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Gas Core Reactors for Deep Space Propulsion

Mikhail V. Shubov

University of MA Lowell, One University Ave, Lowell, MA 01854

E-mail<EMAIL_ADDRESS>
Abstract

The operation of a nuclear power station powering Deep Space transport is
presented in the paper. The power station consists of a Gas Core Reactor (GCR)
with magnetohydrodynamic (MHD) generator. A new model of gas core reactor
which has advantages by comparison with known models is suggested. It is shown
that $200\ ton$ system would generate $200\ MW$ electricity. The gas core
reactor consists of a spherical cavity of $1\ m$ to $1.5\ m$ radius surrounded
by a $50\ cm$ thick liquid beryllium reflector within molybdenum–92 container.
The fuel consists of 235UF4, KF, and K. It is demonstrated that such power
station will work for gaseous fuel pressures of $50\ atm$ to $100\ atm$. The
Monte-Carlo simulations describing paths of thousands of neutrons are
performed. These simulations yield neutronic results, i.e. the fractions of
neutrons causing fission, being absorbed by moderator, and escaping the
reactor. From this information one can calculate minimum needed pressure
inside the reactor, the reactor size, and moderator thickness. The
conductivity of superheated fuel depending on temperature and pressure is also
calculated. It is concluded that space propulsion is the best use of uranium
resources.

Keywords: Nuclear Space Power, Magnetohydrodynamic Energy Conversion, Gas Core
Nuclear Reactors, Space Exploration

Introduction

Deep Space transportation of large payloads would use electrically powered
magnetoplasmadynamic thrusters to provide momentum for maneuvers. These
thrusters use lithium propellant. The thrusters should provide a sustained
acceleration of $10^{-3}\ m/s^{2}$ for the space transport. Although this
value seems incredibly small compared to planetary gravity, it would certainly
be sufficient for most maneuvers of interplanetary travel. As we show in
Subsection 1.3, the minimum electrical power requirement for the thrusters is
$45\ W/kg\cdot M_{\text{tot}}$, where $M_{\text{tot}}$ is the total mass of
the space transport. The space transport includes an electrical power source,
the thrusters, lithium propellant, payload, and the spaceship. Thus, the space
transport must contain an electrical energy source of high specific power.

Several technologies exist for generating electrical energy in outer space in
general and deep space in particular. They are photovoltaic arrays, solid core
fission reactors with turbine generators, and gas core reactors (GCR) with
magnetohydrodynamic (MHD) electrical generators [1]. In this work we describe
a power system consisting of a GCR with a MHD electrical generator. The power
system will generate about $1,600\ MW$ thermal power and convert $200\ MW$
into electric power, which means that thermal efficiency will be 12.5%. The
system with all accessories will have a mass of about $200\ tons$ and thus a
specific electric power of about $1\ kW/kg$. If we scale down such a system,
the specific electric power will have to be much less. All other power systems
have specific power at least 5 times lower [1, p.3]. As we will show in
Section 6, the cost of such system will be $ 4.2 Billion.

The Gas Core Reactor will consist of a spherical or cylindrical cavity
surrounded by liquid beryllium reflector in molybdenum–92 container. The
reactor cavity radius will be from $1.0\ m$ to $1.5\ m$ and beryllium
reflector thickness will be $50\ cm$. The fuel consisting of 235UF4, KF, and K
in gaseous form will enter the reactor cavity at $50\ atm$ – $100\ atm$
pressure. It will be heated by 235U fission to 3,500$\ {}^{o}K$. Then the fuel
gas will expand and generate electric power in a magnetohydrodynamic
generator. Finally, the fuel will be cooled and condensed in radiative heat
rejection system.

The GCR-MHD system was originally suggested in [2, 3]. We introduce several
modifications in the construction and work of the system. First, we lower the
working temperature of the gas from 4,000 ${}^{o}K$ to 3,500 ${}^{o}K$. Since
4,000 ${}^{o}K$ exceeds the melting point of tungsten, reactor working at such
temperature may be worn out very quickly. Second, we suggest a different
neutron moderator. In [2, 3] it is beryllium oxide. We propose to use liquid
beryllium within molybdenum tubing. That would greatly simplify the cooling of
moderator and slightly improve the neutron balance. Unlike the previous works,
we have performed calculations of neutron balance using extensive Monte Carlo
simulations. We have shown that tungsten can not be used as a GCR shell, while
molybdenum can.

Now we are in a position to outline the content of the present work. In
Section 1 we present a brief review of possible means of propulsion for deep
space transportation. First we discuss the electric propulsion. The basic
engine for space propulsion is the magnetoplasmodynamic (MPD) thruster. At
high power, MPD thrusters have many advantages. Namely, they have simple
construction, about 50% efficiency, and high specific thrust. They use
inexpensive propellants like lithium and sodium. Another engine is called
Variable Specific Impulse Magnetoplasma Rocket (VASIMR) [4]. VASIMR is also
efficient, but it is much more complicated and expensive then MPD. It is
useful only for light spacecraft. Second, we describe thermal nuclear
propulsion (TNP). The advantage of TNP is that it uses space-based resources
such as water and requires low specific power. The disadvantage of TNP is low
specific impulse. Third, we calculate the power requirements for electrical
and thermal propulsion. Generally, electric propulsion requires 45 $kW$ per 1
$N$ of thrust. That translates to 230 – 450 $kW$ thermal power per Newton of
thrust. TNP uses at least 100 times less energy, but it is unusable on most
space missions, because it requires a lot of neutral propellant like water
which is not available at most space destinations.

In Section 2 we give an overview of the concept of a gas core reactor (GCR).
GCR consists of a cavity filled with fissile gas and surrounded by neutron
reflector. As fission chain reaction takes place in the reactor cavity, the
gas is heated to 3,500 ${}^{o}K$. Then the hot gas expends and generates
electrical energy via a magnetohydrodynamic generator. The fuel is passed
through the reactor cavity thousands of times. The gas is a mixture of 235UF4,
KF, and K. The neutron reflector is liquid beryllium within molybdenum tubing.
The reactor is presented in Figure 3. Previous models of GCR considered gas at
4,000 ${}^{o}K$. In our opinion, such heat will destroy the generator very
quickly.

In Section 3 we present the calculations of neutron balance. These
calculations determine under which conditions the reactor can achieve
criticality. The varying conditions include different reactor cavity sizes,
tubing materials, reflector wall thicknesses, and fuel pressure. Each neutron
starts its path from a 235U fission, undergoes a number of collisions, and
finally ends its path either by being absorbed or by escaping the reactor.
Each neutron travels through the reactor’s three regions – gaseous fuel,
refractory shell, and reflector. We present mathematical formulae describing
neutron paths. We derive neutron velocity after elastic and inelastic
scattering. Finally, we calculate the neutron balance and reactor criticality.
We calculate the impact of neutron absorbtion by fission products and neutron
multiplication in beryllium. The two aforementioned factors almost cancel each
other.

In Section 4 we present the calculation results. First, we present the
fraction of neutrons reflected by a liquid beryllium reflector. In some cases,
over 90% of neutrons are reflected. Second, we show that a reactor with
tungsten refractory shell can never achieve criticality. Third, we show that a
reactor with molybdenum-92 refractory shell can easily achieve criticality.
The lower melting point of molybdenum would present an engineering problem.

In Section 5 we calculate plasma density and conductivity for different
temperatures and pressures. First, we express plasma conductivity in terms of
electron collision cross-sections. Given that these cross-sections for
$UF_{4}$, $UF_{5}$, and $UF_{6}$ are still untabulated, some experimental work
remains to be done. We show that conductivity sufficient for a
magnetohydrodynamic generator can be achieved by expanding plasma. In some
cases, the expanding gases may have to be mechanically heated in order for
conductivity to stay high. Mechanical heating is accomplished by placing
tungsten obstacles in the path of the high velocity gas stream. Second, we
calculate the generator efficiency. It turns out to be 12.5%. Third, we
calculate the specific power of the radiator heat rejection system.

In Section 6 we discuss economic considerations. First we list the state of
the art deep space launch costs. Falcon by SpaceX is the most cost-efficient
launch system. It can launch payload into Low Earth Orbit (LEO) for $5,400 per
$kg$. Second we estimate the cost of the reactor itself. The most expensive
material used in the reactor is the fuel – $89,000 per kilogram of
${}^{\text{235}}$U. We estimate the cost of the whole reactor at about $4.2
billion. One of the factors lowering reactor cost is the much lower safety
requirement compared to reactors on Earth. Third we calculate the fuel
efficiency of GCR. The reactor will use 40% of the fissile isotope 235U and
discard 60%. This will have to be done due to accumulation of fission products
and 236U in the uranium fuel. As we show in Subsection 6.1, one gram of input
${}^{\text{235}}$U would produce $1,080\ KWh$ electrical energy, which is much
less than $8,400\ KWh$ produced in modern pressurized water reactors [5, p.
111]. Nevertheless, given the cost of delivering materials into deep space
this is an outstanding performance.

Finally, we argue that the best use of uranium resources is for space
propulsion. The energy density of 235U fuel is 68 $GJ/g$ making it perfect for
space applications. For terrestrial applications, coal reserves have 21 times
as much energy as uranium reserves.

##  1 Deep Space transportation

Using conventional chemical rocket fuel delivered from Earth for Deep Space
colonization and exploration is cost prohibitive. The cost of fuel and fuel
tanks in Deep Space is at least $20,000 per $kg$ due to the cost for bringing
these materials from Earth [6]. We explore two options for Deep Space
propulsion – electric propulsion and thermal propulsion. Electric propulsion
uses fuel brought from Earth and produces very high exhaust velocities.
Thermal propulsion uses materials obtained in space – such as water (obtained,
e.g., from ice on asteroids) and has moderate exhaust velocities.

###  1.1 Electric propulsion

The energy and fuel requirements for space missions or maneuvers are given in
terms of $\triangle v$ which is the change in velocity of the spacecraft
needed to accomplish a given mission or maneuver [7, pp.9–11]. To evaluate the
amount of fuel needed for a maneuver with $\triangle v$ we use the
Tsiolkovsky’s rocket equation [7, p.9]:

$\frac{m_{0}}{m_{1}}=\exp\left(\frac{\triangle v}{v_{0}}\right),$ (1.1)

where $m_{0}$ is the mass of the rocket with fuel, $m_{1}$ is the mass of the
vehicle after the fuel has been consumed, and $v_{0}$ is the exhaust velocity
of the rocket engine. The highest value of $v_{0}$ for a storable liquid
rocket propellant, hydrazine and liquid fluorine, is $4.22\ km/s$ [8]. An
obvious way to deliver greater mass over space routes demanding large
$\triangle v$ is to use engines with much greater exhaust velocity. For low
power applications ion thrusters and Hall effect thrusters [9, p.5-6] can be
used. For applications using hundreds of thousands of kilowatts or greater
power, magnetoplasmadynamic thruster and Variable Specific Impulse
Magnetoplasma Rocket (VASIMR) can be used.

Magnetoplasmadynamic thrusters pass current through a propellent and use
Lorentz force to expel it at 15 to 60 $km/s$ [10]. These thrusters generally
have high power of 100 – 500 $KW$. The construction of these thrusters is
quite simple [11, p.15] and schematically shown on Fig. 1 below

Figure 1: Magnetoplasmadynamic thruster

This thruster has not been used in space yet because of requirement of
hundreds of kilowatts of electric power. Sometimes the simplest ideas are the
hardest to implement. Variable specific impulse magnetoplasma rocket (VASIMR)
“is an electromagnetic thruster for spacecraft propulsion. It uses radio waves
to ionize and heat a propellant. Then a magnetic field accelerates the
resulting plasma to generate thrust (plasma propulsion engine)" [4].

