Twenty-One Castle Mike

On the Technological and Market Feasibility of Introducing Martian ores of Elements and Minerals into the General Marketplace

This version is deprecated. A more recent attempt to relate the notes of my Uncle (a nuclear weapons research physicist for USG) can be found here.

Part A: Primary Transit Propulsion Systems

— Reference Design Twenty-One Castle Mike

Click here for schematics

A classic Viking photo of the surface of Mars (color corrected to match appearance as seen by a human being on the surface)

Abstract: [redacted] Marine Construction, LLC dba [redacted], commissioned this study to determine if an investment of less than 25 billion USD would suffice to establish a trading infrastructure under the controlling interest of [redacted] in valuable ore material with the following characteristics:

1.)   A return on investment of at least 3x in 10 years.

2.)   A probability of technical success credibly estimated of greater than 90%

3.)   A trading infrastructure of indefinite duration with extraction occurring within this solar system and with an intended international market.

4.)   To position [redacted] as the sole entity with the capacity to provide transportation within the solar system of both low and high mass payloads at a market tolerable price and risk.

Note:

This document is intended for an audience of business executives and scientists and assumes cursory knowledge of spaceflight and general mathematical literacy. All units used in this document are Systeme Internationale (SI) and wherever units are not specified, SI units should be assumed.

Introduction

[redacted] is a privately held corporation incorporated in the State of DE with its combined, interested conglomerate assets exceeding 50 billion USD. Its primary mode of business is marine [redacted] for the petroleum industry. As of 2011, [redacted] has earned over 12 billion USD in the sale of [redacted]. An innovative paradigm shifter, [redacted] continually seeks heterodox ventures with the potential for large reward.

This is one of several documents whose purpose is to provide the Business and Technical Requirements for a Reference Design model. In this document, the Reference Design discussed regards a direct kinetic fusion power plant for spaceflight. The Reference Design specifies an operational (reusable) fusion reactor that is vacuum transit, VT, certified (known traditionally as space flight certified) and has a maximum rated power of 11 – 21 Terra watts continuous. The design was code-named Twenty-One Castle Mike to denote its power parameters; to wit, that this Reference Design shall specify a maximum power output of 21 Tw continuous (a Case Model), possibly slightly less depending on the design chosen. Due to considerable limitations in the current understanding of fusion power, we’ve opted to employ a novel design based on nuclear weapons research. Specifically, this technology involves the invention of several new composite technologies and is referred to overall as Thin Film Electric-Magnetic Tokamak fusion, or TFET, fusion (also Thin Film Electromagnetic Tokamak fusion, or TFET, fusion). Patents on this technology are now pending.

Chapter 1

Toward defining a goal

The justification in developing this technology was to solve problems identified with transit propulsion between Mars and Earth. The Reference Design was created both for referential purposes in our discussion as well as the nearest known design (along with a fission proposal) intended for actual use in this project. The justification for taking this route must begin with some discussion of the problem of Earth-Mars transit.

What we found was that the mass fraction (that is, the fraction of cargo mass to remaining vehicle mass) of chemical propulsion schemes is workable but does not satisfy the criteria under which this study was commissioned; namely, to develop a project plan with a  high probability of success. The reasons for this are interdependent. Using chemical propulsion, an earth weight of about 10000 mT (the approximate total mission weight we envision in the first step of this project) will require an ungodly amount of LH2 and LOx. These massive fuel loads will increase launch to LEO costs. But more importantly this mass requirement will necessitate a more modest Martian Trans-Insertion (MTI) burn which in turn will yield a longer transit. While not problematic by itself, the combination of this fact with the unknowns surrounding human weightlessness conspire to create serious concern regarding meeting the requirement for a high probability of success using chemical propulsion. A solution to this problem is to perform two separate MTIs under chemical burn. The first is the cargo burn, which is “heavy” and slow, taking a year or more to reach Mars. The second, also laden with fuel, is a personnel burn in which the crew and their life support, and little else, is transited at much higher velocity, arriving at Mars in less than 4 months. We recommend this workaround as a backup if other methods prove intractable. In that vein, we recommend a research and development fund for the alternative propulsion scheme proposed supra with a contingency fund attached to finance the heavier launch and material cost required by a chemical solution, should the former fail to produce.

Space Medicine. The current state of the art in medical knowledge of human weightlessness, and what to do about its effects, is marginal at best. When human beings are removed from the effective proximity of the gravitational field of Earth several physiological abnormalities begin to occur in a cascading phenomenon of biological chaos. First, the cardiovascular system (and we point out that blood circulation occurs not only as a result of the heart’s pumping action but also by assistance of the blood vessels themselves – a systemic phenomenon) is phylogenetically adapted to operate in an accelerated reference frame of about 9.8 m / s². Therefore, when exposed to a much lighter acceleration the cardiovascular system responds systemically by overpumping to those areas of the body which, under normal orientation in the Earth’s gravitational field, experience negative ambient profusion. It does this because it has not adapted to circumstances where this field is significantly altered or variable. It’s a “dumb” pump that pushes blood more forcefully to the head than the feet because it “assumes” gravity is always there. This causes a cascade of problems in spaceflight. First, symptoms generated from too much fluid in the head cause medical problems by themselves. But, in response to this “over profusion” of the head, the cardiovascular system “down regulates” blood volume in an attempt to reduce the amount of fluids in the head. Now, the body lacks sufficient blood volume because the body is confused. We identified this particular physiological problem as the most significant, not solely because of its immediate medical effects, but because of what it means for any attempts to address weightlessness problems generally; in particular, the use of centrifugal force. In the absence of an appreciable gravitational field, a spacecraft component can be rotated such that, if someone locates inside the component at or near the arc swept out by the rotating component, an acceleration is present which produces “artificial gravity”. From a physics perspective such an acceleration is perfectly safe and normal as the body has no way to distinguish between different types of accelerations if all reference cues are removed. But that is a key qualifier. This holds only if all cues are removed. Let us be clear; the physics is more fundamental to nature than the medicine, so it is not possible for such a force to create any problems the same force created on the surface of the Earth by gravity would. But again, this only holds in a pure sense. If the radius of rotation is too short, the human being is aware that they are spinning, even if they can’t “see” any stationary reference by which to make that judgment, because of what we shall refer to here as “circular differential acceleration” (in physics this could be properly characterized by tensor mathematics). But to be precise, we should probably say, a “circular differential acceleration” that is physiologically significant. This is because this differential acceleration also exists at the surface of the Earth. However, the difference in acceleration measured at one’s feet versus at their head, assuming they are standing at sea level, is too small to be of significance or to have any physiological effect. But this is not true under rotations with relatively short radii. These rotational contraptions, called centrifuges, have radii so short that the difference in acceleration at the feet and the head can be felt by the person subjected to it. The only way to overcome this is to make the radius larger. At some unknown radius, the individual can no longer feel this differential force, or “artificial weight”. And the state of the art provides no answer regarding the radius question. But perhaps more troublesome, we don’t know at what radius we must construct a centrifuge such that there is no measurable physiological impact. We are completely in the dark on this crucial question. This concerns these authors because of the distinct possibility that a centrifuge could do more medical harm than good. If, for example, the body is confused upon losing gravity, imagine the potential for confusion if the body is subjected to differential forces acting on the blood stream! While medical experts should be consulted to confirm these findings, we do not recommend any further research into that question by itself as it will not be sufficiently productive if performed on Earth. Such possibilities would only be imaginable under prolonged exposure, but that is what we face with an interplanetary voyage. We feel this is dangerous, guinea pig territory and that relying on centrifuges to solve the weightlessness problem is not a winning strategy. We recommend securing the advice and consultation of a qualified medical expert who can speak to the uncertainty of centrifugal forces acting on the human body (but not necessarily on a solution or how to create a centrifuge). Answering that simple question will help. We recommend experts chosen from the field of space medicine in particular, as well as vascular surgeons and cardiovascular experts.

We recommend that a transit vehicle be equipped with a small centrifuge that the crew can use only when lying down, to reduce the differential forces acting on the cardiovascular system, and that this centrifuge be used on demand so that the crew can stop using it if problems are encountered. Using a centrifuge while the crew sleeps may provide enough gravitational simulation to eliminate other problems such as bone and muscle atrophy. Early data suggests that just a few hours a day at 1G is sufficient to drastically reduce the effects of weightless vis-à-vis atrophy. However, given the paucity of data, the overall time spent in this environment must be taken seriously if a high probability of success is desirable. The reason for this is that human beings subjected to “weightless” environments for extended periods of time have been documented to suffer permanent damage and require considerable recuperation time at 1G to regain full functionality. For a crew whose destiny is a planet with only 40% of the gravitational force of Earth, this is a serious problem as they have no opportunity to fully recover. Given that a crew arriving at Mars must be physically functional, the transit time becomes crucial to the probability of mission success. Spaceflight experience indicates that most space travelers will suffer minimal incapacitation if the time spent in microgravity is 1 month or less. We identify this time duration as a critical threshold. Beyond this limit human beings will require greater recuperation than Mars can provide. A limited centrifuge as suggested allows for flexibility in its use on the basis of sound recommendations from professionals in the medical field. In any case, we suspect that persons living and working on Mars in the weaker gravitational field of Mars – and what can be done to remedy all the adverse medical effects -  will be the object of lengthy studies and bedevil medical science for some time; long after our proposed mission succeeded.

The Medical and Propulsion Technology Connection. Having noted the issues identified as critical with respect to microgravity, we next note its interrelationship to the propulsion mode chosen. Using LOx and LH2 the transit duration will be at least one year. We believe this is unacceptable as we have insufficient information regarding the physiological impacts of centrifuges and believe that 1 month represents the limit of transit time affordable for a mission with a high probability of success. A more potent burn mode is required. But before examining what we learned regarding propulsion options, we must also elaborate on another hazard of long duration spaceflight (here, by long duration we mean a cruise exceeding 1 week outside celestial orbit). Galactic Cosmic Radiation (GCR) is the mother of all radiation and is extremely deadly. It exists freely in and throughout space. At Earth, and at LEO, we are protected from GCR by the Earth’s magnetosphere which deflects the radiation. However, once we pilot outside the magnetosphere we are barraged with GCR. Apollo astronauts noticed that, as they left the magnetosphere’s protection en route to the moon, and when they closed their eyes, they could see mysterious flashes of light. This was GCR. If your exposure is so great you can see it you can bet it is doing biological damage. Though this issue is touted by many in popular “lets go to mars” forums and circles as being a showstopper, it is actually one of the easiest problems to resolve. Regardless of the propulsion choice made, LH2 will likely be a passenger. We’ve sketched a transit vehicle (which, by all counts, is a true “ship”) which is cleverly designed to place the crew inside a protective sleeve, capped on both ends, with LH2 and LOx (as has been proposed by others). These two constituents will, at a thickness of about 1 meter, deflect all GCR to background levels. Though LH2 is used for the transit burns, we envision a project plan in which a small residual amount of LH2 is retained so that, in an emergency, the crew could return. We will explain this “assured return mode” later, but for now, we point out that not all LH2 will be burned at the beginning of the journey and that with proper ballasting, a 1 meter layer covering all directions can be ensured for the crew.

The choice of transit propulsion represents the greatest risk factor for this project.

Chemical propulsion schemes, such as the chemical reaction of LOx with LH2, represents a “known quantity”, a technology for which experience is ample and confidence can, with adequate funding and care, be ensured. However, the conspiring issues of microgravity and transit times for chemical propulsion force us to consider more exotic alternatives for propulsion. We need a propulsion scheme that will reduce the transit time to one month or less. Obviously, we would prefer to choose the emergent technology most developed as the alternative. Indeed, we shall find that to confidently develop the propulsion system we will require, 5 billion USD must be allocated for that purpose alone, the single greatest cost of the entire project. Unfortunately, there is an overdose of science fiction and fantasy thinking in too many areas of discussions about Mars missions and Astronautics generally. Propulsion is no exception and there is much popular confusion about:

1.)   the maturity of alternative propulsion methods

2.)   the economic, or comparative, quality of alternative propulsion methods

Considerable advances have been made, especially in the last 10 years. This is remarkable considering the paucity of public or private funding in such research. Nevertheless, considerable progress has been particularly evident in plasma based fusion propulsion. The free markets have awakened to the emerging impact of fusion power and it has caused the tree of research to diverge considerably. Having said this, all things are relative. The current state of the art is “light years” away from producing a viable fusion power production plant, whether kinetic or electric. We will begin to understand the depth of this problem as we proceed through this document. Various strategies for fusion propulsion have ballooned and it can be a daunting task to sort out which technologies represent the best fit for pursuit in this project. Therefore, we will approach the problem based on the business requirements of this mission, then identify the technology or collection of technologies best suited to this project which we feel can be developed near term for less than 5 billion USD. To do this, we’ll start with the fundamentals of physics. And when we do that we will see that our realistic options diverge in two distinct directions; either the far simpler and more attainable “closed-system gas-core fission reactor” or the more challenging fusion approach. We will start by assuming that the fusion approach is taken.

First, we need a digression to restore the discussion from Fantasy Island and work with numbers and quantities that reflect the real physics involved in interplanetary spaceflight. To do this, we need to deconstruct the notion of “specific impulse”, then discard it for a more useful term in its stead. Specific impulse is simply a ratio that describes the propulsive force generated from a given mass of propellant. This sounds very much like the number we need to focus on in order to identify what type of propulsion system is best suited for a real Mars voyage. The problem however, is that it is misleading. By conservation of momentum, we can characterize rocket thrust in vacuum as an equation relating the product of the overall vehicle mass and velocity to the product of the mass and velocity of the propellant exhaust. Conservation of momentum assures us that these two products will always be equivalent. The shortcoming in using specific impulse as a discussion tool is that it assumes exhaust velocity with no accounting for the mass required to generate it. And the mass required is a function of the design employed. If we increase the exhaust velocity we indeed increase the amount of energy we can extract from the exhaust as a propulsive force over a distance. However, what specific impulse doesn’t tell us is how much and in what manner was mass required to generate that velocity in the first place. Since our spacecraft is a semi-closed (to be technically accurate, but consider it closed for all practical purposes), self-contained system all propulsive energy must come from the ship itself and, if chosen poorly, the propulsion system may add an enormous mass liability in order to generate the assumed exhaust velocities. This is precisely the problem with electric ion propulsion. Therefore, we need to algebraically characterize what we are trying to pin down.

The term generally used to refer to the quantity we seek is called “specific power”. But even this figure must be calculated carefully in order to generate the pragmatic figure we are seeking. More digression is needed.  We’ll start by calculating the propellant mass required.

To calculate propellant mass required we need to take a few steps. Let m_p be the total mass of the propellant at the beginning of the TMI burn. Let m_i be the total mass of the loaded ship at the same time; that is, at the beginning of TMI burn. Now, let Δv be the transit velocity assuming a relative starting velocity of 0 and let v_e be the velocity of the propellant as it exits the nozzle (and relative to the nozzle). Finally, we’ll denote the final mass at transit burn completion as m_f. As it turns out, when we integrate over the transit to determine the fuel and velocity relations we end up with a term like 1 / x; thus resulting in a convenient mathematical relation given by a logarithmic function. Therefore, the definite integral is equivalent to a term involving the exponential function, e, which makes the math convenient. We motivate this discussion by pointing out that a simple calculation of the total mission mass without accounting for power plant weight yields a minimum power output of about 500 Gw, a bridge too far for fission power. Of course, we can rework the transit to include a transit burn for the entire first half of the journey, but we still end up with a power requirement in excess of 100 Gw. Therefore, it can be clear from this why we are exploring propulsion options beyond the most readily available (though one variation of fission propulsion might work – see the second alternative. It has the potential to deliver power levels as high as 2 Tw if adequately developed).

For our purposes, we will frame the discussion by specifying the hard requirements for a Earth to Mars transit:

• It cannot depend on Hohman transfers, or other infrequent events to make a transit in either direction. The ship must have the power to complete the transit regardless of the relative orientation of celestial bodies.
• Its fuel load should be limited to 60000 kg mass. We shall see in what follows that a generous mass is readily attainable; the challenge lies more in how many, if any, crew restarts of the reactor will be desirable during the mission and how many rocket launches would be required to provide the fuel. Based on the Reference Design for this projects rocket, and attempting to keep the number of launches feasible, we set the product of 60000kg mass and ten; that is, 600,000 kg to match the upper bound for one such rocket of mass lifted (at Earth gravity) to a high polar orbit. See the Reference Design mentioned for details on the 41 GLASS Pack which we reference [One of the concerns given me when asked to do this study was to ensure that the environmental impact at Mars be given special consideration above and beyond that typically given by the petorleum industry. I was encouraged to hear this and it is one of the reasons why I have recommended operating high power devices exclusively from high polar orbits, regardless of celestial body].
• The transit time would ideally last about 5 – 10 days but may last as long as 30 days. The less exposure to microgravity the better. In addition, to establish a practical cargo infrastructure relatively short transit times will be necessary.
• Fuel consumption on any one Mars transit should require no more than 1/2 total fuel supply onboard. This means that roughly 30,000 kg mass of fuel is available for Mars transit burns. We shall provide some margin and set the fuel burn at 22280 kg mass round trip.

What we shall see shortly is that the problem of rocket propulsion has an elegant way of reducing to what our common sense might have already suggested to us with not a little clarity already: Given the momentum relation shown it is evident that overall efficiency of a propulsion plant will hinge on two things; how much mass “you push away from you” and “how fast you push it”. Since propellant is basically throwaway and serves no useful mission purpose after the burn we have a strong incentive to reduce the overall propellant mass we have to carry with us for the burn, which in turn increases the amount of energy needed for the transit. This leaves us with only one variable to manipulate (for improved efficiency); exhaust velocity, v_e. So, it reasons that we are seeking the highest possible v_e we can attain with near term technology.

The key to understanding the practical boundary conditions for interplanetary spaceflight is in finding a means within a semi-closed system to produce the greatest value of v_e with the smallest total propulsion system mass. It has nothing to do with specific impulse.

If a selected system can “bleed off” electrical current in a small proportion of thrust to provide ship-wide electrical power, this is acceptable and preferable to having to use an onboard power plant for ship services. We shall see later that the thermal energy available for a Brayton cycle or other similar heat rejection and energy conversion scheme will be enormous during the transit burn; but only during the transit burn. But for now we circle back to our previous comments regarding specific power. We shall need to revisit our assumptions if the propulsive thrust is provided by any means other than a direct, kinetic thrust. Such a scheme requires a conversion of energy which will invariably (and most likely) add an enormous mass penalty. Stated more fundamentally, the energy needed to generate v_e should come predominately from the propellant itself, not an outside source. Thus, such schemes should be avoided if possible because:

1.)   They add unnecessary mass to the transit and are wasteful forms of “double-dipping” fuels.

2.)   They add unnecessary and risk enhancing complexity to the overall system.

We can now state the following technical requirements:

Δv / ve   ≤ 1 / 3300	[eq. 1.07]
Technical   Requirement 1

α ≥    1 Mw / kg	[eq. 1.08]
Technical   Requirement 2

For the purposes presenting, nothing else matters.

The reader may be wondering at this point why we are considering such high mission masses. In what shall be shown infra a one-way, self-sufficient mission will require a staggering scale of hardware to ensure a high probability of success and a six-person crew won’t cut it (the number usually offered in Mars proposals). We defer that conversation until we complete the propulsion analysis. Also, one can reference the other Reference Designs in this project plan for more information on mission mass and a breakdown of cargo.

In seeking an appropriate power plant and ship design we must therefore identify:

1.)   A rocket power scheme that is either flight certified or in near term development which can be produced in a set of three, independent ship power plant systems all for less than 5 billion USD (one propulsion plant will serve as a spare).

2.)   A specific power and exhaust velocity sufficient to produce a mass fraction of approximately 40 to 60% and a Earth to Mars transit duration of 1 month or less (no Hohlmann transfer tricks), conveying a mission mass of 10000 mT Earth weight on each of two ships. A third power plant system will serve as backup.

3.)   If nuclear, a plant that obviates the need for heavy shielding and thermal conversion (or improves the efficiency of conversion to near 100%).

4.)   If nuclear, a plant that allows for a highly efficient conversion of excess propulsive energy for ship power – both for re-circulating energy as needed for propulsion and for shipboard services – in transit.

Our own research shows that only a narrow band of options fits the criteria thus far elucidated. With only one exception, the physics relations will necessitate a propulsion system theoretical power density ≥ to that given by fusion or matter annihilation sources (the latter is the presently known natural limit to specific power) which excludes electric ion or similar “double-dip” schemes and most of fission thermal rocketry. Penn traps used for containing antimatter are several orders of magnitude too massive for use as fuel tankage; requiring a mass-reducing development cost estimated at near 1 trillion USD.  Furthermore, no “ore” source and existing infrastructure to obtain antimatter naturally exists. Given that E=mc², creating the fuel artificially is not a winning strategy.

Therefore, we have identified fusion kinetic thrust and Closed System Gas-Core Fission rockets with sharply limited shielding requirements as the best value proposition. Both are further identified as emitting charged particles as, at least, a partial byproduct and as using a fuel whose consumer is economically cheaper to technologically develop than the cost of accessing any given alternative fuel.

 

1. Thin Film Electric-Magnetic Tokomak Fusion (TFET)

The distance between Earth and Mars, for the purposes of illustration, can be estimated at 3.1 x 10⁹ meters (about 2 A.U.). We set Δv = 10000 m / s, the upper limit velocity before ablative shielding will be required (see later discussion).

3600 * 24 = 86400 seconds are in a day

to satisfy the fuel mass limitation, we shall keep the fuel expense well below the allocated 15000 kg with a value of 11140 kg mass one way (2 burns) and 22280 kg round trip (4 burns). Each burn is 5570 kg. The exact amount of deceleration will be mission dependent, but we allow for a maximum deceleration in any case.

ð  Δv = (3.1 * 10⁹) / 432000 ≈ 10000 m / s

ð  10000 * 432000 =  4.32 * 10⁹ meters transit path length

The well-known Tsiolkovsky rocket equation is:

mf   / mi = e -( Δv / ve)	[eq. 1.01]
Tsiolkovsky   Rocket Equation

ð  m_f  = m_i  * e ^{-( Δv / ve)}

ð  m_f = (2500000) * e ^{-( 10000 / 0.015c)}

ð  m_i =2505570 kg

ð  2505570 – 2500000 ≈ 5570 kg mass fuel.

We assume a total package mass of 2500000 kg. For mission purposes we’ve set the total allowed cargo and hull mass at 1000000 kg (10000 mT Earth weight) and the propulsion system at 1500000 kg (15000 mT Earth weight). then we can set our values, after some re-arranging, as:

Δv = ve [ ln (mi / mf) ]	[eq. 1.02]
Mars transit   baseline minimum delta v required
Reference Design   => (0.015c) * (ln (2505570 / 2510000))≈ 10000 m / s

and

α =    Tk / (mi   – fm – cm – hm)	[eq. 1.03]
Mars transit   baseline minimum specific power required
Reference Design=> ((7*10^12)/(1.5 * 106)) ≈ 6 Mw / kg

and

Tk =  α(mi   – fm – cm – hm)	[eq. 1.04]
Mars transit   baseline minimum kinetic thrust required
Reference Design => (3.5   * 106)( 1500000) ≈ 5.25 Tw
Usable maximum power rating 7 Tw

where

f_m is the fuel mass,  p_i is the propulsion system tare mass, m_i is the total mass of the entire assembly at mission start, c_m  is the total cargo mass, h_m is the remaining assembly weight (superstructure – heat rejection panels are part of propulsion system mass) and T_k is the total kinetic thrust power. The three equations supra are the bunny trail that leads one inexorably to fusion kinetic propulsion for spaceflight. In the case supra we have not considered the burn time and we will address that directly.

We know that the lower bound for fast ⁴He from the reaction we are considering can be conservatively averaged and estimated as 0.015c.

We shall find infra that the fuel consumption of this Reference Design is about 0.087 kg / s. To determine the burn time for Mars transit we have:

T =    fm / c	[eq. 1.05]
MTI burn   duration, 7 Tw kinetic
Reference Design => 5570 / 0.087 ≈ 64022 seconds ≈ 18 hours

Where T is the burn time in seconds, f_m is the mass of fuel burned for the transit and c is the fuel consumption rate at maximum rated power. Therefore, the acceleration imparted to the hull/truss assembly is 10000 / 57471 ~ 0.175 m / s / s for a period not less than 18 hours. This results in two transit burns per transit, one for acceleration and one for deceleration, resulting in an overall Earth-Mars transit time of just under 7 days. The exact elapsed time will depend on relative celestial orientations at the time of transit.

