1. Physics of fusion power Lecture 2: Lawson criterion / Approaches to fusion.

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

3. 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  r 2 (with r the radius of the ball) The cross section of various fusion reactions as a function of the energy. (Note logarithmic scale) 1 barn = 10 -28 m 2

4. Averaged reaction rate One particle (B) colliding with many particles (A) Number of reactions in  t is Both  as well as v depend on the energy which is not the same for all particles. One builds the average The cross section  Schematic picture of the number of reactions in a time interval  t

5. 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)

6. 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 Schematic picture of the calculation of the averaged reaction rate (Integrand as a function of energy) The Maxwellian (multiplied by the velocity) The cross section The product of distribution and cross section

7. Compare the two Cross section as a function of energy Averaged reaction rate as a function of Temperature The averaged reaction rate does not fall off as strongly when going to lower energies

8. 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)

9. 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 10 6 m/s for Deuterium and 6 10 7 m/s for the electrons In a reactor of 10m size the particles would be lost in 10  s.

10. 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).

11. Week five …… Guest lecturer and international celebrity Dr. D. Gericke will give an overview of inertial confinement fusion …..

12. 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

13. 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

14. Fusion power The total fusion power then is Using quasi-neutrality For a 50-50% mixture of Deuterium and Tritium

15. Fusion power To proceed one needs to specify the average of the cross section. In the relevant temperature range 6- 20 keV The fusion power can then be expressed as

16. The power loss The fusion power must be compared with the power loss from the plasma For this we introduce the energy confinement time  E Where W is the stored energy

17. 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

18. 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

19. 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

20. 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

21. Energy Cycle in a Power Plant External Power In Neutrons (80% of fusion power) Energy losses due to imperfect confinement D T He (20% of power) Steam Turbine Electrical power out 3GW 1GW 0.2GW Waste heat 1.8GW

22. n-T-tau is a measure of progress Over the years the n- T-tau product shows an exponential increase Current experiments are close to break- even The next step ITER is expected to operate well above break-even but still somewhat below ignition

23. Some landmarks in fusion energy Research Initial experiments using charged grids tofocus 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 alsoin development. Recently, massive improvements in lasertechnology have allowed ICF to comeclose to ignition: planned for this year.

24. Alternative fusion concepts

25. End of lecture 2

26. 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

27. Force on the plasma For the electric field Define an average velocity Then for the magnetic field

28. Force on the plasma Averaged over all particles Now sum over all species The total force density therefore is

29. 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.

30. 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

31. 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

32. 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).

33. 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

34. Reformulating the Lorentz force Using The force can be written as Then using the vector identity

35. Force on the plasma One obtains Important parameter (also efficiency parameter) the plasma-beta Magnetic field pressure Magnetic field tension