Introduction to Magnetic Fusion
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MIT Plasma Science & Fusion Center Introduction to Magnetic Fusion Dennis Whyte MIT Plasma Science and Fusion Center Nuclear Science and Engineering SULI, PPPL June 2015 Whyte, MFE, SULI 2015 1 Overview: Fusion Energy • Why? ! To meet growing world energy demands using a safe, clean method of producing electricity. • What? ! Extract net energy from controlled thermonuclear fusion of light elements • Who? ! International research teams of engineers and physicists. • Where? ! Research is worldwide ….eventual energy source has few geographical restrictions. • When? ! Many feasibility issues resolved….demonstration in next decades • How? ! This lecture Whyte, MFE, SULI 2015 2 Fusion & fission: binding energy, E = Δm c2 Whyte, MFE, SULI 2015 3 Thermonuclear fusion of hydrogen powers the universe: stars Whyte, MFE, SULI 2015 4 As a terrestrial energy source we are primarily interested in the exothermic fusion reactions of heavier hydrogenic isotopes and 3He • Hydrogenic species definitions: ! Hydrogen or p (M=1), Deuterium (M=2), Tritium (M=3) Whyte, MFE, SULI 2015 5 Exothermic fusion reactions of interest Net energy gain is distributed in the Kinetic energy of the fusion products Whyte, MFE, SULI 2015 6 D-T fusion facts d + t → α + n Q = 17.6 MeV 2 3 4 1 H + 1 H → 2 He + n Q = 17.6 MeV Conservation of energy & linear momentum when Ed,t<<Q mn 1 mα 4 Eα ! Q = QDT = 3.5 MeV En ! Q = QDT = 14.1 MeV mα + mn 5 mα + mn 5 Whyte, MFE, SULI 2015 7 A note about units • In nuclear engineering, nuclear science and plasmas we use the unit of electron-volt “eV” for both energy and temperature ! N.B. Energy and temperature are NOT the same thing! • The eV, rather than Joules, is the natural energy unit when dealing with elementary particles: Energy gained by a particle with unit electron charge falling through 1 Volt of electrostatic potential energy…that’s natural? • Yes! E.g. Electrostatic particle accelerator, beam of protons fall through potential drop of 1 MV x 1 Ampere current. Beam power: V x I =1 MW..so what is the energy per proton? 106 eV! e ⎛ C ⎞ 6 I ⎛ proton ⎞ 6 6 J i I ⎜ ⎟ i10 V = ⎜ ⎟ i10 e V = 10 e ⎝ s ⎠ e ⎝ s ⎠ s Whyte, MFE, SULI 2015 8 Handy dandy unit conversions & ! Typical physical parameters • 1 eV = 1.6 x 10-19 J (conversion between single particle & SI) • 1 e = 1.6 x 10 -19 C • 1 eV ~ 11600 Kelvin ~ 104 K (describing temperatures) • 10 keV ~ 108 K ~ 100 million K (typical fusion T)! 4 1/2 • vth ~ 10 (T / M) m/s {T in eV, M in amu} ! At typical fusion temperature thermal velocity of protons ~ 106 m/s • 1 barn = 10-28 m2 = 10-24 m2 (nuclear cross section) Whyte, MFE, SULI 2015 9 We also use eV to describe mass ! through energy-mass equivalence • E = m c2 " m = E / c2 • Electron: 0.511 MeV / c2 • Atomic mass unit ~ proton ~ neutron: 970 MeV / c2 ~ 103 MeV • So note D-T fusion gain: ~ 20 MeV arises from relative decrease in rest mass energies of products compared to reactants! ~ 20 MeV / (5x1000 MeV) ~ 0.4% Whyte, MFE, SULI 2015 10 Fusion is a nuclear reaction…it occurs due to the strong nuclear force • Fundamental competition: nuclear binding force vs. repulsive Coulomb force for the like + charged nuclei. U rnucleus Potential ! repulsive diagram for Coulomb force Interacting! nuclei If the nuclei get Close enough r