EM   = Tk * dEM /   vf	[eq. 1.06]
Earth to Mars   single transit burn energy expenditure
Reference Design => (5.25   * 1012) * (4.32 * 109) / (10000) ≈ 2.27 * 1018   Joules
Cross check: ½ mv2 = ½ (2.5 * 107)(10000)2  = 1.25 * 1015 Joules

Where E_M is the Earth to Mars single transit burn energy expenditure, T_k, is the kinetic thrust power, d_EM, is the Earth to Mars distance and v_f is the final transit velocity. A final note to make is that we have not considered the energy required to transition from a high polar orbit (which we specify infra) to escape velocity. This is why we used generous fuel load figures. We can also see that our energy requirements are well within the required margin.

All of these calculations are order of magnitude estimates and do not take into account what happens if specific impulse, for example, is varied during the burn. However, margins of error of appropriate magnitude have been used to compensate for these variables.

The oft-recommended fusion fuel is ³He. However, this fuel is so rare on Earth that only the moon serves as a viable source. But extracting such a fuel from the moon considerably exceeds the increased costs of simply developing a propulsion scheme using a different, more technologically challenging fuel. This is the same problem that, in conjunction with other factors already stated, disqualifies antimatter schemes from consideration. The fusion fuel in mind is ²H-¹¹B which meets the requirement that little or no shielding is needed (high kw/kg -  we shall see that at the fantastic power levels we are considering considerable shielding will still be required). It requires no heat conversion cycle (we shall see later that a heat rejection system will be needed) and electrical power can be readily tapped from the electromagnetic deceleration of charged particles. With regard to developmental readiness, the playing field is already so scattered that an amalgam of different schemes will be necessary to leverage the component technologies closest to developmental readiness. Rather than digressing into a lengthy discussion of why other technologies were excluded, we will simply point out that application of the above equations will suffice to explain our choice. What we have quickly learned is that this propulsion issue has dropped us deeply into the fusion power rabbit hole, with all its complex and drawn out research and development angles which must be examined.

The Challenges of Fusion Power. Therefore, we have approached the problem from first principles. Fusion power requires an enormous amount of input power to initiate, making it fundamentally different than just about any power scheme known (including fission). It is this drawback that has retarded development of fusion power for some fifty years now with no end in sight. What one learns as they delve into the subject matter is that fusion power is, fundamentally, a power conversion problem. The power levels, physical properties and other conditions required to initiate and sustain fusion and recover power, are so high that state of the art in materials science is still unable to effect it. And we should note that:

When we talk of materials science limitations, we speak broadly to include both the materials science issues of schemes which involve the interaction of external masses with plasma as well as the materials required to transfer power generally. So, by removing direct physical contact in power transfer schemes we do not magically eliminate this problem. The problem is more fundamental than that. It is the magnitude of difference in energy between two physical systems, and the transferal of same back and forth, that taxes materials science far beyond its current capacity. Practically speaking, we lack the materials science necessary to convert (to electrical power) and contain (the fusion itself) the enormous power such a reactor, by design requirement, would produce. And even fuels that produce product that can be electromagnetically harnessed do not help as the materials science problem is more fundamental than that; it is a physical material that initially absorbs the “force” of the moving particles. However, and fortunately, kinetic propulsion does not necessarily face the latter issue.

Scientists astutely realized over 50 years ago the difficult proposition that fusion presents. In order to ensure continued support and funding for a technological development with such a long development time one would be best advised to speak publicly only of the accomplishments and do their best to minimize awareness of the hurdles. This is exactly what has happened. Examples abound, but the use of figures such as fusion Q, triple products and terms such as “ignition” are laughable. Equally worthless are accomplishments of the enormous power achieved in fusion reactions in which the reaction occurred on the order of peta seconds. We qualify these statements by emphasizing that these terms have pure science merit, but that they are not figures of merit for our purposes. Fusion Q is a quantity that represents the amount of power required to sustain a fusion reaction and nothing else. It is a relatively fundamental value (dependent on the reactor design, of course) and doesn’t take other power losses into account. The triple product is simply a product of three quantities whose product has a known lower limit to allow fusion to occur (regardless of fusion “density”). Labs around the world are just now reaching this magic triple product with enough net power production to run a large lawnmower; provided they smother it first with gobs of electrical power. This will not get you to Mars and it won’t make economic sense as a substitute for fission. And the term “ignition” helps but still doesn’t get us there. Ignition (perhaps outside the strict definition used in the Magnetic-confinement only cheerleading squad) relates by proportion the power fusion plasma loses to the environment with respect to the power the fusion reactions produce. But the real figure of merit, aside from the obvious figure of power density, is the total power input required with respect to the total output power that can be recovered. This is a broad term. First, we must know how much power we lose just trying to kick off the fusion reaction (the Q ratio gives us that). Then, we must know how much power we lose to power transfers from the plasma to the environment (an efficiency loss). Next, we must know what percentage of remaining power (once we subtract these first two losses from the total fusion power output of the reaction) we can actually technologically recover (fusion output can come in multiple energy forms). Finally, we must take account of the efficiencies of each recovery method used.  This process alone can get complicated as it partly depends on the fuel being burned. In any case, the proposed ITER reactor in southern France, the most ambitious long shot for fusion, will not come anywhere near meeting these requirements. Notwithstanding public relations propaganda, it is a pure science project with no practical application.

So, the natural question is, what is/are the stumbling blocks exactly? After fifty years surely success must be imminent? The answer to this is likewise complex, but we’ll try to summarize it here in the context of our suggested solution. Going back to the fundamentals of power conversion, the problem is that we don’t have a single-mode method of adequate material science sophistication to reach the so-called triple product of fusion (a necessary fusion condition of plasma) in a true thermonuclear regime. Indeed, such an advance is likely over 5 decades away, at best. However, the reason why “single-mode” only schemes seem to be favored is deeply political and cultural and we won’t spend too much time on that discussion. But also scientists, as already stated, are interested in pure research and right now tokamaks make the most sense for doing that kind of research. Needless to say, there has been a fail-setting tendency to focus on single-mode power transfer (wittingly or not) to solve the fusion challenge and it simply won’t work with foreseeable advances in materials science. We know this as a fundamental fact of nature. The key transfer issue with fusion is something called “confinement”, which refers to the confinement of fusion plasma so that a barrier between the material world and the high thermonuclear regimes of fusion can be maintained. Inexorably linked to confinement is the actual problem of converting power from outside this environment to inside it and vice versa – the power transfer problem. So, power transfer with fusion plasma is intimately linked to containment and vice versa. If you can adequately contain plasma, you can probably create the triple product regime necessary for fusion. But, the single-mode mentality has not only been resistant to use of multiple confinement methods in one scheme but it has also led to most scientists refusing to realize the opposite corollary: if you can create the conditions of a triple product, can you likely confine it?[1]

Chapter 2

Current Research in Fusion power

The Current State of the Art

One of the important indices for evaluating the performance of fusion plasmas is a product of the plasma density, n, the energy confinement time, τ, and the plasma temperature, T, that is called the “Fusion Triple Product”, nτT. As of June, 2008, high n_i0τ_ET_i0 of  ~1 * 10²¹ m^-3 * s* keV, where n_i0, τ_E and T_i0 are the central ion density, the energy confinement time, and the central ion temperature, respectively, has been achieved in tokamaks, such as JT-60U, JET, TFTR, and DIII-D. Though increases in density almost always carry with them increases in temperature, it is not so intuitively clear that for fusion to occur the greatest increase in density possible with the smallest corresponding increase in temperature yields the highest values for nτT. In typical setups, the density is increased and the total heating power is reduced while density is maintained. This results in the highest values of nτT and is known as the annealing process (as a metallurgical analogue). Required triple product quantities for sustained fusion vary depending on reactor design, but the following values provide ballpark figures:

Deuterium-Tritium 1.24×10²⁴ m^-3s

Deuterium-Deuterium 1.28×10²⁶ m^-3s

D-³He 2.24×10²⁶ m^-3s

²H-¹¹B it is about 3.01×10²⁷ m^-3s

Above is the LANL MTF design. The green cylindrical layer corresponds to an Al sleeve that is symmetrically crushed around the plasma. We recommend replacing this Al sleeve with a confined chamber sleeve that allows the rapid introduction of a highly ionized gas or liquid. The design shown here is another example of the losing proposition of using power transfer schemes that involve contact between plasma and matter.

Once the “dark horse” of fusion development, MTF fusion schemes have been researched at length at Los Alamos National Laboratories. In this scheme both magnetic and inertial plasma confinement strategies are employed. By splitting the confinement between two methods the high cost, scale and mass of magnetic only confinement is avoided. But, as is often the case fusion research that once appeared promising eventually evidences signs of insurmountable and unexpected difficulties as research scales upward. In the case of MTF the problem relates to the inertial confinement method which relies on physical contact between an Al sleeve and the plasma being contained. In order to compress the plasma to thermonuclear regimes the sleeve is rapidly and with narrow symmetrical tolerance collapsed around the plasma inside it. Of course, contact with Al introduces thermal instabilities that spoil the reaction. What is needed is a way to boost the confinement into the thermonuclear regime beyond what can be easily done with magnets and in a manner that doesn’t contaminate the plasma with solid matter.

One variant of inertial confinement schemes is to use a Reactor Pumped Laser (RPL) to generate the temperatures needed. This is a tantalizing idea since it relies on the potential of a fission reaction running at ultra high power. In this sense, it confirms our suspicions that a direct power conversion from a properly shaped critical mass into plasma will allow major rating increases in power transfer for fissile material in the scheme we propose. By taking away most of the thermal heating energy created in fissile reactions, reactor mass is controllable and can be kept within the temperature regimes allowed by the materials (especially the Americium itself). What it neglects to address however involves the same issue we’ve discussed supra: delivering this much power through a noble gas (and even into plasma) has been demonstrated but delivering it through laser equipment and optics has not. Indeed, by using lasers the same materials science limitations to power conversion arise. The most salient weak link in the chain appears to be in the optics used to focus the lasers. Power ratings are extolled in the highest ranges but run only for peta seconds before the optics require a cool down period of several hours. We will propose that the only viable scheme is to convert power directly from the nuclear reaction to the plasma itself; albeit via electrostatic transfer of energy from fissile product to fusion fuel. Doing so can provide, as needed, up to 100 million K to the plasma, a temperature far exceeding even the higher temperature needed for ²H-¹¹B fusion.

Helion Energy and General Fusion are, at least at a fundamental level, making attempts at creating fusion power. In the Helion case, an inertial approach is used by electromagnetically accelerating two plasmas toward each other into a collision along a linear accelerator path. General Fusion uses the most similar approach but suffers from the issue of physical contact between the plasma and matter. We suspect it will fail. Tri-Alpha Energy, a “stealth-mode” company (like us), has revealed very little but is apparently attempting some kind of MTF scheme. None of these combine the three common confinement modes as we’ve suggested (in a manner that eliminates all known experimental limitations).

The Confinement Problem: a Power Conversion interdependency

Current fusion research has had the opportunity to expose some fundamental voids of understanding of nature that represent extreme project risk. One of these is known as magnetic reconnection, which we suspect may be due to a lack of understanding of nature at a deep level. While that may not be the case, it is of sufficient concern to warrant discussion. Magnetic reconnection is the result of multiple magnetic field lines “bleeding” off into other nearby field lines when magnetic field strengths are very strong and confined. Fusion research has encountered, at least in some cases, stumbling blocks due to the fact that this magnetic reconnection phenomenon occurrs several orders of magnitude faster than the current state of the art predicts. This is a red flag. There are many reasons for saying this, but one very important one is the academic denial currently swirling around traditional Tokamak research (magnetic-only confinement). Tokamak research is generating billions in research grants and construction projects and the dirty little secret made so to keep the funds flowing is that they won’t work and researchers have no clue why. The reason is that this phenomenon known as “magnetic reconnection” occurs in magnetically confined plasmas (of moderate energy levels and beyond):

1.)   With limited or no predictability [the second derivative of the reconnection changes without any understood cause]

2.)   Faster than MagnetoHydroDynamic Theory (MHD) algebraically posits (ominous)

3.)   In such a manner as to foul any attempts at confining a plasma for a useful length of time.

4.)   In a space so small that experimental observation of the event is virtually impossible (which suggests serious issues if one tries to resolve this issue by gaining more insight into it).

5.)   Is likely exponentially proportionate to energy output scaling (chasing your tail).

Whether or not this stands as an impediment to fusion research generally depends on the system design and approach and can be assessed using what is called a “Lundquist number”. High Lundquist numbers indicate high conducting plasmas while low Lundquist numbers represent low conducting plasmas. Due to the problem of magnetic reconnection, which is only solved by the current state of the art by using low conducting plasmas or – to state an equivalent – an option less dependent on magnets in the first place, we recommend a reactor development project which anticipates the lowest plasma conductivities or much more limited use of magnetic field lines. What we see in the Lundquist number is a critical issue in which, at some critical value (dependent upon the scheme being used) magnetic reconnection begins to cause losses of plasma energy that occur faster and faster as the Lundquist number rises. This “critical” Lundquist number threshold can vary apparently from 10³ up to 10⁷ (the wide range of values is due to the fact that different schemes – or reactor designs – will result in different critical values).

In the journal Physics of Plasmas A. Bhattacharjee, Yi-Min Huang, H. Yang, and B. Rogers wrote in “Fast reconnection in high-Lundquist-number plasmas due to the plasmoid Instability” that the real October surprise in magnetic-only confinement is that as the plasma confinement scales upward the critical Lundquist number drops. This is a critical point because Tokamaks are driven by Magnetic Induction in which a current is induced in the plasma by strong external magnets. If the conductivity drops, the driving of the entire free body diagram slows. Indeed, the expensive ITER experiment may fail to produce the much touted results of “break even” in which the output energy exceeds the input energy because of this. According to D. Najouks in “Plasma-Material Interaction in Controlled Fusion”, if energy diffusion increases in the ITER experiment (which we know it will because the energy also scales upward in ITER), a fusion burn will not be achieved. Attempts to understand what is going on with magnetic reconnection are frustrated by the fact that in Tokamaks the spatial region in which it occurs is too small to be measurable by any conceivable measuring device. The significance of this cannot be overstated. It means that even trying to understand magnetic reconnection better is likely to require years of research and development. For these reasons we regard all magnetic-only confinement schemes to fall in the pure science realm for the foreseeable future and that they currently have little or no immediately usable value for our project. This conclusion will gain substantial support as the discussion progresses.

In magnetic-only confinement schemes a high value for the ratio of the plasma density to the magnetic field strength (called beta) is desirable. However, it is precisely in these regimes where magnetic confinement instabilities become severe. To summarize the salient points:

1. The conductivity of plasma drops as density increases; but this is the driving force of the entire reactor
2. Magnets lose their capacity to dampen inertial movements when density increases.
3. Much higher plasma densities are absolutely necessary for a practical power density regardless of the fuel used (a point to be proven in what follows)

For a preview of where this is going, the reader might care to think about what might happen if an additional force, much stronger than magnetism, were to complement these magnetic field lines. We will return to this point shortly.

The three most common instabilities in traditional Tokamak designs are known as ideal kink modes, resistive wall modes, and neoclassical tearing modes. This makes sense when one considers what beta means: if plasma density is considerably higher, relatively speaking, than the magnetic field strength, the fields usefulness in controlling the orbits of particles in a torus lessens since the higher density serves to accentuate small perturbations and instabilities such that the magnets cannot dampen them back out. The magnetic field is overwhelmed by the inertia of the higher density plasma motion. Electrostatic assistance (at about 10⁶ bar) would dampen these instabilities out handily. Large density gradients become very hard to control in a magnetic field of limited strength over the same region of space (the gradients of field strength and density do not match in such cases). When the density of plasma varies considerably as one moves outward from the central torus orbit (called the bootstrap value) instabilities occur if a so-called conducting wall is not present. In other words, the plasma literally needs to conduct current through to the external environment (not good for confinement). Again, this arises because variable density through the torus “beam” of particle flow translates, in neutral two-species plasma, to a higher density current which, when stronger at the edges, begs for a nearby conductor. This issue does not present with positively charged, electrostatically stabilized single species plasma. Finally, a so-called “Neo-Classical Tearing Mode” instability occurs when a plasma density which varies as position changes outward from the central orbit has a region in which this variance is non-monotonic. Again, 10⁶ bar of electrostatic force will readily eliminate this (most magnetic-only confinement schemes operate at just a few atmospheres). Operating at six orders of magnitude greater density while keeping the same field strength will cause beta to skyrocket, precisely what is desirable to dampen instabilities. Beta values are a telling clue that supports our contention that the magnetic field is itself causing these instabilities due to the fact that the field is not applied symmetrically throughout the plasma and because Magnetic fields are insufficient, by themselves, to attain stability in fusion reaction regimes. This is because the magnetic field strength on the outer orbits is necessarily weaker than the magnetic field on the inner orbits (due to the intrinsic geometric properties of a torus which bunch the magnets closer together on the inside of the torus).

Thus, increasing magnetic field strength won’t eliminate these problems, but increasing density by the application of a non-magnetic force will. This is so obvious it is frustrating to observe the endless attempts to eliminate instabilities using the very mechanism that is creating it and the apparent unwillingness to acknowledge this simple no-win strategy. Frustrating examples of dancing around this subject without addressing it are numerous, but one is illustrative. Many in the field of research have noted that flattening out the torus to a smaller diameter will help eliminate many of these instabilities. Of course it will. It reduces the asymmetric nature of the toroidal magnetic arrangement. But clearly it is not the solution. Symmetry must be enforced with virtual perfect precision and magnetism alone cannot do this. A symmetric force at least equal to the magnetic field strength must be applied to ensure perfect symmetry. In the Neo-Classical Tearing Mode the instability has been noted as being associated with the helical motion of the particle in its orbit. A symmetric dampening force that flattens out these helices eliminates this issue. Indeed, a symmetrically applied electrostatic force would have the effect of not just flattening out the helical travel, but shaping it symmetrically so that its helical orbit path radius is shorter on the higher charge side. The result is a stable, parabolic helix orbit about the main particle orbit path. This is exactly what an electrostatic force will do when applied symmetrically.

Due to the lengthy and well developed research already conducted in plasma fusion schemes we will propose a design that leans on this long history of research and utilizes a toroidal plasma as typically found in traditional Tokamak research reactors. However, we will also propose a new method for eliminating the difficulties currently encountered in MCF by developing a method that is complementary to MFC rather than divergent.

Chapter 3

The Power Conversion Problem

Power Conversion into a Fusion Fuel Source

To explain what we mean, we start fundamentally by asking, has humankind ever been able to create a fusion reaction? That is, has humankind ever been able to cross the triple product boundary into a true, real fusion event? The answer is obviously yes, but it wasn’t confined. The Hydrogen Bomb gives us a clue as to what a real first generation fusion reactor would look like if politics, culture, money and budgets were put aside and outcomes were preferentially sought.  And how was the triple product boundary crossed? The boundary was crossed with a semi-controlled fission explosion. There are at least 3 very good reasons to start from scratch and re-evaluate the fundamental physics of fusion.

First, a process called magnetic reconnection goes haywire (an issue we introduced supra) and results in a geometric increase in power transfer losses as power transfer input increases. Not only is it not understood in these regimes, the very empirical process for studying it will require decades of instrumentation development due, again, to material science limitations. In torus shaped plasma these events occur in a very small space and there are no conceivable instruments available to even observe it directly. The experimental data developed regarding these events already clearly shows that magnetic-only confinement cannot work – but the beat goes on. Desperate attempts to study it in nature using satellite-born instruments will help but it will be no substitute for the hands-on access one needs in a laboratory to study this. We estimate that an understanding of and magnetic work-around to this issue is at least 20 years distant, at minimum. This problem is addressed in Chapter 3.

Second, the much touted increases in triple product factors over the decades means nothing if the magnetic reconnection issues, and the general thermal loss issues, scale exponentially with increasing power transfers. And that is exactly what is happening. Some relief can be found in physical scaling since the volume of plasma increases as the cube of diameter and its surface by only the square of diameter. However, there are practical limits to how large a volume can be created and we are still skeptical for yet more reasons.

Third, current modes of power transfer into the reaction plasma are dreadfully insufficient. The most effective method has been the acceleration of particles into the fuel using expensive and electrical power hungry particle accelerators, but this requires a massive secondary device not suitable for spaceflight.

More about Thin Film Tokamak Fusion (TFET). The solution to introducing sufficient power to the reaction chamber without direct use of materials is to use a device that delivers considerable levels of power via electromagnetic radiation directly into the plasma. This suggests a fission reaction in the power regimes between a standard fission reactor and an uncontrolled critical mass explosion. There are quite a few clever things one can do with Uranium to achieve this. And it gets even more interesting with an element known as Americium-242m. Ultra fast reactors have been constructed and experiments show them to be viable. Of course, we seek something a little hotter. What we envision is a very hot, very short burn of less than 30 minutes (probably just seconds – we will calculate this much more precisely infra). Fission explosions have positive power output over ranges in the 10^-9 second range. This is obviously too fast to control with any foreseeable technology. Doing the math, for a fission reactor with a 10 year refueling cycle (a little on the long side) suggests a median burn time of 3600 * 24 * 365 * 10 * 10^-9 / 2 ≈ 0.158 seconds. So, we’re a little on the slow side of an uncontrolled reaction. This can be achieved using an ultra fast reactor design using Americium 242 with a thin film deposition designed to preferentially vent critically excess fast nuclei outside the reaction mass. Much experimental work (much of it classified for a long time) shows this to be achievable. Recent research in Israel has proven it. The reaction mass is shaped concavely facing the target. This introduces alpha radiation (positively charged Helium nuclei), gamma radiation and neutrons directly into the plasma. Better still, it can be used to electrostatically accelerate fusion fuel product whose kinetic energy will heat the remaining reaction plasma. Boron and Li-Si shutters can be employed to control plasma destabilization and provide an “annealing” feature for the plasma. The heating effect would be staggering and would follow closely on the well-established principles developed in nuclear weapons research. Americium 242m provides approximately 2 Mw power per square meter of film. Given 1 such “shaped charge” with a surface area of 1000 m², this yields a total power output of about 7 Gw, placing the energy deliverable to the plasma in the thermonuclear explosion range after a 30 second burn. This energy, mostly in the form of fast ⁴He nuclei, can be contained via the electrostatic fields discussed infra. Be reflection material inside the torus can trap neutrons to aid in heating. A further refinement will be discussed in what follows.

The Fissile Fuel Injector; a “slow” shaped charge

We recommend the introduction of a purpose built fission reactor whose product is used to electrically accelerate ²H-¹¹B directly into the plasma at energies proportional to its β radiation power output. This can be achieved by using the fissile β radiation to accelerate the fuel by electrostatic attraction in a cylinder within a cylinder design. Magnets around the cylinders further shape the fissile product as it exits the fission reaction chamber. The fissile product can be vented to vacuum and thus keep the plasma pure. This follows closely the design principles behind so-called “Hydrogen bombs” which use fission “kickers”, called “first stages”, to induce a fusion reaction. Diagrams of the Fissile Kicker invention are included in the Twenty-One Castle Mike Reference Design diagrams.

We have proposed the use of Americium-243m as the First Stage, fissile material. It can heat a small region of fusion fuel in the Gw power regime for periods up to several days (thus, several restarts on one kicker “pack”).

Isotope	Half-Life	Specific Activity (Ci/g)	Decay Mode	alpha (α)	Beta (β)	Gamma (γ)
Am-241	430 yr	3.5	Α	5.5	0.052	0.033
Am-242m	150 yr	9.8	IT	0.025	0.044	0.0051
Am-242 (product of Am-242m decay)	16 hr	820,000	β, EC	0	0.18	0.018
Am-243	7,400 yr	0.2	Α	5.3	0.022	0.055
Np-239 (product of Am-243 decay)	2.4 days	230,000	Β	0	0.26	0.17

Key: IT = isomeric transition, EC = electron capture, Ci = curie, g = gram, and

MeV = million electron volts; a dash means that the entry is not applicable.

Americium-242 decays by two means:

by emitting a beta particle (83%) and by electron capture (17%). Certain

properties of americium-242 and neptunium-239 are included here because

these radionuclides accompany the americium decays. Values are given to

two significant figures.