For an Ion thruster with power $P$, exhaust velocity $v_{0}$, and efficiency
$\eta$, the following formula for the thrust force holds:

$F=\frac{2P\eta}{v_{0}}.$ (1.2)

The characteristics of several ion and Hall effect thrusters which have been
used in space are given in [9, p.57]. Those for magnetoplasmadynamic (MPD)
thrusters are given in [11, p.19]. Those for VASIMR thrusters are given in
[12, p.7]. By measuring the size and estimating the volume of MPD thruster
parts on diagram [11, p.20], we conclude that a $250\ kW$ thruster has a mass
not exceeding $500\ kg$. Hence, MPD thrusters have specific mass of $2\ kg/kW$
or less. In table below we have collected the data for electrical propulsion
systems.

Engine | Engine | Propellant | Power | Thruster |  $v_{0}$ $km/s$ | Efficiency
---|---|---|---|---|---|---
type | name |  | $kW$ | mass (kg) |  |
Ion | T6 [9, p.57] | xenon | 6.8 |  | 46 | .68
Ion | NEXT [13, p.2] | xenon | 6.9 | 5.0 | 41 | .7
Ion | HiPEP [9, p.57] | xenon | 30 |  | 87 | .8
Hall | SPT-140 [9, p.57] | xenon | 5 |  | 49 | .55
VASIMR | VF-150 [12, p.7] | argon | 150 | 1,000 | 29 | .61
VASIMR | VX-200 [14, p.9] | argon | 200 | 620 | 30 – 50 | .6
VASIMR | VF-400 [12, p.7] | argon | 400 | 1,000 | 49 | .73
MPD | Fakel [11, p.19] | lithium | 300 – 500 | 1000 | 34 – 44 | .4 – .6
MPD | Energiya [11, p.19] | lithium | 500 | 1000 | 44 | .55

Table 1: Electric thrusters

Scarcity of propellant presents a serious problem for Ion and Hall thrusters
listed above. Xenon is one of the rarest elements on Earth. VASIMR and MPD
thrusters listed use argon and lithium which are readily available. The
efficiency of VX-200 engine for different propellants is :

Figure 2: System efficiency of the VX-200 engine as a function of the specific
impulse [15, p.8].

Although krypton gives better performance than argon, it is too rare to be
used on massive missions. VASIMR engines are generally expensive e.g., a $200\
kW$ engine costs $308 million and a $400\ kW$ engine costs $371 million [12,
p.7], which means a cost of $930 per $Watt$. Unless VASIMR cost per $Watt$ can
be decreased at least by a factor of 20 for multimegawatt thrusters, MPD
thrusters present the only option available.

###  1.2 Thermal nuclear propulsion

A thermal nuclear rocket consists of a super hot reactor, which heats the
propellant and a nozzle through which the heated propellant is expelled.
Propellant can consist of hydrogen, ammonia, hydrocarbons, or water. These
materials should be obtainable on asteroids.

In order to calculate performance for different propellant combinations at
different temperatures, we have used the Rocket Propulsion Analysis program
[16]. This program calculates exhaust velocity and chamber/exhaust composition
for various propellant combinations. The results are tabulated in Table 2
below. The fourth column is the composition inside rocket chamber. In all
cases the area expansion ratio is 200. The internal energy of propellant is
the thermal energy it takes to heat the propellant to rocket chamber
temperature and to accommodate all the phase and chemical changes. The fifth
column is thermal efficiency, i.e. the portion of thermal energy supplied to
the propellant by heating transferred into kinetic energy.

Propellant | Chamber | Exhaust | Chamber | Internal | Thermal
---|---|---|---|---|---
| temperature | velocity | composition | energy | efficiency
| ${}^{o}K$ | $m/s$ |  | $kJ/g$ |
hydrogen | 1,800 | 7,030 | $H_{2}$ | 27.0 | 0.92
ammonia | 1,900 | 3,530 | 17.7% $H_{2}$, 82.3 % $N_{2}$ | 8.3 | 0.75
55% water | 1,450 | 3,180 | 15% $CH_{4}$, 44% $CO$, | 7.3 | 0.69
45% methane |  |  | 12% $CO_{2}$, 6% $H_{2}$, |  |
|  |  | 22% $H_{2}O$ |  |
90% methanol | 1,450 | 2,860 | 9% $CH_{4}$, 59% $CO$, | 4.45 | 0.92
10% water |  |  | 7% $CO_{2}$, 8% $H_{2}$, |  |
|  |  | 17% $H_{2}O$ |  |
water | 1,100 | 1,960 | $H_{2}O$ | 3.9 | 0.49
water | 1,000 | 1,850 | $H_{2}O$ | 3.7 | 0.46

Table 2: Performance of different fuels in thermal nuclear rocket

Liquid hydrogen gives by far the best performance. One problem with using
liquid hydrogen is that it has to be stored at very low temperatures. Liquid
hydrogen propellant is useful at the Asteroid Belt (far from the Sun). Even
though water has the lowest specific impulse, it can be a useful propellant
within the Asteroid Belt, since it is readily available there. The main
problem with high temperature and high pressure steam applications is that
such steam oxidizes boiler tubing. There is a specific alloy, Inconel 740,
containing nickel, chromium, cobalt, and no rare materials such as rhenium
[17], which has outstanding resistance to steam corrosion. When it is
subjected to high pressure steam at 750${}^{o}C$ for 10,000 hours it loses
only $1.2\ mg/cm^{2}$ [17].

###  1.3 Electrical and thermal power requirements

In this subsection, we calculate the power requirement $P$ for a rocket
engine. We are given the engine thrust $F$, the exhaust velocity $v_{0}$, and
engine efficiency $\eta$. The rate of fuel consumption $(-\dot{m})$ is
obtained from the fact that the force is equal to the change of momentum

$-\dot{m}=\frac{F}{v_{0}},$ (1.3)

where the overdot denotes derivative with respect to time. The engine
expelling exhaust at the rate $(-\dot{m})$ and velocity $v_{0}$ supplies the
power

$P_{0}=-\frac{\dot{m}\ v_{0}^{2}}{2}=\frac{Fv_{0}}{2}$ (1.4)

to the exhaust. Given that the engine has efficiency $\eta$, the total power
consumed by the engine is

$P=\frac{Fv_{0}}{2\eta}.$ (1.5)

The power per unit thrust is calculated from (1.5):

$p_{0}\equiv\frac{P}{F}=\frac{v_{0}}{2\eta}.$ (1.6)

From (1.6), one can see that a typical ionic engine with exhaust velocity of
$50\ km/s$ and efficiency $55\%$ requires $p_{0}=45\ kW/N$ of electric power,
where $p_{0}$ is power per unit thrust. The thermal power needed to produce
such electrical power is $230-450\ kW/N$.

Thermal nuclear propulsion uses much less energy then electrical propulsion .
First, it does not need to convert thermal energy into electrical energy.
Second, the lower exhaust velocity reduces cost per unit thrust, even though
it greatly increases propellant consumption. Finally, with some propellants
thermal nuclear rocket has high efficiency. By (1.6), for a steam rocket with
thermal efficiency $0.49$ and exhaust velocity $1,960\ m/s$, the power per
unit thrust is $2,000\ W/N$. For a 90% methanol 10% water rocket with thermal
efficiency $0.92$ and exhaust velocity $2,860\ m/s$, the power per unit thrust
is $1,560\ W/N$. Even if water tanks carried by the space transport are much
heavier than the transport itself, steam rocket has much smaller power
consumption then the electric rocket. Unfortunately, water may not be
available on many space routes, or $\triangle v$ may be high enough to require
electrical propulsion. Thus, in this work we will focus on energy requirements
for electrical engines.

At this point we calculate power requirements for space missions. For
non–urgent space transports, an acceleration of $a_{0}=1\cdot 10^{-3}\
m/s^{2}$ would meet most mission requirements. Such acceleration would produce
$\triangle v=15.8\ km/s$ within six months. Let $M_{t}$ be the total mass of
the space ship, payload, propulsion system, propellant, and energy generation
system. Then the electric power needed for space ship propulsion can be
calculated as follows:

$P=p_{0}F=p_{0}M_{t}a_{0}=M_{t}\ \cdot 4.5\cdot 10^{4}\cdot\frac{W}{N}\ 1\cdot
10^{-3}\ \frac{m}{s^{2}}=M_{t}\ \cdot 45\ \frac{W}{kg}.$ (1.7)

In our work, the GCR with MHD generator has a specific electric power of
$1,000\ W/kg$. Since the magnetoplasmodynamic thruster has specific power of
$1,000\ W/kg$ as well, the whole transport has a specific power of $500\
W/kg$. As we see from (1.7), our transport can carry payloads 11 times its own
mass, or it can deliver smaller payloads more rapidly. For some missions, like
bringing payloads from Asteroid Belt to Earth, much smaller accelerations like
$a_{0}=2\cdot 10^{-4}\ m/s^{2}$ can be used, thus much greater loads could be
carried.

These power requirements can be fulfilled by space solar power as well, but
the cost would be unreasonable. In 2002, these arrays had specific power of
$115\ W/kg$ and cost $1,000 – $2,000 per _Watt_ [18, p.432]. The specific
power has not improved by 2012 [19]. For a solar power panel working 10 years
or 87,660 hours, the cost of $1,000 – $2,000 per $Watt$ translates to $11.4 –
$23 per _kWh_. Thus, nuclear fission reactors are the best sources of power
for transportation in Solar System.

##  2 Gas core reactor

###  2.1 General concept

The main concern for testing new nuclear reactor types on Earth is safety. In
deep space, even if there are accidents they have no chance of contaminating
human habitat. Thus, deep space is an ideal place for testing innovative
nuclear reactor ideas.

The Gas Core Reactor (GCR) is a reactor in which the fuel generates thermal
energy in gaseous state. The reactor can use either fast or thermal neutrons.
The fuel is constantly circulating between the reactor, power generating
system, and the condensing radiator. The condensing radiator, which radiates
thermal energy into space is used as a heat sink. The fuel suggested in the
previous studies is uranium tetrafloride UF4 [3]. Uranium tetrafloride has a
melting point of 1,309${}^{o}K$ [20, p.39] and a boiling point of
1,710${}^{o}K$ [20, p.87]. Uranium tetrafloride is very stable at high
temperature. At 2,500${}^{o}K$, the equilibrium constant for the reaction

$UF_{4}\to UF_{3}+F$ (2.1)

is $1.2\cdot 10^{-6}\ atm$. Recall, that the equilibrium constant for the
above reaction is

$K=\frac{[UF_{3}][F]}{[UF_{4}]},$ (2.2)

where bracketed formulas stand for partial pressures. The fact that UF3 has
very high fluorine affinity is responsible for stability and noncorrosiveness
of UF4. Uranium tetrafloride is compatible with tungsten up to 3,000${}^{o}K$
and molybdenum up to 2,000${}^{o}K$ [3, p.5]. In order to produce electrically
conductive gas, potassium fluoride, pure potassium, rubidium fluoride and pure
rubidium are added to fuel. Potassium fluoride has melting point
1,130${}^{o}K$ and boiling point 1,775${}^{o}K$ [24, p.4-76]. Rubidium
fluoride has melting point 1,106${}^{o}K$ and boiling point 1,683${}^{o}K$
[24, p.4-79].

The fuel cavity is surrounded by a molybdenum-92 shell which is 1 $cm$ to 2
$cm$ thick.
Molybdenum-92 has much lower neutron absorbtion cross-section then other
isotopes. The next surrounding layer consists of a neutron reflector shell
which is $40\ cm$ to $60\ cm$ thick. The best reflector is liquid beryllium
metal in a pressurized molybdenum container. Due to the fact that reflector
absorbs most of the neutron energy, it will have to be cooled, thus a solid
reflector such as beryllium oxide may present considerable technical problems.
The final layer is a metal pressure vessel cooled to a temperature at which it
retains most of its hardness. This vessel contains the molten beryllium
reflector as well as the gas in the reactor core at 40 – 100 $atm$.

