Download The Physics of Energy sources Nuclear Fusion

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.
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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.
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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
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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
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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>
!
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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
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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.
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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
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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
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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
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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
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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….
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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
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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.
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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
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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.
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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
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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
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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
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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
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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
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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 ?
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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)
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