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The Physics of Energy sources Nuclear Fusion B. Maffei [email protected] www.jb.man.ac.uk/~bm Nuclear Fusion 1 What is nuclear fusion? ! We have seen that fission is the fragmentation of a heavy nucleus into 2 more stable components ! Needs to be triggered by neutron Fission Fusion ! According to the plot B/A versus Mass number A, fusion of 2 light nuclei into a more stable nucleus is also an exothermic reaction ! In this case to need to bring to 2 nuclei close to each other for the strong force to come into action ! Need to overcome the Coulomb barrier due to nuclei repulsion ! Need to give some energy for the reaction to occur. Nuclear Fusion Figure from wikipedia 2 Some interesting reactions (1)! ! (2)! ! (3)! ! (4)! ! (5)! ! (6)! p + d →3He + γ d + d →α +γ 1 2 H + 2H → 4He + γ d + d →3He + n 2 H + 2H → 3He + n 2 H + 2H → 3H +1H d +t →α + n 2 H + 3H → 4He + n d + 3He → α + p 2 H + 3He→ 4He +1H d +d →t+ p or H + 2H → 3He + γ 5.49 MeV! ! 23.85 MeV! ! 3.27 MeV! ! 4.03 MeV! ! 17.59 MeV! ! 18.35 MeV! Figure from wikipedia Most interacting nuclei are isotopes of hydrogen (Z=1) for 2 reasons: ! According to the plot B/A they will release the maximum energy ! That minimises the Coulomb repulsive force We can see that reactions leading to production of α particle (particularly stable) produce a large amount of energy. Note: one reaction is missing – p+p. While it is omitted here, it is the primary astrophysical reaction. We will come back to this one later. Nuclear Fusion 3 Which reaction shall we try to use ? ! The ideal reaction to use would: ! ! ! ! Produce a lot of energy Require a minimum energy to start Have a large reaction rate (high probability) Easy (potentially cheap) to produce (1)! ! (2)! ! (3)! ! (4)! ! (5)! ! (6)! p + d →3He + γ d + d →α +γ d + d →3He + n d +d →t+ p d +t →α + n d + 3He → α + p ! (1) has a small cross-section ! (1) and (2) release energy through γ rays: not efficient to keep the reaction on-going ! (3) And (4) are better suited ! more likely to happen than (1) or (2) ! all the energy is released through kinetic energy ! (5) is even more interesting ! Produces more energy (due to the fact that 4He is tightly bound) ! Same Coulomb barrier than D-D reaction but with a larger cross-section ! Pb: requires tritium which is radioactive and needs to be produced through fusion reaction ! (6) has a high Q released through particles and has no radioactive components ! Disadvantage: higher Coulomb barrier ! But has the advantage on (5) that it releases only charged particles, easier to extract energy from, not like neutrons. Nuclear Fusion 4 How is the energy shared between the fragments? ! Let s assume a reaction leading to products a and b releasing an energy Q. For most applications of fusion, the reacting particles have an energy ~ 1-10keV. This is negligible in comparison to Q. We can then write: va and vb are the velocities of the fragments a and b respectively Conservation of energy Conservation of momentum Giving: ! 1 2 ma v! a m 2 = b ! 1 mb v b2 ma 2 1 1 ma v a2 + mb v b2 " Q 2 2 ma v a " mb v b ! The kinetic energy ratio between fragments 1 Q ma v a2 " # ma & 2 %1+ ( m $ b' 1 Q 2 mv " 2 b b # mb & %1+ ( $ ma ' ! For the D-T reaction (5), products are 4He and!neutron the neutron gets 80% of the energy ! For the D-D reactions (3) or (4), products are either t + p or 3He + n the proton or the neutron get 75% of the energy D-T reaction: most of the energy goes with neutron Nuclear Fusion 5 Necessary energy to initiate fusion ! We need to reach an energy equivalent to the Coulomb barrier in order to initiate the fusion reaction ! If we have 2 reacting particles x and y of radius Rx and Ry, the Coulomb barrier is: Vc = e 2Zx Zy ( 4 "# 0 Rx + Ry ) ! The fusion probability decreases rapidly with Zx and Zy. ! The barrier is the lowest for the hydrogen isotopes e2 1 "= = c = 197.3 MeV.fm being the fine structure constant, and 4 #$ 0!c 137 e2 197.3 = MeV.fm 4 "# 0 137 With 1/ 3 R = 1.2A Zx Zy 197.3 Vc = MeV 1/ 3 1/ 3 137 1.2 Ax + Ay fm ( ! ! ) For the D-T reaction, calculation of Vc gives 0.44MeV Even if it is the lowest, still is above the typical incident particle energy of 1-10keV Nuclear Fusion ! 