{"id":13528,"date":"2017-01-03T12:25:44","date_gmt":"2017-01-03T12:25:44","guid":{"rendered":"http:\/\/sitepourvtc.com\/?page_id=13528"},"modified":"2022-03-06T12:35:23","modified_gmt":"2022-03-06T12:35:23","slug":"reaction-rate","status":"publish","type":"page","link":"https:\/\/sitepourvtc.com\/nuclear-power\/reactor-physics\/nuclear-engineering-fundamentals\/neutron-nuclear-reactions\/reaction-rate\/","title":{"rendered":"Reaction Rate"},"content":{"rendered":"
Knowledge of the neutron flux<\/strong> <\/a>(the total path length<\/strong> of all the neutrons in a cubic centimeter in a second) and the macroscopic cross sections<\/strong><\/a> (the probability of having an interaction per centimeter path length<\/strong>) allows us to compute the rate of interactions (e.g., rate of fission reactions). The reaction rate<\/strong> (the number of interactions taking place in that cubic centimeter in one second) is then given by multiplying them together:<\/p>\n <\/a><\/p>\n where: The reaction rate for various types of interactions is found from the appropriate cross-section type:<\/p>\n Multiplying the reaction rate per unit volume (RR = \u0424 . \u03a3) by the total volume of the core (V) gives us the total number of reactions<\/strong> occurring in the reactor core<\/a> per unit time. But we also know the amount of energy released per one fission reaction<\/a> to be about 200 MeV\/fission<\/strong>. Now, it is possible to determine the rate of energy release<\/strong> (power) due to the fission reaction. It is given by following equation:<\/p>\n P = RR . Er<\/sub> . V = \u0424 . \u03a3f<\/sub> . Er<\/sub> . V = \u0424 . NU235<\/sub> . \u03c3f<\/sub>235<\/sup> . Er<\/sub> . V<\/strong><\/p>\n where: A typical thermal reactor<\/strong> contains about 100 tons<\/strong> of uranium<\/a> with an average enrichment of 2%<\/strong> (do not confuse it with the enrichment of the\u00a0fresh fuel<\/strong>). If the reactor power is 3000MWth<\/sub>, determine the reaction rate<\/strong> and the average core thermal flux<\/strong>.<\/p>\n Solution:<\/strong><\/p>\n The amount of fissile<\/a> 235<\/sup>U<\/a> per the volume of the reactor core.<\/p>\n m235<\/sub>\u00a0[g\/core] = 100 [metric tons] x 0.02 [g of 235<\/sup>U\u00a0\/ g of U] . 106<\/sup>\u00a0[g\/metric ton]\n= 2 x 106<\/sup> grams of 235<\/sup>U<\/strong>\u00a0per the volume of the reactor core<\/p>\n The atomic number density <\/a>of 235<\/sup>U\u00a0in the volume of the reactor core:<\/p>\n N235<\/sub> . V = m235<\/sub> . NA<\/sub> \/ M235<\/sub> \u03c3f<\/sub>235<\/sup> = 585 barns<\/strong><\/p>\n The average recoverable energy per 235<\/sup>U\u00a0fission:<\/p>\n Er<\/sub> = 200.7 MeV\/fission<\/strong><\/p>\n
\n\u0424 – neutron flux (neutrons.cm-2<\/sup>.s-1<\/sup>)<\/strong>
\n\u03c3 – microscopic cross section (cm2<\/sup>)<\/strong>
\nN – atomic number density (atoms.cm-3<\/sup>)<\/strong><\/p>\n\n
\n
Reaction Rate and Reactor Power Calculation<\/h2>\n
\nP – reactor power (MeV.s-1<\/sup>)<\/strong>
\n\u0424 – neutron flux (neutrons.cm-2<\/sup>.s-1<\/sup>)<\/strong>
\n\u03c3 – microscopic cross section (cm2<\/sup>)<\/strong>
\nN – atomic number density (atoms.cm-3<\/sup>)<\/strong>
\nEr – the average recoverable energy per fission (MeV \/ fission)<\/strong>
\nV – total volume of the core (m3<\/sup>)<\/strong><\/p>\nExample – Reaction Rate and Reactor Power<\/h2>\n
\n= 2 x 106<\/sup> [g 235 \/ core] x 6.022 x 1023<\/sup>\u00a0[atoms\/mol] \/ 235 [g\/mol]\n= 5.13 x 1027<\/sup> atoms \/ core<\/strong>
\nThe microscopic fission cross-section of 235<\/sup>U (for thermal neutrons<\/a>):<\/p>\n