###  2.2 GCR with MHD generator

The Liquid-Vapor Core Reactors with magnetohydrodynamic generator has been
proposed by Anghaie [3]. In this reactor, a mixture of uranium tetrafluoride,
potassium fluoride, and potassium acts as both the fuel and the working fluid
[3, p.5]. The fuel mixture suggested by Anghaie consists of 32% UF4 and 68% KF
by mass. In the present work we will show that a mixture of 62% UF4, 30% KF 8%
K by mass can perform better. The uranium is almost pure ${}^{\text{235}}$U.
The fuel passes through the cycle, which is similar to the one proposed by
Anghaie [3, p.8]. Below we present the steps of the cycle which is a
modification of the cycle from [3]. The modification is concerned with the
modified chemical composition as well as lower temperature. The schematic
picture of GCR with MHD generator is shown on Fig. 3.

  1. 1.

Liquid UF4 at $1500\ ^{o}K$ is pumped into tubes within an irradiated region
of reactor surrounded by a neutron reflector. The pumping rate is $310\ kg/s$.
The mixture is heated by fission and evaporates within the tubes.

  2. 2.

The gaseous UF4 enters the reactor cavity at $310\ kg/s$. Gaseous potassium is
also pumped into reactor cavity at $40\ kg/s$.

  3. 3.

The reactor cavity is a spherical chamber surrounded by a $50\ cm$ thick
neutron reflector. The reflector is liquid beryllium within molybdenum–92
container. Within the reactor cavity the fuel mixture is heated to $3,500\
^{o}K$. In Anghaie’s model the temperature is $4,000\ ^{o}K$. Using extremely
high temperature will increase the electrical efficiency of the engine, but it
will greatly increase corrosion.

  4. 4.

Reactor core temperature of $3,500\ ^{o}$K necessitates cooling of reactor
walls. One possibility of a cooling system is introduction of liquid potassium
fluoride through numerous spray nozzles in the reactor walls. KF is pumped at
$150\ kg/s$. In a similar arrangement, liquid propellant is sprayed through
the nozzles in the walls of liquid propulsion rockets to shield the combustion
chamber walls from the heat.

  5. 5.

The gas mixture expands through a nozzle and enters the magnetohydrodynamic
generator. The generator efficiency is 12.5%, while in Anghaie’s model it is
22% [3, p.4]. Our model has lower efficiency due to the lower temperature
inside reactor cavity.

  6. 6.

The gas containing 62% UF4, 30% KF 8% K is cooled by passing through the
condensing radiator, where the gas is passed through tungsten tubes which
radiate energy into the space. The gas condenses into liquid. KF can be
separated from UF4 since it has higher boiling point.

  7. 7.

Electric pumps pressurize the liquid to 40 – 100 atmospheres to feed it into
the reactor.

  8. 8.

On each pass a small amount of fuel may be taken to reprocessing unit, where
some fission products and some fission product fluorides are removed. Then the
processed fuel is returned into the main loop.

The whole system would generate at least 1 $kW/kg$ electrical energy. The
reactor and MHD generator are shown in the figure below:

Figure 3: Gas core reactor with MHD generator

A Magnetohydrodynamic generator converts mechanical energy of ionized gas into
electric energy. The diagram in [21] illustrates its principle of operation:

Figure 4: Magnetohydrodynamic generator

In the figure above the hot ionized gas flows in $x$ direction at a velocity
of 1 $km/s$ to 2 $km/s$. A strong magnetic field points in $y$ direction.
Current in $-y$ direction going through the gas is generated.
Magnetohydrodynamic generators have no moving parts making them much better
suited for space missions than turbine engines.

##  3 Simulation

Neutron balance calculations are necessary to determine under which conditions
the fission chain reaction is sustained. The conditions include gaseous fuel
temperature and pressure, reactor inner chamber length and diameter, moderator
thickness, inner shell thickness and material. _We show that the fission can
be sustained for most reactor and moderator sizes as long as the inner shell
is made out of molybdenum rather than tungsten._ Previous works [2, 3]
constructed models of GCR for space propulsion. Each team used a computer code
for calculating neutron balance. _We have written our own Monte-Carlo
simulation code in MatLab._ The code constructs paths of a large number of
neutrons within the reactor. Path of each neutron begins with neutron
generation, usually contains many collisions, and ends with neutron absorbtion
or leakage from the reactor. Using the information for many neutrons we
generate statistical data. The data is used to determine neutron balance and
criticality of the reactor.

###  3.1 General method

The reactor contains several regions with each one having its own composition.
We perform a Monte Carlo simulation tracking a number of neutrons along their
paths within the reactor. Each neutron undergoes a number of events before
being absorbed or lost. Below we give a list of possible events.

  1. 1.

An elastic or inelastic scattering during which the neutron changes both its
energy and flight direction.

  2. 2.

The event of passage of the neutron from one reactor region to another, when
the neutron keeps its energy and flight direction. The reactor has three
regions, they are internal cavity containing gaseous fuel, refractory material
shell, and the reflector shell.

  3. 3.

Absorption or leakage of the neutron, which ends the neutron’s path. If the
neutron is absorbed within a region containing fissile material, or a region
where it can undergo an (n,2n) reaction, then the place of absorption along
with the expected number of secondary neutrons produced are recorded.

The neutron’s energy, coordinates and flight direction are kept track of
during the steps.

For a neutron traveling in a region $R=1,2,3$ (see Subsection 3.3 below) and
having a given energy $E$, there are two probability constants. The scattering
probability $p_{s}(R,E)$ is the probability of scattering per unit length. The
absorption probability $p_{a}(R,E)$ is the probability of absorption per unit
length. Both probabilities have unites of $m^{-1}$. Below we present some
examples of these probabilities. For $1500\ ^{o}$K thermal neutrons in liquid
beryllium reflector, $p_{s}=70\ m^{-1}$ and $p_{a}=3.4\cdot 10^{-2}\ m^{-1}$.
For 1500oK thermal neutrons in graphite reflector with density $2.0\
g/cm^{3}$, $p_{s}=47\ m^{-1}$ and $p_{a}=1.4\cdot 10^{-2}\ m^{-1}$. For 1500oK
thermal neutrons in 184W, $p_{s}=46\ m^{-1}$ and $p_{a}=3.8\ m^{-1}$. As we
will see in Subsection 4.2, the neutron losses within tungsten are very high.

###  3.2 Tabulation of cross-sections

The cross-sections are obtained from the South Korean source "Table of
Nuclids"[22] and US Government source "Evaluated Nuclear Data"[23]. A nuclide
is a single isotope of a single element – like 235U. For each nuclide, the
data is tabulated for the set of 1,231 energies from $10^{-5}\ eV$ to $20\
MeV$ given by

$\mathbf{E}=\left\\{10^{i/100}\right\\}_{i=-500}^{730}\ eV.$ (3.1)

The data for each nuclide are presented as matrix with 1,231 rows and a few
columns. For all nonfissile nuclei, the columns contain the following: Column
1: Energy; Column 2: Elastic scattering cross-section; Column 3: Radiative
capture cross-section; Column 4: First inelastic scattering cross-section;
Column N+3: N${}^{\text{th}}$ inelastic scattering cross-section. For all
fissile nuclei, the following tabulation rules apply: Column 1: Energy; Column
2: Fission cross-section; Column 3: Radiative capture cross-section; Column 4:
Average number of neutrons per fission.

At this point we look at the three reactor regions described in Subsection
3.3. For each reactor region the following data is tabulated for 1,231 values
of neutron energy:

  1. 1.

EV(E) – total event cross-section per nuclide. An event is any scattering,
absorption or fission initiation.

  2. 2.

CPT(E) – total capture cross-section per nuclide. Includes fission triggering.

  3. 3.

RET(E) – the quotient of the average number of fission neutrons produced per
captured neutron.

The tables described above contain the cross-sections of every possible
neutron interaction for every nuclide present within the reactor and for every
possible neutron energy. The total event cross – section per nuclide for every
reactor region can be used to find total event cross section per unit length.
It is done by the following expression:

$P_{i}(E)=n\cdot EV(E),$

where $P_{i}(E)$ is the total event cross section per unit length, and $n$ is
the total nuclide density for a given reactor region. The availability of
$P_{i}(E)$ enables the program to simulate a neutron segment path length by
(3.14) below. Total capture cross-section $CPT(E)$ enables the program to know
the probability of each link in the neutron path being the last one. The the
quotient of the average number of fission neutrons produced per captured
neutron $RET(E)$ enables the simulation program to calculate the neutron
balance.

###  3.3 Reactor regions

The spherical reactor contains three regions (denoted by $R=1,2,3$) – the
gaseous fuel region, the refractory material shell, and the neutron reflector.
These regions are described below.

  1. 1.

Gaseous fuel region

This region extends for $r\in[0,r_{0}]$, where $r_{0}$ is the radius of
reactor cavity. It contains fuel in gaseous form – 62% UF4, 30% KF and 8% K by
mass. By nuclide abundance this region consists of 10% 235U, 28% Potassium,
and 62% Fluorine. An average nucleon weighs 46.2 amu. As 235U absorbs
neutrons, 236U is formed. When 236U absorbs a neutron it produces 237U which
decays to 237Np, which has high capture cross – section for thermal neutrons.
When 237Np absorbs a neutron it produces 238Np, which rapidly decays to 238Pu,
which also has high thermal neutron absorption cross – section. When 238Pu
absorbs a neutron it produces 239Pu, which is similar to 235U in its fissile
properties.

  2. 2.

Refractory material region

This region extends for $r\in[r_{0},r_{1}]$. It is at most $2\ cm$ thick. An
obvious candidate for this region is tungsten – it has the highest melting
point and it is compatible with UF4 up to 3,000oK [3, p.5].

The isotope of tungsten with the smallest neutron absorbtion cross-section is
184 W. Nevertheless, as it is shown in Section 4, even a very thin shell of
184 W absorbs too many neutrons for reactor to function. Moreover, as 184 W
absorbs neutrons, 185 W forms. 185 W may absorb a neutron to form 186 W, or it
may decay to 185 Re – it’s $\beta$ – decay half – life is 75 days. Further
neutron absorptions and $\beta$ – decays will produce 187 Re, osmium isotopes
186 – 190, as well as iridium. All rhenium, osmium and iridium isotopes have
high thermal neutron capture cross – sections.

According to Anghaie, "for reasons of neutron economy, tungsten and tungsten
alloys cannot be used. Instead, materials such as the alloy TZM (99% Mo, 0.9%
Ti, 0.1% Zr) are necessitated both from a materials and a neutronics" [3,
p.5]. Molybdenum used in this region will consist of isotope 92Mo.

  3. 3.

Reflector region

This region extends for $r\in[r_{1},r_{2}]$. It is is $40\ cm$ to $60\ cm$
thick. It contains a neutron reflector for which there are several choices: i)
molten beryllium in a container made of molybdenum isotope 92Mo, ii) beryllium
oxide, iii) graphite, iv) molten beryllium fluoride BeF2 in a container made
of molybdenum isotope 92Mo. In our simulation we use molten beryllium. As 9Be
absorbs neutrons, the isotope 10Be is produced. The isotope 10Be is metastable
with half-life of $1.5\cdot 10^{6}\ years$ [24, p. 11-199] and neutronic
properties similar to those of 9Be.

A small amount of 9Be will undergo the reaction

$^{9}Be+n\to^{6}Li+^{4}He.$ (3.2)

The isotope 6Li has a thermal neutron absorbtion cross–section of $920\ barn$,
thus beryllium would have to be constantly reprocessed to separate lithium.
Given that beryllium is liquid it should not be difficult.

###  3.4 Path of a single neutron

The simulation will generate the paths of $N_{n}$ neutrons each of which
originates from the gaseous fuel region of the reactor. The energy
distribution of neutrons from 235U fission is [25, p. 34]:

$N(E)=0.453\exp\left(-1.036E\right)\sinh\left(\sqrt{2.29E}\right),$ (3.3)

where the energy $E$ is given in $MeV$.

