seawater separation, distillation and hydrogen isotope separation here on Earth, and can be stored cryogenically indefinitely, given an appropriate refrigeration system. It also isn’t radioactive, as compared to Tritium. Like Hydrogen, Deuterium is a chemical explosive if reacted with oxygen, but that is hardly a problem for a starship. Its major downside is that there are two equally probable DD fusion reactions, one that produces Tritium and a proton, and one that produces He3 and a high energy neutron. The Tritium, with it’s lower fusion temperature, will fuse and create the 14.1 MeV neutron already mentioned. The waste energy of the high energy neutrons cannot be recaptured easily, but some drive designs manage to do it. The D+D mix is harder to ignite than D+T, but easier than D+He3. The cascade fusion of 6 Deuterium atoms (table 1, reaction 11) produces 2 He4 atoms, 2 protons and 2 neutrons. About 40% of the energy release is carried by the neutrons.
p+B11 is another interesting reaction that does not produce any neutrons. And it has the further benefit that there are no secondary reactions, as opposed to D-He3 mixes. It is very difficult to initiate, however, and may lose excessive amounts of energy through electromagnetic (bremstalhung) radiation. Despite this, it is actively studied()(). It’s also important to note that releases about a quarter of the energy per kg of other reactions, so a Boron powered spaceship might require more propellant, depending on the burnup fraction.
The Coulomb Barrier
The Coulomb Barrier
Nuclei repel each other due to electrostatic forces by the inverse of the separation distance squared. This force becomes extremely high as the separation between two nuclei approaches zero. However, past a certain threshold the attractive strong nuclear force in the atomic nucleus becomes dominant. The position in which the attractive nuclear force overcomes the repulsive electrostatic force is known as the Coulomb barrier. This barrier becomes an increasingly difficult engineering problem as atomic nuclei get larger. Consequently, light nuclei are easier to fuse. Once the Coulomb barrier is overcome, the strong nuclear force attracts the two nuclei into fusion, and these nuclei convert the mass difference between their original separate states and their new combined state into energy that can be used for useful work. The Coulomb barrier for a particular nuclei interaction determines the amount of energy that needs to be injected, or recirculated, to initiate the fusion reaction in a drive.
Mathematically, the Coulomb barrier is the nuclei separation distance at which a the maximum in potential energy occurs. The shape of this function is seen in the following image:
Figure 1
In the image, positive potential energy results in repulsion between two nuclei, while negative potential attracts two nuclei. Due to the instantaneous change from repulsion to attraction at the Coulomb barrier radii, low-energy nuclei escape fusion reactions. This is why a high energy input is required.
Based on observation of the potential energy versus separation distance graph alone, one may assume that the input energy required to initiate a fusion reaction is the maximum in the the figure. However, due to the effect of quantum tunneling only ~10% of this maximum is required for a sufficiently large number of nuclei to react. For more information on quantum tunneling, please refer to any introductory quantum mechanics textbook.
Driver and gain
Driver and gain
Driver for the sun is gravity
Driver for ICF tests in 1980s was x-rays from nuclear explosions
Driver for ICF can be lasers or ion beams
Driver for Z-pinch is current
Reaction Rates
Reaction Rates
Fusion reactions in a fusion drive are complex because more than one reaction happens at the same time, and they happen in a complex and unstable plasma environment. DD reactions proceed along two paths with equal likelihood, and their reaction products (T and He3) will in turn react with the deuterium plasma. The rates for all these reactions depend on the plasma temperature (as shown in Figure 2) and the number density of the respective ions, which both change over the course of a fusion burn. Wave effects in the plasma create uneven burn conditions that further complicate matters.
Figure 2- Fusion reaction rates by temperature (reference)
Fusion reaction rates (f) are proportional to density and the cross section of the reaction products, as seen in figure 2. The general equation is:
f = n1 n2 <σv> Vp (4.1)
where n1 and n2 are the number densities of the plasma species, Vp is the volume of the plasma core(cm3), and <σv> is the reactivity (reactions/s/cm3).
The rates for DT, DHe3 and DD fusion can be calculated as:
f = nD nT <σv> Vp
f = nD nHe3 <σv> Vp
f = nD <σv> Vp
The reaction rate for DD fusion is exactly twice the others, since owing to the single ion species nD is twice as high.
The reactivities themselves can be calculated using formulae presented by Atzeni(4):
<σv> = C1 ζ-5/6 ξ2 exp(-3ζ1/3ξ) (4.2)
ζ = 1 - (C2T+C4T2+C6T3) / (1+C3T+C5T2+C7T3) ( 4.3)
ξ = C0 / T1/3 (4.4)
For DD,DT and DH3.
For p+B11, a slightly different formulae is used:
<σv> = C1 ζ-5/6 ξ2 exp(-3ζ1/3ξ)+5.41e-15*T^(-3/2)*exp(-148/T) (4.5)
Constants for these equations are found in the table bellow.
Fusion reaction rates
Fusion reaction rates
It can be seen in figure 2 that the reaction rates for p+B11 are much lower than for others and require higher temperatures.