6 ! How to reach this energy threshold? ! We need to increase the kinetic energy of the reacting particles ! The most “economical” way would be to increase the temperature of the initial gas in order to create a plasma (ionised gas) at temperature T Particles in a gas at a temperature T are in thermal motion. Their velocity spectrum is described by the Maxwell-Boltzmann distribution: $ #mv ' p(v) " v exp& ) 2kT % ( 2 2 p(v) is the probability that the velocity is comprised between v and v+dv k: Boltzmann constant k = 1.38 !10 "23 J K -1 = 0.862 !10 "4 eV K -1 The kinetic energy of a particle corresponding to the most probable speed is kT If we want to heat the plasma in order to reach the Coulomb barrier, in the case of D-T reaction, E=kT=0.44MeV T~ 109 – 1010 K However, QM tunnelling through Coulomb barrier and the fact that we have a distribution of energy for a specific T allows fusion for T ~ 106 – 107K which is quite hot still… Nuclear Fusion 7 Reaction rate ! Let s suppose a reaction between particles 1 and 2 with n1 and n2 being the respective particles volume densities ! v is the relative velocity between the 2 species ! σ is the fusion cross-section between the 2 species If we suppose that particles 2 are stationary, the incoming flux density of particles 1 is: n1.v The reaction rate per unit of volume (see lecture 5) is then R = n1n2σv However we have assumed that there was only one speed v. As seen previously, we have a distribution of speed values. We define the average value of vσ as v" = # p(v)"(v)vdv and R = n1n2<σv> ! Nuclear Fusion 8 Reaction rate variation Ex for T fixed ⎛ − mv 2 ⎞ ⎟⎟ p( v) ∝ v exp⎜⎜ 2kT ⎝ ⎠ 2 For a specific T, R max at vm an effective thermal energy Em. R = n1n2<σv> Variation of the crosssection with v variation of <σv> with T (or E=kT) Practical thermonuclear reactor likely to be between 10-30 keV (T=few 108K) for which D-T reaction rate is much higher (>x10) than the other reactions Nuclear Fusion 9 Energy balance ! The goal is to produce energy ! Initially we give some energy in order to initiate fusion ! Then we need to create enough energy for the fusion to be self-sustained • We also have to take into account the energy losses • The main one is through Bremsstrahlung radiation – Emitted when charged particles interact with each other and decelerate – It can be shown that losses varies as T1/2 and Z2. – We need to maintain T above a certain temperature in order for fusion to be much more efficient compared to losses Break-even point Fusion power generated = power needed to maintain plasma temperature However, due to losses (radiation + some of the neutron energy), even when this point is reached, energy still has to be supplied to maintain plasma temperature Ignition point Fusion power generated can maintain the reactor without external source of energy. In the case of D-T reactor: energy deposited by α particles retained by plasma is enough to compensate for energy losses. Nuclear Fusion 10 Plasma energy ! A plasma can be described by a gas of ions and electrons due to its high temperature and overall electrically neutral ! In such a gas, the average kinetic energy of a particle is 3 E = kT 2 ! If the density of specie 1 in the plasma is n1 then the average plasma kinetic energy density due to these particles “specie 1” is E p1 = 3 n1kT 2 In the plasma we will have 2 kinds of ions with nd and nt (d and t for example). If the electron density is ne then the total plasma energy density is 3 E p = (nd + nt + ne )kT 2 But then ne=nd+nt (each ionised atom is giving a nucleus and an electron) and E p = 3(nd + nt )kT Nuclear Fusion 11 Starting the fusion ! We choose to operate the fusion reactor at a temperature high enough for the power gain from fusion to exceed the Bremsstrahlung losses (above 4keV) ! The breakeven point (and possibly ignition) can be reached if we are able to confine the hot reacting plasma long enough that the nuclear energy produced exceeds the energy required to create the plasma Ep The energy released per unit of volume from fusion is E f = nd nt vσ Qτ τ is the confinement time: length of time the plasma is confined so that the reactions can occur Fusion could be maintained if E f > E p " n d n t v# Q$ > 3(n d + n t )kT Nuclear Fusion 12 Lawson criterion If we have the same quantity of the two species (nd=nt). 