Each path will contain one or more segments described below.

Neutron path segment

We define any segment of the neutron path by the following parameters.
$\textbf{r}_{s}\in\mathbb{R}^{3}$ is the origin of the segment. The velocity
direction $\textbf{u}_{s}$ belongs to a unit sphere. The energy of the neutron
is given in $eV$, and determines absorption, scattering and fission cross –
sections and thus absorption probabilities for the neutron. The infinite line
$\mathbb{L}_{s}\subset\mathbb{R}^{3}$ is the linear extension of the neutron
path segment in both directions. We express $\mathbb{L}_{s}$ as

$\mathbb{L}_{s}=\left\\{\textbf{r}_{s}+x\textbf{u}_{s},\ \
x\in\mathbb{R}\right\\}.$ (3.4)

The neutron path segment is illustrated below:

Figure 5: Neutron path segment

The angle between $\textbf{r}_{s}$ and $\textbf{u}_{s}$ is $\varphi$
determined by

$\cos\varphi=\frac{\textbf{r}_{s}\cdot\textbf{u}_{s}}{\|\textbf{r}_{s}\|\|\textbf{u}_{s}\|}=\frac{\textbf{r}_{s}\cdot\textbf{u}_{s}}{\|\textbf{r}_{s}\|}.$
(3.5)

The point at which $\mathbb{L}_{s}$ is closest to the origin is given by

$\textbf{r}_{s0}=\textbf{r}_{s}-\|\textbf{r}_{s}\|\left(\cos\varphi\right)\textbf{u}_{s}=\textbf{r}_{s}-\left(\textbf{r}_{s}\cdot\textbf{u}_{s}\right)\textbf{u}_{s}.$
(3.6)

The closest approach of $\mathbb{L}_{s}$ to the origin is

$\begin{split}\left\|\textbf{r}_{s0}\right\|&=\left\|\textbf{r}_{s}-\left(\textbf{r}_{s}\cdot\textbf{u}_{s}\right)\textbf{u}_{s}\right\|=\sqrt{\left(\textbf{r}_{s}-\left(\textbf{r}_{s}\cdot\textbf{u}_{s}\right)\textbf{u}_{s}\right)\cdot\left(\textbf{r}_{s}-\left(\textbf{r}_{s}\cdot\textbf{u}_{s}\right)\textbf{u}_{s}\right)}=\sqrt{\|\textbf{r}_{s}\|^{2}-\left(\textbf{r}_{s}\cdot\textbf{u}_{s}\right)^{2}}.\end{split}$
(3.7)

Notice, that if $\cos\varphi<0$, then the point of closest approach
$\textbf{r}_{s0}$ is in the direction of the neutron path, otherwise it is
virtual.

At this point we calculate the value of $x$ for which the neutron path
intersects the sphere centered at the origin with radius $r_{i}$. We have to
solve the equation

$\left\|\textbf{r}_{s}+x\textbf{u}_{s}\right\|=r_{i}.$ (3.8)

Rewrite (3.8) as

$\begin{split}r_{i}^{2}&=\left\|\textbf{r}_{s}+x\textbf{u}_{s}\right\|^{2}=\textbf{r}_{s}\cdot\textbf{r}_{s}+2\textbf{u}_{s}\cdot\textbf{r}_{s}x+\textbf{u}_{s}\cdot\textbf{u}_{s}x^{2}=x^{2}+2\textbf{u}_{s}\cdot\textbf{r}_{s}x+\textbf{r}_{s}\cdot\textbf{r}_{s},\end{split}$
(3.9)

which can be rewritten as a quadratic equation

$\begin{split}x^{2}+2\textbf{u}_{s}\cdot\textbf{r}_{s}x+\|\textbf{r}_{s}\|^{2}-r_{i}^{2}=0.\end{split}$
(3.10)

Thus,

$\begin{split}x_{s}=-\textbf{u}_{s}\cdot\textbf{r}_{s}\pm\sqrt{\left(\textbf{u}_{s}\cdot\textbf{r}_{s}\right)^{2}-\|\textbf{r}_{s}\|^{2}+r_{i}^{2}.}\end{split}$
(3.11)

Notice, that the intersection between the neutron path and the sphere of
radius $r_{i}$ is possible only if two conditions are satisfied. First,
$x_{s}>0$. Second, no other sphere where reactor regions are changing is
crossed for an $x<x_{s}$. Thus, if (3.11) has two positive solutions, we take
only the smaller one. Also notice, that for any point within the reactor and
any direction of the neutron path there is a value $x_{s}$, which is the
smallest positive value for which the neutron changes reactor region or
escapes the reactor.

If the neutron continues its flight in a straight line without undergoing a
collision or absorption event, then the flight segment $\mathbb{S}_{s}$ would
be given by

$\mathbb{S}_{s}=\left\\{\textbf{r}_{s}+x\textbf{u}_{s},\ \
x\in[0,x_{s}]\right\\}.$ (3.12)

Let $P_{i}(E)$ be the probability per unit length of the neutron with energy
$E$ undergoing a scattering, absorption, or triggering a fission in the
reactor region $i$. The probability of an event along $\mathbb{S}_{s}$ is

$P=1-\exp\left(-\frac{x_{s}}{P_{i}(E)}\right).$ (3.13)

In a simulation, the value $x_{e}$ where the neutron either changes the
reactor regions or undergoes scattering, absorption, or triggering a fission
can be obtained in two steps. First choose $u\in[0,1]$ from a uniform random
distribution. Then, set $x_{e}$ as

$x_{e}=\min\left\\{x_{s},P_{i}(E)\ln u\right\\}.$ (3.14)

A new segment starts from the point

$\textbf{r}_{s+1}=\textbf{r}_{s}+x_{e}\textbf{u}_{s}.$ (3.15)

Neutron Scattering

Suppose a neutron of mass $m_{0}$, flight direction $\textbf{u}_{0}$ and
energy $E_{0}$ is scattered by a nucleus of mass $m_{1}$, flight direction
$\textbf{u}_{1}$ and energy $E_{1}$. The velocity of the neutron is

$\textbf{v}_{0}=\sqrt{\frac{2E_{0}}{m_{0}}}\ \textbf{u}_{0}.$ (3.16)

We do not know the energy of the scattering nucleus, thus we will have to
generate a random energy. We do have the distribution of particles at
temperature $T$ given by [26, p.312]:

$P(E)=C\sqrt{E}\exp\left(\frac{E}{kT}\right),$ (3.17)

where $k=1.381\cdot 10^{-23}\ J/^{o}K$ is the Boltzmann constant.

The velocity of the nucleus is

$\textbf{v}_{1}=\sqrt{\frac{2E_{1}}{m_{1}}}\ \textbf{u}_{1}.$ (3.18)

The velocity of the center of mass is

$\textbf{v}_{c}=\frac{m_{0}\textbf{v}_{0}+m_{1}\textbf{v}_{1}}{m_{0}+m_{1}}.$
(3.19)

The velocity of the neutron relative to the center of mass is

$\begin{split}\textbf{v}_{0c}&=\textbf{v}_{0}-\textbf{v}_{c}=\textbf{v}_{0}-\frac{m_{0}\textbf{v}_{0}+m_{1}\textbf{v}_{1}}{m_{0}+m_{1}}=\frac{m_{1}\left(\textbf{v}_{0}-\textbf{v}_{1}\right)}{m_{0}+m_{1}}.\end{split}$
(3.20)

The speed of the neutron relative to the center of mass is

$\begin{split}v_{0c}=\frac{m_{1}\left\|\textbf{v}_{0}-\textbf{v}_{1}\right\|}{m_{0}+m_{1}}.\end{split}$
(3.21)

The velocity of the nucleolus relative to the center of mass is derived
similar to (3.20):

$\begin{split}\textbf{v}_{1c}=\frac{m_{0}\left(\textbf{v}_{1}-\textbf{v}_{0}\right)}{m_{0}+m_{1}}.\end{split}$
(3.22)

At this point we consider three types of scattering: i) Low energy elastic;
ii) High energy elastic; iii) Inelastic.

  1. 1.

Low energy elastic scattering

For energies under $50\ keV$, the angular distribution of scattering neutrons
is spherically symmetric. According to "Evaluated Nuclear Data"[23], for light
scattering nuclei up to oxygen and neutron energies up to $50\ keV$, the
angular distribution of scattered neutrons does not deviate from spherical
symmetry by more then 1%. In this case, the scattering direction
$\textbf{u}_{0}^{*}$ is selected from a spherical uniform distribution via the
following set of two commands in MatLab

            uu=randn(1,3);
        uu=uu/norm(uu);

The neutron speed relative to the center of mass is given by (3.21). In an
elastic collision, the neutron will keep it’s speed relative to the center of
mass, but change it’s direction. Thus the new neutron velocity relative to the
center of mass is

$\begin{split}\textbf{v}_{0c}^{*}=\frac{m_{1}\left\|\textbf{v}_{0}-\textbf{v}_{1}\right\|}{m_{0}+m_{1}}\textbf{u}_{0}^{*}.\end{split}$
(3.23)

Adding the center of mass velocity to the neutron’s velocity we obtain the
scattered neutron’s velocity:

$\begin{split}\textbf{v}_{0}^{*}=\textbf{v}_{0c}^{*}+\textbf{v}_{c}=\frac{m_{1}\left\|\textbf{v}_{0}-\textbf{v}_{1}\right\|}{m_{0}+m_{1}}\textbf{u}_{0}^{*}+\frac{m_{0}\textbf{v}_{0}+m_{1}\textbf{v}_{1}}{m_{0}+m_{1}}.\end{split}$
(3.24)

  2. 2.

High energy elastic scattering

Once we know the scattering direction, the equations (3.23) and (3.24) are the
same as for the low energy scattering. The main difference is that the
scattering direction $\textbf{u}_{0}^{*}$ is neither spherically symmetric nor
independent of the original direction $\textbf{u}_{0}$. Let $\beta$ be the
scattering angle which is the angle between $\textbf{u}_{0}^{*}$ and
$\textbf{u}_{0}$. Then

$\cos\beta=\textbf{u}_{0}^{*}\cdot\textbf{u}_{0}.$ (3.25)

For symmetrical scattering, $\cos\beta$ follows a uniform random distribution
within $[-1,1]$ and thus the expectation value of $\cos\beta$ is $0$:

$E(\cos\beta)=0.$ (3.26)

For high energy neutrons, and especially for heavier nuclei, the scattering is
cylindrically symmetric, but not spherically symmetric. "Evaluated Nuclear
Data"[23] tabulates the distributions of $\cos\beta$ for all scattering
nuclides for different incoming neutron energies.

In practical terms, the expected value of $\cos\beta$ increases with neutron
energy and approaches $1$ as the neutron energy comes into the region of
hundreds of MeV – although such energetic neutrons are not considered in the
present work. That means that the deflection angle is smaller for neutrons
with energies of over $2\ MeV$ than for thermal neutrons. Such neutrons
produced within the reactor core will travel deeper into the neutron reflector
before being stopped than the less energetic neutrons. Thus, those of them
which will be neither absorbed nor lost will be thermalized before returning
to the reactor core.

  3. 3.

Inelastic scattering

In inelastic scattering, part of the neutron’s initial energy is transferred
to the scattering nucleus, which goes into an excited state. The excitation
energy $E_{e}$ of the nuclide has to be lower than the energy of the neutron –
nucleus system with respect to the center of mass, otherwise known as the
center of mass energy. The excitation energy in an inelastic scattering is
taken from the system energy relative to the center of mass.