Formally speaking, so-called “hybrid” fusion-fission systems are technically different in that they are inherently safer than what we propose (which is actually a fission-fusion hybrid, not a fusion-fission hybrid). In these systems a fusion reaction is the “kicker” that provides the neutrons for a fission reaction. Because a sub-critical fission reaction is driven by a fusion process, it is inherently safe (at least it can be made so). However, in our case, we are in fact using a fission reaction (which certainly can runaway) to “kick off” a fusion reaction. In the formal hybrid scheme positive feedbacks are not possible. In our scheme they very much are. Having said that, we are merely pointing out the dangers that have always been inherent in fission reactors and we do not feel that this risk is sufficient to cause concern for this project. In the end, we are simply throwing the most powerful source of heat and pressure we know of at the problem, just as was done when developing uncontrolled fusion reactions for bombs (which do in fact work).

The natural question might arise; why not just use a kicker as the main power plant? The reason is that performing this same magic with Uranium 235 instead of Americium 242m is not feasible and Americium 242m is going to cost us over 50 million USD just for the amounts needed for all the reactor starts the project plan will require. This is because Americium is incredibly difficult (and expensive) to manufacture. Attempts at using Uranium 235 in the same way we are proposing are based on flawed approaches, mistakes and fantasy. Fission fragment reactors are attempts to duplicate what we’ve proposed here (and have been called “shaped charges”). These reactors use small fragments (won’t work, thermal explosion of reaction mass will occur) or a dust plasma of Uranium 235. In the latter case the idea is sound but the primary proponent of this idea mistook volume for mass and misunderstood the orders of magnitude. The reaction chamber for a gaseous Uranium reactor would be inordinately huge (it only works with low mass and will not scale). However, these novel approaches are quite similar to what we are proposing here, at least in principle. These ideas share the common feature that power is converted without a material medium directly from fission product to fusion plasma (or thrust in the case of the other proposals).

We submit that this approach has not been contemplated amongst academicians, much less attempted, partly for cultural and political reasons. For civil power production the only advantage fusion offers over fission in a “kicker” design is raw power, albeit with still considerably less radioactivity. The fission reaction need only occur for startup. Once initiated, the fusion reaction can provide all necessary power. In any case, and though not widely acknowledged publicly, this method is the shortest and most certain path to controlled fusion power.

For more information on the interplanetary medium and how it relates to Twenty-One Castle Mike, see the Reference Design GFS Yuri Gagarin. For our part here, we will only briefly note that the overall system we are specifying will be vacuum transit certified which means that it is designed to travel at very high velocity in the interplanetary medium. This has a few implications for the design of Twenty-One Castle Mike.

The temperature of the interplanetary medium is approximately 100,000 K, though its very low density makes this fact rather misleading since it is so far outside normal human experience.  Current estimates place it between 5 – 100 particles / cm³, the former an observed quantity in the vicinity of Earth. The density is believed to decrease with increasing distance from the sun, in an approximate inverse proportion to the square of the distance. Locally, the density is variable, and may be affected by magnetic fields and events such as coronal mass ejections.

The fuel

We specify in this Reference Design a reaction involving the fusion of ²H (protons thereof) with ¹¹B pure ions. ¹¹B is prepared for tankage as isotopically pure (to prevent unwanted neutron flux). ²H is also isotopically pure. We specify its storage to be in liquid form and the ¹¹B storage to be in solid form. Though we have discussed scenarios involving fuel mixtures of equal parts ²H and ¹¹B for order of magnitude validations we here specifically indicate the requirement as being one part ²H for every eight parts ¹¹B. The reason for this is not at once obvious until we consider nuclear cross sections, which we will take up later. In summary, these proportions ensure the highest reaction rate possible. Since the fuel is loaded to a minimum of 30,000 kg mass over what is to be used for reaction, this does not require additional adaptation of the design.

The fuel injection system: the ignition sequence

The fuel required for the reference round-trip transit is 22,480 kg of ²H and ¹¹B. This fuel will be injected into the plasma at ignition sequence start. Once it is exhausted, a restart will be required. Therefore, we now detail the mechanism for fuel storage, plasma creation and fuel handling.

We specify an integrated and self-contained unit which, in its overall construction as a propulsion system, constitutes Twenty-One Castle Mike. Included in this system are 20 large fission reactor whose design is specified in Reference Design Low power services self contained, mobile fission reactor (MFR), a design derived of the United States SP-100 reactor design. It consists of a fully shielded system (to background) with 6, separate reactor cores individually controllable and providing a combined electrical power of 600 kWe. Included in each system is a Brayton cycle that is VT certified and capable of operating in varying G environments. All MFRs brought online brings the system auxiliary maximum rated power to 12 MWe. Heat rejection is by liquid Li circulation.

Ignition sequence begins when, if not already online, all MFRs are brought to maximum rated power and diagnostics are performed to validate the requirement that at least 60 reactor cores remain functioning. This can be accomplished by the built in Li-Ion battery packs each MFR contains which, by remote command, can retract their control rods to start each of the six reactor cores.

We specify a startup plasma Earth weight of 70,000 kg and a final Earth weight at restart of 60,000 kg.  This yields a total fuel load of ²H (the density of L²H is 68 kg / m³ at 20K.) and ¹¹B of approximately 10,000 kg. We further specify the use of a fission kicker as described supra and infra which, when the coolant/moderator is removed from the reaction chamber the shaped charge “fires”. This moderator removal is performed by auxiliary power discussed supra.

Plasma creation

One of the key reasons for using a thin film, apart from the fact that this is required to take advantage of electrostic forces, is that we can reduce overall system weight significantly and dramatically improve the performance of “inducing” plasma current by the application of alternating charge in laminate layers of the vacuum vessel. This obviates the need for a central solenoid. But most importantly it means that current can be run much higher and therefore the forces created by the fields are consqeuently much greater. In this paper we have not yet calculated the impact this would have on confinement and other factors. We will do that now, TBC.

The kicker is designed with Gd shielding of approximately 10 cm thickness. Inside that shield is a W liner which is honeycombed with coolant conduit passages. This kind of engineering is well established and we won’t dwell on the calculations here. Filling the voids of the W shielding is Nb which, upon kicker firing becomes liquid. Auxiliary power is used to circulate the liquid Nb through the W reactor wall and out to an ¹¹B fuel storage bin.  A thermal energy transfer scheme is once again applied here and the ¹¹B, inside a bin set to 1 Pascal pressure (giving the ¹¹B a boiling point of 2075 C), is heated to its vapor pressure and just beyond.  Once the desired thermal energy transfer from the Kicker to the ¹¹B fuel is complete the liquid Nb is rerouted to transfer thermal energy to the ships heat rejection panels. This is accomplished through the thermal transfer first to the liquid Li coolant used by the fusion reactor (which is circulating at 100 m / s).

Simultaneous to the events described supra, auxiliary power is used to depressurize a stream of L²H and run high voltage through it. The specification is that the electrical power applied shall be sufficient to ionize the ²H. ²H has a total ionization energy of 1312 kJ / mole. Therefore,

Paux = Ei * Mkg   * mH	[eq. 3.01]
Auxiliary power   required to start kicker power up sequence

Where P_aux is the auxiliary power required to initiate the kicker power startup procedure, E_i is the total ionization energy of the fuel used for startup, M_kg moles of said fuel per kg and m_H is the total mass of the fuel to be used in the fueling cycle being initiated.

The ²H is then pumped into a cylinder which runs the along the axis of the magnetic thruster unit at the systems extreme after end. This cylinder contains within it another cylinder, inside which β radiation from the kicker will flow. Magnets outside the cylinders deflect particles to maintain a trajectory along the axis of the cylinders. As the β radiation moves through the inner cylinder it “drags” the ²H with it and transfers momentum to the fusion fuel component. This ²H is then magnetically routed into a torus consisting of numerous electromagnets that provide resistance to the momentum of the each ²H particle flowing into the torus. This transfers kinetic energy to electrical energy. This process of decelerating ions is a well established technology and we won’t dwell on it here.

As the ¹¹B is converted fully to its gaseous state it is pumped “downrange” of the kicker and aligned with the cylinders described supra. The β stream is directed to accelerate another stream of electrons by “pushing” electrons ahead of it (see diagrams). The electrons being “pushed” are outside the Gd shielding surrounding the outer cylinder. This secondary β stream we shall designate β₂. An ionization based on the technique used in Electron Beam Ion Traps is performed by colliding the β₂ stream with the ¹¹B fuel gas. The outer cylinder is divided into at least two sections to allow the flow of both ²H and ¹¹B at the same time. We can now utilize eq. 3.01 again by substituting ¹¹B. The result is the power transferred from the Kickers β stream to ionize the ¹¹B. Existing designs show that at least 5000 A / cm² charge density is required of the β stream to ionize the ¹¹B. The ionization energy for ¹¹B is 64740 kJ / mole.

Fuel injection

Once both fuel components are fully ionized the plasma valves for both fuels are opened and the plasma is directed back into the cylinder. As the Kicker emits β radiation through the cylinder and out toward the exhaust, it “drags” the fuel back up the cylinder. At a terminus point, when the fuel has acquired an energy greater than the potential energy thus presented by the electrostatic forces of confinement in the fusion torus, it is magnetically directed into the fusion reaction plasma. As the torus fills with fuel plasma a thin film develops inside the torus due to the geometry of the electrostatic forces of confinement. In addition, the high kinetic energy of the ²H-¹¹B fuel increases the temperature in the fusion torus. Plasma heating is performed by this process alone, that is, the fueling of the fusion reactor chamber by the forceful emission of β radiation from the Kicker, operating at approximately 5 GW power. Once all fuel has been thus injected, the liquid moderator is forced back into the Kicker reaction chamber and the Kicker shuts down. The circulating Nb is stilled and freezes in place. Once the Kicker shuts down, electrical power from the deceleration of ²H ceases. Auxiliary power is decoupled and the fusion reactor sends product to the same magnetic torus used previously and electrical power is generated by decelerating that product. Available electrical power, depending on the settings applied, can be measured in Tera Watts.

The entire Kicker runtime is about 30 minutes. It can also be run for the duration of a transit burn if necessary; a point we will explore later.

For reference, we note that the vapor pressure of ¹¹B is as follows:

Temperature/Vapor pressure:

2348K – 273.15 = 2075 C @ 1 Pa
2562K  @ 10 Pa
2822K  @ 100 Pa
3141K  @ 1 k Pa
3545K  @ 10 k Pa
4072K  @ 100 k Pa

Confinement of the Fuel Source Undergoing Fusion Reactions

Bremsstrahlung radiation losses are a major concern for the ²H-¹¹B reaction we shall specify. However, these losses only occur in the presence of electrons. The lower the proportion of electrons to plasma is the smaller the energy loss due to “Brems”. Thus, it might be useful to perform a sanity check on the very idea of single species plasma in fusion applications. Those in the magnetic confinement Kool-Aid camp will quickly advise that the Brillouin density limit renders single species, positively charged ion plasmas unworkable in fusion applications. This is false. The Brillouin density limit is a well known limit on the achievable density of plasma as radially confined by magnets. It is not a fundamental density limitation anymore than a black hole is subject to a density limitation. This density limitation is an artifact of the interaction of centrifugal forces, the electric force and the magnetic force. Therefore, compressing single-species ionized plasma up to the Brillouin density limit can be readily achieved with magnetic fields and densities beyond that can be readily achieved by other, non-magnetic means.

The Brillouin limit is irrelevant in a torus in which an electrostatic charge is presented both outside and inside the torus, symmetrically compressing the plasma as it rotates. Given sufficient power the magnets are free to compress the plasma to arbitrary densities beyond the Brillouin limit because the centrifugal forces causing it are stabilized by electrostatic forces. The scheme we propose addresses and solves both the Bremsstrahlung radiation issue and the Brilouin density limit that plagues magnetic confinement-only schemes.

Forces created by Pressure

The ITER project proposes the construction of a monstrosity that weighs more than the Eiffel tower yet is orders of magnitude too weak to contain the pressures necessary for practical fusion. This is yet another aspect of fusion research that gets little coverage. Also related to the power conversion problem generally, plasma pressures will have to be extraordinary if a reasonable power density is to be expected, regardless of the fuel chosen. Indeed, herein lies yet another materials science impediment for which adequate materials for this purpose were available only in the last couple of years. Any viable fusion reactor design in the near future will be totally dependent on nanotube materials in order to contain these pressures. As we shall see, this Reference Design specifies a plasma pressure at maximum rated power of about 1.5 million bar. A blanket approximately 2 meters thick made of nanotube materials will be required for the construction of the reactor vessel.

1 Pascal is 1 N / m² means that the confinement forces must apply x amount of Newtons force over an area of 1 m^2, that is, it is the sum of all confinement forces acting over that area.

1 * 10⁶ bar = 1 * 10¹¹ Pascals. Therefore, the sum of all force acting over any given square meter surface must equal 10¹¹ Newtons force.

Let us begin our inquiry into this problem by tackling some fundamental physics first. The strength of a magnetic field can be expressed in the limit law form as:

dB   = (κm / 4 * π ) / ( qv   ds X r / |r|3 )	[eq. 3.02]
Biot-Savart Law

Which can be rewritten as

Fm12 =   κm q1 q2 * (v1 X (v2   X r12)) / r212	[eq. 3.03]
Magnetic Force   acting on q1 due to q2

κ_m= 4π * 10^-7 T * m / A and is known as the “permeability of free space”, q₁ and q₂ are the charges in Coulombs of each observer 1 and 2, v₁ and v₂ are the velocities of the respective observers and r₁₂ is the right line distance between observers 1 and 2. Notice there is no mention of a magnetic field in that equation. This follows naturally from the fact that magnetic fields exist only in the presence of moving charge; that is, only in consequence of an already existing E field. This will become significant in MCF discussions infra.

Likewise, the electrostatic force acting on the same two observers can be found with a similar relation:

Fe12 =   κe q1 q2 / r212	[eq. 3.04]
Electrostatic   force acting on q1 due to q2

where κ_e = 8.987 * 10⁹ N * m² / C² and is the electrostatic proportionality constant and q denotes the charge of a particle in Coulombs. We can arbitrarily choose a value for |F_e12| and |F_m12| such that these two forces acting on two particles differ in magnitude by a known constant, C_p:

Cp [κe q1   q2 / r212] = κm q1 q2   * (v1 X (v2 X r12)) / r212	[eq. 3.05]
Electrostatic   and magnetic force compared for q1 and q2

which is just (4π * 10^-7 * |v₁| * |v₂|) / (8.987 * 10⁹)

But the values for |v₁| and |v₂| cannot exceed c. If we assume the maximum possible boundary value, then we can say that we have [(4π * 10^-7 * |v₁| * |v₂|) / (8.987 * 10⁹)] * c².

What this shows is that:

[(4π * 10^-7 * |v₁| * |v₂|) / (8.987 * 10⁹)] * c² = 12.567;

That is, ceteris paribus, the Magnetic force of Nature is 12.567 times the strength of the Electric force of Nature. The problem, however, is that this assumes velocities far exceeding what even the most ambitious fusion reactor concepts have proposed. And these velocities would require advances beyond the current state of the art in materials science. Tokamak plasma rotational velocities (the highest velocity medium for this purpose) do not exceed 0.1 % the speed of c (e.g. 0.035% c path speed at TEXT facility, Texas). Nor is this needed for fusion. Any significantly higher velocity would simply be an energy loss. Thus, when we take the more reasonable figure the problem becomes evident. The figure above plummets to:

sv = (4π * 10-7 * (c   * 0.01) 2) / (8.987 * 109) = 1.2567 * 10-5	[eq. 3.06]
Ratio of   strength of Magnetic field to Electric field at velocity of merit

That is, the Electric force of Nature, when operating at the boundaries of a fusion regime, is 79 thousand times stronger than the Magnetic force. This is worse than junk science. This is pathological science. Magnetism, for the purposes of confinement (but not necessarily fuel control), is not a winning strategy. Thus, we define the fusion pathology constant as:

Cp = κe / κm   = 7.958 * 104	[eq. 3.07]
The Pathology   Constant

Clearly, the obvious candidate for a realistic fusion reactor pressure providing force is the electric force. This follows immediately from the findings supra and from the fact that we cannot harness either the Weak or Strong nuclear forces because our materials science (especially in terms of tolerances) cannot operate at the scales over which those forces operate. And Gravity would require far too much mass. The remarkably high pressure we seek, 1.5 million bar, has yet to be justified, but a full treatment is promised in what follows.

In terms of traditional Tokamak design, this can be understood as a significant inefficiency in power conversion into the reactor fuel. It also dramatically undermines the specific power due to the fantastic mass requirements of Induction Magnets.

This outlandish inefficiency presents in the form of large induction magnets for which the iV applied to the Solenoid will have to be 79000 times greater than the magnetic field potential energy produced within the plasma, at minimum. Why not just use iV directly and save 79000 times the traditional cost? That question is what we will now attempt to answer.

Charge density, the density of charge in Coulombs, is expressed:

Q =	[eq. 3.08]
Charge density   of a charged volume

where x, y and z are the coordinates locating the total charge and ρ is the density function for charge as a function of r; a position vector within the total charge volume.

To find the total count of charge carriers in a volume we use:

N =	[eq. 3.09]
Number of charge   carriers in a charged volume

Where N is the number of charge carriers in the volume and n ( r ) is the position dependent charge carrier count density.

In attempting to find a more efficient means of plasma confinement we examined the current state of the art in materials science in the area of conductor materials. In particular, we sought an ion superconductor (for reasons that will soon become clear) because of the power conversion issues that will arise if a conductor were to exhibit any Ohmic resistance greater than 0. As a template for the set of compounds already discovered we chose to examine the Ag₂Se superionic conductor and Nickel based ion superconductors that transport Ag ions. Some possible states for the Ag₂Se superionic conductor material are:

At -4 eV excitation, the peak region of charge density for Ag₂Se, the states for Ag and Se are 4d and 4 p respectively. This is two states for that material.

A charge density of 4 is known experimentally for each state => charge density of 8 per atom per eV. The time average of radial separation for Ag₂Se atoms at peak pair values in the superionic state is about 3 Angstrom. This determines the density of ions but does not speak to the density of electrons contained in the lattice atoms of Ag2Se, which are composed of different ions (non-flowing ions).  Therefore, one meter of material will yield

1 / (6 * 10^-10) = 1.67 * 10⁹ [1 Angstrom = 10^-10 m]

positive ionic nuclei, equally divided between Ag and Se ions. Ag ions consist of 47 protons and Se ions consist of 34 protons. This gives an average of about 40 protons per ion (assuming the ionic charge of conduction is the same as the lattice atomic elements, which it apparently is). Therefore, the integral of charge from x₀ to x is:

= 40 * 1.67 * 10⁹ / m = 6.68 * 10¹⁰ ions per meter or,

N =  = (6.68 * 10¹⁰)³ = 2.981 * 10³² ions / m³

which is (2.981 * 10³²) * 1.602176565 * 10⁻¹⁹ = 4.78 * 10¹³ C.

using eq. 3.03 and assuming plasma charge (which we will justify infra), then, at 10 cm distance this generates a force of:

F_i = [ 8.987 * 10⁹ * (4.78 * 10¹³) * (1.602176565 * 10⁻¹⁹ * 2.598 * 10²⁷)] / (0.1)² = 1.79 * 10³⁴ Newtons. We reduce by one order of magnitude to adjust for the vessel geometry:

1.79 * 10³³ Newtons force applied to the plasma.

This figure assumes a drift velocity of 1 m / s, which is conceivably achievable, especially with the large headroom this value gives us.

This Reference Design specifies a minimum force applied to the plasma of ~10¹¹ Newtons / m², the approximate force required to generate 1.5 million bar. Therefore, current technology in superionic conductor material is sufficient for our purpose.

This calculation was for making an order of magnitude estimate of charge density. This is the charge that passes a given point in the conductor per second per unit area. Because this figure exceeds the figure required for plasma compression, it readily exceeds the figure required for creation of magnetic fields for control of the plasma (this topic will be addressed in a following section). Reducing the Reference Design thickness of the conductor and modifying our drift velocity assumption in order to satisfy the pressure containment requirement while minimizing thickness, we have:

1 cm thickness at 1 mm / s drift velocity

ð  1.79 * 10²⁹ Pascal

This still leaves considerable headroom for materials science adjustments (operating temperature being one possible limitation). We shall examine this in more detail in what follows.

Given the findings supra we specify a reactor vessel of a generally toroidal shape and made primarily of nanotube materials. In order to make extraction of fusion product as technically easy and achievable as possible, we specify a reaction plasma generally located in the center of the torus that is a thin film, no more than 3 cm thick. Calculations we shall perform later will show mathematically the superior fusion product capture that will occur as a result of this geometry. We shall also find later that this provides other notable advantages.

A quick inquiry into the current state of the art in nanotube technology will show that the tensile strengths required of this material are currently sufficient to satisfy our requirements for a pressure of 1.5 million bar. However, limitations and restrictions abound. First, the thermal durability of these materials is still poor. This will require active and aggressive cooling. Second, and as alluded to supra, the tensile strength of the material is not the only concern in assessing its overall strength. Third, the plasma pressure to nanotube normal force interface is problematic. Even the hardest nanotube materials cannot withstand the pressure exerted on the inner surface of the material in the geometry we will require. Intrusion of charged particles into the nanotube material will erode (destroy) it. Fourth, there is as yet no manufacturing process to consistently manufacture the volumes of the material we will require. We specify a design that addresses all four concerns thusly:

A nanotube, laminate blanket is constructed in a toroidal shape and thus has a hollow center. Its inner surface shall be made of a neutron super mirror material (alternating layers of Ni and Ti) to be described later. It shall have a thickness of about 2 cm. It shall also have a negative electric charge applied during operation [layer 1]. Embedded in Layer 3 is a honeycomb of conduits equivalent to about 2 cm thickness into which liquid Li is pumped at high velocity (about 100 m / s). For diagrammatic purposes here, we treat it as a separate layer [layer 2]. Backing that layer is a layer of super pressed Gd about 4 cm thick and weighing about 225 mT [layer 3a]. Backing that layer is a layer of super pressed Os about 2 cm thick and weighing about 240 mT [layer 3b]. Backing that layer is a vacuum about 2 cm thick into which liquid N is pumped at modest velocity (about 10 m / s) [layer 4]. Backing that layer is a layer about 1 cm thick of Boron Nitride Wurtzite Nanotube, an electrical insulator and very hard material that resists N penetration. It also provides normal forces for containing plasma pressure [layer 5]. Backing that layer is an electron superconducting layer. The overall charge in this level is significantly less than that of the ion superconducting layers. It is used to moderate the electrostatic forces and will be described in greater detail in the text [layer 6]. Backing that layer is a repeat of layer 5 [layer 7]. Backing this layer is a layer of ion superconducting material about 10 cm thick to be described later. Ions in this conductor experience an electrostatic force exerted from ions within the plasma approximately 20 cm toward the center of the torus [layer 8]. Backing this layer is a layer identical in all respects to layer 5 [layer 9]. Backing this layer is a layer about 4 cm thick made of carbon nanotube material which acts as the primary stiffener and normal force to the electrostatic forces of the plasma [layer 10]. This lamination cycle repeats starting with a layer identical in all respects to layer 4. The total number of layers is approximately 90. Layers consisting of liquid coolants are bridged by carbon nanotube perpendicular “walls” assembled in a helical pattern about the torus. Superconductors have similar shaped insulators running the same helical course (the mathematical description of this helix will be provided later). Each “wall” is a V-shaped hollow vane to support electrostatic symmetry. Layers 2 and 4 receive several hundred very narrow return lines through the V-shaped walls. Return lines for layer 2 are lined with thermal insulators. Due to limitations in manufacturing tolerances, parallel processors are employed to anticipate out of symmetry charge densities and issue commands to correct these anomalies in real time. We can illustrate the layers thusly:

Internal layers (not repeated)

[layer 1]: alternating layers of Ni and Ti: 1 cm

[layer 2]: liquid Li: 2 cm (active Neutron and thermal absorber – flow rate: 100 m / s)

[layer 3a]: Gd: 3 cm (primary Neutron Transport backstop) flux is to background far side

[layer 3b]: Os: 1 cm (primary gamma backstop); is to background far side

Repeating layers

[layer 4]: liquid N: 1 cm

[layer 5]: Boron Nitride Wurtzite Nanotube: 1 cm

[layer 6]: electron superconducting material:1 cm

[layer 7]: Boron Nitride Wurtzite Nanotube: 1 cm

[layer 8]: ion superconducting material: 10 cm

[layer 9]: Boron Nitride Wurtzite Nanotube: 1 cm

[layer 10]: carbon nanotube material: 4 cm:

[layer 11]: Repeat pattern starting with layer 4.

