Fusion Principles Jef ONGENA Plasma Physics Laboratory Royal Military Academy Brussels

Transcription

1 Fusion Principles Jef ONGENA Plasma Physics Laboratory Royal Military Academy Brussels 4 th SIF-EPS International School on Energy Villa Monastero Varenna, Lago di Como 25 July 2017

2 Outline Fusion reactions In the sun On earth Two possible options to realize fusion on earth Inertial Fusion Magnetic Fusion Two possible options to realize magnetic fusion on earth Using a plasma current : tokamak Avoiding a plasma current : helical devices, stellarator Lawson Criterion

3 Fusion: energy source of the universe

4 Energy gain in fusion reactions Results from the difference in binding energy between light nuclei and fusion products Binding energy per nucleon (MeV) Nucleon mass number A Maximum at ~ 62 Ni : tremendous consequences for heavy stars

5 Energy gain in fusion reactions 4 He has a particularly large binding energy Binding energy per nucleon (MeV) Nucleus Total Binding Energy (MeV) D = 2 H T = 3 H He He Nucleon mass number A Large gain in energy when 4 He is one of the reaction products

6 Fusion in the sun

7 Some facts about our sun Temperature at edge From Stefan-Boltzmann law and measured Luminosity L L = 4πσ R 2 sun T 4 edge T edge = 5780K (σ = Stefan-Boltzmann constant = J m 2 K 1 s 1 ) Temperature in centre: Proton thermal energy in centre (= 3/2 kt) equal to potential energy from gravity per proton: 1.5k T centre = Gm p M sun /R sun T centre = K (G=gravitational constant= Nm 2 kg -2 k=boltzmann s constant= J K -1 m p = mass of proton = kg.

8 Interesting recent reference on our sun

9 Fusion from hydrogen to helium

10 Helium from protons only? +4 +2??? Involves conversion proton to neutron Very difficult and slow reaction (which is good for us...) Sun : Every second : 4 million tonnes transformed Energy

11 Fusion on earth

12 Easiest fusion reactions Fusion Temperature Reaction Reaction Needed Energy (in Million Degrees) (in kev) ~700 ~400 ~400 Extensive database on fusion reactions :

13 The simplest fusion reaction on earth

14 Comparison: fusion reaction on earth and in the sun On earth (D-T) D-T reaction has times larger reactivity (cm 3 /s) than the p-p reaction In the sun (p-p) Reactivity (cm 3 /s) Ion temperature (K)

15 Tritium Breeding inside the reactor D + T 4 He + n + ΔE n + 6 Li 4 He + T n + 7 Li 4 He + T + n D + Li 4 He x 10 6 kwh/ kg

16 Ash is 4 He no radioactivity chemically inert : no ozone depletion, no acid rain,... no greenhouse effect Excellent environmental compatibility Does not imply long term storage of radioactive waste part of fuel is active (tritium), but consumed in reaction choice of structural materials to reduce long lived activity Offers prospect to recycle radioactive waste in 1-2 generations Inherently safe malfunction of control system does not lead to runaway gas burner : shutting down gas supply stops reactor Tchernobyl like accident EXCLUDED Inexhaustible fuel consumption is minimal, reaction releases lots of energy Energy source for thousands/millions of years Energy independence no geographical dependence for fuel Avoid geopolitical difficulties Advantages of fusion

17 Energy Coulomb needed to Repulsion: initiate fusion key problem reactionof fusion Height of the Coulomb barrier V C V C = q 2 Z x Z y 4πε 0 (R x + R y ) Joule (R =1.4A 1/3 fm) =1.44 Z x Z y 1.4(A x 1/3 + A y 1/3 ) MeV For the D-T reaction we find V c = 0.38MeV = 380 kev. The corresponding gas temperature is ~ K However the maximum fusion density is reached at kev or K (Note : 1 ev ~ K, see appendix)

18 Most fusion Coulomb reactions Repulsion: occur through key problem tunneling of fusion

19 Fusion Cross-Sections Fusion cross-sections and Reactivities and power Fusion cross-section (in barn = m 2 ) Fusion Reactivity in cm 3 s -1 (averaged over Maxwellian velocity distribution)

20 Optimal Fusion temperature cross-sections for fusionand power D-T reaction : Maximum for T ~ kev Fusion power: P fusion ~ n 2 <σv> E fus At fixed pressure p = cte = nkt -- > n 1/T Thus : P fusion ~ <σv> E fus / T 2 P fusion normalized to max of D-T

21 How to confine matter at very high temperatures?

22 Brief history of fusion research ITER construction start ITER location OK MW in JET 1991 JET first D-T exp worldwide Gorbachev, Reagan : ITER 1984 JET, TFTR, JT60 operation start Start of INTOR later ITER 1972 JET (EU), TFTR(USA), JT60 (Japan) proposed; reactor study First presentation of Inertial fusion systems (Frascati) 1971 Tokamak fever; Conceptual design of reactor, Erice (Brunelli, Martellucci) 1968 Novosibirsk IAEA Conference : Succes of tokamak Failure of pinches: temperature not as high as expected (large thermal losses) st IAEA conference : first plasmas at 1-10 million degrees (ZETA, pinch) nd UN conference : stellarator, pinch, tokamak presented Lawson criterion : performance of fusion systems st UN conference - Bhabha Statement : "in 20 years fusion; water as fuel" 1951 Sakharov and Tamm (Kurchatov Institute, Moscow) : Propose Tokamak Lots of 'unsurmountable' difficulties have been solved But still a lot of challenges ahead

