NEEP 536: Feasibility of Controlled Fusion Energy Spring semester 2003 Prof. D. Whyte #331 ERB [email protected] Tel: 2-4854 NEEP 536, Spring 2003, D. Whyte Overview: Fusion Energy (cid:127) Why? – To meet growing world energy demands using a safe, clean method of producing electricity. (cid:127) What? – Extract net energy from controlled thermonuclear fusion of light elements (cid:127) Who? – International research teams of engineers and physicists. (cid:127) Where? – Research is worldwide (including UW-Madison)….eventual energy source has few geographical restrictions. (cid:127) When? – Many feasibility issues resolved….demonstration in next decades (cid:127) How? – This course NEEP 536, Spring 2003, D. Whyte Course outline 1. Fusion reactor physics – Fusion reactions, fuels, Coulomb collisions, power balance 2. Magnetic confinement concept – Configurations, MHD, pressure limits, power exhaust – Burning plasma experiments (ITER) 3. Inertial confinement concept – Ignition physics, Rayleigh-Taylor, drivers – National Ignition Facility 4. Electrostatic confinement concept 5. Technology of fusion – Neutronics, tritium, activation – MFE: plasma-surface interactions, magnets – ICF: drivers, chambers 6. Economics and social aspects of fusion – Cost of electricity estimates, scalings, safety NEEP 536, Spring 2003, D. Whyte Expected work for course (30%) (cid:127) 5 homework assignments (45%) (cid:127) Written final exam: Friday May 16, 7:25 p.m. (25%) (cid:127) Paper on related topic – Written report will be due 1 week before…. – Brief oral presentation (~15 minutes) to the class near end of semester – Before starting…please let me advise you on topics NEEP 536, Spring 2003, D. Whyte Fusion & fission: binding energy, ∆ E = m c2 NEEP 536, Spring 2003, D. Whyte Exothermic fusion reactions of interest (cid:127) Hydrogenic species definitions (A=1): – Hydrogen or p (M=1), Deuterium (M=2), Tritium (M=3) D + T --> 4He + n + 17.6 MeV D + D --> (50%) T + p + 4 MeV or (50%) 3He + n + 3.3 MeV D + 3He --> 4He + p + 18.3 MeV T + T --> 4He + 2 n + 11.3 MeV NEEP 536, Spring 2003, D. Whyte Exothermic fusion reactions of interest Fusion reaction cross sections 10 1 ) 2 m D+D = T + p 8 0.1 D+D = 3He + n 2 - e D +T 1 ( D + 3He 0.01 s T + T n r a b 0.001 0.0001 1 10 100 1000 Particle energy (keV) NEEP 536, Spring 2003, D. Whyte Estimating fusion reaction rates (cid:127) Fundamental competition: nuclear binding force vs. repulsive Coulomb force U r nucleus repulsive Coulomb force r attracting nuclear force NEEP 536, Spring 2003, D. Whyte Estimating fusion reaction rates (cid:127) Repulsive force acting on two particles with same charge (ions here), U r nucleus Z Z e2 repulsive Coulomb force F = 1 2 ε, dielectric const ~8.9x10-12 F/m 4πεr2 r attracting r Z Z e2 Z Z e2 nuclear force ∫ ∫ U ≡ F(r)dr = − 1 2 dr = 1 2 4πεr2 4πεr ∞ So for protons (taking U=0 at infinity), r ~ 4x10-15m, e=1.6x10-19 C, we get U~5.5x10-14 J nucleus Or U ~ 350 keV NEEP 536, Spring 2003, D. Whyte Some useful units & conversions (cid:127) 1 eV = 1.6 x 10-19 J (particle energy) (cid:127) 1 eV ~ 104 K (describing temperatures) (cid:127) 10 keV ~ 108 K ~ 100 million K (typical fusion T) (cid:127) 1 unit ~ 5x1011 J (per capita energy consumption, U.S.) (cid:127) 16 kW/per capita power consumption --> ~ 5 TW (cid:127) 1 barn = 10-28 m2 (cross section) (cid:127) V ~ 1.4x104 (T / M)1/2 m/s {T in eV, M in amu} th NEEP 536, Spring 2003, D. Whyte