A Few Comments About Fusion

1. A Few Comments About Fusion

2. Binding Energy Energy released when nucleus created from protons and neutrons Larger binding energy per nucleon means more stable nucleus

3. Fusion vs. Fission Fission Fusion

4. Relevant fusion reactions

5. 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

6. Calculation of released energy Energy then follows from Einstein’s formula Physicist’s unit of energy is electron volt (eV) (kilo-electron volt, keV; mega-electron volt MeV)

7. 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

8. 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 years Deuterium is also very easy to separate (i.e., cheap) Tritium is unstable with a half age of 12.3 years There is virtually no naturally occuring Tritium

9. 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

10. 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

11. 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...

12. 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%

13. …. Is the Coulomb barrier Key problem of fusion

14. 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 Reaction cross section of relevant fusion reactions as function of energy. 1 barn = 10-28 m2

15. 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

16. The cross section must be averaged over energies of the particles. Assuming a Maxwell distribution Averaged reaction rate …..

17. Number of fusion reactions as function of average T The product of distribution and cross section (proportional to reaction rate) Particle energy for average T (from Maxwell distribution) The reaction cross section

18. 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

19. 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 Current fusion reactor concepts

20. Implications from high temperature Temperature 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

21. 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

22. Fusion power The 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)

23. Fusion power At the relevant temperature range 6-20 keV the average cross section is Plugging in, the fusion power can then be expressed as

24. The power loss 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

25. 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)

26. Break-even and Ignition The 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%)

27. 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

29. Magnetic confinement: low n high tE In a plasma, all particles are charged If strong magnetic field applied, Lorentz force can be used to trap charged particles Force 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

30. Tokamak / Stellarator

31. Large Helical Device (LHD,Japan)

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

33. 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

34. 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

35. ITER Goals Achieve 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)