23 Realizing Fusion A. Inertial Fusion Using powerful laser or particle beams to compress a tiny pellet Surface Heating Compression Ignition Fusion

24 Realizing Fusion A. Inertial Fusion Very powerful laser systems needed High requirements for isotropic illumination of target

25 Two options: direct and indirect drive direct drive with lasers Better efficiency but: less stable and less symmetric implosion indirect drive by X-rays Less efficient but: more stable and more symmetric implosion Inertial fusion facilities: NIF (USA) and LMJ (France) Planned (EU) : HiPER

26 Targets for Inertial Fusion ablator DT ice Δ DT gas R R ~ 1 mm Δ ~ 0.2 mm 26

27 USA : National Ignition Facility (NIF) Livermore, California Experiments ongoing 4 MJ laser 192 laser beams in operation from March 2009 ~ One experiment / week

28 Realizing Fusion: B. Magnetic Fusion a real challenge Sun ITER (France, in construction) T ~ K R ~ m Heat flux at edge ~ 60MW/m 2 T ~ K R ~ 6 m Heat flux at edge ~ 10MW/m 2

29 Fusion research in Europe

30 Principle of magnetic fusion Neutral gas Plasma Low temperature / High temperature

31 Principle of magnetic fusion Charged particles 'stick' to magnetic field lines (Lorentz force) No magnetic field With magnetic field

32 Principle of magnetic fusion Particles follow magnetic fields how to limit losses at the end of cylinder? Two possible solutions close' magnetic field at ends BUT : too high losses at ends 'close' magnetic fields on themselves toroidal configuration, BUT..

33 Pure Toroidal field does not work : Charge Separation! Fundamental Reason: Gyroradius varies with magnetic field and particle speed Pure toroidal magnetic field Charges Separate Electric Field

34 Pure Toroidal field does not work : Charge Separation! Fundamental Reason: Gyroradius varies with magnetic field and particle speed Pure toroidal magnetic field Charges Separate Electric Field Magnetic field + Electric Field ALL particles move outward! Electron (-) Ion (+)

35 Realizing a helicoidal magnetic field : Option 1 Tokamak Large current induced in plasma (~100kA - 10MA) Poloidal Magnetic Field Induced Plasma Current (by Transformer) Toroidal Magnetic Field Poloidal Magnetic Field

36 Tokamak Final Configuration A torus with a large flux of high energetic neutrons 14.1MeV 10 kev 3.5MeV 10 kev 14.1MeV Nota : 1 kev = C

37 Tokamak Summary Tokamak, from the Russian words: toroidalnaya kamera, s magnitnami katushkami meaning toroidal chamber with magnetic coils Invented by : Andrei Sacharov and Igor Tamm (both Noble Prize Winners) at the Kurchatov Institute in Moscow in 1950 Essentially a tokamak consists of : large transformer coils for magnetic fields plasma ring with large plasma current

38 Two tokamak configurations in use Limiter Configuration Divertor Configuration Magnetic Surfaces Magnetic Surfaces Magnetically Confined Plasma Vacuum Vessel Divertor Plates

39 Realizing a helicoidal magnetic field : Option 2 Stellarator Complex 3D coils create directly a helical field No plasma current no transformer continuous operation

40 Wendelstein 7-X: Overview Physics goals Fusion product should be about 1/50 of a fusion reactor Major, average minor radius: R=5.5 m, <a>=0.53 m Magnetic field on plasma axis: B=2.5 T Test magnetic field optimization physics experiment: H, D plasmas only, additional planar coils heating systems 10MW ECRH, 20 MW NBI, ICRH mimic α-particle heating: ICRH, NBI low impurity content, heat removal divertor Technological goals Reactor feasability of stellarators steady-state operation 30 minute plasma heating with ECRH superconducting coils active cooling of plasma facing components

41 Wendelstein 7-X : Largest Stellarator in the world

42 Wendelstein 7-X : November 2011

43 Wendelstein 7-X : January 2013

44 Stellarator W7-X: Completion of assembly 2014 Configuration and Schedule Start of Commissioning of W7-X: 20 May (~ 2 months ago)

45 How to create the ultra high temperatures needed? In a future fusion reactor: α-particle heating

46 The main difficulty of magnetic fusion: keep a huge T gradient Two positive nuclei (D + and T + ) at short distance strong repulsion EXTREMELY HIGH temperatures needed to bring the nuclei close enough together : ~ K Special methods needed to heat and confine the fuel Very large gradient in temperature (~ K/m) gradients limited by turbulence TURBULENT medium : very complex physics

47 Neutral beam injection Principle: Ion Source Ion Acceleration Neutralisation Neutral beam Ions (losses) Neutral Ionisation Collisional Slowing-down Dump beam PLASMA

