8. Calculations

As aforesaid, the nuclear fusion fuel for this disclosure can be composed of light atomic nucleus like hydrogen, deuterium, tritium, helium, lithium, beryllium, boron, and their various isotopes. Some isotopes like hydrogen-1, helium-3, lithium-6, lithium-7 and boron-11 are the interest for aneutronic nuclear fusion (low neutron radiation hazards), as example: [1]

1H + 2 6 Li → 4He + (3He + 6Li) → 3 4He + 1H  +  20.9  MeV( 153  TJ/kg ≈  42  GWh/kg)
1H + 7 Li → 2  4He + 17.2  MeV ( 204  TJ/kg ≈ 56  GWh/kg)
3He  + 3 He → 4He + 2 1H + 12.9  MeV ( 205  TJ/kg ≈ 57  GWh/kg)
1H + 11 B → 3 4He + 8.7  MeV ( 66  TJ/kg ≈ 18  GWh/kg)

Boron and helium-3 are special aneutronic fuels, due to its primary reaction produce less than 0.2% of the total energy as fast neutrons, meaning that a minimum of radiation shield is required for a spacecraft, and the products kinetic energy is directly convertible to electricity with a high efficiency, more than 95%, as previously described.

Borax is available in the Earth's crust and helium-3 is available in the lunar regolith, both are relatively plentiful if in comparison to tritium.

With hydrogen, boron forms a series of chemical compounds called borane or boron hydrides, as example, decaborane (B10H14) which have low toxicity and high density (950kg/m3), and relatively inexpensive taking account that it is clean and its energy density is higher than the fossil fuels (18×106 kWh/kg versus 13 kWh/kg).

The following calculations take decaborane (B10H14) as example:
1H + 11B + 123keV → 3 4He + 8.68MeV (66 TJ/kg ≈ 18 GWh/kg)
There are 10 × (1H + 11B) reactions and a rest of 4 × (1H)

Electronvolt (eV) is a unit of energy and a Volt (V) is a unit of electric voltage.
Electronvolt to Joule: 1 eV = 1.60218×10-19J
Electronvolt to temperature: 1 eV = 11604.505 Kelvin → 1 eV = 11604.505 K -273.15 = 11331.355 °C
Electronvolt to mass: 1 eV = 1.782662×10-36 kg → 1 MeV = 1.782662×10-30 kg

Charge: proton= +1.60218×10-19 C, electron= -1.60218×10-19 C
Particles mass: proton=1.67262×10-27 kg, neutron=1.67493×10-27 kg, electron=0.00091×10-27 kg
11B mass= 5 protons + 5 electrons + 6 neutrons = 5×1.67262×10-27 + 5×0.00091×10-27 + 6×1.67493×10-27 =
         18.41723×10-27 kg
1H mass= 1 proton + 1 electron = 1×1.67262×10-27 + 1×0.00091×10-27 = 1.67353×10-27 kg
Decaborane (B10H14) mass: 10×18.41723×10-27 + 14×1.67353×10-27 = 207.60172×10-27 kg

Specific energy of decaborane (eV/kg):
10 × (8.68MeV-123keV) / (207.60172×10-27 kg) = 4.12183×1032 eV/kg
Specific energy of decaborane (J/kg):
4.12183×1032 × 1.60218 ×10-19 = 66.03921×1012 J/kg
Specific energy of decaborane (GWh/kg):
66.03921×1012 / (3.6×106) = 18.34422×106 kWh/kg = 18.34422 GWh/kg

Extracting 14 electrons from decaborane to produce positively charged particles:
207.60172×10-27 -14×0.00091×10-27 = 207.58898×10-27 kg
Charge-to-mass ratio of decaborane(C/kg) after extracting 14 electrons:
14×1.60218×10-19 / 207.58898×10-27 = +10.80525×106 C/kg

The specific energy and charge-to-mass ratio are essential parameters to define the magnetic flux and electrical potentials.

Using the specific energy to find velocity of products from nuclear reaction:
E=½mv2 → v= ((E/m) × 2)0.5 → v= ((66.03921×1012) ×2) 0.5 → v=11.49254×106 m/s

Specific impulse: 11.49254×106 / 9.80665 = 1.17191×106 s

Defining the magnet bore about 1 meter (0.5 meter of internal radius) and using the charge-to-mass ratio to find magnetic flux:
r=mv/qB → r= (v/B) × (m/q) → r= (v/B) / (q/m) → B=v/(r × (q/m)) →
B=11.49254×106 / (0.5×10.80525×106) = 2.12721 Tesla
A superconducting magnet of 4.5 Tesla or higher and about 1 meter of bore is sufficient to confine radially the plasma (reactants and products).

The reactants (1H + 11B) needs at least 123keV of kinetic energy for fusing, however 600keV is considered the best, nevertheless, in theory, only 123keV is consumed by the reaction. Losses caused by electromagnetic radiation (bremsstrahlung) are considered a fail of the coating of the magnet bore responsible to reflect the electromagnetic radiation back to plasma.

Calculation of a negative electric potential (first electric potential) for electrostatic acceleration of the positively charged particles to gain enough kinetic energy to fuse:
E = q×V → V=E/q → V= (E/m)/ (q/m) →
V= ((10×600keV×1.60218×10-19)/207.58898×10-27)/ 10.80525×106 = 428.57165×103 Volts
Temperature: 600×103× (11604.505 K -273.15) = 6.79881 billion °C
An electric potential (first electric potential) of -430 kV is enough to the charged particles gain the required kinetic energy of about 7 billions °C.