The graph also shows that while at lower temperatures D-D reactions will dominate a D-He3 mix, at higher temperatures the D-He3 reaction will become more common, reducing the number of side reaction in the mix.
We can also see that if temperatures get too high, the reaction rates start going down for some of the reaction, since the nuclei now have too much energy to “stick together”.
The fusion reactions have been simulated numerically by many authors()(). The results of the two reactions used for this report are
Nuclei+Neutrons Bremstalhung total MeV MeV
D+He3 95% 5%
D+D 60% 40%
Table xx, fusion results
Neutron Capture
Neutron Capture
The neutrons shown in table 1 carry a lot of energy that cannot be manipulated by magnetic fields and are therefore not usable directly for propulsion. However, at the very high densities required for most fusion drive designs, the probability of interactions between the neutrons and other particles is high. With these interactions (impacts with nuclei, mostly), some neutrons lose velocity and become ‘thermal’ neutrons. The ‘thermal’ cross section of most elements is much higher than the high energy cross section, and these neutrons can more easily be captured in fusion reactions, as there is no Coulomb barrier to overcome. This adds their mass to the reaction products and lowers neutron radiation from the drive. The average distance that a particle can travel before having a collision is the Mean Free Path. If the mean free path is shorter than the thickness of a material, most of the neutrons will be absorbed.
Mean free path equation
mfp=M/ρNA𝜎 (4.6)
where:
M= Molar mass (g/mole)
ρ= density (g/cm3)
NA= Avogadro’s number (6.023e-23 atm/mole)
𝜎=cross section (barns)
The notions of mean free path and of neutron absorption can be integrated by the following equation, a variation of the Beer Lambert law ( I/Io=e-(μ/ρ)ρl ):
N/No= e-(nHe3*σHe3*r/2) (4.7)
where, for a pellet rich in Helium3:
nHe3 = Number of He3 nuclei in the pellet
𝜎He3 = Cross section of He3 in Barns
r = compressed pellet radius
For Daedalus, the drive increased density of the propellant between 1000 and 2000 times, and average neutron capture cross section chosen was 30 Barns. Using the equation for the mean free path of neutrons in a plasma the Daedalus team showed that the neutrons never escape the fuel pellets and were entirely absorbed in the reaction, though this conclusion has since been disputed(32). In particular, fusion reactions taking place close to the surface of the pellet would not see their neutrons absorbed and a cross section of 30 barns seem very optimistic.
In fact, for any cross section larger than about 8 Barns, neutron emission from any type of fusion reactions taking place in an ICF pellet will be practically zero. If the high energy cross sections in table 3 are used, then about 15% of the neutrons manage to escape from Daedalus type pellets.
The following table shows the mean free path of neutrons in various materials. The first four elements are in their liquid state. The table also shows the mean free path when the material is compressed 1000 times, as in ICF fusion and its variations. Although not calculated, the high cross section for thermal neutrons leads to even shorter mean free paths.
Triple Product
Triple Product
The triple product of density, time and temperature must be at least 1021 sKeV/m3 to reach the minimum conditions for break-even fusion reactions(). This is the same numerical value as the similar Lawson’s criteria(), though the latter omits the time component.
The triple product equation is:
ntT=triple product(sKeV/m3) (4.8)
where:
n=number of nuclei (density)
te= confinement time (s)
T= Temperature (K) or KeV
te=w/ploss energy density/volumetric nergy loss of the plasma
Formulation from Perakis Paper
n*te*T ≥ 12*kB*T2 /Ech* (σv)
te= confinement time (s)
n=number density of the plasma (density)
T=Temperture (K)
kB=Botltzmann constant
Ech=Energy of the reaction products (excluding neutrons)(J)
(σv)=average product of velocity and cross section
σ=cross section
v=velocity
The Triple Product is more of a guideline than a law. In particular, reaction rates actually decrease at much higher temperatures, so a plasma with very high temperature but low density, for instance, may not ever reach break-even conditions. The more rigorous formula for break-even involves energy calculations using the reaction rates at the target temperature.
There are three paths to achieving the triple product:
1- Increasing Density (n)
1- Increasing Density (n)
In stars, density is a consequence of gravity. This doesn’t seem practical for starship engines, but some very speculative designs use the gravity increase at the surface of a miniature black hole(33)() to create fusion conditions.
In most designs, density is increased using pressure from light or electron beams, from impact shock or from electromagnetic fields. Often a combination of these is proposed.
In ICF, the pressure from the beams compresses the pellet to densities thousands of times higher than their original state. This is the mechanism proposed for Icarus Endeavour, augmented with shock ignition.
Using ultradense deuterium() as proposed by Milos Stanic in his UDD vessel concept () is another possible way of increasing density.
2- Increasing Confinement Time (s)
2- Increasing Confinement Time (s)
This is the main mechanism in stars, since their confinement time is tremendously long. The very long magnetic Z-pinch (+40 m) proposed for Icarus Firefly allows more time to create a more complete burn.
