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Nuclear fusion energy can be most commonly released from fusion fuels such as hydrogen, deuterium, tritium, helium, lithium, beryllium and boron. The isotopes having potential for third generation fusion fuel are hydrogen-1, helium-3[1], lithium-6, lithium-7 and boron-11:[2][3]
| Reactants | Products | Energy Density | ||||||||||
| 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) |
The aneutronic reactions showed above are of notable interest due to low emission of neutrons, production of charged particles in the primary reactions, that can be directly convertible into electricity.[4][5]
Basic Calculation
editExamples of boron hydrides are diborane B2H6, pentaborane B5H9, and decaborane B10H14.
The following example of calculation use pentaborane (B5H9):
Electronvolt (eV) is a unit of energy and 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
| particle | charge | mass |
| proton | +1.60218×10-19 C | 1.67262×10-27 kg |
| neutron | 0 C | 1.67493×10-27 kg |
| electron | -1.60218×10-19 C | 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
Pentaborane (B5H9) mass: 5×18.41723×10-27 + 9×1.67353×10-27 = 107.14792×10-27 kg
Specific energy of pentaborane (eV/kg):
- 5 × (8.68MeV-123keV) / (107.14792×10-27 kg) = 3.99308×1032 eV/kg
Specific energy of pentaborane(J/kg):
- 3.99308×1032 × 1.60218 ×10-19 = 63.97633×1012 J/kg
Specific energy of pentaborane(GWh/kg):
- 63.97633×1012 / (3.6×106) = 17.77120×106 kWh/kg = 17.77120 GWh/kg
Extracting 3 electrons from pentaborane to produce positive ions:
- 107.14792×10-27 -3×0.00091×10-27 = 107.14519×10-27 kg
Charge-to-mass ratio of pentaborane(C/kg) after extracting 3 electrons:
- 3×1.60218×10-19 / 107.14519×10-27 = +4.48600×106 C/kg
The specific energy and charge-to-mass ratio are essential parameters to define the magnetic flux and electric voltages.
Using the specific energy to find the velocity of products from nuclear reaction:
- E=½mv2 → v= ((E/m) × 2)0.5 → v= ((63.97633×1012) ×2) 0.5 → v=11.31162×106 m/s
Specific impulse: 11.31162×106 / 9.80665 = 1.15346×106 s
Defining the magnet bore about 0.9 meter (0.45 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=11.31162×106 / (0.45×4.48600×106) = 5.60341 Teslas
A superconducting magnet of 6 Teslas or higher and about 0.9 meter of bore is sufficient to confine radially the plasma (reactants and products).
Calculation of a negative voltage for electrostatic acceleration of the positive ions to gain enough kinetic energy, at least 123keV, hence 550keV should be enough:
- E = q×V → V=E/q → V= (E/m)/ (q/m) →
- V= ((5×550keV×1.60218×10-19)/107.14519×10-27)/ 4.48600×106 = 916.66667×103 Volts
- Temperature: 550×103× (11604.505 K -273.15) = 6.23224 billion °C
A negative voltage of -920 kV is enough for the positive ions gain the required kinetic energy, equivalent to 6.2 billions °C.
Calculation of a positive voltage to trap longitudinally the reactants allowing the charged products to escaping. A kinetic energy choice between reactants 550keV and products 8.68MeV could be something about 1.5MeV:
- E = q×V → V=E/q → V= (E/m)/ (q/m) →
- V= ((5×1.5MeV×1.60218×10-19)/107.14519×10-27)/ 4.48600×106 = 2500×103 Volts
- V = 2500×103 - 920 kV = 1580×103 Volts
A positive voltage of 1580 kV is enough to trap the reactants allowing the products to escape.
The consumption of a fusion reactor at power of 500MWatts using a fuel with specific energy of 63.97633×1012J/kg:
- 500MW = 500×106 J/s → 500×106 J/s / 63.97633×1012 J/kg = 7.81539×10-6 kg/s
A fuel consumption of 7.82 milligrams per second is enough for producing 500MWatts.
- Ion source current: 7.81539×10-6 kg/s × 4.48600×106 C/kg = 35.05989 C/s
The ion source must provide a current of at least 35.1 Amperes for producing 500MWatts.
Cyclotron frequency: f= qB/ (2πm) = (q/m) × (B/2π) = 4.48600×106 × 6/ (2×3.14159) = 4.28382 MHz
Magnetic pressure: pm = B2/2µ° = 62/ (2×4π×10-7) = 14.32394×106 J/m3
- 14.32394×106 / 101325 = 141.36634 atmospheres
See also
editReferences
edit- ^ E. N. Slyuta (2007). "The estimation of helium-3 probable reserves in lunar regolith" (PDF).
{{cite web}}: Check date values in:|date=(help) - ^ Atzeni S., Meyer-ter-Vehn J (2004). "The Physics of Inertial Fusion: Beam Plasma Interaction, Hydrodynamics, Hot Dense Matter" (PDF).
{{cite web}}: Check date values in:|date=(help) - ^ S. Son , N.J. Fisch (2004-06-12). "Aneutronic fusion in a degenerate plasma" (PDF).
{{cite web}}: Check date values in:|date=(help) - ^ Ralph W. Moir (1997). "Direct Energy Conversion in Fusion Reactors" (PDF).
{{cite web}}: Check date values in:|date=(help) - ^ "Electricity Conversion by Neutralization Process" (Flash video). 2008-12-16.
{{cite web}}: Check date values in:|date=(help)
External links
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