The overall layer pattern is a spiral that works its way outward from the center, providing a continuous, helical conductor surface from the interior of the reactor vessel to the outermost layer.

The key to this arrangement is subtle, so we will explicate it here. Charge density and possibly current in layer 6 is adjusted so that electrostatic forces between the plasma and layer 6 are virtually equal to the total force exerted by the plasma pressure (in other words, the electrostatic forces are adjusted so that they are almost, but not entirely sufficient to support the plasma pressure). If the force exerted on a square meter of surface in layer 6 consequent to the interaction of the charges located in the reaction plasma and layer 6 is denoted F₆ then the Reference Design requires that F_p - F₆ = W_f ≪ F_p where F_p is the totalforce exerted (as an electrostatic potential energy) on the same square meter by the total plasma pressure. And W_f is the normal force exerted over that same square meter by the nanotube layer 7; which is in turn supported by layer 8.  Thus, equilibrium is given by:

Fn – 1= Fn+ Wf@n ∀ n in blanket       [ Wf@n   ≪ Fn   – 1 ]	[eq. 3.10]
Reactor Vessel   Wall Equilibrium Condition (simplified)

where W_f ≪ F_p

And now; the denoument: the force F₆ experiences a normal force of equal magnitude from the next analogous layer backing it; that is, layer 12. And at that layer the same equilibrium condition is repeated. The difference however, is that the electrostatic normal forces decrease with each pattern repetition. The effect is to distribute the plasma pressure force evenly throughout the nanotube material such that the total force created by the pressure is not exerted on a single surface. In each case of electrostatic normal forces, the forces are acting on charged particle to charged particle and no material is involved. The pressure of the plasma, therefore, is transmitted via fields only. After a repetition of some 90 cycles over approximately 2 meters of blanket, the final material layer absorbs the remaining force delivered by plasma pressure by presenting a normal force to it. We will pause here with this overview and describe this pressure confinement system in much more detail in what follows. For now it is only important to understand that the strategy here is to complement and reinforce magnetic field lines that will be created as a consequence of this arrangement. This allows the Reference Design to follow the well established experimental data already amassed in traditional Tokamak fusion.

It has been stated supra that “The choice of transit propulsion represents the greatest risk factor for this project”, and we can now reduce that statement to this:

The ability to create specific classes of carbon and Boron-Nitride nanotube material in quantity represents the greatest risk factor for this project. Once this problem is solved, the remainder is relatively short term development work.

Mass of an atom of ¹¹B is about 11 times that of a proton.

To determine speed of ²H -¹¹B product, we take 2 of the 3 helium nuclei produced by this reaction (because the 3^rd is a “slow” product) and calculate their velocity:

⁴He fast nuclei ²H-¹¹B fusion product v = 3.28 % c or 0.0328 * c; e.g.

(((4 * 10^6[energy of fast ⁴He in MeV]) * (1.60217646 * 10^-19 [conversion to Joules]) / (5 * 10^-27[mass of ⁴He]))^(1/2)) / 299792458[speed of light].

A quantity of 2 of these fast ⁴He are generated with velocity vectors 155 degrees from each other. Coordinate system orientation is random.

Therefore, plasma velocity is best limited to about 0.001 c maximum. This yields:

0.001 * c ≈ 299,792 m/s

The ideal geometry for capturing product for kinetic thrust now appears to be a torus reduced to a cylinder in which fusion occurs in a thin film at the cylinder wall. It is possible to use cylinders with curvature in their walls along the plane of their radius but the math is made more complex. This configuration works best with the geometry of the ²H-¹¹B reaction due to the 155 degree separation between product vectors. An examination of this geometry shows that reaction product can be recovered from a little less than one-half of all reaction product kinetic energy.

We begin by first noting the percentage of total output lost due to re-absorption into the plasma and annealing. This we can estimate to be about 10% based on our “critical mass” calculations and a thin film of no more than six inches thick. In addition, neutron flux will account for as much as 0.1 % of total power output. At Tera Watt power levels this is substantial. Finally, we will need to take into account the loss of the slow ⁴He nuclei which is assumed in the ideal geometry. The total energy release from a ²H -¹¹Bo reaction is 8.7 MeV, of which 8 are accounted for by the fast nuclei. The remainder, 0.7 MeV is the kinetic energy of the slow nucleus. This energy is lost to the aggregate reaction rate power output by re-absorption and annealing.

The strategy is to find a way to capture the useable product mentioned by capturing it in a neutral E orbit outside the plasma. Once there, it can be redirected into accumulators.

An examination of the ideal geometry shows that probabilistically only about 32.5 percent of the reaction product can be recovered for kinetic thrust (this assumes a 10% loss of product by nuclei quantity as a result of product being captured by plasma and remaining therein). These calculations assume a successful deflection of product nuclei whenever incident upon the reactor vessel wall with an angle in the interval (π / 4, 0). Deflections at steeper angles will result in an electrical potential being produced in the electrostatic confinement field; thereby releasing approximately 32.5 % of the power output in the form of electric current. The remaining 35% will be power in the form of significantly reduced velocity of alpha particles and in the re-heating (and annealing) of the plasma; thereby providing the heating necessary to sustain the fusion reaction once initiated. These proportions are estimates that mathematically should be close to the real experimental values one would likely obtain. Some of the power recovered as current can be used to power the electromagnets and other reactor power devices. In addition, some of this power can be tapped for direct electrical within the spacecraft.

Distance between reaction plasma surface and reactor wall must be adjusted, with other related dependencies, to accommodate this optimum geometry. A curved cylinder wall, instead of a straight one, will improve the spatial tightness of the deflection. The final adjustment to make is to note that neutron flux may steal as much as 0.1 % of the total energy output available, therefore, adjusting for that, we can estimate recoverable power at 32.5 – (32.5 * 0.1) = 29.25 % as a very close figure for characterizing reactor efficiency. For our purposes, this alone will require a total reaction rate yielding a power production of 65 TW, enough to ensure the required 21 TW kinetic thrust we need and also take into account the power required to run the magnets and compensate for system energy losses (which, relative to these figures, will be tiny).

Heating of the plasma should be described to preclude any confusion. Heating is the process whereby the fuel is “vibrated” about its orbital axis as it spins in a helical orbit within the torus. The very strong electrostatic confinement coupled with the fine control of orbits given by the electromagnets ensures that even at very high temperatures the vibrations will result only in motions about the orbital path (a circular helix), but vibrations insufficient in strength to break free of this orbital path.

One can view this as a key requirement that the density and orbital confinement be tight enough to overcome the vibrations of heating.

Thus, when temperature is expressed in KeV what we are referring to is the kinetic energy, and hence velocity, of the fuel particles in vibration, not their velocity in their orbit, which is an entirely different energy measure. Thus, the total energy of a given particle is the sum of its kinetic energy due to heating and its kinetic energy imparted by the electromagnets and electrostatic confinement.

Given our previous considerations, we can calculate that our maximum plasma temperature will necessarily be about 122 KeV because we restricted the upper velocity of our fuel to a fraction of the product velocities, as described supra.

The nuclear binding energy of ¹¹Boron is 6.9277 MeV. This value corresponds to the average kinetic energy of collision particles that optimizes fusion reaction rates.

---

[1] But again, this is largely because scientiests are not genuinely concerned with these issues. They are doing pure research and a single-mode is the better approach for their purpose.

TBC in Part II

Part II

A further dive into reaction rates

To determine the “reaction rate” we need to first develop the concept of a cross section. This term gets much play in the fusion fan crowd with equal confusion and controversy.  Many state that a “formula” for cross sections does no exist or that it is just too complicated to work out. This is certainly false. Some of the latest work shows that there are indeed “formulas” that can be used to arrive at very good estimates of experimental data, so good in fact that they exceed by far our requirements for order of magnitude estimates. Here, we use the plan of R.K. Tripathi who showed that a general formula for cross sections can indeed be found. It is a clumsy equation but by no means difficult. It has excellent agreement with experimental results and we provide it infra.

The cross section for a given reaction is usually determined experimentally. However, equations have been developed to approximate these values. The experimentally determined value for the cross section for the p-¹¹B reaction is 1.2 * 10^-28 m³.

The nuclear cross section for particle interaction as a function of the particle energy (due only to relative motion in the system considered; and which is usually just the temperature expressed as an energy in joules) is given by the general formula:

σ_R (E_p) = σ_R = σ_T – σ_el

If we write this value explicitly for the p-¹¹B reaction as a function of incident particle energy (with target energy presumably 0) we have:

σ_p-11B(E_p) = σ_R (E_p)

and is the quantity we seek ∴

σp-11B   (Ep) = [πr02   (AP1/3 + AT1/3 + δE)2   * (1 – (B / Ecm)) ]  / (1028)whose B factor   expands to:B = [1.44*ZPZT/ R]R = [rP   + rT + (1.2*(AP1/3 + AT1/3)/Ecm1/3))]ðσR (Ep) = {πr02 (AP1/3   + AT1/3 + δE)2 * (1 – ([1.44*ZPZT/   [rP + rT + (1.2*(AP1/3 + AT1/3)/Ecm1/3))]]   / Ecm))} / (10-28)andδE =   {1.85S + (0.16S / Ecm1/3) – CE + [0.91 * (AT   - 2ZT) * ZP/(ATAP)]}andS = {AP1/3AT1/3   / (AP1/3 + AT1/3)}andCE = {Dα * [1 - e(-Ek/40)] – cos (0.229Ek0.453)   * 0.292e(-Ek/792)}andfor alpha –   nucleus reactions we use this form of D:Dα    = {2.77 – (AT (8 * 10-3)) + (A2T   (1.8 * 10-5)) – [0.8 / [1 + e[(250 – Ek)/75]]]}	[eq. 3.11]
Nuclear Cross   Section
ð  Reference   Design Twenty-One Castle Mike σp-11B   (Ep) ≈ 1.2 * 10-28   m2

*********************************************************************

Where

σ_p-11B (E_p) = is the cross section as a function of thermal energy. This is the effective geometric plane exposed to an inbound, colliding particle. It is “effective” because its magnitude does not equal a literal area, but is what the area would be ceteris paribus.

A_T = target mass number (sum of all neutrons and protons in nucleus)

A_P = projectile mass number (sum of all neutrons and protons in nucleus)

Z_T = target atomic number (sum of protons in nucleus)

Z_P = projectile atomic number (sum of protons in nucleus)

r₀ = 1.1 * 10^-15 m

E_cm = is the colliding system center of mass energy in MeV

r_P = is the equivalent sphere radius of the projectile (1.29*r, where r is sometimes called charge rms radius)

r_T = is the equivalent sphere radius of the target (1.29*r, where r is sometimes called charge rms radius)

{

The equivalent spherical diameter (or ESD) of an irregularly-shaped object is the diameter of a sphere of equivalent volume

}

E_k = is the collision kinetic energy in AMeV

Let the atomic mass number of a nucleus be denoted A. Then A * MeVs = AMeVs.

A simple, crude approximation can be found using:

σ_R = πr₀² (A_P^1/3 + A_T^1/3 -δ)²

An explicit calculation for the ²H-¹¹B reaction

E_cm = is the colliding system center of mass energy in MeV

E_k = is the collision kinetic energy in AMeV

Let m_P and m_T be the mass of the projectile and target respectively. And let:

v_Ti be the initial velocity of the target

v_Pi be the initial velocity of the projectile

v_Tf be the final velocity of the target

v_Pf be the final velocity of the projectile

v be the final system velocity

m_P v_Pi  + m_T v_Ti = (m_P + m_T) v and

v = ( m_P v_Pi  + m_T v_Ti  ) / (m_P + m_T)

m_P  = (1.67 * 10⁻²⁷) kg

m_T = 11 * (1.67 * 10⁻²⁷) = (1.84 * 10^-26) kg

550000 * (1.60217646 * 10^-19) = 8.81 * 10^-14 joules

E_P = ½ mv² and letting v_Ti = 0

ð  √ 2E_P / m = v

ð  √ 2 * (8.81 * 10^-14) / (1.67 * 10⁻²⁷) = (1.027 * 10⁷) m / s or 0.037c

ð  v = (1.67 * 10⁻²⁷) * (1.027 * 10⁷) / ((1.67 * 10⁻²⁷) + (1.84 * 10^-26)) = 8.55 * 10⁵ m / s

ð  E_cmp = (½ ((1.84 * 10^-26) + (1.67 * 10⁻²⁷)) *(8.55 * 10⁵)²) = 7.33 * 10^-15

ð  (7.33 * 10^-15) / (1.60217646 * 10^-19) = 4.58 * 10⁴ or 46 KeV

ð  E_cm= 0.046 MeV (AKA reduced mass energy)

ð  E_k = 12 * [(8.81 * 10^-14) / (1.60217646 * 10^-19)] / (10⁶) = 6.6 AMeV (AKA collision mass energy)

ð  E_k = [(8.81 * 10^-14) / (1.60217646 * 10^-19)] / 12 * (10⁶) = 0.0458 AMeV (AKA collision mass energy)

ð  E_k = [(8.81 * 10^-14) / (1.60217646 * 10^-19)] / 1 * (10⁶) = 0.549 AMeV (AKA collision mass energy)

and for 580 KeV

580000 * (1.60217646 * 10^-19) = 9.29 * 10^-14 joules

E_P = ½ mv² and letting v_Ti = 0

ð  √ 2E_P / m = v

ð  √ 2 * (9.29 * 10^-14) / (1.67 * 10⁻²⁷) = (1.05 * 10⁷) m / s or 0.037c

ð  E_cmp = (9.29 * 10^-14)

ð  E_cm= 0.480 MeV (AKA reduced mass energy)

ð  E_k =  0.480 / 1 = 0.48 AMeV (AKA collision mass energy)

and for 650 KeV

650000 * (1.60217646 * 10^-19) = 1.0414 * 10^-13 joules

E_P = ½ mv² and letting v_Ti = 0

ð  √ 2E_P / m = v

ð  √ 2 * (1.0414 * 10^-13) / (1.67 * 10⁻²⁷) = (1.12 * 10⁷) m / s or 0.037c

ð  v = (1.67 * 10⁻²⁷) * (1.12 * 10⁷) \ ((1.67 * 10⁻²⁷) + (1.84 * 10^-26)) = 9.32 * 10⁵ m / s

ð  E_cmp = (½ ((1.84 * 10^-26) + (1.67 * 10⁻²⁷)) *(9.32 * 10⁵)²) = 8.72 * 10^-15

ð  (8.72 * 10^-15) / (1.60217646 * 10^-19) = 5.44 * 10⁴ or 54 KeV

ð  E_cm= 0.054 MeV (AKA reduced mass energy)

ð  E_k = 12 * [(1.0414 * 10^-13) / (1.60217646 * 10^-19)] / (10⁶) = 7.8 AMeV (AKA collision mass energy)

ð  E_k = [(1.0414 * 10^-13) / (1.60217646 * 10^-19)] / 12 * (10⁶) = 0.054 AMeV (AKA collision mass energy)

ð  0.0541

and

E_cm=| 0.054| MeV

E_k = |7.8| AMeV

A_T = |11|

Z_T = |5|

r_T = |1.29 * 1.17 * 10^-10| m (fully ionized)

r_P = |1.29 * 0.8775 * 10^-15| m

A_P = |2|

Z_P = |1|

r₀ = |1.1 * 10^-15| m

The cross section of ¹¹B with an incident proton is supposedly 1.2 * 10^-28 m². But at least one of the equations for solving for it comes up with a number in the correct order of magnitude but that is negative. Errors such as this are common in attempts to analytically determine cross sections and experimental data tends to be more reliable. We have examined the record and the figure of 1.2 * 10^-28 m² at 580 KeV appears to be very close to the observable value. Explicit experiments designed for the validation of this Reference Design will be required to both narrow the error range in this quantity and to validate the peak energy as a function of cross section which has been observed to be 580 KeV. This decision is in line with our general practice of accepting the most conservative figures (580 KeV peak will result in the lowest reaction rate).

Machine computation can be performed with the explicit expression below:

{ Attaching a carefully placed  (-1)* resets the sign to a positive value }

pi*((1.1 * 10^(-15))^2) * ( (( (1)^(1/3) ) + ( (11)^(1/3)) + (1.85* ( ((1)^(1/3)) * ((11)^(1/3)) / ( ((1)^(1/3)) + (11)^(1/3) )) + (0.16*( ((1)^(1/3)) * ((11)^(1/3)) / ( ((1)^(1/3)) + (11)^(1/3) )) / ((0.48)^(1/3))) – (2.77 – (11 * 8 * 10^-3) + (((11) ^2) * 1.8 * 10^-5) – (0.8 / (1 + exp((250 – 0.48) / 75) ) )) * (1 – exp(-0.48/40) ) – 0.292*(exp(-0.48/792)) * cos (0.229 * ((0.48)^(0.453))) + (0.91 * ((11) – 2*(5)) / ((11)*(1)))))^2) * (-1)*(1 – ( (1.44*(1)*(5)) / ((1.29*(1.1 * 10^-15)) + (1.29*(1.17 * 10^-10)) + 1.2 * (((1)^(1/3)) + ((11)^(1/3))) / (((0.48)^(1/3)) / (0.48 )))))

= 1.5 * 10^-28

Pressure and density

The first quantity we need to determine is the number of particles in a cubic meter of plasma. From that we can use the reaction rate to calculate the power density, our ultimate goal in the present case. We will attempt to perform a first-order approximation using the Ideal Gas Law, but we will next calculate it based on the actual material involved, a single species plasma. In the latter case, we will need to calculate the density in a much different manner. There, only electrostatic forces at the required pressure will determine density.

Density determined from the Ideal Gas Law

The final quantity we need to begin calculating for different design approaches is the plasma density as a function of temperature and pressure. Values in nuclear fusion reactors vary from 10⁷ particles m³ up to 10³² particles m³ in inertial confinementschemes. Our design requires a pressure of 10⁶ bar, therefore, we can use this as a starting point to determine density. Temperature is given by the known optimization energy for the ²H-¹¹B reaction, which is about 650 KeV of heat energy. Therefore, the question is, at T=650 KeV and P=1000000 bar, what is the density of ²H-¹¹B? Only then can we determine the reaction rate per unit volume per second, and it is thus not solely dependent on the cross section and the proportions of reaction fuels (this seems to be a common misconception in the fusion crowd).

The so-called Ideal Gas Law does not apply to a plasma but we’ll do some calculations that will be useful later:

PV   = nRT	[eq. 3.12]
Ideal Gas Law

n = # grams / MM (is molar mass in g/mol)
PV = (mass in g / MM) x RT
P = (mass in g / MM) x RT / V

But, d = mass in g / V
P = dRT / MM
d = P x MM / RT

To find the density of a gas, take the P in atm times the molar mass, then divide by R (0.08206 L atm / K mol) and T (in kelvins)

ρ =   P*MM / R * T{in Kelvin}	[eq. 3.13]
Density of a gas

R is the universal gas constant and is  8.314472×10⁻⁵ m³ bar K⁻¹ mol⁻¹  OR

R = 8.314472 m³ * Pa /  K / mole

molar mass is the number of particles of the substance per unit mass. One mole is a dimensionless number (a quantity of “things”) and is 6.02214×10²³

Molar Mass of an element alone is the product of the Atomic weight of that element and the mole constant Mu = 1×10⁻³ kg/mol = 1 g/mol.

Atomic “weight” is not a weight but a hokey quantity (a ratio of vague masses). Anyway,

Atomic weight of ¹¹B = 10.811

Atomic weight of ²H = 1.008

Atomic weight of an electron

[at the end of the day, using the ideal gas law for a plasma and atomic “weights” will inevitably render this a rough approximation]

There is this stupid chemistry number called “Dalton”;

A Dalton is another hokey unit of mass equal to 1.660538921 * 10⁻²⁷ kg. Atomic weight is defined as the ratio of the average mass of all isotopes of an element to ½ Dalton‼ wtf? Anyway, we can get an Atomic weight for an electron thusly:

mass of an electron = 9.10938291 * 10⁻³¹ kg

Therefore;

(Atomic “weight”)e = 2 * 9.10938291 * 10⁻³¹ / 1.660538921 * 10⁻²⁷ = 0.00109

And we see that the effect of stripping electrons in plasma will likely be negligible as an approximation. Therefore,

If the fuel is half and half, then,

MM_11B = 1×10⁻³ kg/mol * 10.811 = 0.010811

MM_2H = 1×10⁻³ kg/mol * 1.008 = 0.001008

MM_plasma= ½ * 0.001008 + ½ * 0.010811 = 0.0059095 kg/mol

ρ_plasma = 1 * 10⁶ (bar)* MM_plasma / R * T(Kelvin)

ρ_plasma = 1 * 10⁶ (bar)* (0.0059095) / R * T(Kelvin)

converting energies of 200 KeV and 650 KeV to Kelvin yields:

200,000 * 11604.5 = 2.3209 * 10⁹ Kelvin and

650,000 * 11604.5 = 7.5429 * 10⁹ Kelvin and using this figure and

1 * 10⁶ bar = 1 * 10¹¹ Pascals; continuing,

ρ_plasma = (5.9095 * 10¹¹) / (8.314472) * 7.5429 * 10⁹

ρ_plasma = 9.4227 kg / m³

Finally, the masses of ¹¹B and ²H are,

First, we start with the mass of the proton, which is

1.672621777 × 10⁻²⁷ kg

and the mass of a neutron, which is

1.674927351 × 10⁻²⁷ kg

And multiply and sum the masses for ¹¹B and ²H:

¹¹B_m = (5 * 1.672621777 × 10⁻²⁷) + (6 * 1. 674927351 × 10⁻²⁷) = 1.841267 * 10^-26 kg

²H_m = 1.672621777 × 10⁻²⁷ + 1. 674927351 × 10⁻²⁷ = 3.3475 * 10^-27 kg

ð  ½ * (3.3475 * 10^-27) + ½ * (1.841267 * 10^-26) =  1.0880085 * 10^-26 kg average fuel mass per particle

Density and pressure as a function of electrostatic forces

But for our purposes, gas laws will have essentially no relevance. If we assume a design pressure requirement of 1.5 * 10¹¹ Pascal then we need to calculate the particle count in a thin film whose density decreases as we move outward in accordance with the electrostatic equation for force:

F_p = φ Q_P / (ℓ₀ - ℓ₄)²

Where we let φ = κ_e q_m1 where κ_e = (8.987 * 10⁹ N * m² / C²) is the electric constant and q_m1 is the charge of the particle placed under operating pressure (a “test” charge). Q_P is the charge of the rest of the plasma. (see Figure 1)

∴ as an order of magnitude estimate we know that the force will decrease as 1 / ℓ₀ - ℓ₄)² and the potential energy is:

F_p ⨀ (ℓ₀ - ℓ₄)= – φ Q_P / (ℓ₀ - ℓ₄)²

and F and ℓ are perpendicular to the plane tangent to the plasma surface so that cosine ((π / 4) – ϕ) = 1 ∴

F_p  = – φ Q_P / (ℓ₀ - ℓ₄)

=> F  = – φ Q_P                         ℓ (ℓ)^-1

=> F  = – φ Q_P ln (|ℓ₀ - ℓ₄|)

and if the magnitudes of ℓ and F are known constants (design requirements) then, letting ω = ln (|ℓ₀ - ℓ₄|):

F  / ω= – φ Q_P

We next treat our test particle as a sheet of charge with A = 1. Then q_m1 and Q_P are related by:

q_m1 = Q_P / (ℓ₀ - ℓ₄). Casting the problem as a test charge a distance (ℓ₀ - ℓ₄) from a charged disk (r < 10 in this approximation since curvature will reduce fidelity):

e = F / q_m1 = { Q_P (1 – ([ln(ℓ₀ - ℓ₄)] / √ [ [ln(ℓ₀ - ℓ₄)]² + (ri)²])) } /  [ 8ϵ₀π(ri)² ]

ð  Q_P = F * [ 8ϵ₀π(ri)² ] / { q_m1 (1 – ([ln(ℓ₀ - ℓ₄)] / √ [[ln(ℓ₀ - ℓ₄)]² + (ri)²])) }

And the particle count, N₁, in a “charge” area q_m1 = Q_P / (|ℓ₀ - ℓ₄|) is:

q_m1  / (ϵ)

where ϵ = 4.806 * 10^-19, the averaged charge per particle (at 50/50 mix). In a later section we will calculate the required reactor vessel wall charge.