In order to determine the energy of the system relative to the center of mass,
we recall the neutron and nucleus velocity in the center of mass frame of
reference:

$\begin{split}v_{0c}=\frac{m_{1}\left\|\textbf{v}_{0}-\textbf{v}_{1}\right\|}{m_{0}+m_{1}},\qquad
v_{1c}=\frac{m_{0}\left\|\textbf{v}_{0}-\textbf{v}_{1}\right\|}{m_{0}+m_{1}}.\end{split}$
(3.27)

Given that the neutron energy of inelastic scattering is at least $50\ keV$
for all nuclei considered, the velocity of the nucleus is negligible with
respect to neutron velocity. Thus,

$\begin{split}v_{0c}=\frac{m_{1}}{m_{0}+m_{1}}v_{0},\qquad
v_{1c}=\frac{m_{0}}{m_{0}+m_{1}}v_{0},\end{split}$ (3.28)

where $v_{0}=\|\textbf{v}_{0}\|$ is the speed of the neutron in the reactor
frame of reference. The total kinetic energy of the neutron and the nucleus in
the center of mass frame of reference is

$\begin{split}E_{c}&=\frac{1}{2}\left(m_{0}v_{0c}^{2}+m_{1}v_{0c}^{2}\right)=\frac{1}{2}\left[m_{0}\left(\frac{m_{1}}{m_{0}+m_{1}}v_{0}\right)^{2}+m_{1}\left(\frac{m_{0}}{m_{0}+m_{1}}v_{0}\right)^{2}\right]\\\
&=\frac{1}{2}m_{0}v_{0}^{2}\left[\frac{m_{1}^{2}}{\left(m_{0}+m_{1}\right)^{2}}+\frac{m_{1}m_{0}}{\left(m_{0}+m_{1}\right)^{2}}\right]=E_{0}\frac{m_{1}}{m_{0}+m_{1}}.\end{split}$
(3.29)

After the scattering event, the center of mass kinetic energy will be

$E_{c}^{*}=\frac{m_{1}E_{0}}{m_{0}+m_{1}}-E_{e}.$ (3.30)

As a result of the inelasticity, the center of mass energy is decreased by the
factor of

$\frac{E_{c}^{*}}{E_{c}}=\frac{\frac{m_{1}E_{0}}{m_{0}+m_{1}}-E_{e}}{\frac{m_{1}E_{0}}{m_{0}+m_{1}}}=1-\frac{E_{e}}{E_{c}}\frac{m_{0}+m_{1}}{m_{1}}.$
(3.31)

Therefore, in the inertial frame of reference, both the speed of the scattered
neutron and the scattered nuclei will decrease by the factor of

$\sqrt{1-\frac{E_{e}}{E_{c}}\frac{m_{0}+m_{1}}{m_{1}}}.$ (3.32)

Substituting (3.32) into (3.23), we obtain the velocity of an inelastically
scattered neutron relative to the center of mass:

$\begin{split}\textbf{v}_{0c}^{*}=\frac{m_{1}\left\|\textbf{v}_{0}-\textbf{v}_{1}\right\|}{m_{0}+m_{1}}\sqrt{1-\frac{E_{e}}{E_{c}}\frac{m_{0}+m_{1}}{m_{1}}}\
\textbf{u}_{0}^{*}.\end{split}$ (3.33)

Adding the center of mass velocity to the neutron’s velocity we obtain the
scattered neutron’s velocity:

$\begin{split}\textbf{v}_{0}^{*}=\textbf{v}_{0c}^{*}+\textbf{v}_{c}=\frac{m_{1}\left\|\textbf{v}_{0}-\textbf{v}_{1}\right\|}{m_{0}+m_{1}}\sqrt{1-\frac{E_{e}}{E_{c}}\frac{m_{0}+m_{1}}{m_{1}}}\
\textbf{u}_{0}^{*}+\frac{m_{0}\textbf{v}_{0}+m_{1}\textbf{v}_{1}}{m_{0}+m_{1}}.\end{split}$
(3.34)

###  3.5 Neutronics

Neutronics refers to the balance of neutrons which has to be maintained within
reactor. We introduce criticality, otherwise known as multiplication factor,
which is the ratio of the number of neutrons produced by fission to the number
of neutrons absorbed or lost [27, p.3]:

$k=\frac{\text{neutrons produced in current fission
generation}}{\text{neutrons absorbed in previous fission generation}}.$ (3.35)

If $k<1$, the reactor is subcritical and fission stops very quickly. If $k>1$,
the reactor becomes supercritical and the neutron flux multiplies until the
criticality is restored to 1 or there is an accident. Thus for a working
reactor, $k=1$ with possible oscillations of very small magnitude.

Define the fuel utilisation factor [27, p.4] as

$f=\frac{\text{number of neutrons absorbed by the
fuel}}{\left\\{\begin{split}&\text{number of neutrons absorbed by the fuel,
}\\\ &\text{refractory sphere, reflector, fission products,}\\\ &\text{ and
leaked neutrons}\end{split}\right\\}}.$ (3.36)

Define regeneration factor [27, p.4] as

$\eta=\frac{\text{neutrons produced from fission}}{\text{neutrons absorbed in
fuel}}.$ (3.37)

As we will see in simulations, for the reactor we are considering,
$\eta=1.96$. The relationship between criticality, fuel utilisation factor and
regeneration factor is

$k=f\eta=1.$ (3.38)

It follows, that the proportion of the neutron flux absorbed by the fuel is

$f=\frac{1}{\eta}=51\%.$ (3.39)

The total fraction of neutrons leaked from reactor, absorbed by fission
products, the 92Mo shell, and reflector and lost in other ways has to be 49%.

In the gas core reactor, the density of fuel will be much lower than that of
solid or liquid core reactors. Thus, ensuring that 51% of the neutrons are
absorbed by the fuel would be a challenge. In order to do so we need the
following:

  1. 1.

Maximize the fuel density. To do so we maximize the pressure within the limits
presented by the pressure vessel.

  2. 2.

Maximize the size of reactor cavity within the limits imposed by reactor mass.
Even though reactors existing on Earth have almost no mass limit, the cost of
placing material into deep space is at least $20,000 per kg.

  3. 3.

Use reflector of sufficient thickness to minimize neutron leakage. Generally,
$40\ cm$ to $60\ cm$ reflectors will be used.

  4. 4.

Use the least amount of tungsten or no tungsten at all in the refractory
shell. Tungsten is compatible with UF4 vapor up to 3,000oK, while molybdenum
is compatible with UF4 vapor up to 2,000oK only [3, p.5].

  5. 5.

Separate fission products with high neutron absorbtion cross – section from
the fuel stream. This should be possible, since the fuel is constantly
recirculating in and out of reactor zone. These products can be discarded into
space.

Most metal fission products will appear as fluorides. Products with boiling
point significantly different from that of uranium tetrafluoride (1417 oC) can
be easily separated by distillation. The worst neutron absorber 135Xe cannot
be separated as it absorbs neutrons very fast.

  6. 6.

Discard the fuel as it is strongly contaminated by 236U. This will mean that
some 235U will also be discarded – in our model 60%.

Neutron absorbtion by fission products

As 235U undergoes fission, fission products will form. Metallic products will
be present mostly in the form of fluorides. Neutron absorption by fission
products is called poisoning [28]. Some important neutron absorbers or neutron
poisons are presented in Table 3. Fission yield denotes the number of nuclei
produced per fission.

Isotope | Fisson | Thermal neutron | Comment
---|---|---|---
| yield | cross section |
|  | at $0.025\ eV$ |
135Xe | 0.065 |  $2.6\cdot 10^{6}$ barn | Absorbs neutrons very fast
149Sm | 0.011 |  $4.0\cdot 10^{4}$ barn | Absorbs neutrons fast, produces 150Sm
|  |  | with 109 barn thermal neutron
|  |  | absorption cross section
151Sm | 0.008 |  $1.5\cdot 10^{4}$ barn | Absorbs neutrons fast, produces 152Sm
|  |  | with 2,760 barn neutron resonance integral

Table 3: Important neutron poisons

According to Kotov [29, p. 190], the total fission product absorbtion (FPA) is
equivalent to $FPA=0.11$ per each fission event. In order to account for
fission product absorbtion, we multiply the number of neutrons produced per
fission by $(1-FPA)$. V. Kotov has developed a regime for which $FPA=0.07$. In
this regime the fuel is used in a cycle for 5 to 10 hours and then is rested
for 35 – 50 hours during which most 135Xe decays [29, p. 191]. Given that fuel
can be stored in liquid form, possibly this cycle can be used.

Neutron multiplication by beryllium

Neutrons are multiplied in beryllium via the reaction

$^{9}\text{Be}+n\to^{8}\text{Be}+2n-1.665\ MeV\to 2^{4}\text{He}+2n.$ (3.40)

This reaction is possible only for neutrons with energy exceeding 1.8 $MeV$.
The scattered neutrons have much lower energy than the original neutron. Only
the most energetic neutrons can produce tertiary neutrons. According to a 1959
calculation [30], beryllium moderator increases the number of neutrons by 5.1%
to 7.9%. According to a 1963 Soviet calculation [31], number of neutrons is
increased by 10% in beryllium and 8% in beryllium oxide. In our calculations,
the number of neutrons is increased by 9% to 11%. For beryllium moderator, the
neutron gain almost offsets the absorbtion loss.

##  4 Simulation Results

In this section we present the results of numerical simulations concerning
neutron balance. These simulations show that the reactor can not achieve
criticality if the shell surrounding reactor cavity is made of tungsten over 1
$mm$ thick. The reactor can easily achieve criticality if it has a
${}^{\text{92}}$Mo shell 1 $cm$ thick.

###  4.1 Neutron reflector

At this point we determine the optimal thickness of neutron reflector. To this
end we look at the paths of the neutrons which are reflected by liquid
beryllium shell and returned into the reactor cavity. Let $p(x)$ be a fraction
of reflected neutrons which have achieved depth of at least $x$ within the
neutron reflector. In all cases, the function $p(x)$ is strictly decreasing
and such that $p(0)=1$, and

$\lim_{x\to\infty}p(x)=0.$ (4.1)

Let $R(r)$ be the neutron reflectance of an infinite liquid beryllium shell
surrounding a reactor cavity of radius $r$. Then a liquid beryllium shell of
thickness $x$ surrounding a reactor cavity of radius $r$ would have
reflectance $(1-p(x))R(r)$.

Liquid beryllium reflector

We estimate the function $p(x)$ for liquid beryllium reflectors with reactor
cavities of different radii. In each case, we have used 20,000 neutrons. For
these simulations, we have used the programs _InfiniteCilynderReflector_Be.m_
and _SphericalReflector_Be.m_ which we have written ourselves. The results are
tabulated below:

Reactor cavity |  | Penetration depth $x$, cm
---|---|---
radius $cm$ |  | 30 | 35 | 40 | 45 | 50 | 55 | 60 | 65 | 70
50 | $p(x)=$ | 3.5% | 2.3% | 1.7% | 1.1% | 0.8% | 0.6% | 0.36% | 0.27% | 0.17%
70 | $p(x)=$ | 3.7% | 2.6% | 1.9% | 1.4% | 1.0% | 0.7% | 0.48% | 0.33% | 0.17%
100 | $p(x)=$ | 3.3% | 2.3% | 1.5% | 1.1% | 0.8% | 0.5% | 0.35% | 0.20% | 0.14%
150 | $p(x)=$ | 3.3% | 2.2% | 1.5% | 1.1% | 0.7% | 0.5% | 0.38% | 0.26% | 0.18%
200 | $p(x)=$ | 3.5% | 2.3% | 1.6% | 1.2% | 0.8% | 0.5% | 0.39% | 0.30% | 0.25%

Table 4: Reflected neutrons penetrating liquid beryllium reflector to depth
$x$

As it will be seen from Table 5 below, the _optimal thickness_ of molten
beryllium neutron reflector should be $40\ cm$ to $60\ cm$ for most reactors.

