A deuterium-tritium plasma with equal deuterium and tritium densities is run at

PHYSICS · Multi-part · Fermi University Physics · Plasma Physics · Earth, Climate & Energy Physics · Fusion plasmas

Energy from Physics, Ch Fusion as a power plant: gain, beta, wall loading and breeding

Problem

A deuterium-tritium plasma with equal deuterium and tritium densities is run at T=20 keVT = 20\,\mathrm{keV}, where the Bosch-Hale rate coefficient is ⟨σv⟩=4.33×10−22 m3 s−1\langle\sigma v\rangle = 4.33\times10^{-22}\,\mathrm{m^3\,s^{-1}}. Use the zero-dimensional power balance with Eα=3.5 MeVE_\alpha = 3.5\,\mathrm{MeV}, Efus=17.6 MeVE_{\rm fus} = 17.6\,\mathrm{MeV} and the bremsstrahlung coefficient CB=5.35×10−37 W m3 keV−1/2C_B = 5.35\times10^{-37}\,\mathrm{W\,m^3\,keV^{-1/2}}. ((a)) Find the density-confinement product nτEn\tau_E and the triple product nTτEnT\tau_E this plasma needs to ignite. ((b)) Find both again for a plasma gain Q=5Q = 5. ((c)) At 14 keV14\,\mathrm{keV} the ignition triple product is 3.0×1021 keV s m−33.0\times10^{21}\,\mathrm{keV\,s\,m^{-3}}, with nτE=2.1×1020 s m−3n\tau_E = 2.1\times10^{20}\,\mathrm{s\,m^{-3}}. A colleague argues that 20 keV20\,\mathrm{keV} is the easier operating point because its nτEn\tau_E is lower. In a magnetic plasma whose pressure 2nkBT2nk_BT is capped by the beta limit at a given field, which product fixes the confinement time needed? By what factor must τE\tau_E be longer at 20 keV20\,\mathrm{keV} than at 14 keV14\,\mathrm{keV} for ignition?

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