The Physics of Energy sources Nuclear Fusion

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1 The Physics of Energy sources Nuclear Fusion B. Maffei Nuclear Fusion 1

2 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 Fusion Fission 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

3 Some interesting reactions (1) (2) () (4) (5) (6) p + d d + d d + d d + d d + t α + n d + He He + γ α + γ He + n t + p α + p or H + H + H + H + H + H + 2 H H H H H He He + γ 4 4 He + γ He + n H + H He + n 4 1 He+ 1 H 5.49 MeV 2.85 MeV.27 MeV 4.0 MeV MeV 18.5 MeV 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. Figure from wikipedia 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

4 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) () (4) (5) (6) p + d d + d d + d d + d t + p d + t α + n d + He + γ α + γ He + n He α + p (1) has a small cross-section (1) and (2) release energy through γ rays: not efficient to keep the reaction on-going () 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 4 He 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

5 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: v a and v b are the velocities of the fragments a and b respectively Conservation of energy Conservation of momentum Giving: 1 2 m v 2 a a = m b 1 2 m v 2 m a b b 1 2 m v 2 a a m v 2 b b " Q m a v a " m b v b The kinetic energy ratio between fragments D-T reaction: most of the energy goes with neutron Nuclear Fusion m v 2 a a " 1 2 m bv 2 b " For the D-T reaction (5), products are 4 He and neutron the neutron gets 80% of the energy For the D-D reactions () or (4), products are either t + p or He + n the proton or the neutron get 75% of the energy Q # 1+ m & a % ( $ ' m b Q # 1+ m & b % ( $ ' m a

6 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 R x and R y, the Coulomb barrier is: V c = e 2 Z x Z y ( ) 4"# 0 R x + R y The fusion probability decreases rapidly with Zx and Zy. The barrier is the lowest for the hydrogen isotopes " = With e 2 4#$ 0 c = 1 17 R =1.2A 1/ fm being the fine structure constant, and e 2 = "# 0 17 MeV.fm V c = =197. Z x Z y ( ) MeV 1.2 A x 1/ + A y 1/ MeV.fm For the D-T reaction, calculation of V c gives 0.44MeV Even if it is the lowest, still is above the typical incident particle energy of 1-10keV Nuclear Fusion 6 c

7 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: $ ' p(v) " v 2 exp & #mv2 ) % 2kT ( p(v) is the probability that the velocity is comprised between v and v+dv k: Boltzmann constant k =1.810 "2 J K -1 = "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~ 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 ~ K which is quite hot still Nuclear Fusion 7

8 Reaction rate Let s suppose a reaction between particles 1 and 2 with n 1 and n 2 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: n 1.v The reaction rate per unit of volume (see lecture 5) is then R = n 1 n 2 σv 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 = n 1 n 2 <σv> Nuclear Fusion 8

9 Ex for T fixed Reaction rate variation p(v) v v exp 2kT 2 m 2 For a specific T, R max at v m an effective thermal energy E m. R = n 1 n 2 <σv> Variation of the crosssection with v variation of <σv> with T (or E=kT) Practical thermonuclear reactor likely to be between 10-0 kev (T=few 10 8 K) for which D-T reaction rate is much higher (>x10) than the other reactions Nuclear Fusion 9

10 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 T 1/2 and Z 2. 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

11 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 E = kt 2 If the density of specie 1 in the plasma is n 1 then the average plasma kinetic energy density due to these particles specie 1 is E p1 = 1 2 n kt In the plasma we will have 2 kinds of ions with n d and n t (d and t for example). If the electron density is n e then the total plasma energy density is Ep = ( nd + nt + ne ) kt 2 But then n e =n d +n t (each ionised atom is giving a nucleus and an electron) and E p = ( nd + nt ) kt Nuclear Fusion 11

12 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 E p The energy released per unit of volume from fusion is E f = n d n t 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$ > (n d + n t )kt Nuclear Fusion 12

13 Lawson criterion If we have the same quantity of the two species (n d =n t ). n d τ > 6kT vσ Q If we call n the total ion density then n d =n t =n/2 then: nτ > 12kT vσ Q Lawson criterion Shows how large the product density x duration of the plasma must be before we achieve break-even condition Nuclear Fusion 1

14 Examples D-T reaction (Q=17.6MeV), ref to plot t(d,n) 4 He for <σv> vs kt Suppose n=10 20 m - Operated at kt=1kev <σv> = 6x10-27 m s -1 nτ > 1.1x10 24 s m - τ > 10 4 s The confinement time must exceed nearly hours, far too long Operated at kt=10kev <σv> = m s -1 nτ > 0.7x10 20 s m - τ > 0.7s Operated at kt=20kev <σv> = 4.5x10-22 m s -1 nτ > x10 19 s m - τ > 0.s D-D reaction (Q=4MeV) Suppose n=10 20 m - Operated at kt=10kev <σv> = 5x10-25 m s -1 nτ > 6x10 22 s m - 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 = 6x10 20 s kev m - D-D at 100keV nτt ~ x10 2 s kev m - Nuclear Fusion 14

15 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+ 6 Li t+ 4 He 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~ K. Nuclear Fusion 15

16 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

17 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 v F F qv = B B v = If B and v are not perpendicular, the trajectory follow a helical path 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. Nuclear Fusion 17

18 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

19 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 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 Resulting B field However, a toroidal magnetic field is non-uniform. It becomes weaker at large radius plasma tends to go towards the walls. Addition of 2 fields Resulting field lines Nuclear Fusion 19

20 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

21 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 50MJ With about 10 micro explosions per seconds.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 A D-T pellet - Figure from wikipedia Nuclear Fusion 21

22 ICF phases Irradiation of pellet by lasers Shock wave compressing core Formation of plasma atmosphere Ignition of core Absorption of laser beam by atmosphere Fusion energy produced Material violently ejected from surface resulting in imploding shock wave Nuclear Fusion 22

23 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 s which will then be the confinement time τ. Applying Lawson s criterion for a D-T reaction (nτ > 0.7x10 20 s m - ) we need n of at least m -. To heat a spherical pellet of 1mm diameter to a mean energy of 10keV per particle we need E = 4 π ( ) = ev 10 5 J We need to supply this energy in about 10-9 sà W 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 W for short intervals of time Nuclear Fusion 2

24 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 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

25 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 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

26 Summary We have seen that D-T reaction is much more efficient than the D-D reaction 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 = ( nd + nt ) kt The energy released per unit of volume from fusion is Fusion could be maintained if τ is the confinement time vσ Qτ 12kT Break even condition when: nτ > vσ Q All we have to do now, is just be able to produce a plasma with T~ K. Nuclear Fusion 26 E f = n E f > E p " n d n t v# Q$ > (n d + n t )kt Two main research fields pursued to reach fusion: Magnetic and Inertial confinements Could you describe briefly their principle? d n t

27 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

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