The Physics of Energy sources Nuclear Fusion

The Physics of Energy sources

Nuclear Fusion

B. Maffei

Bruno.maffei@manchester.ac.uk www.jb.man.ac.uk/~bm

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.

Some interesting reactions

p

He

d

n

t

d

p

t

d

d

n

He

d

d

d

d

He

d

p

+

→

+

+

→

+

+

→

+

+

→

+

+

→

+

+

→

+

α

α

γ

α

γ

3 3 3

H

He

He

H

n

He

H

H

H

H

H

H

n

He

H

H

He

H

H

He

H

H

1 4 3 2 4 3 2 1 3 2 2 3 2 2 4 2 2 3 2 1

+

→

+

+

→

+

+

→

+

+

→

+

+

→

+

+

→

+

γ

γ

5.49 MeV! ! 23.85 MeV! ! 3.27 MeV! ! 4.03 MeV! ! 17.59 MeV! ! 18.35 MeV!

or

(1)! ! (2)! ! (3)! ! (4)! ! (5)! ! (6)!

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.

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 d He p

n t d p t d d n He d d d d He d p + → + + → + + → + + → + + → + + → + α α γ α γ 3 3 3 (1)! ! (2)! ! (3)! ! (4)! ! (5)! ! (6)!

!   (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

How is the energy shared between the fragments?

!   Let s assume a reaction leading to products a and b releasing an energy Q.

! 1 2 mava 2 + 1 2 mbvb 2 "Q ! m_av_a " m_bv_b ! 1 2 mava 2 " Q 1+ ma m_b # $ % & ' ( ! 1 2 mbvb 2 " Q 1+ mb m_a # $ % & ' (

For most applications of fusion, the reacting particles have an energy ~ 1-10keV. This is negligible in comparison to Q. We can then write:

Conservation of energy

Conservation of momentum

D-T reaction: most of the energy goes with neutron

v_a and v_b are the velocities of the fragments a and b respectively

! 1 2 mava 2 1 2 mbvb 2 = mb m_a

!   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

Giving:

The kinetic energy ratio

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

!

e

2

4

"#

₀

=

197.3

137

MeV.fm

!

R

=

1.2

A

1/ 3

fm

V

c

=

197.3

137

Z

_x

Z

_y

1.2

(

A

1/ 3_x

+

A

1/ 3_y

)

MeV

!

"

=

e

2

4

#$

₀

!

c

=

1

137

being the fine structure constant, and



c

=

197

.

3

MeV.fm

With

For the D-T reaction, calculation of Vc gives 0.44MeV

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

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 _{– 10}10 _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 _{– 10}7_{K which is}

quite hot still…

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

#

m

v

2

2kT

$

%

&

'

(

)

p(v) is the probability that the velocity is comprised between v and v+dv

k: Boltzmann constant

The kinetic energy of a particle corresponding to the most probable speed is kT

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: n₁.v

The reaction rate per unit of volume (see lecture 5) is then R = n₁n_2σv 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 v_σ as

v

"

=

#

p

(v)

"

(v)vdv

Reaction rate variation

⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ − ∝ 2kT v exp v ) v ( 2 2 m p Ex for T fixed R = n₁n₂<σv>

For a specific T, R max at v_m 

an effective thermal energy E_m.

Variation of the cross-section with v 

variation of <σv> with

T (or E=kT)

Practical thermonuclear reactor likely to be between 10-30 keV (T=few 108_K)

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 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

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

!   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

kT

E

2

3

=

kT

n

E

_{p1 1}

2

3

=

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

kT

n

n

n

E

_p

(

_{d t e}

)

2

3

+

+

=

But then n_e=n_d+n_t (each ionised atom is giving a nucleus and an electron) and

kT

n

n

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

τ

σ

Q

n

n

E

_f

=

_{d t}

v

The energy released per unit of volume from fusion is

τ is the confinement time: length of time the

plasma is confined so that the reactions can occur

E

_f

>

E

_p

"

n

_d

n

_t

v

#

Q

$

>

3(

n

_d

+

n

_t

)

kT

Lawson criterion

If we have the same quantity of the two species (n_d=n_t).

Q

kT

n

_d

σ

τ

v

6

>

If we call n the total ion density then n_d=n_t=n/2 then:

Lawson criterion

Shows how large the product density x duration of the plasma must be before we achieve

break-even condition

Q

kT

n

σ

τ

v

12

>

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 n_τT to measure the difficulty of meeting a particular target criterion

D-T at 20keV  n_τT = 6x1020 _{s keV m}-3

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 break-even 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

He

t

Li

n

+

6

→

+

4

The neutrons will come from the D-T reactions

The lithium is contained in a breeding blanket placed around the reactor

How to get there ?

!  

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

A bit of Electromagnetism

B q

F ₌ v_× 

A charged particle q moving at speed v in a uniform magnetic field B experience a Lorentz force

B

v

F

If B is perpendicular to v the

trajectory of the charge is circular If the trajectory follow a helical path B and v are not perpendicular,

B q F ₌ v_{⊥ ×}  ⊥

v



v



=

v



B

v can be decomposed in a parallel and a perpendicular component relatively to B. The acting force will change the direction of the perpendicular

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

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

However, a toroidal magnetic field is non-uniform.

It becomes weaker at large radius 

plasma tends to go towards the walls.

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



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

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

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

ICF phases

Irradiation of pellet by lasers

Formation of plasma atmosphere

Absorption of laser beam by atmosphere

Material violently ejected from surface resulting in imploding shock wave

Shock wave compressing core

Ignition of core

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 (n_τ > 0.7x1020 _{s m}-3 _{) we need n of at}

least 1029 _{– 10}30 _m-3_.

To heat a spherical pellet of 1mm diameter to a mean energy of 10keV per particle we need

J

eV

E

(

0

.

5

10

3

)

3

10

29

10

4

5

.

24

10

23

10

5

3

4

≈

×

=

×

×

×

=

π

−

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

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?

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

Summary

All we have to do now, is just be able to produce a plasma with T~ 107- ₁₀8 _K….

d

+

t

!

!

+

n

17.59 MeV!

We have seen that D-T reaction is much more efficient than the D-D reaction

D-T reaction: most of the energy goes with neutron

kT

n

n

E

_p

=

3

(

_d

+

_t

)

•  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:

τ σ Q

n n

E_{f = d t} v

The energy released per unit of volume from fusion is

τ is the confinement time

!

E_f > E_p " n_dn_t v# Q$ > 3(n_d + n_t)kT

Fusion could be maintained if

Q kT n σ τ v 12 >

Break even condition when:

Two main research fields pursued to reach fusion: Magnetic and Inertial confinements Could you describe briefly their principle ?

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)