3- Increasing Temperature(T)
3- Increasing Temperature(T)
The temperature of a gas is the measure of the average velocity of the particles in the gas. The particles do not have ‘heat’ and velocity, for a gas or a plasma heat is velocity.
In stars, the temperature self-regulates as the star expands under light pressure until the radiation losses are equal to the gains from the fusion reactions.
In fusion drives, increased temperatures can be obtained by shocks (Endeavour, Zeus), by electrical resistance (Firefly) and is generally accomplished by the ignition system.
Antimatter can also be used. Injected in small quantities, the antimatter reacts with matter to create energy that heats the plasma to fusion conditions.
RMF: This whole section could be fleshed out some more.
Fusion Propulsion
Fusion Propulsion
An internal study carried out by Icarus Interstellar() listed no less than 20 technologies that could potentially be used to achieve nuclear fusion. This is an embarrassment of riches, that shows that fusion is far from a mature field, rather like the period when steam cars still competed with electrical and gas powered vehicles in the early 20th century. The following table shows some of the fusion technologies or sources.
General Physical Characteristics of Fusion Propulsion Systems
General Physical Characteristics of Fusion Propulsion Systems
Fusion rockets use a propellant, which in most systems is comprised of the reaction products, unspent fuel, and other material directly involved in the underlying fusion reaction. These materials are contained within a reaction chamber and ejected via a magnetic nozzle. The nozzle typically includes an energy recovery system that is used to power the ignition system, called the ‘‘driver’’. A secondary energy production system is usually required to enable start-up of the propulsion system.
In some variations, the driver may be powered by an external source(american navy).
The power flow within a fusion propulsion system is shown in the following graph. The graph also shows the physical systems that are required to produce fusion propulsion system.
Figure 2- Generic fusion propulsion system
In some designs the energy recovery system is replaced with an external source
For each type of fusion propulsion there are variations from this general arrangement. Specific implementations of fusion systems are explored in the pages on the various vehicles.
⭐ Fusion drive example
Pf= Fusion power, Pf=Fe*ṁ*bf
Pp= Plasma power, Pp=Pf-Pn-Pb
Pn= Neutron losse s
Pb= Bremsstrahlung losses
Pt= Thrust power, Pt=(Pp-Pr-Pnl)*ηn
Pnl= Nozzle (reaction chamber) power losses
Pr= Recovered power, Pr=Prl-Ps
Prl= Energy recovery system power losses
Ps= Power to ship services
Pi= Driver power (ignition power)
Es= Energy stored in energy recovery system
Pth= Thermal power loads from radiation losses, Pth=(Pn+Pb)*area ratio
ηn= Nozzle efficiency
bf= Burnup fraction
ṁ= Mass flow of propellant
G= Gain, G=Pf/Pi
Fusion power = fusion energy release*mass flow*burnup fraction, Pf=Ef*ṁ*bf
Plasma power = Fusion power-Neutron losses-bremsstrahlung losses, Pp=(Pf-Pn-Pb)ηn
Thrust power = (Plasma power-Recovered power-Nozzle power losses)*nozzle efficiency, Pt=(Pp-Pr-Pnl)ηn
Ignition power = Power recovery-Power recovery system losses- Ship services, Pi=Pr-Prl-Ps
Recovered power may be replaced by an external source. Project Longshot() used a separate nuclear reactor to power ignition.
Fusion gain = Fusion power/ Recovered power, G=Pf / Pr
According to Newton’s laws of motion the vehicle thrust (F) is equal to the mass flow rate (ṁ) times the exhaust velocity(ve) and the efficiency of the nozzle (ηn):
F=ṁ*ve*ηn
Since for a rocket the power (P) can be related to the thrust by the following equation:
Pt=1/2ṁve2 or
ve =√(2Pt/ṁ)
It follows that the thrust is related to the power by:
F=ṁ*√(2Pt/ṁ)*ηn or
F=√(ṁ*2Pt)*ηn
The rocket equation uses the exhaust velocity Ve to find the mass ratio of the vehicle and the final rocket velocity:
Rocket equation:
Vf=Ve*ln(mo/mf)
If the mass flow is constant (as per most of these fusion designs) the distance crossed while under power is:
s=Ve*(t-(m0/ṁ))*ln(m0/mf)+t(Vo-Ve)
where (t) is the time under power and Vo the initial velocity.
Using our earlier mass flow (ṁ) of 1 kg/s and power (P) of 35 300 GW, and supposing an overall efficiency of 0,8, including fusion losses, the thrust from the fusion drive will be F=√(1kg/s*35300GW*2)*0,8= 8 403 000 N or 857 tonnes. This is about ¼ of the thrust of the Saturn 5 rocket.
Fusion technology checklist
Fusion technology checklist
Adam Crowl
For an analysis of a fusion drive design, here is a handy checklist:
Bremsstrahlung radiation
Bremsstrahlung radiation
References
References
1-Sun power reference
2-Orion references
3-ICF reference
4- Disturbing the universe, Freeman Dyson
x- Icarus Intestellar paper on various fusion schemes, Kelvin
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