 Figure 1

The number of particles present must obey the relation:

[Q_P / (4.806 * 10^-19)]  = (2.791 * 10²⁷)^2/3 * (|ℓ₄ - ℓ₅|)

ð  [Q_P / (4.806 * 10^-19)]  = (2.791 * 10²⁷) * (0.1)

ð  Q_P   = (4.806 * 10^-19) * (2.791 * 10²⁷) * (0.1) = 1.34 * 10⁸ Coulomb per plasma 2D sheet.

This yields 1.34 * 10⁸ Coulomb per square meter of plasma and ((1.34 * 10⁸)^1/2)³

= (1.55 * 10¹²) Coulombs / m³. ∴

An explicit machine calculation gives:

Q_P  = {|F| * (|ℓ₀ - ℓ₄|) * ( 8ϵ₀π(ri)² ) / q_m1(1 – (|[ln(ℓ₀ - ℓ₄)]| / √ [|[ln(ℓ₀ - ℓ₄)]|² + (ri)²]))}

|F| = (1.34 * 10^8)  / ((0.1) * ( 8*(8.85 * 10^(-12))*π(5)² ) / (   ((4.806 * 10^(-19)) * (2.791 * 10²⁷) / (0.1) ) * (1 – (|[ln(0.1)]| / √ [|[ln(0.1)]|² + (5)²]))})

|F| = (1.34 * 10^8) / ((0.1) * ( 8*(8.85 * 10^(-12))*pi *(5)^2 ) / ( ((4.806 * 10^(-19)) * (2.791 * 10^27)^(2/3) / (0.1) ) * (1 – ((ln(0.1)) / sqrt ((ln(0.1))^2 + (5)^2)))))

= 3.26 * 10¹⁸ Pascal

{

The W-80 thermonuclear bomb generates about 6.4 * 10¹⁵ pascal pressure. Ivy Mike was 2 to 4 percent fuel fusion.

1 Megaton = 4.184 * 10¹⁵ Joules

A good estimate of the energy release duration is about 50 microseconds, or 1 * 10^-6

∴

the power output is ~ (4.184 * 10¹⁵) / (1 * 10^-6) = 4.18 * 10²¹ W

We will be getting only ~10⁸ W / m³ at ~10¹⁶ Pascal. Notice the dramatic impact the electrostatic repulsion has on pressure. A nuclear explosion is considerably more powerful than what we are proposing here, but the pressures we will have to accommodate are higher. This mismatch is due to the large influence that electrostatic charge in a single species, or highly charged, plasma has on pressure.

}

This is seven orders of magnitude above our target pressure of 10¹¹. We set that value because it represents the current technological limit of materials science.

The plasma total volume per film layer of 5,500 m³ cannot be increased without putting the specific power too low and creating an overall reactor vessel weight > 15000 mT. If we add 10 thin film layers to the design, we can recover an order of magnitude thusly:

Q_P   = (4.806 * 10^-19) * (2.791 * 10²⁶) * (0.1) = 1.34 * 10⁷ Coulomb per plasma 2D sheet.

|F| = (1.34 * 10⁷) / ((0.1) * ( 8*(8.85 * 10^(-12))*pi *(5)^2 ) / ( ((4.806 * 10^(-19)) * ((2.791 * 10²⁶))^(2/3) / (0.1) ) * (1 – ((ln(0.1)) / sqrt ((ln(0.1))^2 + (5)^2)))))

= 7.02 * 10¹⁶ Pascal

This remains five orders of magnitude above the maximum sustainable pressure. The only remaining solution is to dope the plasma, or otherwise moderate its electrostatic pressure, with an electron charge sufficient to decrease the electrostatic forces to a resultant pressure of:

3.9 * 10¹¹ Pascal

while maintaining the same number of fuel ions in the plasma.

First, we recommend working the geometry of the reactor vessel in a more suitable fashion to generate the same target plasma volume. That is, we’ll consider now a diameter of 75 m, a height of 90 m and 25 layers of thin film plasma. This better balances the geometry required while keeping the same volume specified supra.

This could be done by adding a conductive layer in the nanotube laminate blanket such that a negative conductor lies alongside the positive conductor. This negative conductor would spiral outward from the inside to the outside, just as all the other layers do. The negative charge would be just sufficient to relieve the additional pressure applied to the blanket from the ~10¹⁶ Pascal plasma pressure; creating an effective plasma pressure of about 3.9 * 10¹¹ Pascal. The reason for this approach is that the layering of the electron charge carriers will distribute the forces exerted between them and the positive charge in the same way we suggested for the positive charge carriers.

Based on the information thus discovered, we specify the following parameters:

Plasma charge +1.34 * 108 Coulomb with spin 100K m / s arc speed	[eq. 3.14]
Resultant   pressure maximum = 7.02 * 1016 Pascal
25 layers of thin film plasma
Maximum rated Gross power = 11 Tw
Plasma toroid height = 90 m
Plasma toroid diameter = 75 m
Plasma film thickness = 0.1 m
Plasma circumference ≈ 179 m
Plasma total volume = (((pi * 75) *   90) * 0.1) * 25 = 53000 m3

The rationale for these specifications is continued infra.

We will mitigate this high pressure using a negative charge carrier conducting layer mixed with the nanotube laminate layers.  The resultant pressure will then be ~10¹¹. By doing this, we will be able to introduce ~10²⁷ particles / m³ / s for reaction. We shall see the implications of this choice in the continuing discussion.

Description of the multi-layered thin film

[TBC]

This is why polywell fusion and other electrostatic schemes like it cannot work. Using the Ideal Gas Law is extremely deceptive when dealing with a single species, or highly charged, plasma. Gas and plasma have completely different properties. Therefore, another innovation will be required to squeeze (8.84 * 10²⁸) particles/ m³ / s into the reaction plasma. There now exists only one option remaining: doping of the reactor wall with negative charge carrier layers to mediate the electric forces and reduce the forces transferred to materials.

The strongest nanotubes to date have a specific strength of about 48000000 N·m·kg⁻¹. Using the maximum feasible weight for the reactor of 1000 kg / m³ gives:

1000 * 48000000 = 4.8 * 10¹⁰

which is ≈ 1.5 * 10¹¹ Newtons. This represents a current materials science limit.

[TBC]

Obviously, we will need to determine, by at least an order of magnitude, if the Fissile Kicker will have sufficient power to inject the fuel as required supra.

The Reaction Rate

Ncm   = ρplasma \ 1.0880085 * 10-26 kg	[eq. 3.15]
Number of fuel   particles per m3 of plasma
ð    Reference Design   Twenty-One Castle Mike Ncm   = 2.791 * 1026 particles / m3

The reactor Reference Design has a total fusion reaction volume of approximately 5525 m³  (assuming a thin film laminate with a total of 3 inches thickness in reaction material).

The rate at which fusion reactions occur in a volume of fuel is called the reaction rate. And it is given by the equation:

μ= n1n2       σR(Ep) vT ai √ (N2(ai))	[eq. 3.16]
Fusion Reaction   Rate
Reference Design Twenty-One   Castle Mike => μ = 2.6 * 10-8/ atom / m3 / s and
Reference Design Twenty-One   Castle Mike => μ = 7.26   * 1018 / m3 / s

where n₁ is the reactant number density of the projectile/s and n₂ is the reactant number density of the target. n₁₊₂ is one if no non-fuel species are present. σ_R(E_p) is the “cross section” as a function of E_p, E_p is the energy of the incident particle, and v_T is the relative particle velocity, all of which are due to relative motion only. E_r is the energy released from each reaction (8.7 MeV). N₂ is the particle density function of the “incident” particle stream, and N₁ is the total number of particles available for reaction. Power density follows directly with:

ρP / m3 = ErN1n1n2       σR(Ep) vT  * |(ai)| * √ (N2(ai))	[eq. 3.17]
Fusion Power   Density
Reference Design Twenty-One   Castle Mike => ρP ~108 W / m3

μ= n₁n₂  σ_R(E_p) v_T ai √ (N₂(ai))

μ= (0.2)*(0.8) * (ln(0.1)) * (1.2 * 10^-28) * (1.12 * 10^7)* ((2.791 * 10^27) ^(1/2))

= 2.6 * 10^-8

and (2.791 * 10²⁶) * (2.6 * 10^-8) = 7.26 * 10¹⁸ reactions / m³ / s (we justify 10²⁶ later)

Burn to exhaustion elapsed time is:

(2.791 * 10^27) / ((2.791 * 10^27) * (2.6 * 10^-8)) = 3.85 * 10⁷ seconds or 445 days at maximum power

If the probability of interaction is given by:

P =  σ_R * v* x* √ (N₂(x)) dx

(1.60217646 * 10^-19) * (8.56 * 10⁶) = 1.37 * 10^-12

Where N₂ is the particle density function of the “incident” particle stream, and if N₁ is the total number of particles available for reaction, then power density is given by:

ρ_P / m³ = E_rN₁n₁n₂  σ_R(E_p) v_T * |ai| * √ (N₂(ai))

ð  E_r * n₁n₂ * N₁ * σ_R(E_p) * v_T * |ln(a)| * √N₂

And as discussed previously, the particle count, like power density, will decrease as the square of the distance from the center of the plasma. Thus “a” becomes ln a (which is the solution to the integral above).

2* (0.2)*(0.8) *(8.6605021 * 10^26) * [|ln(0.1)|] * (1.2 * 10^-28) * (1.12 * 10^7)* ((8.6605021 * 10^26) ^(1/2)) * (1.37 * 10^-12) ~ 3.46 * 10⁷ watts / m³

ρ_P / m³ = N₁ *

ð  N₁ * σ_R * v * √ N₂

Using the velocity calculated in our cross section discussion, we can write an explicit machine calculation for a distribution of notable N₁ values as:

1.)   2* (0.2)*(0.8) *(8.6605021 * 10^25) * [|ln(0.1)|] * (1.2 * 10^-28) * (1.12 * 10^7)* ((8.6605021 * 10^25) ^(1/2)) * (1.37 * 10^-12) = 1.09 * 10⁶ watts / m³

2.)   2* (0.2)*(0.8) *(8.6605021 * 10^26) * [|ln(0.1)|] * (1.2 * 10^-28) * (1.12 * 10^7)* ((8.6605021 * 10^26) ^(1/2)) * (1.37 * 10^-12) = 3.46 * 10⁷ watts / m³

3.)   2* (0.2)*(0.8) *(5 * 10^27) * [|ln(0.1)|] * (1.2 * 10^-28) * (1.12 * 10^7)* ((5 * 10^27) ^(1/2)) * (1.37 * 10^-12) = 4.8 * 10⁸ watts / m³

4.)   2* (0.2)*(0.8) *(2.791 * 10^27) * ((-1)*ln(0.1)) * (1.2 * 10^-28) * (1.12 * 10^7)* ((2.791 * 10^27) ^(1/2)) * (1.37 * 10^-12) = 2 * 10⁸ watts / m³

5.)   2* (0.2)*(0.8) *(8.6605021 * 10^27) * [|ln(0.1)|] * (1.2 * 10^-28) * (1.12 * 10^7)* ((8.6605021 * 10^27) ^(1/2)) * (1.37 * 10^-12) = 1.09 * 10⁹ watts / m³

6.)   2* (0.2)*(0.8) *(8.6605021 * 10^28) * [|ln(0.1)|] * (1.2 * 10^-28) * (1.12 * 10^7)* ((8.6605021 * 10^28) ^(1/2)) * (1.37 * 10^-12) = 3.46 * 10¹⁰ watts / m³

The reaction vessel contains 5500 * x = y m³ plasma. ∴

(1.09 * 10^6)*(55000) = 60 Gw total power output

(3.46 * 10^7)*(55000) = 1.9 Tw total power output

(2 * 10^8)*(55000) = 11 Tw total power output

(1.09 * 10^9)*(55000) = 60 Tw total power output

(3.46 * 10^10)*(55000) = 1.9 Pw total power output

And we’ve specified

(2 * 10^8)*(55000) = 11 Tw total power output

as the design requirement. This requirement works out to be 11 Tw maximum total output, 7 Tw usable maximum total output and 5.25 Tw guaranteed continuous; which is the requirement we sought to satisfy. The difference of 7 – 5.25 is a function of how the reactor is fueled and at what pressure range it operates since as it proceeds through a full burn it will consume some of that fuel and the internal pressure would be variable. At burn initiation there would be 7 Tw power slowly decreasing monotonically to a minimum power of 5.25 Tw before the burn ends.

We can now see that as we increase the number of atoms in each m³ we increase the cross section. That in turn increases the reaction rate. The values for n₁ and n₂ are the concentration percentages for projectile specie and target specie in the plasma (for the ²H-¹¹B reaction it is recommended to be something like 8 to 1, however, by this equation alone, the obvious optimization is half and half).

To determine the fuel consumption rate we use:

c =   rVmp	[eq. 3.18]
2H-11B fuel consumption at   maximum rated power
Reference Design   => (7.26 * 1019)*(55000)*( 2 * 1.0880085 * 10-26) ≈ 0.087 kg / s
18 hour burn consumption: (18 * 3600) *   (0.087) = 5638 kg mass

where c is the fuel consumption rate in kg per second, r is the number of particles interacting per second per m³ (which is 7.26 * 10¹⁹), V is the volume of the reaction plasma  and m_p is the mass of the two particles colliding (which is 2 * 1.0880085 * 10^-26).

This results in both a commercially and scientifically sensible alternative, at an entry level, to a fission power source. At a pressure of 1 million bar the plasma has the consistency of light Styrofoam, about 22 pounds per cubic meter.

We are now in a position to consider the effect of plasma current driven by alternating current. We believe an achievable goal would be to increase the toroidal magnetic force operating on the plasma by a factor of ten by increasing the current of the plasma. We recommend increasing the rotational velocity of the plasma to 0.01c, or about 3 million m/s.

This will result in an order of magnitude increase for values of N, therefore,

5.)   2* (0.2)*(0.8) *(8.6605021 * 10^27) * [|ln(0.1)|] * (1.2 * 10^-28) * (1.12 * 10^7)* ((8.6605021 * 10^27) ^(1/2)) * (1.37 * 10^-12) = 1.09 * 10⁹ watts / m³

With a vacuum/reaction vessel containing a much more feasible 550 * x = y m³ plasma. ∴

(1.09 * 10^9)*(5500) = 6 Tw total power output

Allowing for some variance in initial and final power output by varying the pressure over the burn, we have a range of values of:

5 to 7 Tw, which is exaclty what we sought.

Recasting our previous considerations yields

Plasma charge +1.34 * 108 Coulomb with spin 3 * 106 m / s arc speed	[eq. 3.14]
Resultant pressure maximum = 7.02 * 1017 Pascal
25 – 100 layers of thin film plasma
Maximum rated Gross power = 11 Tw
Plasma toroid height = 20 m
Plasma toroid diameter = 10 m
Plasma film thickness = 0.1 m
Plasma circumference ≈ 31.4 m
Plasma total volume = (((pi * 10) *   18) * 0.1) * 25 = 5654 m3

Which results in weights that are feasible for placing in orbit.

Finally, we must be careful to note that to create the pressures discussed here, a force equal to the sum of this pressure and the electrostatic repulsion of the E field within must be present.

Next, we consider the boundary limitations required for the plasma geometry and its magnetic and electric fields. The first thing we want to understand is whether or not thermal perturbations of toroidal orbits will be annealed within the plasma without destabilizing it. For magnetic only confinement this is a problem that cannot be easily calculated. However, in our scheme, the presence of an E field makes this much easier.

We can frame the problem thusly. We seek a geometry and electrostatic force such that, whenever a particle of ¹¹B or ²H is perturbed from its orbital path, it’s deviation does not place the particle outside the electrostatic potential well.  We can diagram it thusly:

Figure 1 (a) and (b) below

Let us begin by drawing a line segment in Figure 1 (b) in order to evaluate the electrostatic forces involved. We will mark each point with the letter ℓ_n∈ ck. U(p₀) is located at the point charge provided by the charge contained in the nanotube blanket’s outer layers (the total charge in the outer superconducting layers), call it ℓ₀. U(p₁) is located at the reactor vessel’s outer (meaning outer toroidal) interior wall. This is the ultimate orbital boundary which fusion product should never reach. We denote that point ℓ₁. Continuing to work our way inward from the outside of the torus toward the plasma, the next point is E_nout, call it ℓ₂. This is the point at which a locally, electrically neutral orbit can be found. This orbit exists because a modest negative charge exists on the solid wall surface at ℓ₁. It is considerably weaker than the primary charges we will consider, but is strong enough at ℓ₂ to have a local effect whereby any particle in the immediate vicinity (i.e. between any demarcated points in ℓ) will “feel” a force significant to the particles perturbations in this area which we will examine shortly. Continuing through the vacuum, we next consider point ℓ₃ (h_out) which is the outer horizon, or outer surface of the plasma. The next point, ℓ₄, is the component location at ck where the particle being acted upon is located. Here we will consider bj = 0 and ai > 0. The next point, ℓ₅, is the electrically neutral orbit at the center of the plasma. This is the orbit whereby the forces applied from Q_o and Q_i on a particle at this orbit cancel. It is a surface and has a toroidal shape. In the diagram of Figure 1 (b) it is given that ℓ₄ = ℓ₅ but this need not be so. However, the upper boundary we wish to test is for a reaction that occurs at the plasma surface, that is, ℓ₃, and we seek the condition ℓ₃ = ℓ₄. At this orbit the electrostatic potential energy to acquire in order to reach ℓ₁ is minimized. Continuing along the line segment we next arrive at ℓ₆ which is the inner horizon and is analogous to ℓ₃. Points ℓ_6-9 have like analogies to the outer points. Therefore, the condition we must enforce is:

Assuming ℓ₀ > ℓ₉ the boundary condition is met at:

-U(ℓ1)m1   = [( Fℓ0 -  Fℓ4   ) ⨀ (ℓ0 - ℓ4   )] – [(Fℓ4 -  Fℓ9)⨀ (ℓ4 - ℓ9 )]	[eq. 3.19]
Electrostatic   potential energy

where F denotes a force vector.

And it is required that:

|-650 KeV| = [( F_ℓ0 -  F_ℓ4 ) ⨀ (ℓ₀ - ℓ₄ )] – [(F_ℓ4 -  F_ℓ9)⨀ (ℓ₄ - ℓ₉ )]

ð  1.0414 * 10^-13 = [( F_ℓ0 -  F_ℓ4 ) ⨀ (ℓ₀ - ℓ₄ )] – [(F_ℓ4 -  F_ℓ9)⨀ (ℓ₄ - ℓ₉ )]

ð  1.0414 * 10^-13 / cos( θ ) =  [( F_ℓ0 -  F_ℓ4 ) * (ℓ₀ - ℓ₄ )] – [(F_ℓ4 -  F_ℓ9)* (ℓ₄ - ℓ₉ )]

If we set s = 0.01 (1 cm) and cos( θ ) = 1 ( a good choice as this is the upper bound condition) then:

1.0414 * 10^-13 =  [( F_ℓ0 -  F_ℓ4 ) * (ℓ₀ - ℓ₄ )] – [(F_ℓ4 -  F_ℓ9)* (ℓ₄ - ℓ₉ )]

ð  1.0414 * 10^-13 + [(F_ℓ4 -  F_ℓ9)* (ℓ₄ - ℓ₉ )] = [( F_ℓ0 -  F_ℓ4 ) * (ℓ₀ - ℓ₄ )]

Let φ = κ_e q_m1 where κ_e is the electric constant. Then the equations for potential energy, starting with forces, are:

F_o = φ Q_o / (ℓ₀ - ℓ₁)² – φ Q_o / (ℓ₀ - ℓ₄)²

F_i = φ Q_i / (ℓ₄ - ℓ₉)² – φ Q_i / (ℓ₁ - ℓ₉)²

F_p = φ Q_p / (ℓ₄ - ℓ₅)² – φ Q_p / (ℓ₁ - ℓ₅)²

F_R = F_o – F_i – F_p

ð  U( Δℓ )_m1  = F_R ⨀ (ℓ₁ - ℓ₄) ∴

We can expand thusly:

U( Δℓ )_m1  =

[φ Q_o / (ℓ₀ - ℓ₁)² - φ Q_o / (ℓ₀ - ℓ₄)²  -  φ Q_i / (ℓ₄ - ℓ₉)² - φ Q_i / (ℓ₁ - ℓ₉)²  - φ Q_p / (ℓ₄ - ℓ₅)² - φ Q_p / (ℓ₁ - ℓ₅)²]

⨀ (ℓ₁ - ℓ₄)

ð  U( Δℓ )_m1  =

φ [Q_o / (ℓ₀ - ℓ₁)² - Q_o / (ℓ₀ - ℓ₄)²  - Q_i / (ℓ₄ - ℓ₉)² - Q_i / (ℓ₁ - ℓ₉)²  - Q_p / (ℓ₄ - ℓ₅)² - Q_p / (ℓ₁ - ℓ₅)²] ⨀

(ℓ₁ - ℓ₄)

And since equilibrium requires that Q_i = Q_o we can simplify with

Q = Q_i = Q_o

ð  U( Δℓ )_m1  =

φ Q[1 / (ℓ₀ - ℓ₁)² - 1 / (ℓ₀ - ℓ₄)²  - 1 / (ℓ₄ - ℓ₉)² - 1 / (ℓ₁ - ℓ₉)²] – φ Q_p [1 / (ℓ₄ - ℓ₅)² - 1 / (ℓ₁ - ℓ₅)²] ⨀

(ℓ₁ - ℓ₄)

And cosine returns 1 when examining the upper bounds so that,

U( Δℓ )_m1  =

φ Q(ℓ₁ - ℓ₄)[1 / (ℓ₀ - ℓ₁)² - 1 / (ℓ₀ - ℓ₄)²  - 1 / (ℓ₄ - ℓ₉)² - 1 / (ℓ₁ - ℓ₉)²] – φ Q_p (ℓ₁ - ℓ₄) [1 / (ℓ₄ - ℓ₅)² - 1 / (ℓ₁ - ℓ₅)²]

And the boundary condition of interest is solving for Q or Q_p for given values in ℓ and the constraint that the electrostatic potential well must be greater than or equal to the highest thermal energy of any fuel particle (we examine fusion product separately). For the ²H-¹¹B reaction this value will be constant, namely;

T_e = 1.0414 * 10^-13

ð  [ T_e / φ(ℓ₁ - ℓ₄) ]+ Q_p [1 / (ℓ₄ - ℓ₅)² - 1 / (ℓ₁ - ℓ₅)²] =

Q [1 / (ℓ₀ - ℓ₁)² - 1 / (ℓ₀ - ℓ₄)²  - 1 / (ℓ₄ - ℓ₉)² - 1 / (ℓ₁ - ℓ₉)²]

So that, solving for Q_p we have:

Q_p  =  Q[1 / (ℓ₀ - ℓ₁)² - 1 / (ℓ₀ - ℓ₄)²  - 1 / (ℓ₄ - ℓ₉)² - 1 / (ℓ₁ - ℓ₉)²] – [T_e / φ(ℓ₁ - ℓ₄) ] / [1 / (ℓ₄ - ℓ₅)² - 1 / (ℓ₁ - ℓ₅)²]

Or for Q:

Q = [T_e / φ(ℓ₁ - ℓ₄) ]+ Q_p[1 / (ℓ₄ - ℓ₅)² - 1 / (ℓ₁ - ℓ₅)²] /  [1 / (ℓ₀ - ℓ₁)² - 1 / (ℓ₀ - ℓ₄)²  - 1 / (ℓ₄ - ℓ₉)² - 1 / (ℓ₁ - ℓ₉)²]

where κ_e = 8.987 * 10⁹ N * m² / C². We can state this equation as a function whose domain includes; T_e, ℓ₀, ℓ₁, ℓ₄, ℓ₅, ℓ₉, and q_m1. However, the general form would be:

Q_p (T_e, ℓ₀, ℓ₁, ℓ₄, ℓ₅, ℓ₉, φ(q_m1) )

Q (T_e, ℓ₀, ℓ₁, ℓ₄, ℓ₅, ℓ₉, φ(q_m1))

Assuming ℓ₀ > ℓ₉ the boundary condition is met at:

Qp  =Q[1 /   (ℓ0 - ℓ1)2 - 1 / (ℓ0 - ℓ4)2  - 1 / (ℓ4 - ℓ9)2   - 1 / (ℓ1 - ℓ9)2] –[Te / φ(ℓ1   - ℓ4) ] / [1 / (ℓ4   - ℓ5)2   - 1 / (ℓ1 - ℓ5)2]	[eq. 3.20]
Minimum total   plasma charge required for thermal stability	{Ep ≥ 650 KeV}

And

Q =[Te / φ(ℓ1   - ℓ4) ]+ Qp[1   / (ℓ4 - ℓ5)2 - 1 / (ℓ1 - ℓ5)2] /[1 /   (ℓ0 - ℓ1)2 - 1 / (ℓ0 - ℓ4)2  - 1 / (ℓ4 - ℓ9)2   – 1 / (ℓ1 - ℓ9)2]	[eq. 3.21]
Minimum total   wall charge required for thermal stability	{Ep ≥ 650 KeV}

The pressure of the plasma is just the sum of the forces acting on a test particle located at the point where pressure is measured; i.e.