The reflectance $R(r)$ has also been estimated for different radii of reactor
cavity. We have also performed several tests for reactors shaped as infinite
cylinders rather than spheres. In each case, the simulation ran a set of
20,000 neutrons. For all simulations, we have used the programs
_InfiniteCilynderReflector_Be.m_ and _SphericalReflector_Be.m_ mentioned
above. The reflectances are tabulated below:

Cavity | Neutron | Neutron | Neutron | Neutron
---|---|---|---|---
radius $r$ | Reflectance $R$ | Reflectance $R$ | Reflectance $R$ | Reflectance $R$
in $cm$ | for fission | for thermal | for fission | for thermal
| neutrons | neutrons | neutrons | neutrons
| spherical | spherical | cylindrical | cylindrical
| reactor | reactor | reactor | reactor
50 | $85.9\pm 0.2\%$ | $91.6\pm 0.2\%$ | $88.5\pm 0.2\%$ | $93.4\pm 0.2\%$
70 | $87.1\pm 0.2\%$ | $92.6\pm 0.2\%$ | $88.9\pm 0.2\%$ | $93.8\pm 0.2\%$
100 | $88.6\pm 0.2\%$ | $93.3\pm 0.2\%$ | $90.0\pm 0.2\%$ | $94.2\pm 0.2\%$
150 | $89.8\pm 0.2\%$ | $93.6\pm 0.2\%$ | $90.8\pm 0.2\%$ | $94.4\pm 0.2\%$
200 | $90.0\pm 0.2\%$ | $93.7\pm 0.2\%$ | $90.8\pm 0.2\%$ | $94.6\pm 0.2\%$

Table 5: Infinite beryllium sphere or cylinder neutron reflectance

As follows from Table 5, reflectance $R$ for fission neutrons in a spherical
reactor is a little bit lower than the reflectance for fission neutrons in a
cylindrical reactor. A similar conclusion can be made about the reflectance
for thermal neutrons.

In operation of an actual reactor, a fission neutron is usually reflected
several times before being lost or absorbed. As we will see below, 27% to 36%
of neutrons will be lost or absorbed by the reflector. For the reactors with
criticality of 1, reflector loss would be 31% to 33%. Thus, an average neutron
is reflected about three times.

###  4.2 Reactor neutron balance

We have performed several simulations for a spherical reactor surrounded by a
neutron reflector shell. Here we make calculations for a reactor with cavity
surrounded by a tungsten shell. The tungsten shell initially contains 184W.
Eventually, as 184W absorbs neutrons it will contain 185W, 185Re, 186Re,
187Re, 188Re, 186Os, 187Os, 188Os, 189Os, 190Os, and iridium. All of these
isotopes have large thermal and resonance neutron absorbtion cross-sections.
It is convenient to introduce the effective thickness of tungsten shell which
is thickness of non-irradiated shell of 184W, which has the same neutron
absorbtion properties as an irradiated 184W shell. Depending on irradiation,
the effective thickness of 184W shell should be up to 4 times greater than the
actual thickness of the 184W shell.

We have performed tests for reactors with 50 $cm$ molten beryllium reflector
shells. We use reactors with cavity radii of 1.0 $m$ and 1.5 $m$. We use fuel
gas densities of 20 $kg/m^{3}$ and 40 $kg/m^{3}$. We use tungsten shells of
effective thickness of 0 $mm$, 1 $mm$, 2 $mm$, and 5 $mm$. In each case the
sample size of 5,000 neutrons have been used. In all working reactors, the
multiplication factor $k$ is 1. If our calculations show that $k>1$, it means
that there is room for relaxing reactor conditions – like reducing fuel
density, or reducing reflector thickness, or using a thicker tungsten shell.
If our calculations show that $k<1$, it means that the reactor will not work
until adjustments are made. For all simulations, we have used the program
_SphericalReactor_Be.m_ which we have written ourselves. The program
calculates average number of fission neutrons produced per neutron absorbed by
fuel. This number is always $1.90$ to $1.96$. By multiplying this number by
the fraction of the neutrons absorbed by fuel we obtain the multiplication
factor. The results are tabulated below.

Reactor cavity radius $cm$ |  Tungsten shell effective thickness $mm$ |  Beryllium reflector thickness |  Fuel density $kg/m^{3}$ |  Percent neutrons absorbed by fuel |  Percent neutrons absorbed by W |  Percent neutrons absorbed by reflector |  Percent neutrons lost |  Multi- plication factor
---|---|---|---|---|---|---|---|---
100 |  0 |  50 |  20 |  59.1 |  0 |  20.8 |  20.1 |  1.15
100 |  0 |  50 |  40 |  64.4 |  0 |  18.5 |  17.1 |  1.23
150 |  0 |  50 |  20 |  65.0 |  0 |  18.2 |  16.8 |  1.24
150 |  0 |  50 |  40 |  67.5 |  0 |  17.1 |  15.4 |  1.28
100 |  1 |  50 |  20 |  49.5 |  9.1 |  21.3 |  20.1 |  0.97
100 |  1 |  50 |  40 |  56.6 |  6.8 |  18.8 |  17.8 |  1.09
150 |  1 |  50 |  20 |  55.6 |  8.0 |  19.5 |  16.9 |  1.07
150 |  1 |  50 |  40 |  59.6 |  7.2 |  18.3 |  14.9 |  1.14
100 |  2 |  50 |  20 |  47.8 |  12.9 |  19.8 |  19.5 |  0.93
100 |  2 |  50 |  40 |  52.6 |  11.5 |  19.0 |  16.9 |  1.01
150 |  2 |  50 |  20 |  53.7 |  11.7 |  18.0 |  16.7 |  1.03
150 |  2 |  50 |  40 |  58.1 |  10.6 |  15.6 |  15.8 |  1.11
100 |  5 |  50 |  20 |  39.4 |  22.5 |  19.1 |  19.1 |  0.77
100 |  5 |  50 |  40 |  45.1 |  19.6 |  17.8 |  17.5 |  0.87
150 |  5 |  50 |  20 |  46.4 |  20.7 |  17.4 |  15.5 |  0.90
150 |  5 |  50 |  40 |  50.6 |  17.8 |  16.3 |  15.3 |  0.96
Table 6: Neutron balance for a reactor with beryllium reflector and tungsten
inner shell

The above table shows that a tungsten shell of even 1 $mm$ effective thickness
absorbs 7% – 9% of neutrons. (See “Percent neutrons absorbed by W" column and
take data for Tungsten shell effective thickness equal to one.) The neutrons
absorbed by the tungsten shell are almost completely reflected neutrons, thus
a 1 $mm$ effective thickness lowers criticality by .14 – .18. These numbers
are obtained by subtracting the multiplication factors for reactors with 1
$mm$ tungsten shell from multiplication factors for similar reactors with no
tungsten shell. Given that even a 0.5 $mm$ tungsten shell will soon attain an
effective thickness of 2 $mm$, it follows, that very little tungsten can be
used as the inside shield. For lighter reactors, no tungsten at all should be
used for the internal shell.

Instead of tungsten, an alloy consisting almost entirely of molybdenum – 92 is
used. Once again, we perform tests for reactors with with 50 $cm$ molten
beryllium reflector shells. We use reactors with cavity radii of 1.0 $m$ and
1.5 $m$. We use fuel gas densities of 20 $kg/m^{3}$ and 40 $kg/m^{3}$. We use
92Mo shells of effective thickness of 1 $cm$ and 2 $cm$. The molybdenum shell
should be thick since it contains many tubes carrying potassium and potassium
fluoride. Molten potassium and potassium fluoride are sprayed from many
micronozzles on the reactor walls, thus shielding the walls from the extreme
heat.

Reactor cavity radius $cm$ |  92Mo shell effective thickness $cm$ |  Beryllium reflector thickness |  Fuel density $kg/m^{3}$ |  Percent neutrons absorbed by fuel |  Percent neutrons absorbed by Mo |  Percent neutrons absorbed by reflector |  Percent neutrons lost |  Multi- plication factor
---|---|---|---|---|---|---|---|---
100 |  1 |  50 |  20 |  53.9 |  7.0 |  20.0 |  19.1 |  1.03
100 |  1 |  50 |  40 |  57.6 |  5.7 |  19.0 |  17.7 |  1.09
150 |  1 |  50 |  20 |  58.7 |  6.9 |  18.2 |  16.2 |  1.11
150 |  1 |  50 |  40 |  61.3 |  6.0 |  17.5 |  15.3 |  1.15
100 |  2 |  50 |  20 |  49.1 |  10.3 |  20.2 |  20.4 |  0.94
100 |  2 |  50 |  40 |  54.2 |  9.8 |  18.0 |  18.0 |  1.02
150 |  2 |  50 |  20 |  54.7 |  10.6 |  18.3 |  16.5 |  1.03
150 |  2 |  50 |  40 |  57.8 |  9.8 |  17.4 |  15.0 |  1.08
Table 7: Neutron balance for a reactor with beryllium reflector and molybdenum
inner shell

From Table 7 we conclude that a reactor with 1.0 $m$ cavity radius, 50 $cm$
reflector shell thickness, 1 $cm$ molybdenum inner shell can function with
fuel density of 20 $kg/m^{3}$. A reactor with 1.0 $m$ cavity radius, 50 $cm$
reflector shell thickness, 2 $cm$ molybdenum inner shell can function with
fuel density of 40 $kg/m^{3}$.

##  5 Density and chemical equilibrium calculations

The primary goal of these calculations is determining the plasma conductivity
at different temperatures and pressures. In order for the magnetohydrodynamic
generator to work, plasma conductivity must be sufficiently high.

###  5.1 Fuel density inside the reactor

We would like to find out the pressure and molecular composition of the fuel
with mass composition of 62% UF4, 30% KF and 8% K inside the reactor. We are
given the fuel density of $20\ kg/m^{3}$ to $40\ kg/m^{3}$ and its temperature
of $3,500\ ^{o}$K. We use the Rocket Propulsion Analysis (RPA) program [16] to
solve this problem. Here, Rocket Propulsion Analysis is not used to calculate
anything about the propulsion system itself. Instead, RPA is used to calculate
the chemical equilibrium in the reactor chamber depending on composition and
pressure.

We have calculated the pressures for reactor fuel at $3,500\ ^{o}$K and
different densities. For $20\ kg/m^{3}$, the pressure is $52.5\ atm$, for $30\
kg/m^{3}$, the pressure is $78\ atm$, and for $40\ kg/m^{3}$, the pressure is
$104\ atm$. The size of the pressure vessel required to withstand these
pressures would depend on the temperature at which it would be kept during the
work of reactor.

###  5.2 Plasma expansion

As the plasma leaves the reactor, it expands in a nozzle, with some of the
thermal energy being converted into mechanical energy. Then it enters the
magnetohydrodynamic (MHD) generator where its kinetic energy is converted into
electrical energy. Along the way and inside the MHD generator, the plasma is
subject to the neutron irradiation from the main reactor. The main purpose of
irradiating the expanding plasma is to keep its temperature high enough for
conducting electricity. Additional benefit of keeping the plasma at high
temperature is the increased kinetic energy it will acquire during expansion.

The extra ionization of gas provided by fission fragments also increases
conductivity [32], but in our case the fluxes up to $10^{16}\
n/\left(cm^{2}\cdot s\right)$ affect conductivity very slightly [32, p.120].
Only the highest flux reactors can achieve neutron fluxes above $10^{15}\
n/\left(cm^{2}\cdot s\right)$, [33, p.3], thus all of ionization would have to
come from thermal equilibrium. The main reason why electrons released by
fission product ionization do not contribute to conductivity is the fact that
they are absorbed by UF4, UF5 and UF6 as soon as they become free. The
molecules of UF4, UF5 and UF6 have electron affinities of $1.27\ eV$, $3.52\
eV$ and $5.63\ eV$ respectively.

