Chemical, Biological and Environmental Engineering

A Few Comments About Fusion

Advanced Materials and Sustainable Energy LabCBEE

Binding Energy

Energy released when nucleus created from protons and neutrons

Larger binding energy per nucleon means more stable nucleus

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Fusion vs. Fission

Fusion

Fission

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Relevant fusion reactions

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Calculation of energy released Released energy follows from the mass deficit.

Consider the reaction

Masses of products are

The mass deficit (Total mass before minus total mass after) for reaction is

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Energy then follows from Einstein’s formula

Physicist’s unit of energy is electron volt (eV)

(kilo-electron volt, keV; mega-electron volt MeV)

Calculation of released energy

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Energy released by 1kg of D-T mixture

1 kg of a Deuterium/Tritium mixture would allow for a number of fusion reactions N

This would generate

If released over 24h, this is around 4 GW

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Availability of the fuel Natural abundance of D is 0.015% of all H (1 in 6700)

However, at current rate of energy use there is enough H in the ocean for 1011 yearsDeuterium is also very easy to separate (i.e., cheap)

Tritium is unstable with a half age of 12.3 yearsThere is virtually no naturally occuring Tritium

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Availability of the fuel: T Tritium can be bred from Lithium

Note that the neutron released in the D-T fusion reaction can be used for this purpose

Enough Lithium on land for 10k to 30k years at low cost If the oceans included, enough Li for 107 years

Advanced Materials and Sustainable Energy LabCBEE

Why fusion …. A large amount of fuel is available, at a very low cost

The fuel is available in all locations of the earth.

Like fission, fusion is CO2 neutral

Fusion would yield only a small quantity of high level radioactive waste.There is only a small threat to non-proliferation of weapon material

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But... An energy producing working concept is yet to be

demonstrated.

The operation of a fusion reactor is hindered by several difficult (and rather interesting) physics phenomena

Also bear in mind that the cost argument thus far focuses on the fuel only

However, the cost of the energy is largely determined by the cost of the reactor...

Advanced Materials and Sustainable Energy LabCBEE

Distribution of energy over the products

Energy released as kinetic energy of products

Kinetic energy is not equally distributed:

Since both energy and momentum are conserved

You can solve for the energy in He and n

Therefore n has 80% of energy and He has 20%

2 21 1 and 0

2 2fusion He He n n He He n nE m m m m

21 and

2n He

He He He fusion n fusionHe n He n

m mE m E E E

m m m m

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Key problem of fusion

…. Is the Coulomb barrier

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Reaction Cross Section

Cross section is the effective area connected with the occurrence of the reaction

If you are playing billiards, the cross section is pr2

(with r the radius of the ball)

Reaction cross section of relevant fusion reactions as function of energy.

1 barn = 10-28 m2

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Averaged reaction rate Imagine particle B bombarded by many particles A

Number of collisions in Dt is

Bear in mind that s and v both depend on the energy (which is not the same for all particles)

Cross section s

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Averaged reaction rate …..The cross section must be averaged over energies of

the particles. Assuming a Maxwell distribution

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Number of fusion reactions as function of average T

Particle energy for average T (from Maxwell distribution)

The reaction cross section

The product of distribution and cross section(proportional to reaction rate)

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Compare the two

Cross section as a function of energy

Averaged reaction rate as a function of Temperature

Averaged reaction rate has lesser dependence on energy

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Current fusion reactor concepts Based on a mixture of Deuterium and Tritium

Designed to operate at around 10 keV

(10 keV is equivalent to 100,000,000 K)

Matter is in the plasma state (fully ionized)

Both decisions are related to reaction cross section

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Implications from high temperatureTemperature expresses an averaged energy.

You can convert between K and eV

(so 10 keV is 100 million Kelvin)

The average thermal velocity at 10keV can be estimated as

This is 106 m/s for Deuterium nuclei in plasma

In a reactor of 10 m size the particles would be lost in 10 ms...

1eV = 11605 K 1K=8.616x10-5 eV

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Lawson criterion Derives conditions where production of fusion energy is

possible

We derived reaction rate of particle B due to particles A as

In the case of more than one particle B we could get

Remember we derived <su> for a given temperature

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Fusion powerThe total fusion power then is the reaction rate times energy

Using quasi-neutrality (Deuteriums and Tritiums are indistiguishable)

For a 50-50% mixture of Deuterium and Tritium (nD=nT=1/2n)

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Fusion powerAt the relevant temperature range 6-20 keV the average cross

section is

Plugging in, the fusion power can then be expressed as

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To examine power economy if devices, power produced must be compared with power loss from the plasma

For this we introduce the energy confinement time tE

Ratio of energy content and power loss (e.g. Thermal conduction)

Where W is the stored energy density

The power loss

Eheat

WP

13

3heat E heat EW nTV P nT PV

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Fusion Power to Heating Power ratio

Combine this with the fusion power derived earlier

This is called the “n-T-tau product”

We can get two strategies for fusion energy from here:

High n, low tE

Low n, high tE

(remember, temperature is fixed by cross section at 10 keV)

2 27.7fusionP n T V

0.16fusionE

heat

PnT

P

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Break-even and IgnitionThe break-even condition is defined as the state in which the

total fusion power is equal to the heating power

Note that some power could be externally supplied...

Ignition is defined as the state in which the energy produced by the fusion reactions is sufficient to heat the plasma

Remember that neutrons (80% of the energy) escape reactor; energy in He remains for plasma heating (20%)

1 0.16 Break even when 6fusionE E

heat

PnT nT

P

5 0.16 Ignition when 30fusionE E

heat

PnT nT

P

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Inertial Containment Fusion: high n low tE

Rapid compression and heating of a solid fuel (high n) pellet using laser or particle beams.

Fusion occurs for a few mS... (low t)

Idea is to obtain a sufficient amount of fusion reactions (Pfusion) to generate energy (Pheat) before the material flies apart

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Magnetic confinement: low n high tE

In a plasma, all particles are chargedIf strong magnetic field applied, Lorentz force can be used to

trap charged particlesForce causes charged particles to 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

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Tokamak / Stellarator

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Large Helical Device (LHD,Japan)

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Stellarator

• Inside the device it looks something like this • Picture from LHD in JAPAN

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Tokamak progress as n-T-tau

Current experiments are close to break-even

The next step ITER is expected to operate above break-even but still below ignition

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ITER: International Thermonuclear Experimental Reactor

1985 partnership between EU, Russia (started by Soviet Union…), USA (left in 1999, returned in 2003), Japan, Canada (left in 2003), RoK (2003), India (2005), PRC (2007)

• Budget about G€10… (as in 10 billion euros)• 50% from host “nation” (EU), remainder shared by others

Advanced Materials and Sustainable Energy LabCBEE

ITER GoalsAchieve steady-state plasma with Q > 5 (5x break even)

Momentarily achieve Q > 10 (ten times more thermal energy from fusion heating than is supplied by auxiliary heating

Maintain fusion pulse for up to eight minutes.

Develop technologies needed for fusion power plant

Verify tritium breeding concepts.

Refine neutron shield/heat conversion technology

(most of energy in the D+T fusion reaction is released in the form of fast neutrons)