The force acting at a given point can be solved by just choosing a point:

f_o = φ Q_o / (ℓ₀ - ℓ₁)²

f_i = φ Q_i / (ℓ₄ - ℓ₉)²

f_p = φ Q_p / (ℓ₄ - ℓ₅)²

^OR

f_p = φ Q_p / (ℓ₀ - ℓ₄)²

f_R = f_o + f_i + f_p

|f_R| = 1.5 * 10⁶ Bar

ð  |f_R| = |f_o| – |f_i| – |f_p| = 1.5 * 10⁶ Bar = 1.5 * 10¹¹ Newtons

And the field is:

e_o = κ_e Q_o / (ℓ₀ - ℓ₁)²

e_i = κ_e Q_i / (ℓ₄ - ℓ₉)²

e_p = κ_e Q_p / (ℓ₄ - ℓ₅)²

The rate change in these forces can be given by taking the ratio of any F to a unit distance lying along a boundary condition vector, say ℓ_bc. That is,

∇F_o = φ Q_o   { ∂² / ∂ } [1 / (m₀ - m₁)² - φ Q_o / (m₀ - m₄)² ] / (m₀ - m₄)

∇F_i = φ Q_i     { ∂² / ∂ } [1 / (m₄ – m₉)² - φ Q_i / (m₁ - m₉)² ] / (m₀ - m₄)

∇F_p = φ Q_p   { ∂² / ∂ }  [1 / (m₄ - m₅)² - φ Q_p / (m₁ – m₅)² ] / (m₀ - m₄)

∇F_R = ∇F_o + ∇F_i + ∇F_p

ð  ∇U( Δℓ )_m1  = ∇F_R ⨀ (m₁ - m₄)

where we use m to denote position vectors (all components).

Let:

κ_e = 8.987 * 10⁹

q_m1 = 4.806 * 10^-19

And (1.5 * 10¹¹) is the force required over one square meter, so q_m1 must be the charge present on a square meter surface of thin film; thus q_m1 becomes q_p.

Let:

φ = (8.987 * 10⁹) (4.806 * 10^-19).

ℓ₀ = 0.55

ℓ₁ = 0.50

ℓ₄ = 0.45

ℓ₅ = 0.40

ℓ₉ = 0.25

We found supra that the number of particles per cubic meter in the plasma is about (6.66 * 10²⁷). Since the average charge for each particle was estimated at 4.806 * 10^-19 we will assume that value. And the electric field we denote e:

e_o = κ_e Q_o / (ℓ₀ - ℓ₁)²

e_i = κ_e Q_i / (ℓ₄ - ℓ₉)²

e_p = κ_e Q_p / (ℓ₄ - ℓ₅)²

We will cast the problem by expressing Q as a charge density using the “force of a charged disk on a test charge” form:

e = ρ_Qo / 2ϵ₀ (1 – Δℓk / √ [Δℓk² + (ri)²])

ð  e = F / q_p = { ρ_Qo(1 – (Δℓk / √ [Δℓk² + (ri)²])) } / 2ϵ₀

where ϵ₀ = 8.85 * 10^-12 and ρ = Q / 4π(ri)² so,

e = F / q_p = { Q₀ (1 – (Δℓk / √ [Δℓk² + (ri)²])) } /  [ 8ϵ₀π(ri)² ]

ð  Q₀ = F * [ 8ϵ₀π(ri)² ] / { q_p (1 – (Δℓk / √ [Δℓk² + (ri)²])) }

ð  F = { q_p (1 – (Δℓk / √ [Δℓk² + (ri)²])) } / [ Q₀ 8ϵ₀π(ri)² ]

We can set |F| to |F_dr|, the Reference Design required pressure by constraining our evaluation to a square meter surface and casting F as the force required per particle:

Q₀ = { (3.26 * 10¹⁸) / ((2.791 * 10^27)^(2/3)) } * [ 8ϵ₀π(ri)² ] / { q_p (1 – (Δℓk / √ [Δℓk² + (ri)²])) }

ð  Q_i = { (3.26 * 10¹⁸) / ((2.791 * 10^27)^(2/3)) } * [ 8ϵ₀π(ri)² ] / { q_p (1 – (Δℓk / √ [Δℓk² + (ri)²])) }

ð  Q_p = { (3.26 * 10¹⁸) / ((2.791 * 10^27)^(2/3)) } * [ 8ϵ₀π(ri)² ] / { q_p (1 – (Δℓk / √ [Δℓk² + (ri)²])) }

Q0 = { (3.26 * 1018) /   ((2.791 * 10^27)^(2/3)) } * [ 8ϵ0π(ri)2   ] / { qp (1 – (Δℓk   / √ [Δℓk2 + (ri)2]))   }	[eq. 3.22]
Required charge   at reactor vessel wall

q_p is a sheet of charge tangent to the plasma surface and located within it. Previously, when we calculated the required plasma charge, we used the symbol Q_P to denote the same thing. It corresponds to the horizon (h_out) in Figure 1.

Explicitly, we can set:

ℓ₀ - ℓ₁ = 0.19 m

ℓ₄ - ℓ₉ = 0.24 m

ℓ₄ - ℓ₅ = 0.1 m

(ri) = 5

q_p  = plasma surface = ((s)*(1.34 * 10⁷)) [see calculations infra]

s = plasma spin speed = 100000 m / s

Q₀ = ( (7.02 * 10¹⁶) / ( (2.791 * 10^26)^(2/3) ) ) * ( 8 * ( 8.85 * 10^-12) * pi * (5)^2 ) / ((((1)*( (1.34 * 10⁷)) * ((4.806 * 10^-19)))) * (1 – ((0.01) / sqrt((0.019)^2 + (5)^2))) )

= 142.2 Coulomb

142.2 / (4.806 * 10^-19) = 2.96 * 10²⁰; the number of charge carriers / m³ in the wall and

(7.02 * 10¹⁶) / (2.96 * 10²⁰) = 2.37 * 10^-4 Newtons / particle / m²

100000 * ((2.791 * 10^26)) * (4.806 * 10^-19) * 2*(ℓ₄ – ℓ₅) = q_m1 = 2.68 * 10¹²

=> (ℓ₄ – ℓ₅) = q_m1  / 100000 * ((2.791 * 10^26)) * (4.806 * 10^-19) * 2

=> (ℓ₄ – ℓ₅) = (2.68 * 10¹²) / (100000 * ((2.791 * 10^26)) * (4.806 * 10^-19) * 2)

=> (ℓ₄ – ℓ₅) = 0.1 m

Is the thickness required of the plasma sheet and is partly the basis for characterizing this approach as thin film fusion.

[TBC]

At this point a brief digression is in order. Previously we calculated the density of our plasma; however, this was an approximation using the Ideal Gas Law. What we see here is that the density we specified will not be the most preferable. The density will, in fact, be driven by the electrostatics we are now discussing. The above calculation assumes a pressure of 1.5 * 10¹¹. This is an “electrostatic” pressure. If we specify a particular density at that pressure we will necessarily affect the total charge of the plasma. The higher the charge density of the plasma the further the values of charge for wall and plasma diverge; which also follows logically from the fact that we have a required resultant pressure we are seeking. But that same density affects the reaction rate. These are the interdependencies, qualitatively described, that have been lurking in the equations.

The key Reference Design question is “what is the lowest wall charge we can allow and still be able to transfer the pressure, or force, of 1.5 * 10¹¹ Newtons to the torus structure”?

We can address this numerically by solving for the force exerted by a charge in the superconductor layer onto the Boron Nitride insulator layer. The key question is, does the particle have enough force to penetrate the very hard Boron Nitride surface? This now exposes the wisdom in choosing Boron Nitride Wurtzite nanotube material:

BNW begins to show noticeable deformation at around 1 GPa. But this is just beyond the pressures we have specified. While this cannot constitute an exact statement of the forces involved and considerable experimentation will be required, we can see that BNW has an order of magnitude sufficient hardness to resist penetration by conducting ions from the adjacent conductor layer. However, these hardness ratings are based on cross sections of about 1 mm. Taking the cross section of AgSe as an example, we know that one such ion occupies a space of very roughly:

(3 * 10^-10)²

And hardness tests are conducted by applying pressure over a region of about:

(1 * 10^-3)²

ð  (1 * 10^-3)² / (3 * 10^-10)² = 1.1 * 10¹³

142 Coulomb wall charge:

(142) / 2π (ri)² = 0.9 Coulombs m²

0.9 / (4.806 * 10^-19) = 1.87 * 10¹⁸ charge carriers / m² ∴

(1.87 * 10¹⁸) / (10⁶) = (1.87 * 10¹²) and if we specify ~100 nanotube layers in the torus:

(1.87 * 10¹²) * (10²) / (1.1 * 10¹³) = 17.34 that is;

The BNW material is just within an order of magnitude, assuming 100 nanotube layers, of being able to maintain structural integrity under the force of a charge of only 142Coulombs. The reason for this is that the fewer charge carriers present the more force each has to impart to the surrounding material to obtain the normal force they require. Thus, the lowest wall charge that can likely be fabricated with current materials and development is about:

1 * 10^-3 Coulomb

Obviously, as a development risk, and given the very broad approximations we are making regarding this untested material, we specify an effective lower limit of:

1 * 10^-1 Coulomb

f_p = φ Q_p / (ℓ₀ - ℓ₄)²

Q_o = [(8.987 * 10⁹) (4.806 * 10^-19)] / [{(1.5 * 10^11)  / ((2.791 * 10^26)^(2/3))} * (.2)²]

The force acting to push plasma apart (destabilize it) is:

ð  F = { q_p (1 – (Δℓk / √ [Δℓk² + (ri)²])) } / [ Q₀ 8ϵ₀π(ri)² ]

Which we find to be negligible under our thin film specification, but we must include it to accommodate changes.

We have validated that this plasma can possess the charge necessary to provide a plasma pressure of 1.5 million Bar. We now turn our attention to the question of whether or not the current materials state of the art can provide a material that can hold sufficient charge for the requirements derived infra.

Average charge of fuel particle in plasma:

(5+1/2 = 3) * 1.602 * 10^-19 = 4.806 * 10^-19 and

Earlier we noted that a m³ of plasma at 1.5 * 10⁶ Bar contained 2.791 * 10²⁶ of fuel particles mixed half and half. To get the charge counts, we can average over that value thusly:

5 +1=6

ð  6 / 2 = 3 is the average charge per particle and

ð  3 * 8.6605021 * 10²⁶ is the charge carrier count present at a material density of 9.4227 kg / m³

But since we are looking for a charge contained in the laminate solid layers (proton superconductors) we must modify this density to be more realistic. Let us pick a density for an order of magnitude approximation by assuming the density of copper. This is around 9000 kg / m³. And 9000 / 9.4 ~ 10³. Therefore, for a laminate layer we could realistically expect to squeeze in as much as

3 * (2.791 * 10²⁶) * (4.806 * 10^-19) * 10³ * 5500= 2.2 * 10¹⁵

yielding results similar to what we obtained for the plasma. At this point we are now in a much better position to begin describing the magnetic fields and the field interactions.

As we can see, thermal instabilities are easily excluded from this plasma. Ironically, the challenge we will face here won’t be instabilities in plasma. This plasma will follow B lines like solid rock. However, the intense pressure of the plasma is what makes this challenging technologically. We will next calculate whether or not this many charge carriers will prevent the fuel product from escaping to the neutral E orbit.

A related question however, is will the product overshoot the neutral surface and impinge on the vessel wall? To prevent this the interesting relation must hold:

Fs   > Fp + 4 MeV / ℓ	[eq. 3.23]
Boundary for   separation of product from reactor vessel wall

The product plasma is also sufficient by itself to deflect all particles incident upon it due to heating. This means that should the reaction plasma and electrostatic forces from solid materials be insufficient to fully correct and stabilize a particle, the product plasma can force any particle of any heat energy back into the reaction plasma where the electrostatic forces acting therein are sufficient to stabilize its orbit.

To explicitly verify that our Reference Design lies within an order of magnitude from what is achievable we can use a simple construction. In Figure 1, h_in denotes the plasma inner surface and U(p₁) denotes the potential energy of a particle reaching the outer reactor vessel wall (by “outer” we mean the interior surface of the reactor vessel furthest from the center of the torus). To establish the boundary condition a little more precisely, we seek the condition such that a nucleus of ⁴He product at 4MeV discharging from h_in with a velocity vector, v_{0 ,} pointed directly at U(p₁); that is, perpendicular to the tangent plane of the reactor vessel wall. This is the most energy efficient path a reaction product can take to reach the reactor vessel wall. In this mode the particle has an initial component magnitude of velocity in this direction that is equal to its speed. In order to prevent it from striking the wall and to possibly contain it, we need a force of deflection perpendicular to v_i with a force sufficient to compel an acceleration which in turn is sufficient to change the vector v₀ by π / 2 while | v₀ | remains constant. Furthermore, this angular change must place v₀ parallel to the orbital path defined by magnetic field lines for a particle with its initial conditions.

The line lying along the points denoted at h_out – U(p₁) in Figure 1 is the vacuum gap through which the particle in question will take a parabolic path around the torus. After a deflection of π / 2 the particle will promote to orbit.

Applying some trigonometry to the problem, we first note that deflections do not occur in one basis axis only, but occur on the i’th and j’th bases (see Figure 1). Therefore, there are two magnetic fields in play here (we will look at this more closely later) and the deflection of fusion product will occur across a plane, not an axis. Therefore, the acceleration required of each deflected particle is determined by calculating the trigonometric component values for v₀. If s is the arc length of the parabolic path traveled and | v₀ | remains constant, the i’th acceleration component required across the i-j plane is given by:

| ai | = ½ (| v0 | cos (α))2 / s	[eq. 3.24]
Reactor vessel   geometric dependency; α = π / 4
|ai|= ½ (1.3836   * 107) * cos (π / 4)   / (1/4 * π * 75) = 8.3 * 104 m / s / s

Where α is the angle between the i’th and j’th components of v₀. Note that the circumference is divided by 4 because the interval, s, extends only for ¼ rotation. But what we would like to know is how much force must act on the particle to impart this acceleration to it. That will finally tell us how strong of a B field must be present (we will describe the B fields in detail infra). Of course, we know that F = dp / dt, where p is a classic momentum vector, which just works out to be F = ma. ∴

|F| / m = ½ | v₀ |² / s                   [because F ⨀ s = ½ mv²]

where m is the mass of the deflected particle. Recall the mass of the proton and Neutron, respectively,

1.672621777 × 10⁻²⁷ kg,

1.674927351 × 10⁻²⁷ kg

In this Reference Design the mass of the ⁴He product is

m = 2(1.672621777 × 10⁻²⁷) + 2(1.674927351 × 10⁻²⁷)

and at 4 MeV we have:

| v₀ | = √ [2 * (4000000 * (1.60217646 * 10^-19)) / (6.695 * 10^-27)]

ð  | v₀ | = 1.3836 * 10⁷ m / s and

ð  | v₀ | / c = 0.04615

With a Reference Design specification of 75 meters diameter for the circle marking the thin film plasma, we set s = 75*π meters. This is a conservative figure as it only represents circumference, not a helical path. Finally, since F = ma we get:

F = (6.695 * 10^-27) * (8.3 * 10⁴) = 5.56 * 10^-22 Newtons per reaction.

The amperage of the plasma is calculated by solving for the amount of charge passing a cross section of the torus in one second. For now we will naively assume a velocity for the plasma which we may refine later, but one based on typical plasma speeds in today’s tokamaks; about 100,000 m / s.

The total number of charge carriers in the Reference Design plasma is, as indicated above, 2.791 * 10²⁶ * 53000 = 1.48 * 10³¹. For a plasma constant pressure gradient (this gradient is only constant for vectors lying in the tangent plane to the plasma surface. The vector v₀ must by definition lie in this plane) the charge is (4.806 * 10^-19) * (1.48 * 10³¹) = 7.11 * 10¹² Coulombs.

With a circumference of about 75*π m the torus is traversed 100,000 / 75*π = 4.19 * 10³ times per second.  ∴, the total plasma current is:

i = (4.806 * 10^-19) * (1.48 * 10³¹) * (4.19 * 10³) = 2.98 * 10¹⁶ Amperes

Since μ = (2.6 * 10^-8), the number of reactions occurring per unit length is:

((2.6 * 10^-8) * (2.791 * 10^26))^(1/3)  = 1.94 * 10⁶

=> (5.56 * 10^-22) * (1.94 * 10⁶) = 1.08 * 10^-15 Newtons

And to get the required Tesla for this deflection we multiply:

(2.98 * 10¹⁶)*(1.08 * 10^-15) = 32.18 Tesla

We recall from eq. 3.02 (magnetic force acting on q₁ due to q₂) that,

F_m12 = κ_m q₁ q₂ * (v₁ X (v₂ X r₁₂)) / r²₁₂  ∴

7.315 * 10^-22 = (1.2567 * 10^-6) q₁ q₂ * (v₁ X (v₂ X r₁₂)) / r²₁₂

As an order of magnitude estimate we will choose a point charge representing one cubic meter of “local” reaction plasma. This results in a charge, q₂ = (2.791 * 10²⁶) * (1.60217646 * 10^-19) = 4.47 * 10⁷ Coulombs. We can set q₁ to the charge of ⁴He product; i.e. (2 * 1.60217646 * 10^-19), however, the figure we seek is the product of this value and the total number of product nuclei in the cubic meter we are considering which undergo deflection. Therefore, q₁ = (μ = 7.26 * 10¹⁸) * (2 * 1.60217646 * 10^-19) = 2.33 Coulombs. r is a bit trickier for estimation purposes. Since the electrostatic force drops with the square of the distance, we need an “average” distance in a 1 m³ volume, assuming q₁ lies on that averaged surface within the cubic meter evaluated and q₂ is at the center. Without delving into the mathematical theorems that justify this claim, for now we will note that a distance of ½ * 1/3 = 1/6 meter from q₂ is a sufficient approximation. ∴

F_m12 = 1.4 * 10^-24 = (1.2567 * 10^-6) * (2.33) * (4.806 * 10^-19) * (v₁ X (v₂ X r₁₂)) / (1/6)²

which just reduces to:

(1.4 * 10^-24) / (2.1709 * 10^-23) = 0.0645 = (v₁ X (v₂ X r₁₂))

0.0645 = (|v₁| sin (ω) (|v₂| sin (δ) / 6))

However, the upper boundary condition we seek to validate consists of a geometry ∃

sin (ω) = sin (δ) = 1 so,

6 * 0.0645 = |v₁| * |v₂|

ð  0.388 = |v₁| * |v₂|

There is little need to go further as we can quickly see that the deflection forces are orders of magnitude more than sufficient to deflect product from the reactor vessel wall. Please note that this likewise means that the confinement forces which force fuel particles into their orbits upon thermal heating to 650 KeV are yet stronger; by several orders of magnitude, than the thermal energies. And this is just the magnetic field. But there is a related question that must be addressed. How much energy does it require to deflect these particles? This is a pertinent question since the product possesses the very energy we seek to harness and we would not want deflection to reduce its energy by any appreciable amount.

Recall that |v₀| = 1.3836 * 10⁷ m / s. Thus the energy of the particle at creation is,

k = ½ m|v₀|² = 4 MeV and the energy expended to deflect the particle is,

U = – F ∙ s

The problem, of course, is that the generous magnetic field created allows for a wide range of values for s. Therefore, we will pick a reasonable value for what it is worth and try to validate this better once a smaller range for s can be determined (in what follows). Let s = 0.001 m. This is a very steep deflection but appears to be consistent with the forces we are seeing (we shall see that our choice was modest). And again, cosine (φ) = 1 in the geometry of the boundary conditions we are validating. Therefore,

-U = (7.315 * 10^-22) * (0.001) = 7.315 * 10^-25 Joules

and k = 4 MeV or 4000000 * (1.60217646 * 10^-19) = 6.408 * 10^-13 Joules and

even at a steep deflection (which will give us plenty of margin to adjust electrostatic forces for other purposes) the ratio of energy expended to energy gained is:

(7.315 * 10^-25) / (6.408 * 10^-13) = 1.142 * 10^-12; that is,

the free body system is gaining 10¹² times the energy it is losing for this needed deflection and the energy loss is virtually zero.

ed   = ( F ∙ s ) / 6.408 * 10-13 Joules (or   4 MeV)	[eq. 3.25]
Ratio of   deflection energy expended to product kinetic energy
Reference Design   Twenty-One Castle Mike => (7.315 * 10-25) / (6.408 * 10-13) = 1.142 * 10-12

Derivation of Mean Free Path in thin film plasma

A simple form of mean free path can be estimated by the relation:

ℓ = 1 / (σ * n)	[eq. 3.26]
Mean Free Path   (parallel to average density surface)
Reference Design   Twenty-One Castle Mike=>1/((1.2 * 10^-28)*(2.791*10^26)) =   29.86 m

ℓ = 1/ (σ * n)

where ℓ is the mean free path, n is the number of particles present per cubic meter in the volume considered and σ is the effective cross-section of the particle whose mean free path is sought.

From previous calculations we know that σ = (1.2 * 10^-28)and that n = (2.791*10^26) so,

ℓ = 1/((1.2 * 10^-28)*(2.791*10^26)) = 29.86 m

Of course, this won’t suffice since the plasma pressure is variable as we pass from the outer surface to the inner surface of the plasma. ∴ we need a term that describes the mean free path, ℓ, as a function of height “above” or “below” E_n  (see Figure 1). Refining our approach, we get:

ℓ = 1/ ( (1.47 * 10^-11) * N(ℓ_k) = 29.86 m,

where N(ℓ_k) is a function that describes the particle count as a function of ℓ_k, that is, the k’th component of ℓ ∀ ℓ ∈ k. This function is proportional to the electrostatic force as the square of ℓ_k; i.e. N_average = c / ℓ_k², where c is some constant. In order to solve for the mean free path one must integrate over that path to count the number of particles present. ∴

N(ℓ_k) =  cℓ_k^-1

ð  N(ℓ_k) = c   ℓ_k^-1

ð  N(ℓ_k) = c [ ln (ℓ_k)_f - ln (ℓ_k)_i ]

ℓ = 1 / (σ c [ ln (ℓk)f   - ln (ℓk)i ] / 1015)	[eq. 3.27]
Mean Free Path   (general form)

Reverting again to Figure 1, let the mean free path be denoted m₁, and let us solve for ϕ,

|m₁|² = δ²i + ε²k

ð  ϕ = tan^-1(ε / √ [|m₁|²  - ε²])

ð  ϕ = tan^-1(ε / √ [ [ 1 / (σ c [ ln (ℓ_k)_f - ln (ℓ_k)_i ] / 10¹⁵) ]²  – ε²])

Which suggests that, on average, for any ϕ < tan^-1(ε / √ [ [ 1 / (σ c [ ln (ℓ_k)_f - ln (ℓ_k)_i ] / 10¹⁵) ]²  – ε²]) the fusion product is re-absorbed into the plasma and:

Te = Pt {ϕ < tan-1(ε / √ [ [ 1 / (σ c   [ ln (ℓk)f - ln (ℓk)i   ] / 1015) ]2  – ε2])}	[eq. 3.28]
Thermal plasma   input due to fusion

where T_e is the thermal energy introduced to the plasma from fusion reactions and P_t is the total power output of the reactor. The above equation is the reason for choosing a thin film plasma since the ratio of mean free path to plasma thickness determines the proportion of fusion energy output discharged as thermal heating of the plasma to power output remainder. And as we have seen, the mean free path is on the order of 30 m.

A color rgraphic showing the helical layering scheme. Working outward from the plasma this pattern is repeated about 100 times.

This verifies the efficacy of the geometry chosen for thin film plasma.