Together (~ nuclei radius)! Distance of separation then the binding nuclear force Dominates --> fusion attracting nuclear force Whyte, MFE, SULI 2015 11 The Coulomb repulsion must be overcome with very high particle energies • Repulsive force acting on two U particles with same charge (ions), rnucleus 2 repulsive Z Z e Z Z e Coulomb force U[J] = 1 2 [J] → U[eV ] = 1 2 4πε0r 4πε0r r attracting Z Z nuclear force U[eV ] = 1.4[eV i nm] 1 2 r [nm] -6 Z=1 " rnucleus ~ 4 fm = 4x10 nm " U ~ 350,000 eV Compare to average atom energy at room temperature Ethermal ~ 300 K ~ 0.03 eV ! Whyte, MFE, SULI 2015 12 Despite Ucoulomb~350 keV, ! we can obtain D-T fusion at T~10 keV • Three important effects allow this:! 1. Quantum-mechanical tunneling (wave-particle nature of the interacting nuclei_ provides finite probability for energies below Ucoulomb! 2. Large low-energy nuclear resonance for d-t 3. Fusion achieved in thermalized population distribution with significant population at E > T Whyte, MFE, SULI 2015 13 Fusion cross-sections: Peak of D-T reaction at 0.06 MeV " x 300 energy gain! http://www.psfc.mit.edu/research/MFEFormulary/ Whyte, MFE, SULI 2015 14 Cross-section has units of area but is actually an expression of normalized probability of a reaction occurring per target particle Linear reaction coefficient in homogenous medium dP µ [m−1] ≡ d−t dx µ [m−1] m2 d−t σ ⎣⎡ ⎦⎤ ≡ 3 ntarget[#/ m ] For nuclear reactions, nuclei typically must approach to nuclear distances for strong-force to come into play. Thus “cross-section” -14 2 -28 2 of (rnuclesu ~10 m) ~ 10 m = 1 barn is quite large… Whyte, MFE, SULI 2015 15 Fusion cross-sections: Peak of D-T reaction at 0.06 MeV " x 300 energy gain! Whyte, MFE, SULI 2015 16 Major conceptual challenge #1:! Nuclei have much higher probability of undergoing non-fusion Coulomb collisions ) 2 Cross-section (m Whyte, MFE, SULI 2015 17 Major conceptual challenge #1:! Nuclei have much higher probability of undergoing non-fusion Coulomb collisions Coulomb collision occurs “at distance”, unlike nuclear reactions By carefully integrating over appropriate impact parameters MeV ions are dominated by energy loss to electrons in matter This “competes” against fusion reactions because it removes energy of needed fast particles Whyte, MFE, SULI 2015 18 Fusion is easy, produced all the time in accelerator labs. Fusion energy is hard because it requires thermalized population • Let’s try to make a fusion energy reactor by producing a 1 MeV beam of deuterons streaming into a tritium target • Mass density solid hydrogen ! ~ 0.1 g/cm3 • From previous graph: Stopping power for the deuteron! S ~ 103 MeV cm2/g " S ! ~ 102 MeV / cm! • Distance 1 MeV deuteron travels in solid tritium before coming to rest due to slowing down by Coulomb collisions ! 2 -2 E0 / (S !) ~ 1 MeV / 10 MeV / cm ~ 10 cm (100 microns) • Probability of fusion reaction? Average cross-section ~ 2 barn (generous!) = 2 x 10-24 cm2 continued on next slide Whyte, MFE, SULI 2015 19 Fusion is easy, produced all the time in accelerator labs. Fusion energy is hard because it requires thermalized population -3 • n ~ ! NA / A = 0.1 * 6e23 / 1 = 6e22 cm : this is the volumetric number density of tritons in target • Probability of fusion per unit distance in tritium target! = 2x10-24 cm2 * 6e22 cm-3 = 0.1 fusion reactions / cm-1 • Approximate probability that ion fuses before full stopping:! 