Calculation of a positive potential (second electric potential) to confine longitudinally the reactants allowing the charged products (helium-4) escaping. A kinetic energy choice between reactants 600keV and products 8.68MeV would be something about 1.4MeV (suitable for lithium hydride cross section too):
E = q×V → V=E/q → V= (E/m)/ (q/m) →
V= ((10×1.4MeV×1.60218×10-19)/207.58898×10-27)/ 10.80525×106 = 928.57191×103 Volts
V = 928.57191×103 - 430 kV = 498.57191×103 Volts
A positive electric potential (second electric potential) of +500 kV is enough to confine the reactants allowing the products to escape.

As aforesaid, fusing positively charged particles represents a normal energy production and low bremsstrahlung radiation, otherwise fusing negatively charged particles represents a high energy production and high bremsstrahlung radiation, however, for highest energy production, the charge-to-mass ratio must keep as low as possible, that is the plasma charged particles must be a quasi-neutral plasma resulting in a high density, which implies in a higher magnetic flux and a higher acceleration and confinement voltage.

The consumption of the reactor at power of 200MWatts using a fuel with specific energy of 66.03921×1012J/kg:
200MW = 200×106 J/s → 200×106 J/s / 66.03921×1012 J/kg = 3.02850×10-6 kg/s
A fuel consumption of 3.03 milligrams per second is enough for producing 200MWatts.
Ion source current: 3.02850×10-6 kg/s × 10.80525×106 C/kg = 32.72374 C/s
The ion injector belt must provide a current of at least 32.8 Amperes for producing 200MWatts.

Cyclotron frequency: f= qB/ (2πm) = (q/m) × (B/2π) = 10.80525×106 × 4.5/ (2×3.14159) = 7.73869 MHz
Magnetic pressure: pm = B2/2µ° = 4.52/ (2×4π×10-7) = 8.05721×106 J/m3
8.05721×106 / 101325 = 79.51848 atmospheres

Fuel consumption for a spacecraft of 500000kg (500 tons) to reach an acceleration(a=∆v/∆t) of 20m/s2:
1) m1v1 + m2v2 = 0 → 500×103×20 + m2×11.49254×106 = 0 → m2 = 0.87013 kg
  or by Tsiolkovsky rocket equation: ∆v=ve×ln(mi/mf) → 20 = 11.49254×106×ln(500×103/(500×103-m2)) → m2 = 0.87013 kg
2) ½m1v1² + ½m2v2² = E → ½500×103×20² + ½0.87013×(11.49254×106)² = E → E = 56.46282×1012 J
A fuel consumption of 0.87013 kg/s is enough for a spacecraft of 500000kg (500 tons) reach an acceleration of 20m/s2 (2 g-force).

A travel between Earth and Moon (perigee=348 200 km and apogee=402 100 km) at an acceleration of 20 m/s2 (2 g-force):
Midway: s= (103× (348200 + 402100)/2)/2=187.575×106m
s=s0+v0t+½at2 → 187.575×106=0+0+½×20×t2→ t=4.33099×103 s → t×2 ≈ 3 hours
v2=v02+2a∆s→ v2 = 0+2×20×187.575×106→ v=86.61986×103m/s→ v=311.83149×103km/h
Fuel consumption: 0.87013× (2×4.33099×103) = 7537.04865kg
The travel between Earth and Moon, including acceleration and deceleration, will take 3 hours and a decaborane consumption of 7537.04865kg, reaching a maximum velocity of 311.83×103km/h at the midway.

A travel between Earth and Mars (closest=55 758 006 km and farthest=400 000 000 km) at an acceleration of 20 m/s2 (2 g-force):
Midway: s= (103× (55758006 + 400000000)/2)/2=113.93950×109m
s=s0+v0t+½at2 → 113.93950×109=0+0+½×20×t2→ t=106.74244×103 s → t×2 ≈ 3 days
v=v0+at → v=0+20×106.74244×103 → v=2.13485×106m/s → v=7.68546×106km/h
v2=v02+2a∆s → v2 = 0+2×20×113.93950×109 → v=2.13485×106m/s → v=7.68546×106km/h
Fuel consumption: 0.87013× (2×106.74244×103) = 185.75959×103kg
The travel between Earth and Mars, including acceleration and deceleration, will take 3 days and a decaborane consumption of 185.76×103kg , reaching a maximum velocity of 7.68×106km/h at the midway.


World energy consumption per year is about 500EJ (500×1018Joule ≈ 5708 TWh):
Energy density of the fossil fuels: 13 kWh/kg = 46.9×106J/kg
Fossil fuel consumption: 500×1018/46.9×106=10.66098×1012kg
That is about 10.66 billion tons of carbon dioxide (CO2) and other toxic gases going to atmosphere each year increasing the greenhouse effect.

Specific energy of the decaborane: 18×106 kWh/kg = 66×1012J/kg
Decaborane consumption: 500×1018/66×1012=7.57575×106kg
That will be only 7576 tons of clean, inert, safe and light helium gas ascending above the ozone layer per year. Some helium gas may escape to the outer space and be swept by the solar wind.


This nuclear fusion reactor evolves an improved fusion energy apparatus, that can be used to generate electricity at high efficiency; to thrust a spacecraft at very high performance levels, at inexpressive radiation hazards, requiring insignificant shielding; relatively inexpensive and abundant fuel supply; having a scalability of size and power, easier engineering and maintainability.