The conductivity of plasma is given by the formula [20, p.136]

$\begin{split}\sigma&=1.44\cdot 10^{7}\left(\Omega\cdot
m\right)^{-1}\left(\frac{Q_{e}}{\text{\AA}^{2}}\right)^{-1}\left(\frac{T}{1,000\
^{o}K}\right)^{-0.5}\frac{n_{e}}{n_{g}}\\\ &=1.44\cdot 10^{7}\left(\Omega\cdot
m\right)^{-1}\left(\frac{Q_{e}}{\text{\AA}^{2}}\right)^{-1}T_{3}^{-0.5}x_{e},\end{split}$
(5.1)

where $\sigma$ is the conductivity, $Q_{e}$ is the mean collision cross-
section between electrons and atoms or ions, $T_{3}$ is the temperature in
$1,000\ ^{o}$K, and $x_{e}=n_{e}/n_{g}$ is the molar fraction of free
electrons. According to Klein, [20, p.136], $Q_{e}\thickapprox
10\text{\AA}^{2}$. The exact value of $Q_{e}$ remains to be determined by more
theoretical and possibly experimental research.

At $2,400-2,900\ ^{o}$K, the conductivity can be approximated by

$\sigma=.85\left(\Omega\cdot m\right)^{-1}f_{e},$ (5.2)

where $f_{e}$ is the molar fraction of free electrons in parts per million.

Below we tabulate free electron abundance for two fuel compositions at
different temperatures and pressures of 2 $atm$ and 0.4 $atm$. Composition A
contains 62% UF4, 30% KF and 8% K by mass. Composition B 62% UF4, 38% KF by
mass. The results have been obtained by a Rocket Propulsion Analysis
calculation [16].
T, oK Free electron abundance, ppm for composition and pressure Composition A
Composition B 2 $atm$ 0.4 $atm$ 2 $atm$ 0.4 $atm$ 1,800 0 0.1 0 0 1,900 0 0.5
0 0 2,000 0.3 1.5 0 0 2,100 0.9 4.2 0 0 2,200 2.3 11 0 0.2 2,300 5.5 27 0.1
0.5 2,400 12 58 0.3 1.4 2,500 26 119 0.6 3.5 T, oK Free electron abundance,
ppm for composition and pressure Composition A Composition B 2 $atm$ 0.4 $atm$
2 $atm$ 0.4 $atm$ 2,600 51 225 1.4 8.7 2,700 95 395 3.2 21 2,800 166 645 6.9
50 2,900 277 991 15 111 3,000 437 1,440 30 233 3,100 658 2,020 61 453 3,200
947 2,740 120 814 3,300 1,310 3,630 223 1,370

Table 8: Free neutron content of gaseous fuel at different temperatures

As it can be seen from the above table the composition without free potassium
performs much worse than the composition containing free potassium. In order
for the MHD generator to work, we need the gas entering it to have resistivity
of $0.016\ Ohm\cdot m$ at most [3, p.4]. Thus, using formula (5.2), one can
evaluate that the free electron concentration must be at least $50\ ppm$.

For pressures of $2\ atm$ and higher, the expanding fuel can be heated by a
neutron flux. The region where expansion accompanied by nuclear heating takes
place is called the inner expander. From the data presented in Table 8, it
follows that the gaseous fuel at $2\ atm$ pressure must have a temperature of
$2,600\ ^{o}$K. According to a Rocket Propulsion Analysis calculation [16], if
the fuel expands adiabatically from $78\ atm$ pressure (corresponding to $30\
kg/m^{3}$ density) to $2\ atm$ pressure, then its final temperature would be
$2,090\ ^{o}$K. Thus, heating by neutron irradiation and fission must provide
enough energy to increase the temperature by extra $510\ ^{o}$K.

###  5.3 Additional expansion and heating of gaseous fuel

The gaseous fuel leaves the inner expander with velocity $1.2\ km/s$, internal
pressure $2\ atm$, temperature $2,600\ ^{o}$K, and density $1.02\ kg/m^{3}$.
Previous models discussed in the literature [2, 3, 20] suggest reducing
internal gas pressure to $0.65\ atm$. In our model, the pressure is reduced to
$0.4\ atm$. As we see from Table 8, in order for the gaseous fuel to have
electron density of at least $50\ ppm$, it has to have a temperature of
$2,400\ ^{o}$K at $0.4\ atm$ internal pressure.

According to a Rocket Propulsion Analysis calculation [16] if gaseous fuel
adiabatically expands from $2\ atm$ to $0.4\ atm$, its temperature will
decrease from $2,600\ ^{o}$K to $2,050\ ^{o}$K. Thus, in order for MHD
generator to work, the gas would have to be heated by at least $350\ ^{o}$K.
Neutron and fission heating at densities under 1 $kg/m^{3}$ is impossible, and
the use of generated electricity for heating is wasteful. One option, which
should be explored, is mechanical heating. Mechanical heating uses some of the
the kinetic energy of the gas to heat it. It is achieved by placing obstacles
like tungsten or tantalum carbide rods in the path of the gas stream. These
obstacles will cause some of the gas kinetic energy to be transferred into
heat.

The MHD generator will also convert some of the gaseous fuel’s kinetic energy
into thermal energy by means of resistive heating of the gas. The gaseous fuel
will enter the MHD generator as a turbulent flow with velocity $1.26\ km/s$,
internal pressure $0.4\ atm$, temperature $2,400\ ^{o}$K, and density $0.22\
kg/m^{3}$. The MHD generator will extract about 50% of the gas kinetic energy
and convert it into electricity [2, p.102]. The gas will also expand to
internal pressure of $0.08\ atm$ in the generator itself [3, p.8]. The gas
will slow down and retain 15% of its kinetic energy [3, p.8]. Since 50% of the
gaseous fuel’s kinetic energy will turn into electrical energy and 15% of
kinetic energy will remain, 35% of initial kinetic energy will become thermal
energy of the gas. The thermal energy supplied to the gaseous fuel in MHD
generator will thus be $.35\cdot v^{2}/2$ which is $280\ J/g$. According to a
Rocket Propulsion Analysis calculation, the fuel gas will leave the MHD
generator at $2,920\ ^{o}$K. Some of the waste heat may be used to heat liquid
UF4 as it begins the next cycle.

###  5.4 Gaseous fuel condensation

The gaseous fuel containing 62% UF4, 30% KF and 8% K by mass enters condensing
radiator at $0.1\ atm$ pressure. As the mixture cools it begins to condense.
As it condenses, there will be equilibrium between UF3, UF4, K and KF given by
the following diagram:

$K+UF_{4}\leftrightharpoons KF+UF_{3}.$ (5.3)

According to Rocket Propulsion Analysis calculation, as the temperature falls
to $1,490\ ^{o}$K, 34.5% of the fuel condenses as UF3 crystal. As the
temperature drops to $1,300\ ^{o}$K, less than a quarter of potassium remains
unbound to F. The exact phase diagram of solution containing UF4, UF3, KF, and
K remains to be _determined experimentally_.

We would suggest dissolving the fuel in liquid MgF2 and/or 88SrF2. Most of the
solvent would remain liquid throughout the cycle and never enter the reactor
cavity. Finding the best solvent and accounting for all fission product
effects remains _an open problem._

###  5.5 Electrical efficiency estimation

Now we give approximate estimate on the thermal energy which is supplied to
fuel in each cycle and on the amount of thermal energy which is converted to
electric energy by the MHD generator. As the fuel begins the cycle, the 62%
UF4 fraction and the 30% KF fraction enter the reactor as liquids. The heat of
vaporization of KF is $4.2\ kJ/g$ and the heat of vaporization of UF4 is $1.0\
kJ/g$ [16]. Thus for the whole fuel mass, vaporization takes 1.9 $kJ/g$.
According to a Rocket Propulsion Analysis calculation, heating the gaseous
fuel to $3,500\ ^{o}$K takes another $1.0\ kJ/g$. As the fuel expands, it is
heated by fissions caused by neutron flux raising the final temperature from
$2,090\ ^{o}$K to $2,600\ ^{o}$K, which takes another $0.3\ kJ/g$. The total
amount of thermal energy supplied to the fuel during each cycle is thus about
$1.9+1.0+.3=3.2\ kJ/g$.

According to a Rocket Propulsion Analysis calculation as the fuel expands from
$78\ atm$ pressure corresponding to $30\ kg/m^{3}$ density to $2\ atm$
pressure, it obtains $0.75\ kJ/g$ in kinetic energy. Additional expansion from
$2\ atm$ pressure to $0.4\ atm$ pressure will bring only $0.05\ kJ/g$ in
kinetic energy, since most of the energy of expansion will be returned as
thermal energy by mechanical heating. Previous studies [2, p.102] mention
isentropic efficiency of 50% for a Space based MHD generator. Thus, $0.8\
kJ/g$ kinetic energy will yield $0.4\ kJ/g$ electrical energy. The overall
efficiency of the power plant consuming $3.2\ kJ/g$ thermal energy per cycle
and producing $0.4\ kJ/g$ electrical energy per cycle is 12.5%.

###  5.6 Radiator: the heat sink

Any heat engine needs a heat sink. A heat engine with electrical power output
$P_{e}$ and efficiency $\eta$ needs a heat sink with thermal power

$P_{s}=\frac{1-\eta}{\eta}P_{e}.$ (5.4)

For a 12.5% ($\eta=0.125$) efficient system, the heat sink has to reject
almost 7 times more power than the electric generator produces. In our case,
the heat sink is a radiator. A radiator is a structure composed of metal tubes
linked by metal bends. The working fluid passes through the tubes and radiates
its thermal energy. In our case, the working fluid composed of UF4, KF, and K
changes phase from gas to liquid within the radiator. The structure also has
many faucets which are robotically controlled – if one or more tubes leak,
than the working fluid flow is shut from them.

The main parameter for radiator performance is the physical weight of the
system needed to reject a unit of thermal power measured in $kg/kW$. Due to
the Stefan-Boltzmann Law, “the absolute measure of the heat rejection
effectiveness is the $T^{4}$ dependence" [3, p.7]. Since UF4 has much higher
condensation point then other working fluids, it offers an “order-of magnitude
improvement in radiator performance" [3, p.7]. The plot showing radiator
performance is given in Fig. 6 below [3, p.17]:

Figure 6: Radiator performance

##  6 Economic Considerations

###  6.1 State of the art deep space launch costs

Overall, from 1957 to 2015, there have been 5,510 space launches in the world
of which 5,046 were successful. The decade with most (1,231) space launches
was 1970s [34]. During the years 1990 – 2010, 16,200 tons of payload have been
launched into space –mostly Low Earth Orbit (LEO) and 65% of that mass
returned to Earth with Space Shuttle [35].

Saturn V Rocket used in Moon expedition costs $110 million in 1970 dollars
[36, p.406], which is $680 million in 2015 dollars. It could have placed 108
tons into LEO [37, p.43]. Thus, by 1969, very heavy launch vehicles could
place payload into orbit at $6,300 per kg in 2015 dollars. Most experts in
early 1970s believed that in a few years, the cost of placing payload into
orbit would be $2,200 per kg in 2015 dollars [38, p.35]. By 1980s space launch
should have cost $400 – $600 per kg payload in 2015 dollars [39, p.viii].
Unfortunately, the launch costs have not improved since 1969. In 2002, the
cost of LEO delivery has been $6,900 per kg for Delta4 Heavy and $18,400 per
kg for Titan 4B [40, p. 211]. In 2016, Ariane 5 ECA delivers payload to LEO
for $8,500 per kg [6, p.74], while Falcon Heavy delivers payload to LEO for
$5,100 per kg [6, p.94]. The combination of bright predictions in early 1970s
and quite slow progress by 2016 makes the time frame for future development
uncertain.

###  6.2 GCR cost estimation

We would like to estimate the cost of a Liquid-Vapor Core Reactors with MHG
described in the previous section. The concept reactor [2] has a mass of $200\
tons$ and electrical power of $200\ MW$. Given that our system has 12.5%
efficiency, the thermal power would have to be $1.6\ GW$. We would like to
estimate material inventory needed for the reactor.

  1. 1.