6kT ndτ > vσ Q If we call n the total ion density then nd=nt=n/2 then: 12kT nτ > vσ Q Nuclear Fusion Lawson criterion Shows how large the product density x duration of the plasma must be before we achieve break-even condition 13 Examples D-T reaction (Q=17.6MeV), ref to plot t(d,n)4He for <σv> vs kT Suppose n=1020 m-3 ! Operated at kT=1keV <σv> = 6x10-27 m3s-1 nτ > 1.1x1024 s m-3 τ > 104s The confinement time must exceed nearly 3 hours, far too long ! Operated at kT=10keV <σv> = 10-22 m3s-1 nτ > 0.7x1020 s m-3 τ > 0.7s ! Operated at kT=20keV <σv> = 4.5x10-22 m3s-1 nτ > 3x1019 s m-3 τ > 0.3s D-D reaction (Q=4MeV) Suppose n=1020 m-3 Operated at kT=10keV <σv> = 5x10-25 m3s-1 nτ > 6x1022 s m-3 τ > 600s This is about 100 times larger than D-T mainly due to the poor cross-section and low Q We would need to heat the plasma at a much higher temperature kT~100keV Temperature, plasma density and confinement time all have to be attained simultaneously. Designers of the reactors will refer to the triple product nτT to measure the difficulty of meeting a particular target criterion D-T at 20keV nτT = 6x1020 s keV m-3 D-D at 100keV nτT ~ 3x1023 s keV m-3 Nuclear Fusion 14 So what do we need for a reactor ? ! The Lawson criterion is for break-even condition ! In order to get to the ignition point when we can switch off the external heating of the plasma, we need roughly 6 times the breakeven condition ! So far, the largest current experiment (JET) has achieved slightly less than break-even producing an output of 16MW for a few seconds We have seen that D-T reaction is much more efficient than the D-D reaction While deuterium is a naturally occurring isotope and fairly available, tritium is not. Consequently, the D-T fuel requires the breeding of tritium from lithium using n+ 6Li → t + 4He The neutrons will come from the D-T reactions The lithium is contained in a breeding blanket placed around the reactor All we have to do now, is just be able to produce a plasma with T~ 107- 108 K…. Nuclear Fusion 15 How to get there ? There are 2 main branches of research in order to get to a practical solution ! Magnetic confinement fusion ! The plasma consists of charged particles. ! By applying a specially configured magnetic field it is possible to confine the plasma in a region thermally insulated from the surroundings. ! Internal confinement fusion ! A small pellet of fuel is caused to implode so that the inner core reaches such a temperature that it undergoes a mini thermonuclear explosion. ! This is using the radiation of several very powerful lasers Nuclear Fusion 16 A bit of Electromagnetism A charged particle q moving at speed v in a uniform magnetic field B experience a Lorentz force F = qv × B If B is perpendicular to v the trajectory of the charge is circular B F Nuclear Fusion If B and v are not perpendicular, the trajectory follow a helical path B v F = qv ⊥ × B v⊥ v= v v can be decomposed in a parallel and a perpendicular component relatively to B. The acting force will change the direction of the perpendicular component only. 17 Magnetic confinement fusion (MCF) ! Because we need to heat up the plasma at very high temperature, we have to thermally insulate it from the walls of the container confinement ! Two possibilities with magnetic field • Using a magnetic mirror to trap the plasma within a section of the magnetic field • Using a closed-field geometry: toroidal field Magnetic mirror Higher field strength At point P the force direction is towards the central axis Going from P to Q the field strength is changing, making the field lines converging at point Q and changing the direction of the force. Under the influence of this force, the particle is reflected back towards the region of weaker field. By having a zone of higher field strength at each end we can contain the plasma in the weak field zone Nuclear Fusion 18 More practical: closed-field geometry Here we use a toroidal geometry (like a doughnut). This is based on the TOKAMAK design. Transliteration of the Russian word (toroidal chamber with magnetic coils). Invented in the 50s by I. Yevgenyevich Tamm and A. Sakharov (original idea of O. Lavrentyev). A toroidal field is created by passing a current through a solenoid Solenoid Resulting B field In order to correct the deviation a second field (poloidal) is introduced. Field is generated by passing a current in either • External coil windings • Along the axis of the toroid, through plasma Addition of 2 fields Resulting field lines However, a toroidal magnetic field is non-uniform. It becomes weaker at large radius plasma tends to go towards the walls. Nuclear Fusion 19 Experimental assembly JET (Joint European Torus) Current generating the poloidal field is induced by