The next step is to verify the efficacy of propellant capture and exhaust. In the case of fusion product we consider a significantly higher energy than that experienced from thermal instabilities. Maximum fusion product will be about 4 MeV. Recalling that 1 electron volt = 1.60217646 × 10^-19 joules we have:

(4 * 10⁶) * 1.60217646 × 10^-19  = 6.408 * 10¹³ Joules. Based on our previous calculations we can see that the reaction plasma will have no chance of confining this and virtually 100% of the fusion product (including the “slow” product) will easily reach either the α – ρ neutral E orbit surface or the β – ρ neutral E orbit surface, depending on its velocity components. But once it arrives, unless its angle of incidence is rather steep, it will also readily become circularized into this E neutral orbit. Note that the coordinate system in which the product emanates is already in relative motion in consequence of the orbits induced by the magnetic fields around the torus. This means that regardless of the products vector, a tendency to circularize its orbit and find its way to an accumulator at the top or bottom is inevitable. Manipulation of the force dot product as done supra will show that cos θ will indeed have to be less than 1 in order to capture product, however, the lost product will be mostly re-circulated back into the reaction plasma or slowed and captured by the vessel wall as mentioned previously.

Vessel capture is not as serious of an issue as it may sound like at first. Looking at the equations already worked for thermal heating, we can see that product reaching the wall from steep incident angles will have a reduced energy and be captured by an Osmium-Beryllium-¹⁰Boron vessel sleeve. This sleeve can be replaced on regular maintenance cycles and due to the type of fusion reaction involved, neutron fluxing of the sleeve will not render it too “hot”. Because product moving within the vessel is in contact with a strong electrostatic field, unstable movements therein will cause voltage potentials to oscillate within the torus and this can be used to recapture much of that energy. At first blush this would appear to result in a statistical cancelling effect. However, if the torus is shaped … like a torus, as opposed to a straight cylinder, the differences in electrostatic forces can be separated by computer control.

See the related image files for more details regarding the Reference Design.

Confinement materials science at pressures up to 10⁶ Bar

Another daunting challenge in this reference design was in what to do about pressure. Straightforward calculations quickly revealed that any commercially and practical alternative to fission power would require enormous operating pressures. This is because the reaction rate depends so heavily on it and plasmas are not normally very dense. This simple observation has escaped many fusion cheerleaders who still don’t seem to realize that there is little point in designing a fusion reactor that does not outperform fission reactors by orders of magnitude. And this requires massive pressures. It is simply impractical to build a reaction vessel with the volume of stadium. And if you don’t have that much reaction going on you are better off with fission.

We will now continue the discussion we initiated supra by deriving the essential equations related to the nanotube blanket there described. In particular, we wish to characterize the proportional relationships between the E and B fields created and how they affect the plasma.

Derive equations of motion for:

1. magnetic force lines

[TBC]

Control of the Fuel Source Undergoing Fusion Reactions

Lampe and Manheimer [2] (L and M) have emphasized that the collisional rate νpB for

(p-B) momentum exchange scattering is 37 times faster than the fusion reaction rate νF ,

and that the collisional (p-B) drag rate although slower than νpB by a factor of twice the

mass ratio mp/mB is still faster than νF by a factor 7. They discussed a variety of collisional

processes which would destroy the colliding beam equilibrium on time scales much faster

than the fusion reaction time. They considered an explicit equation for the time evolution

of the transverse spread of the proton beam, based on the Fokker-Planck kinetic equation,

and their calculations clearly showed the heating of the proton beam due to p-B collisions

in the limit νpB > νpe, where νpe is the collisional rate for proton-electron (p-e) momentum

transfer. They concluded that the required colliding beam equilibrium “cannot be sustained

for long enough to provide fusion gain.”

(d_scF_R  / dL)² = (_cF_E )² + (dF_s / dL)²

dα = tan^-1((d_scF_r  / (dF_s / dL))² )

Wherever

∫ tan-1 ((_cF_E)_i  / (_cF_E))_f  dL = ∫ tan^-1(d_scF_r  / (dF_s) dL

is taken over the line representing the inner lining of the torus shell the plasma is confined at any arbitrary pressure and/or temperature provided sufficient power is present. This is a field equation and there is no dependency on mass-bearing material in this relation. This relation requires two magnetic fields, one primary created by a solenoid and the other the result of moving positive charge in a proton superconductor induced by the electrostatic repulsion of positive ions in the plasma. An ultimate materials science limitation will present if a sufficiently large current in the proton superconductor forces the material into a non-superconducting regime. At that point, increases in current will be limited to the capacity of active cryogenic cooling and the thermal behavior of the superconductor. This is a materials science power conversion limitation and not a direct pressure/temperature relation.

To find Tesla, B, Required to confine a fusion reaction:

B = 2ϵ₀nk(T_e – T_i)

where ϵ₀ is the permeability of the medium, T is the temperature in Kelvin and n is the density (k boltzman’s constant?).

To explain why electrostatic stabilization is needed, and to illustrate its remarkable value to magnetic confinement, we explicate the condition we seek thusly:

q₁ * q₂ / 4πϵ₀r² ≥ T*A*m

where q₁ * q₂ are the charge sums of opposing electrostatic forces from the outside (q₁) of the torus and the inner core (q₂) of the torus, evaluated at some point x in the plasma. T is the Tesla of the magnetic force at x. And A*m is Amp * meters. This condition guarantees an elimination of all magnetically induced instabilities of particle orbits in a torus throughout the thermonuclear regime. As T increases Capacitance can be slowly adjusted (creating a negligible magnetic field) by modifying potential. Because the charge positions symmetrically, the change is also symmetric about the torus. Obviously, large electrostatic forces presenting in the vicinity of the torus create considerable problems that must be addressed. To understand this, we first note that an electrostatic neutral surface exists about the torus, call it the y surface. This surface is the preferred location for the placement of all electrical current requirements and electronic equipment. This greatly reduces or eliminates the “effective” resistance that current would encounter outside it (actually, depending on the geometric location of an electrical current, it may be the case that the current is accelerated by the electrostatic forces as well). For the central solenoid, where the highest currents appear, we can eliminate the electrostatic problem by creating one of two separate circuits on the central solenoid axis – one “above” the torus or one “below” the torus, depending on the direction of plasma flow chosen.  The current then travels around the girth of the central solenoid such that its radial component remains neutral (from the current’s perspective) but its zenith winding component is favored, or “pulled” by the positive electrostatic force acting on it from the plating above (“pulling” electrons toward it). In other words, the electrostatic force alone will “energize” the magnets and serve as a booster to the electromagnetic power. But a better arrangement (and what we propose) is to place the current coils on both ends of the central solenoid such that one is boosted and the other is retarded. A circuit outside the coulomb barrier (a ceramic insulator) reconciles these current energies to neutralize the power gain/loss. Because this “effective” resistance is not a material science phenomenon, no thermal or ohmic heating occurs and superconductors can still function as usual. In order to reach the condition stated above, the central solenoid current must be equal to or less than the electrostatic forces.

A final note regarding the case where a fission reactor is employed to boost temperature and pressure, we should note that the ability to confine these pressures and temperatures must still exist. Therefore, should the difference be too great, the fusion reactions will have to be pulsed over short time intervals to prevent the plasma from expanding to the torus shell and making contact with it. In this mode of operation, the fission reactor will require reflectors that provide a periodicity to the heat and pressure giving the plasma a time to “relax” back into full confinement.

The subject of confinement in a steady state of fusion activity deserves greater treatment. There is a kind of Fantasy Island mentality regarding this issue as well. Current tokamaks have no means of maintaining plasma stabilities once a significant number of fusion reactions begin to occur. The reason for this, beyond the lack of magnet tolerance symmetry that prevents steady state in the first place, is that simply adding heat to the plasma is not a winning strategy. Heat results in a Maxwellian distribution of more or less random particle motions. To visualize the problem, it helps if we take another ride on a particle in the plasma and experience what is occurring in the reactor.

As heat is added it tends to make particles in the plasma move with vectors of essentially random direction and magnitude. The thermal vector realized must be vector summed with the vector of an impinged particles initial motion at the instant of collision. This is because this initial motion, in stable plasma, is the vector (or, path integral, actually) enforced by the magnetic field and whose time durability is necessary to have stable plasma. The components lying parallel to this motion are additive (and if a proper choice in heating is used, will result in larger absolute values) and will serve to increase the particle’s energy in the instantaneous direction dictated by the magnetic field. However, the components not parallel with this vector’s components are essentially 100% destabilizing components that the magnets must suppress and “correct”. If it doesn’t, this heat energy is ultimately realized as heat delivery to the outer walls of the torus. The “suppression” to prevent this requires energy. Therefore, heat energy cannot be added faster than the magnets can compensate for this change. This means that power levels of magnets and heat generators are tightly coupled. RF and particle injection heating will destabilize plasma if injected at a power level exceeding the power of the magnets to suppress or “turn” random kinetic energies in the correct direction. Current tokamaks are light years away from being able to do this, due primarily, once again, to lack of symmetric forces applied and excessive rates of magnetic reconnection. This is why greater fusion performance is seen during the annealing period, since this is the time during which the magnets can recover stability of the plasma.

The design we propose addresses this problem thusly. In plasma the unique problem is how to force collisions between fuel particles in the plasma only along the intended flight path required for stable plasma; i.e. controlled fusion. In single species plasma, the solution is to force all kinetic gains from heating into a statistical vector sum that favors the direction of plasma orbital paths.

A substitute for Magnetic Induction

The primary driver of the system is plasma current. But without an inductor this cannot be accomplished in a traditional Tokamak. This Reference Design specifies a new procedure for generating the rotational velocity (rads / s) required of the plasma. The imparting of rotational motion of the plasma and the combination of controlling charge density and rotational velocity is a necessary condition for the TFET system to function.

[TBC]

Thermal management of the plasma

Obviously, if all of the ⁴He product is being prejudicially scooped out of the plasma the instant it is created there will be no intrinsic heating mechanism to maintain operating temperature. However, not all ⁴He can be removed. Therefore, the key question is whether or not that portion that cannot be removed would possess sufficient summed energy to heat the plasma. This can be estimated by examining the geometry of the plasma. First, the plasma is only about 3.3 cm thick. By taking the neutral orbit within the plasma as the average reaction locus some simple trigonometry will give us an idea of how much ⁴He is captured by the plasma, as previously shown in the derivation of mean free path.

Power Conversion out of a Fusion Fuel Source

One of the previously noted problems with Tokamak reactors is that no matter how we may design them, and especially when using super light fuels which are technically easier for other reasons, tokamaks invariably create too many necessary interfaces with physical matter. This means that thermal losses will skyrocket when any attempt is made to actually harness the power from fusion reactions. Use of thermal power transfer, as is so common with fission reactors, is simply unrealistic with fusion as it will invariably run up power transfer losses. Put another way, the materials science needed to control power levels when tapping them thermally is decades in the future. The current ITER plan is to use Li-Si blankets to “capture” energetic product from fusion reactions which is embedded round the torus. The very plasma destabilizing events that magnets are good at controlling will skyrocket when fusion product is allowed to physically interact, by physical contact, with the chamber. This unworkably high proportion of power transfer out of the system by Neutron Transport will quickly become a heat sink sucking all the heat out of the plasma. And there is the more basic question of, what will happen to fusion plasma in a “dirty” deuterium or tritium fuel scenario? Ash from these reactions will easily foul the plasma because, as in the case of the Li-Si blankets, the ash will form a heat sink by which all heat will, with high preference, conduct out of the plasma. To understand the magnitude of the heat sink problem, note that for the reactions being considered at ITER, temperature regimes of something like 150 million K are involved (at these thermal variances between system and external, does an exact number really matter for a heat sink?). A small reliance on common sense would tell you that any such system must be almost completely isolated from potential heat sinks (when we speak of a “heat sink” we are loosely referring to the fact that the outer materials are super frigid compared to the plasma and act as a heat sink whenever any power transfer path presents). Tokamaks, as practical reactors and by current design, have no way of achieving this.

Whatever power output scheme is used, it must be one that bypasses materials altogether in order achieve power output rates in the TeraWatt range. The only way to do that is by clever use of the forces of nature, the only two to which we have practical access being the Electric and Magnetic forces of nature. For some background on this subject, Neutron Transport is the study of how Neutrons from nuclear reactions interact with matter. It is important because this is the only known way in which Neutrons can interact with the universe. They do not “respond” to the Electric or Magnetic forces. But this necessarily makes any power transfer from or to Neutrons a materials science issue. And it is precisely the “easy” fusion fuels in consideration by most current research efforts that produce their power output in such large proportions as Neutron energy; also called Neutron flux. Any undergraduate physics student should quickly see this problem and ignore talk of any use of such fuels; unless, of course, some truly eye-popping development in materials science is just around the corner. But ITER isn’t about solving the fusion riddle. It is about funding. This should be an encouragement to the reader, because it suggests that fusion power is not as far off as it may seem. We shall see in what follows that, even using a supposedly aneutronic fuel source will result in problems of considerable difficulty in thermal management. And we shall provide its solution.

The ²H-¹¹B reaction

The ²H-¹¹B reaction involves several individual reactions which need to be examined. We will first look at neutron production:

Neutrons come primarily from the reaction

¹¹B + α → ¹⁴N + n + 157 keV

The reaction itself produces only 157 keV, but the neutron will carry a large fraction of the alpha energy, which will be close to E_fusion / 3 = 2.9 MeV. Another significant source of neutrons is the reaction

¹¹B + p → ¹¹C + n − 2.8 MeV

These neutrons will be less energetic, with energy comparable to the fuel temperature. In addition, ¹¹C itself is radioactive, but will decay to negligible levels within several hours as its half life is only 20 minutes.

Since these reactions involve the reactants and products of the primary fusion reaction, it would be difficult to further lower the neutron production by a significant fraction. A clever magnetic confinement scheme could in principle suppress the first reaction by extracting the alphas as soon as they are created, but then their energy would not be available to keep the plasma hot. The second reaction could in principle be suppressed relative to the desired fusion by removing the high energy tail of the ion distribution, but this would probably be prohibited by the power required to prevent the distribution from thermalizing.

Before we proceed, we need to explain what we mean by “branching probability”. One can think of a branching process as a random walk. Let S_i denote the state in period i, and let X_i be a random variable that is iid over all i. Then the recurrence equation is

Si   + 1 = Si   + Xi+1 – 1 =	[eq. 3.29]
Recurrence   Equation

with S₀ = 1. To gain some intuition for this formulation, one can imagine a walk where the goal is to visit every node, but every time a previously unvisited node is visited, additional nodes are revealed that must also be visited. Let S_i represent the number of revealed but unvisited nodes in period i, and let X_i represent the number of new nodes that are revealed when node i is visited. Then in each period, the number of revealed but unvisited nodes equals the number of such nodes in the previous period, plus the new nodes that are revealed when visiting a node, minus the node that is visited. The process ends once all revealed nodes have been visited.

The following reaction is the primary source of the neutron flux we seek to manage. It is caused by ⁴He product being re-absorbed back into the plasma. Since this is the product, it is desirable that most of it is removed:

¹¹B + ⁴He → ¹⁴N + n + 157 keV

Where the neutron produced will carry about 2.9 MeV. The rest of the neutrons come from the reaction:

¹¹B + p → ¹¹C + n − 2.8 MeV

Where the neutron produced will carry about 650 KeV (or, the thermal energy of the plasma).

¹¹B + p → ¹²C + γ + 16.0 MeV

with a branching probability relative to the primary fusion reaction of about 10⁻⁴.  This just means that there are, probabilistically averaged, about 0.04 of the above reactions for every one primary reaction. This introduces another issue that will be addressed infra.

Finally, isotopically pure fuel will have to be used and the influx of impurities into the plasma will have to be controlled to prevent neutron-producing side reactions like these:

¹¹B + d → ¹²C + n + 13.7 MeV

d + d → ³He + n + 3.27 MeV

Other reactions include:

¹¹B + p→_0 + ⁸Be + 8.586 MeV,

¹¹B + p→_1 + ⁸Be* + 5.65 MeV,

⁸Be*→_12 + _12 + 3.028 MeV.

The kinetic energy interval for this reaction product is [3.4 MeV, 10 MeV]; though 90% of the total power produced is from particles with kinetic energy < 5 MeV.

Experimentally determined reactions and values (4 channels total):

¹¹B + p → ¹²C → α₀ + ⁸Be → α₀ + α₀₁ + α₀₂           μ = 4.4 mb/s     E = 8.583 MeV  ϕ = 154 deg

¹¹B + p → ¹²C → α₁ + ⁸Be → α₁ + α₁₁ + α₁₂           μ = 6.8 mb/s     E = 5.683 MeV ϕ = 165 deg

¹¹B + p → ¹²C  → α₂ + α₃+ α₄                              μ = ? barns       E = [2.0 MeV, 2.7 MeV]

¹¹B + p → ¹²C → ¹²C + γ                                     μ = 10 barns      E = [2.0 MeV, 2.7 MeV]

see ExperimentallyDerivedCrossSectionsForBoron.pdf for

Nuclear reaction analysis cross sections measurements for Boron and Carbon

by Senzo Simo Miya

To understand thermal losses we begin by an examination of Neutron reflectors. These are surfaces, also known as Neutron super mirrors, that reflect Neutron flux rather than absorbing it. This, obviously, is the preferred method of managing thermal losses due to Neutron flux. Whatever incident radiation is reflected is not only managed but fed back into the reaction plasma.

Ion sputtered (polished) neutron super mirrors made of pure isotope (with ⁵⁸Ni and ⁶²Ni) NiC/Ti laminates have very high reflectivities (nearly 90% incident). However, for an 11 Tw Reference Design we would need a reflectivity of something like this since:

11 Tw @ 1% neutron flux => 0.11 Tw to the wall => 90% reflectivity is therefore 11 Gw. This is still a monstrous amount of thermal energy to manage.  The good news is that neutrons from the presenting fusion reaction are weak; with energies insufficient to produce fission reactions. Therefore, a reflector normally used for fast neutrons would likely reach or exceed a 99% reflectivity.

Use of ion sputtering and laminate structures of pure isotopes ⁵⁸Ni and ⁶²Ni, as well as surface oils and other “tricks”, could raise the reflectivity yet higher.

This results in a highly likely achievable specification of 99% reflectivity which yields 11 Tw * 0.01 ≈ 110 Gw and 110 Gw * 0.01 = 1.1 Gw. This is the power level that must be thermally managed in our reference design. This is still considerable and will require new innovations in thermal management never before used in fission reactors.

But we must point out one important fact: our assumptions here are worse case in that we are assuming that ⁴He product is not removed from the plasma. Of course, we know it must be removed to make use of the power output. However, product vectors lying parallel or near parallel to the tangent plane of the thin film will be re-absorbed. The safe bet therefore, is to specify the thermal management herein until experimental data can settle the issue more precisely.

Additionally, we noted supra that gamma radiation will be emitted (at 4, 12 and 16 MeV) which we can average as having an energy of 12 MeV and which occurs 0.04 times each time the primary reaction, which releases 8.7 MeV, occurs. Thermal (and radiation) management of this must also be considered. This results in an effective contribution to total power on the order of 0.04 / 12000000 ≈ 3.33 * 10^-9 eV for every 8.7 MeV primary reaction. Taking a ratio of the two shows that:

(3.33 * 10^-9) / 8700000 ≈ 3.83 * 10^-16

which delivers (31 * 10¹²) * (3.83 * 10^-16) = 0.0118 watts to the reactor wall. While this may seem like very little radiation, and it is in some respects, the effect over time could be harmful to persons or equipment. Therefore, a thin super pressed Os shell outside the Gd shell is indicated to absorb the gamma radiation thus emitted. This shell is described infra.

Energy reaching the wall through slow neutron flux:

Let

r = ratio of neutron flux reflected to total incident flux

p = full power output of reactor

n = ratio of neutron flux power to total reaction product power for the ²H ¹¹B fusion reaction.

and

f_i = incident flux power = p*n

f_w = neutron flux power that is absorbed by wall

Then,

fw =   fi (1 – r)	[eq. 3.30]
Power absorbed   by reactor wall via Neutron Transport, 2H 11B

We recommend that a thin neutron super mirror made of pure isotopes of Ni, C and Ti form the inner reactor vessel liner. We specify a required reflectivity of slow neutrons be 99% or better. Additionally, we specify a proportion of Carbon to Nickel (to be an alloy in one layer of the mirror laminate) adequate to raise the melting point of the Ni alloy up to about 2000 deg C; the approximate melting point of Ti. We suspect this will result in a mix of about 80% Ni and 20% C. Directly behind this we recommend a 2 cm layer of liquid pure isotopes of Li coolant. The maximum coolant temperature we specify at or about 1450 deg C (the boiling point for Li is 1615 deg C and its melting point is about 450 deg C). The upper bound is low enough that the Nickel-Carbon and Titanium can tolerate the temperatures while still providing a large enough temperature gradient to make use of heat rejection panels highly efficient. For the reflector to maintain structural integrity, this leaves a temperature gradient of approximately 1000 deg Celsius. This thermal energy would be dissipated over heat rejection panels with a minimal differential temperature between near absolute zero and 450 deg Celsius. This leverages Wien’s displacement law to the extent feasible with current material science and is essential for tackling the problem of waste heat in a spacecraft application.

Finally, behind the Li coolant a 4 cm solid layer of super pressed Gadolinium is employed. We will justify this thickness infra. It is pressed to a density > 8,000 kg / m³. The higher this value the longer its service life; and it should be designed to be replaceable on a regular maintenance schedule, as should the super mirror. For this Reference Design we assume 10000 kg / m³. This yields a total vessel weight due to the Gadolinium shell of about 2,200,000 kg, or 2200 mT. This (and the Os layer) is one of the heaviest objects requiring orbital delta v. Based on these calculations, the Reference Design can expect a Gadolinium shell useful service life of about 15 months at reactor full power. At that point excessive embrittlement will require a full replacement. This is the reactor’s primary neutron flux absorber and safety backstop for neutron flux. And it absorbs only what remains of the flux that was not reflected by the super mirror.

Backing the Gd shell we specify a super pressed Os shell of 4 cm thickness. We specify a pressing of the Os elemental solid to create a density not less than 25,000 kg / m³. This is the reactor’s primary gamma radiation absorber. This is one of the heaviest components to be lifted to Earth orbit; weighing approximately 5,500,000 kg.

Let ρ_Os = the density of the pressed Os liner. Then the mass is,

MOs =   (ρAT)Os / 9.8 m / s / s ≈ 56 mT	[eq. 3.31]
Mass of the   Osmium reactor vessel liner

Where A and T are the liner Area and Thickness, respectively.

Neutron transport through circulating liquid Li and Gd liner

Let ρ_Gd = the density of the pressed gadolinium liner. Then the mass is,

MGd =   (ρAT)Gd / 9.8 m / s / s ≈ 22.45 mT	[eq. 3.32]
Mass of the   Gadolinium reactor vessel liner

Where A and T are the liner Area and Thickness, respectively.

The thermal values are achieved at thermal equilibrium. The actual energy carried away from the reactor vessel walls depends on the rate flow of the fluids. Behind the Gadolinium layer we specify a liquid Nitrogen layer followed finally by the first, Boron Nitride Wurtzite nanotube layer of the reactor vessel.

We specify a coolant flow rate of at least 100 m / s for the Li coolant. This results in the ability to remove the equivalent energy of 100,000 deg C / m² / s. Unfortunately, this requires, at minimum, an equal drop of energy over the heat rejection panels. As per reference design, this suggests the rejection by electromagnetic transfer to vacuum of about 100,000 deg C for each of some 700 m² of reactor wall surface. And that

ð  700 * 100,000 = 70 million deg C / s

This places the reference design loosely in the right order of magnitude with a drop of 0.01% of total reactor power output. That is;

70 * 10⁶ / 0.0001 = 700 billion deg C / s and, as per reference design, if the heat rejection panels have a total area of;

A = 100 meters * 33 meters * 48 then

A = 100 * 33 * 48 = 158000 m²

which yields a minimum required rejection rate at the surface of each panel of:

70 * 10⁶ / 158000 = 443 deg C / m² / s

If rejecting equally from both sides it is 443 / 2 ≈ 220 deg C / m² / s per side

This satisfies the minimal temperature differential indicated supra of 450 deg C.

let b = 2.8977685 * 10⁻³ m·K (Wien’s proportionality constant) then the optimum wavelength for thermal rejection is:

λmax   = b / T	[eq. 3.33]
Wien’s  Displacement Law

where T is the absolute temperature in Kelvin and λ_max is the wavelength at which emissivity is maximum. At thermal equilibrium the electromagnetic emission temperature is the maximum coolant operating temperature; e.g. about 1500 deg C. Therefore if [K] = [°C] + 273.15

=> 1773.15 K

=> λ_max = (2.8977685 * 10⁻³) / 1773.15 ≈ 1.63424893 * 10^-6 m

[TBC]

Derive equations of motion for:

1. finish neutron flux thermal management
2. Determine required thickness of Gd liner

Revisiting the fuel injection system and heat rejection panels

We are now in a position to specify the requirements for heat rejection.