48 Neutral beam injection in practice Source: Positive / Negative ions

49 Example of Neutral beam injection (in JET) One of the neutral beam injectors in JET Four of the 8 PINIs (Positive Ion Neutral Injector) are visible: 120keV 2MW D 0 beam Total power available From NBI in JET: max. ~ 36MW

50 How do radiowaves heat plasmas? Particles turn around magnetic field ion cyclotron frequency : ω ci = 2π f ci = q m B Z A B Z=ion charge, A=ion mass f ci different for D ions, H ions, electrons Hydrogen: 42 MHz B magnetic field = function of radius f ci depends on the position in the plasma Example in plasma centre at R ~ 3m and for B ~ 2.7 Tesla in JET Electron: 77 GHz Deuterium: 21 MHz

51 How do radiowaves heat plasmas? If RF Wave frequency f ICRF = ion cyclotron frequency f IC These ions see a constant electric field These ions absorb the energy RF wave they become very energetic Energy gained transferred by collisions to background electrons and ions general plasma heating These ions are said to be resonant with the wave

52 How do radiowaves ICRH heat in tokamaks plasmas? Resonance layer ω c (R) ω c (R) ω 1 ω = ω c R ω c =(q/m)b

53 Example of RF heating systems (in JET) 3.7GHz / 6MW 24MW / m MHz ~4MW / 0.9m MHz 1-3MW / 2.9m MHz 1-3MW / 2.9m 2

54 Power Amplification Factor Q, Break-even, Ignition Q = P fusion P external heating Breakeven Q=1 when P fusion = P external heating Ignition Q = when P external heating = 0: no external heating needed Selfsustained fusion reaction Note: Q relates to the balance between fusion and external heating power only It is not representative for the balance between total power consumption (magnetic fields, additional systems) and fusion power output

55 Plasma Energy and Energy Confinement Time W p = 1.5 k [n e (r) T e (r)+n i (r) T i (r)] dv V dw p dt W p = P If P = 0 W p (t)= W e -t/τ E p (0) τ E Transport losses by conduction and convection τ E measures how fast the plasma looses its energy τ E is a measure for the thermal insulation of the plasma Under stationary conditions (dw/dt=0) : τ E = W p P

56 Lawson Criterium for D-T reaction Heating power must be large enough to compensate for the losses P heat + P α P transport + P Bremsstrahlung P heat Q = P fusion = 5 P α Q + 5 P α W p + C B T e Q τ E 1 ee 4 n2 a < σ v > Q + 5 3n ekt e Q τ E 1/2 n 2 e Plasma neutrality : n i = n e W p = 3 n e kt e 50%-50% D-T reaction : n T = n D = ½ n e + C B T 1/2 e n 2 e

57 Lawson Criterium for Breakeven (Q=1) n e τ E = 3kT e < σ v > E α (Q + 5) 4Q C B T e 1/2 Condition for Breakeven (Q=1) at 15 kev in D-T n e τ E m 3 s

58 Lawson Criterium for Ignition (Q= ) n e τ E 3kT e < σ v > E α 4 C B T e 1/2 Condition for Ignition (Q= ) at 15 kev in D-T n e τ E m 3 s

59 Present tokamak devices close to Break-Even Present machines are close to produce fusion energy comparable with the energy required to sustain the plasma (breakeven : Q=1) Next step devices (ITER) are expected to produce significantly more fusion energy than the energy required to sustain the plasma (close to Ignition Q= ) TEXTOR

60 A P P E N D I X

61 Some parameters of our sun (for ref.) Ref. Kenneth R.Lang, The sun from space Springer Editions, ISBN

62 Particle Coulomb energies Repulsion: and temperatures key problem of fusion Particles in a gas at temperature T Velocity described by a Maxwell Boltzmann distribution p(v) v 2 exp( mv2 kt ) p(v) is the probability that the velocity is located in the interval [v, v+dv] The kinetic energy E of a particle with the most probable speed is kt (k is the Boltzmann constant : J/K Conversion factor between ev and T E = 1eV = J T = / ~ K

63 Fusion reactions and energy release Energy released (MeV) Reactions involving H or isotopes Aneutronic fusion reactions Reactions in the stars: p-p, CNO, Carbon burn, Reference: S. Atzeni, J. Meyer-ter-Vehn (2004). "Nuclear fusion reactions". The Physics of Inertial Fusion. University of Oxford Press. ISBN

64 Fusion Cross-Sections Fusion cross-sections and Reactivities and power

65 Energy of Fusion fusion reaction cross-sections products and power A + B C + D Momentum Conservation : m A v A + m B v B = m C v C + m D v D Energy ~ kev <<< Energy ~ MeV To a very good approximation: m C v C + m D v D = 0 or m C v C = - m D v D Thus also : m C v C = m D v D and therefore m 2 C v2 C = m2 D v2 D Ratio of energies of products C and D: E C E D = 1 2 m C v2 C 1 2 m Dv 2 D = m D m C Conclusion : Lightest particle get largest share in energy Examples : D+T 4 He + n: neutron gets 80% of the energy D+D t + p or 3 He + n: proton or neutron gets 75% of the energy