transformers action on plasma. A current pulse in the primary winding induces a large current of up to 7MA in the plasma The current going through the plasma: • Creates the poloidal field • Provide resistive heat to the plasma JET - Figure from wikipedia Additional heating is needed to raise the temperature of the plasma RF heating with radio/micro-wave radiation (~25-55MHz) Neutral beam heating: accelerate beam of H or D ions then neutralisation + collision with plasma Nuclear Fusion 20 Inertial confinement fusion ! Principle ! A pulse of energy is directed from several directions on a small pellet of fusible material ! Energy from pulses is heating the material until fusion occurs ! 1 pellet will contain ~ 1mg of D-T liberating 350MJ ! With about 10 micro explosions per seconds 3.5GW The pulse of energy can be delivered with a laser. The beam can be separated in several beams in order to illuminate the target from several directions Several synchronised lasers could also be used Nuclear Fusion A D-T pellet - Figure from wikipedia 21 ICF phases Irradiation of pellet by lasers Formation of plasma atmosphere Absorption of laser beam by atmosphere Shock wave compressing core Ignition of core Fusion energy produced Material violently ejected from surface resulting in imploding shock wave Nuclear Fusion 22 What do we need for ICF ? In order to reach a energy per particle of kT~10keV, we estimate that the compression of the pellet will take about 10-9 – 10-10 s which will then be the confinement time τ. Applying Lawson s criterion for a D-T reaction (nτ > 0.7x1020 s m-3 ) we need n of at least 1029 – 1030 m-3. To heat a spherical pellet of 1mm diameter to a mean energy of 10keV per particle we need 4 E = π (0.5 ×10 −3 )3 ×10 29 ×10 4 = 5.24 ×10 23 eV ≈ 10 5 J 3 We need to supply this energy in about 10-9 sà 1014W This is without considering losses that will exist. Power conversion from electrical to radiation in laser is not very efficient: 10% at best This means that we require a minimum electrical power of 1015W for short intervals of time Nuclear Fusion 23 Status ! Most of the progress have been done through MCF ! Several facilities have been developped ! The most recent and promising results have been achieved with JET • Has managed just below breakeven in 1997 with 16MW for a few seconds ! Construction of a new experimental reactor has been decided in 2006: ITER • • • • International Thermonuclear Experimental Reactor First plasma operation is expected in 2016 ? 5 Billion €, one of the most expensive techno-scientific project Designed to produce ~ 500MW for 400 sec ! Followed by DEMO as a first production of net electrical power ! Concept of ICF has been proven ! Most of the development have been performed at Lawrence Livermore Lab. • 1978 Shiva laser – proof of concept • Nova laser ~ 10 times the power of Shiva but failed to get ignition due to laser instability ! With progress in laser development several projects are planned • National Ignition Facility with possible ignition in ~ 2011? • HiPER and Megajoule in Europe Nuclear Fusion 24 Fusion power plant concept Reaction between Lithium and neutron given through D-T reaction will produce the necessary tritium. As a best estimate we can imagine to have the first ignition in the 2015-2020 horizon. Further development will need ~ 10-20years Commercial power plant will take another ~ 10-20years No commercial fusion reactor is planned before ~2050 Nuclear Fusion 25 Summary We have seen that D-T reaction is much more efficient than the D-D reaction 17.59 MeV! d +t !! +n D-T reaction: most of the energy goes with neutron • Fusion needs to be triggered to overcome the Coulomb barrier • We need to reach a very temperature and create a plasma Energy of the plasma: E p = 3(nd + nt )kT The energy released per unit of volume from fusion is τ is the confinement time Fusion could be maintained if Break even condition when: E f = nd nt vσ Qτ E f > E p " n d n t v# Q$ > 3(n d + n t )kT nτ > 12kT vσ Q ! All we have to do now, is just be able to produce a plasma with T~ 107- 108 K…. Two main research fields pursued to reach fusion: Magnetic and Inertial confinements Could you describe briefly their principle ? Nuclear Fusion 26 References ! Most of the material of this lecture is coming from ! Lilley, J. Nuclear Physics Principles and Applications (Wiley 2006) ! Kenneth S. Krane, Introductory Nuclear Physics (Wiley 1988) Nuclear Fusion 27