0.1 cm-1 * 10-2 cm ~ 10-3 (oh man that doesn’t sound good) ! Note this probability is independent of the target number density.! • Power balance?! - Power supplied by you: 1 MeV / particle - Power gained from fusion: ! ~20 MeV / fusion * 10-3 fusion/ particle ~ 0.02 MeV/particle Whyte, MFE, SULI 2015 20 Thermonuclear fusion is necessary! Coulomb collisions in thermalized population Interior of sun ~ 15 MK Fusion energy needs confinement Needs Temperature ASSURES FUSION ENERGY MUST BE IN PLASMA STATE Whyte, MFE, SULI 2015 21 Major conceptual challenge #2:! There are no natural sources of tritium -4 • Deuterium is so abundant on earth…fD/H ~ 1.5x10 … as to make an essentially infinite fuel supply for fusion. ! Think of the # water molecules in the ocean!! • But tritium does not occur naturally due to its short half-life, ! T " 3He + e- 1/2 life = 12.3 years.! • So Tritium must be bred. How? ! D + n --> T + (e.g. CANDU fission reactor) ! This is the present “seed” source for fusion tritium amounting to ~20 kg Whyte, MFE, SULI 2015 22 Major conceptual challenge #2:! Fusion must be a breeder reactor • For fusion we have another important set of nuclear reactions involving the neutron product from the D-T fusion reaction and lithium that can produce tritium.! 6Li + n " T + 4He + 4.8 MeV! 7Li + n " T + 4He + n - 2.3 MeV • Lithium is highly abundant: 93% Li-7, 7% Li-6 Whyte, MFE, SULI 2015 23 Major conceptual challenge #2:! Fusion energy system is a very simple breeder reactor d + t → α + n Q = 17.6 MeV n + 6 Li → t +α Q = 4.8 MeV d + 6 Li → 2α Q = 22.4 MeV Whyte, MFE, SULI 2015 24 A fusion energy device heats itself and recycles neutrons internally for tritium hydrogen fuel Energy Before (MeV) Energy After (MeV) ~ 0.01 14.1 ~0.01 3.5 Plasma physics: T=10 keV Whyte, MFE, SULI 2015 25 A fusion energy device heats itself and recycles neutrons internally for tritium hydrogen fuel Energy Before Energy After ~0.01 14.1 + 3.5 ~0.01 + Plasma physics, MAlfven >1 Alphas heat plasma through scattering Whyte, MFE, SULI 2015 26 A fusion energy device heats itself and recycles neutrons internally for tritium hydrogen fuel Escapes plasma 14.1 MeV Nuclear Physics + + Whyte, MFE, SULI 2015 27 A fusion energy device heats itself and recycles neutrons internally for tritium hydrogen fuel 14.1 " 0 + Heat + Nuclear Engineering Electricity Whyte, MFE, SULI 2015 28 A fusion energy device heats itself and recycles neutrons internally for tritium hydrogen fuel 6-Li + n " He + T + + Nuclear Engineering, Radiochemistry Whyte, MFE, SULI 2015 29 A fusion energy device heats itself and recycles neutrons internally for tritium hydrogen fuel Surrounding materials Nuclear Material Science sheath + + Low-T Plasma Material Science Whyte, MFE, SULI 2015 30 The lure of fusion • Nearly limitless fuel:! Deuterium + Lithium-6 ! 2 Helium-4 + 22 MeV • No radioactive waste in fuel cycle • Million times power density of “chemical” energy ! Minimal fuel / waste stream ! Minimized environmental footprint • Inherently safe because it requires T > 5 keV Whyte, MFE, SULI 2015 31 The basic idea of electricity generation from fusion • For electricity generation we absorb the power of the fusion reactants in a “blanket” and use this heating to run a turbine/ generator cycle.! • Tricky bit? ! The fusion core.