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# Introduction to Reactor Physics and Analysis PHGN 590

Colorado School of Mines

GPA 3.94

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This 12 page Class Notes was uploaded by Donato Hoeger Jr. on Monday October 5, 2015. The Class Notes belongs to PHGN 590 at Colorado School of Mines taught by Staff in Fall. Since its upload, it has received 19 views. For similar materials see /class/219611/phgn-590-colorado-school-of-mines in Engineering Physics at Colorado School of Mines.

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Date Created: 10/05/15

Nuclear Power I Fission process review 11 Reactor design HI Fission bombs IV Fusion process V Gamow peak VI Lawson Criteria VII Fusion devices Tokamak Laser implosion Nuclear Power I Fission process review n 235U gt 139La 95Mo 2n Energy released Semiempirical mass formula gives the following energy released for symmetrical ssion Q 2BEA2 22 BEAZ 1 223 as A23 213 1 For 235U this gives 183 MeV Nuclear Chain Reaction Review neutron multiplication including role of moderating and controlling material v fast neutrons produced per ssion 8 additional neutrons produced due to fast neutron ssion gt 1 p fraction of neutrons surviving moderat10n and control f fraction of neutrons captured in fuel SfItot fraction of captured neutrons which induce ssion Neutron multiplication factor for reactor kV8pf SfStet For natural U 072 235U and 12C I 247 102 p 89 f 88 GfGtot 54 k 19 for in nite reactor FIGURE 535 Lowenergy neutron ssion cross section anf of U235 For com parison the total neutron cross section am of U235 is shown also The difference between the two cross sections is mainly any By permission from D J Hughes and J A Harvey Neutron Cross S ctions lst ed and J R Stehn et al Neutron Cross Sections 2d ed suppl no 2 vol 3 Sigma Center Brookhaven National Laboratory Upton New York 1955 and 1965 Ir r I T I I I I I I I 1 I I I I I I p j 92W 5x102 gt o E 102E 39 b I 2 g p b 39 C 50 9 8 8 3 h quot 8 U CI 10 4 5 1 I l I I I I l I I 14 l I I I l I I I 01 05 l 5 10 Neutron lab energy I ev Reactor Kinetics Prompt neutrons 99 Delayed neutrons 1 Delayed neutrons arise from quotlonglivedquot ssion products which decay via neutron emission quotprecursor nuclidequot Let C be the precursor density dC k I C dt 4 3 T where B is the fraction into C nT it the thermal neutron density k is the neutron multiplication number and 7 is the C decay constant How long does the neutron live Determined by the time it takes to slow down thermal diffusion time td 10394s p This the characteristic time to produce the next generation of neutrons Let Rft be the ssion rate per volume Bit E Rft dt td tk1td Exponential Rft R0 e Reactor period T Id k1 Fork101 T 1 sec Thus in one second the reactor power would increase by a factor of em 22000 Inclusion of delayed neutrons If B is the fraction of neutrons arising from delayed neutronsthen Average lifetime lB td 7v For 235U X 0065 s gt T 80 s mgt 1013 factor instead of 22000 can39lwlla le Neutron diffusion equation d k nI HTt 7C dt td k HT KC td 39 ltk1gt k 39Reactivityquot p Measures how far away from critical For p 3 gt quotprompt criticalquot pB gt quotdollarsquot Nuclear Explosives Get a super critical mass together before it blows itself apart UBU mantle 215U or 1 VI u Subocrihul 5 quot395 Fuel Electrical trigger Chumvcul cxploswc It takes about 50 generations of neutron multiplication to produce enough energy to initiate explosion This takes about 05 ps in which time about 1020 neutrons are produced Fusion Devices Fusion Reactions d d gt 3He n 4 MeV d d gt 3H p 4 MeV d t gt 4He n 176 MeV 3He d gt 4He p 18 MeV To overcome the Coulomb repulsion requires a temperature in eV of kT oclO fm 140 keV However due to quantum mechanical tunneling Gamowfactor fusion can occur at signi cantly lower energy Elwingad39m z n Sax Gamow Peak Maxwell distribution Gamow tunneling factor Ykl 1 E Gamow Peak Most likely energy for the fusion reaction to occur at a given temperature Plasma Fusion Reaction Rate ab fusion reaction rate per volume in a plasma at temperature T Thermal 4 average Rab 11a nb Oab Vab Multiplying by the Q of the reaction gives the power density Powervolume Rab Qab For dt plasma with n 1015cm3 and kT 100 keV one gets 103 Wcm3 Lawson Criterion Since it costs 3nkT to heat the plasma up the breakeven point for na nb n is n2 Q 5x7 1 3 n kT where 1 is the plasma confinement time This gives the Lawson breakeven criterion n 1 gt 3kTQ 0V N 1014 scm3 d t at 10 keV

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