[TBC]

Fusion Product Capture for Outbound Power Conversion

Even with the most advanced, clever design, MFC or not, these realizations suggest that the super light fuels are simply not workable with first generation fusion reactors. It is so far beyond materials science to convert 15 TeraWatts of thermal power from Neutron Transport to useable power that discussion of it is borderline pathological. But this is exactly the “path” ITER is following. But the practicably nearest supposedly “aneutronic” fuel alternative, ²H-¹¹B, which can afford power transfer without physical contact, has other conditions that make it a tough proposition. First and foremost, the triple product for this reaction is higher. But once that is solved, one has to consider that such a plasma of ²H-¹¹B fuel will have to be put in a state that is not, technically speaking, plasma. Known as single species plasma, ²H-¹¹B would have to be stripped clean of all or most electrons so that all its fusion power isn’t produced in a form for which no conceivable power conversion exists: Bremsstrahlung radiation. This problem is similar to the Neutron Transport issue, but not as challenging. The simplest answer is not to use electrons in the reaction plasma, which are the source of it. As positively charged, single species plasma, the ²H-¹¹B fuel would orbit in helical paths around a torus in one direction only.

However, another problem is encountered in that, for magnetic confinements of this type, a density limit is realized (known as the Brillouin density limit). Density is one of the factors of the triple product. We have proposed stabilizing the magnetic confinement using electrostatic positive charge to collapse the plasma into its magnetically defined orbits (relying on a multi-confinement scheme). This allows the Brillouin density limit to be exceeded because this limit is an artifact of how magnets force orbits.

Let us assume then, for the sake of discussion, that a setup in which power transfer can occur without mass contact (no thermal conduction) is employed by using the ²H-¹¹B reaction. Then we know that electrostatic, symmetric forces will have to be applied to the plasma to “assist” the magnetic forces. Otherwise, the exponentially growing instabilities of the plasma will cause exponentially increasing power transfer losses as we try to pass the triple product (due to the limitations of magnetic reconnection in this regime and, most generally, due to the fact that the locating of magnets in a torus is inherently asymmetric). Once a fusion reaction is reached (beyond ignition) the fusion energetic product will be two fissile products that immediately decay into three ⁴He nuclei and spread at very high velocities with positive charge. So, the power form this reaction gives us is moving electric charge, something we can tap without thermal contact. Having said this, we are still not in the clear. Mass to mass contact is unavoidable. When the fusion reaction ramps up to full power density a nasty quantity of power will present. The highly energetic, ⁴He ions (the primary ²H-¹¹B reaction product) cannot all be subjected to negative accelerations to tap off their energy into a useable current. There simply isn’t enough space and, statistically speaking and perhaps more significantly, not all particles will have velocity vectors that can be exploited. Let us take a walk inside the torus to explain what we mean. Remember, we have to keep the plasma stable, so trying to perform this deceleration outside the tokamak is not a winning strategy. When a fusion reaction occurs a more or less random direction of flight for the Helium product is “decided”. Some of these particles will travel with the plasma rotation, more or less. Some will travel in the opposite direction. But some will be ejected at right angles to the orbits. Once again, electrostatic confinement helps out. Let us set an electrostatic neutral zone in the center of the plasma, a special, neutral orbit path. As we move outward from that path a massive (and it will need to be about 10⁶ Bar at between stable and neutral E) electrostatic positive charge will push us back toward the neutral orbit. This helps. But the Helium nuclei can easily overcome this electrostatic force. But, the pressure from the electrostatic force will tend to influence the flight vector of the particles into a torus orbit in which, given a sufficient component at right angles to the neutral orbit, will cause the Helium product to arc either out toward the torus shell or in toward its center. The Helium nucleus will then depart the plasma and head for the wall. It is in this wall that the Helium product, with carefully designed symmetry, can be captured, but not fully stopped (not yet), into a stream of Helium nuclei collected at the torus edge and central axis. A helical collector can then force these nuclei into a deceleration using inductive stators to generate current. At this point, we are outside the plasma so no destabilization occurs as a result. A selected remainder can be vented outside the apparatus for thrust. But these are the happy paths.

Some statistical fraction of ⁴He will invariably impinge on the walls due to unfavorable initial kinetic vectors, straight on collisions with the vessel wall or by simply being too energetic to capture. Though we can minimize these “losses”, they will generate heat when the power levels skyrocket. Minimizing this kind of loss will be crucial but is far, far easier than the challenges faced by today’s tokamaks (like ITER) as regards the same issue. And as the fusion reaction heats up, this tiny power loss still incurred even in the “best” design here suggested will be enormous. In other words, a surplus of heat energy will be needed; one far beyond the surplus capable of being generated by neutral particle injection or RF heating (with foreseeable materials science technology). Given a limited power loss as described, cooling can probably be practically achieved, but the loss will still force us to reconsider the existing heating schemes now in use. What is needed is a heating mechanism that allows us to extend the confinement time indefinitely. Even at the theoretical level there is only one known, certain power transfer device that can do this: A fission reaction.

Fg;12 = ¡G

m1m2

r2

12

^r12

~Fe;12 = ke

q1q2

r2

12

^r12

~Fm;12 = km

q1q2

r2

12

~v1 £ (~v2 £ ^r12) :

An accounting of magnetic dipole interactions in a TFET System

Particle	Magnetic   dipole moment in SI units   (10−27 J/T)	Spin   quantum number (dimensionless)
electron	-9284.764	½
proton	+14.106067	½
neutron	-9.66236	½
muon	-44.904478	½
deuteron	+4.3307346	1
triton	+15.046094	½

 

Notice the diagram:

F ( r, m1, m2)   =[3*μ0 \ 4π r5] *[ (m1   ̇ r) m2 + (m2   ̇ r) m1 + (m1   ̇ m2) r – ( { 5(m1 ̇ r) (m2 ̇ r) }  / r2) / r ]	[eq. 3.34]
Force on   particle 1, two particles with magnetic moments

and torque, τ, is

τ  = [3 μ0 m1m2   / 4 π r3] * [3 cos (ϕ - α ) * sin(ϕ – β ) + sin(β - α )	[eq. 3.35]
Torque on   particle 1, two particles with magnetic moments

where r and θ are measured from the m₁ coordinate system with m₁ lying on the x axis and μ₀ is the fundamental constant called the “permeablility of free space”, (1.25663706 × 10^-6 m kg s^-2 A^-2) in SIunits. m₁ and m₂ are the respective magnetic moments for the two particles.

FYI:

To calculate nuclear magnetic moments:

The values of g^(l) and g^(s) are known as the g-factors of the nucleons, thus:

μj   = g(ℓ) * ℓ + ½ g(s)	[eq. 3.36]
Nuclear Magnetic   Moment; s = ½ , j = ℓ + ½

and

μj   = [j \ (j + 1)] (g(ℓ) (ℓ +1) – ½ g(s))	[eq. 3.37]
Nuclear Magnetic   Moment; j = ℓ – ½

where total angular momentum is denoted j, orbital angular momentum ℓ, spin s, and μ_j is the nuclear magnetic moment contributed by j.

But magnetic moments for various nuclei are known and can be looked up. Magnetic moment for ¹¹B is 2.682 (assuming spin 3/2). g^(ℓ) = 1.788 in units of e/2Mc.

and for Helium, ⁴He, it is 1 *10^-6

For ²H it is

And the potential energy of such a system is:

H =   – (μ0 / 4 π r3jk) * (3 (mj * ejk)( mk * ejk) – mj   * mk)	[eq. 3.38]
Potential energy   in Dipole-Dipole interaction

where e_jk is a unit vector parallel to the line joining the centers of the two dipoles. r_jk is the distance between two dipoles, m_k and m_j.

For two interacting nuclear spins:

H =   – (μ0 γj γk / 4 π r3jk)   * (3 (Ij * ejk)( Ik * ejk)   – Ij * Ik)	[eq. 3.39]
Potential energy   in interaction of nuclear spins

γ_j, γ_k and r_jk are gyromagnetic ratios of two spins and spin-spin distance respectively.

[TBC]

{

Derive equations of motion for:

1. validate effect of nuclear magnetic moments

}

Chapter 4

Research and Development phases

Phases 1 to 5

Chapter 5

Interdependencies, Derivations and proofs

Basic equations of TFET Fusion: a derivation from Magnetohydrodynamics

Based on our considerations supra we now specify the performance and characteristics of Reference Design Twenty-One Castle Mike:

Torus Height (outer shell):                                                                           90 meters

Torus Diameter (outer shell):                                                                       75 meters

Total structural weight (locates all components):                                        100mT

Reactor vessel weight:                                                                                12000mT

Magnet weight:                                                                                            1000mT

Total system Earth weight:                                                                     15000 mT

Total Fusion output power:                                                                          11 Tw

Total thermal power recirculated to sustain reaction: 3 Tw

Total electric power:

electromagnetic systems power requirement to sustain reaction: 0.25 Tw

coolant and heat rejection system electric power requirement: 0.001 Tw

total power available to Brayton cycle (onboard ship services): 0.01 Tw

[for each of 3 units, all redundant]

Total useable kinetic power (at maximum rated power): 7 Tw

Total thermal power loss to reactor vessel wall via fast ⁴He: 0..5 Tw

Total thermal power loss due to inefficiency of recycling thermal losses due to Neutron Transport: 100 Gw

Total power loss to Brensstrahlung: ~1 Mw

Specific Power: 3.5 Mw / kg

Am-242m film power: 2Mw / m²

Total Kicker power output: 7 Gw

Kicker runtime at rated maximum power per fuel emplacement: 10 days

Gd reactor vessel liner lifetime at maximum rated power: 15 months

Hours of operation at maximum rated power between overhauls: 160 hours

Neutron flux at Gd liner exterior surface at maximum rated power: ~0

Plasma reaction mass volume per layer: 5525 m³

Number of plasma layers: 25

Plasma operating density at maximum rated power: 9.4427 kg / m³

Plasma Reaction mass pressure at maximum rated power: 1.5*10⁶ Bar

Mars expedition proposal total mass ²H-¹¹B, fast burn + 1/2 slow burn:     30,000 kg

These numbers were derived from the reaction rates and masses discussed infra. The Space Shuttle Main Engine turbo pumps, by comparison, had a specific power of about 153 kW / kg. It is often stated that the vast distances of space make interplanetary travel a challenge. This is not entirely correct. Movement in space is demonstrably more efficient than in or on just about any medium. The issue isn’t distance, its mass. It takes precious little force to impart a noticeable velocity to several thousand tons of bulk mass in LEO. There are professional drivers who drive a million miles in just a few years. And missions to Jupiter using chemical propulsion are measured in similar time frames. But since Jupiter is some 930 million miles distant the added efficiency of propulsion in space is made up for in distance. But getting a truly powerful power plant, something far more powerful than, say, a diesel engine, into orbit, is costly. For example, an 800 horsepower diesel engine (very large) is about 600 Mw power. Multiplying by a factor of ten (6 Gw) with a rocket, travel times to Jupiter make sense if light weight components and virtually no mass to speak of is carried as cargo. The problem, however, is when we contemplate moving large, worthwhile masses across the solar system. So, we can easily see why power outputs can climb into the Tw range. And that is the issue facing the Mars proposition.

We recommend that [redacted] secure the employ of key figures working on the LANL MTF project. Key figures with the most experience available should be retained. Individuals most knowledgeable of the construction of tokamaks, especially the ITER experiment, should be retained. We further recommend the establishment of a test facility in the American southwest where fusion firings can be triggered under regulatory approval. Involvement of the Nuclear Regulatory Agency will be necessary for the development of the fission kicker. Due to the size of such a propulsion system and its inherent complexity, we recommend the construction of a full size version of this engine as a first pass with the intent of redesigning and modifying as a process toward a one-off product and as experimental work dictates in each step of validation. This first engine would be a spare. Once the design specs are verified and the one-off complete, LEO insertion will be needed to test fire it in full kinetic mode. This should culminate in an automated round trip to the Moon with at least three auto sequence stops and starts. Initial startups of kickers in orbit should be unmanned until shutter operating parameters are comfortably understood. Once these tests and experiments are validated (and we expect more one-off changes to be necessary in orbit), two copies of the final one off should be manufactured and placed in LEO for validation testing and later mission use.

Even if this particular approach proves unworkable, we believe that some kind of similar two stage scheme is the winning strategy. Without it, Mars under the requirements here given becomes a hard proposition to such an extent that we prefer to await a leadership response before entertaining any alternatives. Having said that, we point out that this engine has broad application beyond space flight and that [redacted] could easily position this technology for very impressive margins in other markets, including commercial power generation and use by the U.S. military for powering naval vessels. We believe this program can be fully funded for 5 billion USD or less and be completed in three years. If government run, we scale by a factor of ten on both quantities.

1. Confined Plasma-Core Tokomak (CPT) Fission

We’ve developed for this Reference Design a derivative of the Open System Gas-Core Fission Reactor Rocket; a technology that has undergone pre-production software validation and is considerably closer to development than a fusion power source. We have denominated our derivative Confined Plasma-Core Tokomak Fission, or CPT Fission.

As shown supra, the distance between Earth and Mars, for the purposes of illustration, can be estimated at 3.1 x 10⁹ meters (about 2 A.U.). We set Δv = 3600 m / s, the lower limit velocity required to reach Mars in thirty days or less (see later discussion). We do not recommend atmospheric braking at Mars given the masses involved. These are broad estimates that simply include sufficient TMI burn times, deceleration burn times and transit burns, which obviously leaves headroom.

3600 * 24 = 86400 seconds are in a day

to satisfy the Hydrogen fuel mass limitation, we shall keep the fuel expense well below the allocated (this is one order of magnitude greater than that required by the fusion approach) 1000000 kg with a value of 488000 kg mass one way (2 burns) and 9660000 kg round trip (4 burns). Each burn is 249000 kg. The exact amount of deceleration will be mission dependent, but we allow for a maximum deceleration in any case.

ð  Δv = (3.1 * 10⁹) / (86400 * 10) ≈ 3600 m / s

{We are limited to a 30 day transit, therefore, we’ll need two burns of a maximum duration of 15 days}

ð  3600 * (1.3 * 10⁶) =  4.7 * 10⁹ meters transit path length

The well-known Tsiolkovsky rocket equation is:

mf   / mi = e -( Δv / ve)	[eq. 1.01]
Tsiolkovsky   Rocket Equation

ð  m_f  = m_i  * e ^{-( Δv / ve)}

ð  m_f = (2500000) * e ^{-( 3600 / 50000)}               {25,000 mT initial total, loaded vehicle earth weight}

ð  m_i =2505570 kg

ð  ((2500000 + (4 * 249000)) * (exp-( 3600 / 50000))) ≈ 140000 kg

ð  249,000 kg per burn; approx. 1 million kg total fuel load

We leave approximately 3000 kg allowance for the Uranium fuel that will be required, though much more can be found in the mass savings of employing this power plant over the fusion power plant. In any case, this increased fuel consumption will require at least 13 additional launches over and above the fusion option in order to bring this much fuel into LEO (assuming a launch cargo weight of 600000 kg). As with the fusion option the total mass is about the same, though the masses are proportioned differently. The structural mass of the fusion reactor is much higher than the structural mass of the CPT fission reactor, but the fuel load under CTP power adds back about the same amount of mass. We assume a total package mass of 1800000 kg. For mission purposes we’ve set the total allowed cargo and hull mass at 1000000 kg (10000 mT Earth weight) and the propulsion system at 700000 kg (7000 mT Earth weight), 10000 kg for Uranium fuel and 60000 kg for Hydrogen propellant. Then we can set our values, after some re-arranging, as:

Δv = ve [ ln (mi / mf) ]	[eq. 1.02]
Mars transit   baseline minimum delta v required
Reference Design   => (0.0028c) * (ln (2505570 / 2510000))≈ 1500 m / s

and

α =    Tk / (mi   – fm – cm – hm)	[eq. 1.03]
Mars transit   baseline minimum specific power required
Reference Design=> ((2*10^12)/(1.5 * 106)) ≈ 1.33 Mw / kg

and

Tk =  α(mi   – fm – cm – hm)	[eq. 1.04]
Mars transit   baseline minimum kinetic thrust required
Reference Design => (3.5   * 106)( 1500000) ≈ 5.25 Tw
Usable maximum power rating 7 Tw

where

f_m is the fuel mass,  p_i is the propulsion system tare mass, m_i is the total mass of the entire assembly at mission start, c_m  is the total cargo mass, h_m is the remaining assembly weight (superstructure – heat rejection panels are part of propulsion system mass) and T_k is the total kinetic thrust power. The three equations supra are the bunny trail that leads one inexorably to fusion kinetic propulsion for spaceflight. In the case supra we have not considered the burn time and we will address that directly.

We know that the lower bound for fast ⁴He from the reaction we are considering can be conservatively averaged and estimated as 0.015c.

We shall find infra that the fuel consumption of this Reference Design is about 0.087 kg / s. To determine the burn time for Mars transit we have:

T =    fm / c	[eq. 1.05]
Mars transit   burn duration, 7 Tw kinetic
Reference Design => 5570 / 0.087 ≈ 64022 seconds ≈ 18 hours

Where T is the burn time in seconds, f_m is the mass of fuel burned for the transit and c is the fuel consumption rate at maximum rated power. Therefore, the acceleration imparted to the hull/truss assembly is 10000 / 57471 ~ 0.175 m / s / s for a period not less than 18 hours. This results in two transit burns per transit, one for acceleration and one for deceleration, resulting in an overall Earth-Mars transit time of just under 7 days. The exact elapsed time will depend on relative celestial orientations at the time of transit.

EM   = Tk * dEM /   vf	[eq. 1.06]
Earth to Mars   single transit burn energy expenditure
Reference Design => (5.25   * 1012) * (4.32 * 109) / (10000) ≈ 2.27 * 1018   Joules
Cross check: ½ mv2 = ½ (2.5 * 107)(10000)2  = 1.25 * 1015 Joules

Where E_M is the Earth to Mars single transit burn energy expenditure, T_k, is the kinetic thrust power, d_EM, is the Earth to Mars distance and v_f is the final transit velocity. A final note to make is that we have not considered the energy required to transition from a high polar orbit (which we specify infra) to escape velocity. This is why we used generous fuel load figures. We can also see that our energy requirements are well within the required margin.

All of these calculations are order of magnitude estimates and do not take into account what happens if specific impulse, for example, is varied during the burn. However, margins of error of appropriate magnitude have been used to compensate for these variables.

Derive equations of motion for:

1. interdependent relationships between all the above.

[TBC]

References

Appendix

Kelvin   temperature conversion formulae
	from   Kelvin	to   Kelvin
Celsius	[°C] = [K] − 273.15	[K] = [°C] + 273.15
Fahrenheit	[°F] = [K] × 9⁄5 − 459.67	[K] = ([°F] + 459.67) × 5⁄9
Rankine	[°R] = [K] × 9⁄5	[K] = [°R] × 5⁄9
For   temperature intervals rather than specific temperatures, 1 K = 1 °C = 1.8 °F = 1.8 °R Comparisons among various temperature   scales

Pressure   units
	Pascal (Pa)	Bar (bar)	Technical atmosphere (at)	Atmosphere (atm)	Torr (Torr)	Pound-force per square inch (psi)
1   Pa	≡ 1 N/m2	10−5	1.0197×10−5	9.8692×10−6	7.5006×10−3	145.04×10−6
1   bar	100,000	≡   106 dyn/cm2	1.0197	0.98692	750.06	14.5037744
1   at	98,066.5	0.980665	≡ 1 kgf/cm2	0.96784	735.56	14.223
1   atm	101,325	1.01325	1.0332	≡ 1 atm	760	14.696
1   torr	133.322	1.3332×10−3	1.3595×10−3	1.3158×10−3	≡ 1   Torr; ≈ 1 mmHg	19.337×10−3
1   psi	6.895×103	68.948×10−3	70.307×10−3	68.046×10−3	51.715	≡ 1 lbf/in2

At current U.S. consumption rates; if one Twenty-One Castle Mike were to tap into the U.S. electrical power grid, the plant would have a load of about ½ Tw on average year round and generate 43 billion USD per year in fees based on the average cost of electrical power in the U.S. At wholesale rates and after paying extortion fees to the distributors you would likely draw about 20 billion a year. To make it attractive, you could cut the cost in half, making 10 billion a year in the U.S. alone.

And yes, you read that correctly. One Twenty-One Castle Mike could supply all the electrical power used in the United States year-round about 20 times over. Peak Oil is solved.

Common Conversions

electron volt = 1.60217646 * 10^-19 joules

1 Barn = 10⁻²⁸ m²

1      amu = 1.660538921 * 10⁻²⁷ kg

The Proton

Classification   Baryon

Composition    2 up quarks, 1 down quark

Statistics          Fermionic

Group               Hadron

Interaction        Gravity, Electromagnetic, Weak,Strong

Symbol(s)         p, p+
, N+

Antiparticle       Antiproton

Theorized         William Prout (1815)

Discovered       Ernest Rutherford (1919)

Mass    1.672621777(74)×10⁻²⁷ kg^[1]

938.272046(21) MeV/c^2[1]
1.007276466812(90) u^[1]

Mean lifetime    >2.1×10²⁹ yr (stable)

Electric charge +1 e
1.602176565(35)×10⁻¹⁹ C^[1]

Charge radius   0.8775(51) fm^[1]

Electric dipole moment        <5.4×10⁻²⁴ e·cm

Electric polarizability           1.20(6)×10⁻³ fm³

Magnetic moment        1.410606743(33)×10⁻²⁶ J·T^−1[1]

1.521032210(12)×10⁻³ μ_B^[1]
2.792847356(23) μ_N^[1]

Magnetic polarizability         1.9(5)×10⁻⁴ fm³

Spin                 ¹⁄₂

Isospin             ¹⁄₂

Parity               +1

Condensed       I(J^P) = ¹⁄₂(¹⁄₂⁺)

Boron

General   properties
Name, symbol, atomic number	boron, B, 5
Pronunciation	/ˈbɔrɒn/
Element   category	metalloid
Group, period, block	13, 2, p
Standard   atomic mass	10.811(7)
Electron configuration	[He] 2s2 2p1
Electrons   per shell	2, 3 (Image)
Physical   properties
Phase	solid
Liquid density at m.p.	2.08 g·cm−3
Melting   point	2349 K, 2076 °C, 3769 °F
Boiling   point	4200 K, 3927 °C, 7101 °F
Heat of fusion	50.2 kJ·mol−1
Heat of vaporization	480 kJ·mol−1
Molar heat capacity	11.087 J·mol−1·K−1
Vapor   pressure
P     (Pa) 1 10 100 1     k 10     k 100     k at     T (K) 2348 2562 2822 3141 3545 4072
Atomic   properties
Oxidation   states	3,   2, 1[1] (mildly acidic oxide)
Electronegativity	2.04 (Pauling scale)
Ionization energies (more)	1st: 800.6 kJ·mol−1
	2nd: 2427.1 kJ·mol−1
	3rd: 3659.7 kJ·mol−1
Atomic   radius	90 pm
Covalent   radius	84±3 pm
Van der Waals radius	192 pm
Miscellanea
Magnetic   ordering	diamagnetic[2]
Electrical resistivity	(20 °C) ~106 Ω·m
Thermal conductivity	27.4 W·m−1·K−1
Thermal expansion	(25 °C) (ß form) 5–7 [3]µm·m−1·K−1
Speed   of sound (thin   rod)	(20 °C) 16,200 m·s−1
Mohs hardness	~9.5
CAS registry number	7440-42-8
Most   stable isotopes
Main   article: Isotopes of boron
iso NA half-life DM DE (MeV) DP 10B 19.9(7)%* 10B     is stable with     5 neutrons[4] 11B 80.1(7)%* 11B     is stable with     6 neutrons[4] *Boron-10     content may be as low as 19.1% and as high as 20.3% in natural samples. Boron-11 is the remainder in such cases.[5]
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