Uranium ${}^{\text{235}}$U:
The reactor should work for $5$ years at the thermal power of $1.6\ GW$,
producing $2.5\cdot 10^{17}\ J$ of thermal energy. Each ${}^{\text{235}}$U
fission produces $190\ MeV$ [27, p.10], thus ${}^{\text{235}}$U produces

$1.9\cdot 10^{8}\frac{eV}{\text{atom}}\cdot\frac{1.602\cdot 10^{-19}J}{1\
eV}\cdot\frac{6.022\cdot
10^{23}\text{atom}}{\text{mole}}\cdot\frac{1\text{mole}}{235\ g}=78\cdot
10^{9}\ \frac{J}{g}=78\ \frac{GJ}{g}.$ (6.1)

The total amount of ${}^{\text{235}}$U consumed by the reactor would be

$\frac{2.5\cdot 10^{17}\ J}{78\cdot 10^{9}\ J/g}=3.2\cdot 10^{6}\ g=3.2\ ton.$
(6.2)

As we have mentioned earlier, as much as 60% of ${}^{\text{235}}$U would be
discarded with the growth of ${}^{\text{236}}$U and fission product
contamination. Thus, the reactor would need $8.0\ tons$ of ${}^{\text{235}}$U.
If the uranium used in the reactor is 99% enriched, and the fuel consists of
62% UF4 and 30% KF and 8% K, then ${}^{\text{235}}$U makes up 46% of fuel by
mass. Thus the total mass of the fuel is $17\ tons$. In our system, one gram
of input ${}^{\text{235}}$U would produce $1,080\ KWh$ electricity.

  2. 2.

Beryllium:
The reactor gas core is a cylinder surrounded by a 2 $cm$ thick molybdenum
shell. The shell is surrounded by a 50 $cm$ thick liquid beryllium shell. The
spherical cavity inside neutron reflector shell would have radius $r_{0}=1.2\
m$. The outer radius of the shell is $r_{1}=1.72\ m$. The total volume of
beryllium oxide shell denoted by $V_{Be}$ is given by

$V_{Be}=\frac{4\pi}{3}(r_{1}^{3}-r_{0}^{3})=13.7m^{3}.$ (6.3)

The density of liquid beryllium is $1.69\ g/cm^{3}$, thus the mass of
beryllium shell is $23.2\ tons$.

  3. 3.

Tungsten:
Most of the rest of the engine as well as the heat radiation system is made
from tungsten – thus total tungsten mass is about $140\ tons$.

Now we can look at the prices of materials used. Beryllium costs up to $500/kg
[41]. Tungsten costs up to $40/kg [42]. The price of ${}^{\text{235}}$U has
two components – the price of natural uranium and the price of isotope
separation. The separation work is measured in separative work units (SWUs)
[43, p.2]. In order to separate the feed mass $M_{f}$ of uranium with
${}^{\text{235}}$U fraction $x_{f}$ into product with mass $M_{p}$ of uranium
with ${}^{\text{235}}$U fraction $x_{p}$ and waste with mass $M_{w}$ of
uranium with ${}^{\text{235}}$U fraction $x_{w}$, we need to apply the
separative work given by [43, p.2]:

$\text{Separative
Work}=M_{p}V\left(x_{p}\right)+M_{w}V\left(x_{w}\right)-M_{f}V\left(x_{f}\right),$
(6.4)

where

$V\left(x\right)=(1-2x)\ln\left(\frac{1-x}{x}\right).$ (6.5)

Thus, in order to separate $M_{f}=200\ kg$ of natural uranium containing
$x_{f}=7.1\cdot 10^{-3}\ {}^{\text{235}}$U into $M_{w}\sim 199\ kg$ of
depleted uranium containing $x_{w}=2.1\cdot 10^{-3}\ {}^{\text{235}}$U and
$M_{p}=1.01\ kg$ enriched uranium containing $x_{p}=0.99\ ^{\text{235}}$U, we
need $262\ kg$ separation work, which is 262 SWU’s. In 2014, one SWU costed
$140 [43, p.2]. At present, natural uranium costs $112 per $kg$ [43, p.2], but
the prices are subject to fluctuation. In 2007, uranium prices reached $300
per $kg$ [44]. Uranium can be extracted from sea water at $260 per $kg$, [45]
which should stabilize the price. In order to obtain $1\ kg$ of
${}^{\text{235}}$U, we need $200\ kg$ of natural uranium at $260 per $kg$ and
262 SWU at $140 each – $89,000 total. The total material cost is $850 million.

The capital costs of nuclear power systems in the year 2000 was $2.4 per
$Watt$ in 2015 dollars. The price in 2015 was $4.6 per $Watt$ [46, p.98] due
to new safety regulations. Given that in the deep space, safety is less
important we can assume the system cost of $2.4 per $Watt$ which is $480
million total.

The reactor can be launched into deep space by Falcon 9 rocket, or its future
derivative. Given that nuclear fuel must be enclosed in special capsules, the
total payload should be about 250 tons. The cost of launching 250 tons of
payload into deep space by modern Falcon 9 rocket is $3.2 billion [6, p.19].

The grand cost of the 200 $Megawatt$ system is $850 million plus $240 million
plus $3.2 billion, which is $4.2 billion total or $21 per $Watt$. Solar power
systems which cost at least $1,000 per $Watt$ can not compete with nuclear
power.

###  6.3 The best use of uranium resources

During the 20${}^{\text{th}}$ century, nuclear technology has played an
important role in development of Humankind. The existence of nuclear weapons
was a major factor during Cold War and in prevention of major wars after WWII.

During the early years of nuclear energy use, it was expected that nuclear
power would provide inexpensive electricity, which did not become the reality.
Nuclear power production has experienced exponential growth between 1959 and
1989, then slow growth in 1990s and has stabilized in 2000s [47, p.109]. In
USA, the total nuclear power stabilized at about $100\ GW$ by the end of 1980s
and has not grown up to 2017. According to the leading experts in 1972, the
total installed nuclear power should have been $508\ GW$ by 1990 and $1,200\
GW$ by 2000 [5, p. 14].

In thermal reactor based electric power stations, which use almost exclusively
the rare isotope 235U, one kilogram of the natural uranium produces about
40,000 $kWh$ electricity [47, p.77]. Fast breeder reactors which could utilize
all of the uranium are extremely expensive [29, p.179]. Reasonably assured
resources of uranium consist of 4.6 million tons of uranium [47, p.21] from
which 27,000 tons of 235U can be separated. In terms of global energy
reserves, 235U contains 21 times less energy than coal [48, p.17]. Thus,
nuclear energy hardly has future perspectives on the Earth.

Nevertheless, fission energy can be a key factor in the initial stages of
Solar System colonization. As we have shown in this work, nuclear energy is
much less expensive than solar energy in deep space. Moreover, it can be used
far from the Sun, where solar power is very weak. Delivery of chemical fuel
into Deep Space is prohibitively expensive, yet 235U has thermal energy
potential of 68 $GJ/g$, thus it can store compact energy for deep space
missions. Finally, even though fast breeder and thorium based reactors have
been extremely expensive on the Earth due to safety concerns, they may be
inexpensive in deep space.

##  7 Summary

Several options have been considered in the past for deep space propulsion.
Even though chemical fuel rockets are very well developed, the cost of
bringing enough chemical fuel into deep space is prohibitive. The best option
for deep space propulsion would be magnetoplasmodynamic thrusters which use
electric energy to eject lithium plasma at 15 to 60 $km/s$. These thrusters
have power of $500\ kW$ each and electrical efficiency of 40% to 60%. The only
realistic option for supplying these thrusters with electric power would be
nuclear reactors and nuclear fuel from the Earth. The nuclear reactors and
spaceships used in deep space must be as compact as possible, since it costs
at least $20,000 per kilogram to deliver any payload into deep space.

The power requirement for magnetoplasmodynamic thrusters is at least $45\ W$
per kilogram of loaded spaceship mass. Thus, nuclear electric power stations
with maximal possible specific power (power per unit mass) are needed. The
system we describe in this work is the Gas Core Reactor (GCR) with
magnetohydrodynamic (MHD) generator. One such system proposed to power a heavy
space transport has electrical power output of $200\ MW$ and a mass of $200\
tons.$ It would operate 5 years using $9.3\ tons$ of 235U, which costs $89 per
gram. The overall cost of such station is estimated at $4.2 Billion.

The Gas Core Reactor (GCR) is a reactor in which the fuel generates thermal
energy in gaseous state. In our concept, the fuel consists of a mixture of
235UF4, KF, and K. The mixture enters the reaction chamber in gaseous state at
$40\ atm$ to $60\ atm$ pressure. The reaction chamber is surrounded by a shell
of liquid beryllium neutron reflector in a molybdenum – 92 container. Within
the reaction chamber, the working fluid is heated by fission to 3,500oK. Then
the working fluid expands through a nozzle and enters a MHG. Finally, the
working fluid is cooled and liquefied in a condensing radiator. Then it enters
the reactor, where it is vaporized at a high pressure. Then it returns to
reaction chamber and the cycle repeats itself.

We have performed a numerical simulation describing neutronic performance of
the gas core reactor. Each simulation consisted of tracing the simulated paths
of 5,000 to 20,000 neutrons within the reactor. From these paths, criticality
as well as the rates of neutron absorption and fission for different reactor
regions have been estimated. The simulation error is proportional to the
inverse of the square root of the number of neutrons: Err$\sim
1/\sqrt{N_{n}}$.

Each neutron path is simulated from emission to absorption or escape. A
neutron is emitted from a random point within the reactor cavity. It is
emitted in a random direction. It has a random energy chosen from an energy
distribution (see formula (3.3)). As the neutron travels within the reactor,
it undergoes events. The following constitutes an event: 1) neutron moving
from one region of reactor to another, 2) neutron scattering, 3) neutron
absorbtion, 4) neutron escape from the reactor. The last two events end the
neutron path. The reactor is divided into three regions. The first and
innermost region is a sphere of gaseous fuel. The second region is a thin
shell of refractory material region. The third region is a thick shell of
Beryllium reflector.

The probabilities of absorbtion or scattering taking place within a region of
reactor are determined by neutron absorbtion and scattering cross – sections.
These cross – sections are energy dependent. Both inelastic and elastic
scattering events are considered in the simulation. A typical neutron
undergoes several hundred collisions within the Beryllium reflector region
before being absorbed or escaping the reactor. About 90% of neutrons
thermalise before escaping or being absorbed.

After simulating the paths of the neutrons, we calculate the reactor
criticality. The regeneration factor $\eta$ is the quotient of the number of
neutrons produced from fission to the number of neutrons absorbed in fuel. In
the gaseous fuel region, $\eta=1.96$, thus by multiplying the fraction of the
neutrons which end up absorbed in the fuel region by $1.96$, we obtain
criticality. The criticality has to be corrected for two effects which almost
cancel each other out: neutron absorbtion by fission products and neutron
multiplication in Beryllium.

The first simulation deals with performance of Beryllium neutron reflector.
The fractions of neutrons which leak from reactor for different values of
reactor cavity radius and reflector thickness are presented in Table 4.
Reflector thickness of $50\ cm$ produces leakage of about 0.9%. The values of
reflectance for an infinite Beryllium medium surrounding a cavity of a given
radius are presented in Table 5. These reflectances are 86% to 90%. The second
simulation shows that tungsten can not be used as refractory material, since
it’s neutron absorbtion is too high. The results are in Table 6. The third
simulation shows that 92Mo is a good choice for refractory material.

Next, we calculate the density and resistivity of the gaseous working fluid
entering the MHD generator. We show that in order for the plasma to be
conductive it has to be kept at a temperature of at least 2,600 oK. We
estimate the efficiency of the power station at 12.5%. Then we describe the
radiation heat sink and show that it has to be kept at high temperature of
about 1,500oK. A colder heat sink would enable the station to generate
electricity with greater efficiency, but it would take too much mass.

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