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Transcript
Physics of fusion power
Lecture 2: Lawson criterion /
Approaches to fusion.
Key problem of fusion

…. Is the Coulomb barrier
Cross section


The cross section is
the effective area
connected with the
likelihood of
occurrence of a
reaction
For snooker balls the
cross section is pr2
(with r the radius of the
ball)
1 barn = 10-28 m2
The cross section of various
fusion reactions as a function of
the energy. (Note logarithmic
scale)
Averaged reaction rate



One particle (B) colliding
with many particles (A)
Number of reactions in Dt is
Both s as well as v depend
on the energy which is not
the same for all particles.
One builds the average
The cross section s
Schematic picture of the number
of reactions in a time interval Dt
Averaged reaction rate …..

The cross section
must be averaged over
the energies of the
particles. Assume a
Maxwellian
Averaged reaction rates for
various fusion reactions as a
function of the temperature (in
keV)
Sizeable number of fusion reactions
even at relatively low temperatures

Even for temperatures
below the energy at
which the cross
section reaches its
maximum, there is a
sufficient amount of
fusion reactions due to
the number of particles
in the tail of the
Maxwell distribution
The Maxwellian (multiplied by the
velocity)
The cross section
The product of
distribution and
cross section
Schematic picture of the calculation
of the averaged reaction rate
(Integrand as a function of energy)
Compare the two
The averaged reaction rate does not fall off as strongly when going to
lower energies
Cross section as a function of
energy
Averaged reaction rate as a
function of Temperature
Current fusion reactor concepts



are designed to operate at
around 10 keV (note this is
still 100 million Kelvin,
matter is fully ionized or in
the plasma state)
Are based on a mixture of
Deuterium and Tritium
Both decisions are related
to the cross section
Averaged reaction rates for
various fusion reactions as a
function of the temperature (in
keV)
Limitations due to the high
temperature




10 keV is still 100 million Kelvin (matter is fully
ionized, i.e. in the plasma state)
Some time scales can be estimated using the
thermal velocity
This is 106 m/s for Deuterium and 6 107 m/s for the
electrons
In a reactor of 10m size the particles would be lost
in 10 ms.
Two approaches to fusion

One is based on the rapid
compression, and heating
of a solid fuel pellet through
the use of laser or particle
beams. In this approach
one tries to obtain a
sufficient number of fusion
reactions before the
material flies apart, hence
the name, inertial
confinement fusion (ICF).
Week five ……

Guest lecturer and
international celebrity
Dr. D. Gericke will give
an overview of inertial
confinement fusion …..
Magnetic confinement ..


The Lorentz force connected with a magnetic field
ensures that the charged particles cannot move
over large distances across the magnetic field
They gyrate around the field lines with a typical
radius
At 10 keV and 5
Tesla this radius of
4 mm for
Deuterium and
0.07 mm for the
electrons
Lawson criterion




Derives the condition under which efficient
production of fusion energy is possible
Essentially it compares the generated fusion power
with any additional power required
The reaction rate of one particle B due to many
particles A is
In the case of more than one particle B one obtains
Fusion power

The total fusion power then is

Using quasi-neutrality

For a 50-50% mixture of Deuterium and Tritium
Fusion power

To proceed one needs to specify the average of the
cross section. In the relevant temperature range 620 keV

The fusion power can then be expressed as
The power loss



The fusion power must be compared with the power
loss from the plasma
For this we introduce the energy confinement time
tE
Where W is the stored energy
Ratio of fusion power to
heating power

If the plasma is stationary

Compare this with the fusion power

One can derive the so called n-T-tau product
Break-even

The break-even condition is defined as the state in
which the total fusion power is equal to the heating
power

Note that this does not imply that all the heating
power is generated by the fusion reactions
Ignition condition


Ignition is defined as the state in which the energy
produced by the fusion reactions is sufficient to heat
the plasma.
Only the He ions are confined (neutrons escape
magnetic field and plasma) and therefore only 20%
of the total fusion power is available for plasma
heating
n-T-tau

Difference between inertial confinement and
magnetic confinement: Inertial short tE but large
density. Magnetic confinement the other way
around

Magnetic confinement: Confinement time is around
3 seconds

Note that the electrons move over a distance of
200.000 km in this time
Energy Cycle in a Power Plant
External Power In
Electrical power out
0.2GW
1GW
Steam Turbine
D
He (20% of power)
3GW
T
1.8GW
Waste heat
Energy losses due to imperfect confinement
Neutrons (80% of fusion power)
n-T-tau is a measure of
progress



Over the years the nT-tau product shows
an exponential
increase
Current experiments
are close to breakeven
The next step ITER is
expected to operate
well above break-even
but still somewhat
below ignition
Some landmarks in fusion
energy Research






Initial experiments using charged grids to
focus ion beams at point focus (30s).
Early MCF devices: mirrors and Z-pinches.
Tokamak invented in Russia in late 50s: T3
and T4
JET tokamak runs near break-even 1990s
Other MCF concepts like stellarators also in
development.
Recently, massive improvements in laser
Alternative fusion concepts
End of lecture 2
Force on the plasma

The force on an individual particle due to the
electro-magnetic field (s is species index)

Assume a small volume such that

Then the force per unit of volume is
Force on the plasma

For the electric field

Define an average velocity

Then for the magnetic field
Force on the plasma

Averaged over all particles

Now sum over all species

The total force density therefore is
Quasi-neutrality

For length scales larger than the Debye length the
charge separation is close to zero. One can use the
approximation of quasi-neutrality

Note that this does not mean that there is no
electric field in the plasma.
Under the quasi-neutrality approximation the
Poisson equation can no longer be used directly to
calculate the electric field


The charge densities are the dominant terms in the
equation, and implicitly depend on E.
Divergence free current

Using the continuity of charge

Where J is the current density

One directly obtains that the current density must
be divergence free
Also the displacement current
must be neglected

From the Maxwell equation

Taking the divergence and using that the current is
divergence free one obtains the unphysical result:

To avoid this, the displacement current must be
neglected, so the Maxwell equation becomes
Quasi-neutrality





The charge density is assumed zero (but a finite
electric field does exist)
One can not use the Poisson equation to calculate
this electric field (since it would give a zero field)
Valid when length scales of the phenomena are
larger than the Debye length
The current is divergence free
The displacement current is neglected (this
assumption restricts us to low frequency waves: no
light waves).
Force on the plasma

This force contains only the electro-magnetic part.
For a fluid with a finite temperature one has to add
the pressure force
Reformulating the Lorentz
force

Using
The force can be written as

Then using the vector identity

Force on the plasma

One obtains
Magnetic field pressure

Magnetic field tension
Important parameter (also efficiency parameter) the
plasma-beta