Homework 24 Solutions Due Friday 4 9 You are given a bare spherical fast reactor of pure fissile material a By equating the geometrical buckling to the material buckling derive expressions for i The critical radius ii The critical volume iii The critical mass b Calculate these values for a U235 spherical reactor with the following parameters f 5 297neutron b 18 75 cmg 3 tr 8 246b a 2 844b Compare your result to the actual critical mass of the Godiva Experiment composed of 93 9 enriched U235 where M critical 48 8kg c Calculate these values for a Pu239 spherical reactor with the following parameters 2 98neutrons f 1 85b 19 74 cmg 3 tr 6 8b c 0 26b Compare your result to the actual critical mass of the Jezebel Experiment composed of pure Pu239 where M critical 20 53kg d Discuss the criticality situation for a sphere made out of U238 with the following data 2 6neutrons f 0 095b 19 05 cmg 3 tr 6 9b c 0 16b Discuss the results of your calculations for the attainable critical masses for these materials a Start by equating expressions for geometric and material buckling 2 k 1 Rc L L 2 2 Bg Bm Rc 2 L k 1 k 1 Rc 3 4 3 4 L 4 4 L3 Vc Rc 3 3 3 k 1 3 k 1 2 4 4 L3 M c Vc 3 3 k 1 2 Vc 4 4 L3 3 3 k 1 2 Mc 4 4 L3 3 3 k 1 2 F D For these equations the following relationships hold L2 and k pf f 1 1 aT a a a But all three reactors are in question are bare reactors so k f since Ta aF a Given these expressions we can now calculate critical radius volume and mass b For the bare U235 spherical reactor calculations are as follows 5 297 k pf f 1 1 1 1 86252neutrons 2 844 a g 23 neutrons Av 18 75 cm 3 6 02 10 mol N 4 803 10 22 neutrons g cm 3 M 235 mol 4 803 10 8 246 10 a N a 4 803 10 22 neutrons 2 844 10 24 cm 2 0 1366 cm1 cm 3 tr N tr L D a 1 3 tr a 2 48cm Rc 1 86252 1 22 neutrons cm 3 1 3 0 396 0 1366 24 cm 2 0 396 cm1 2 48cm 8 396cm 4 3 4 3 Rc 8 396cm 2479 44cm 3 3 3 g M c 18 75 cm 3 2479 44cm 3 46 49kg Vc Error 48 8 46 49 4 8 48 8 When compared to the experimentally determined critical mass the error is not due to the fact that neutron leakage was ignored Taking a non zero extrapolation distance will actually decrease the critical radius since the flux profile would reach a zero value outside the core volume The error is due to the fact that the Godiva experiment used 93 9 enriched fuel while we have assumed a sphere of pure U235 The extra mass in the Godiva experiment was due to the non fissile material present c For a bare Pu239 spherical reactor the calculations are as follows 2 98 1 85 k 2 6128neutrons 1 85 0 26 g 23 neutrons Av 19 74 cm 3 6 02 10 mol N 4 972 10 22 neutrons g cm 3 M 239 mol 4 972 10 6 8 10 a N a 4 972 10 22 neutrons 2 11 10 24 cm 2 0 1049 cm1 cm 3 tr N tr L Rc D a 1 3 tr a 3 065cm 2 6128 1 22 neutrons cm 3 24 1 3 0 338 0 1049 cm 2 0 338 cm1 3 065cm 7 58cm 4 3 4 3 Rc 7 58cm 1826 68cm 3 3 3 M c 19 74 cmg 3 1826 68cm 3 36 06kg Vc Error 20 53 36 06 75 65 20 53 The reason for the large error here is the approximation often made for calculating the 1 1 diffusion coefficient D is not a good one to make since the 3 a tr 3 tr absorption cross section for Pu239 is not significantly small enough to ignore Assuming that assumption is not made the results of the above calculations change significantly L Rc D a 3 tr a a 2 679cm 2 6128 1 1 3 0 338 0 1049 0 1049 1 2 679cm 6 626cm 4 3 4 3 Rc 6 626cm 1218 61cm 3 3 3 g M c 19 74 cm 3 1218 61cm 3 24 06kg Vc The error while still large can now be explained by the assumptions inherent in neutron 20 53 24 06 17 19 diffusion theory Error 20 53 d For a bare U238 spherical reactor the calculations are as follows 2 6 0 095 k 0 968neutrons 0 16 0 095 L Rc not real 0 968 1 It is not possible to create a critical assembly with pure U238 The smallest mass required to create a critical assembly is for Pu239 Homework 25 Solutions Due Friday 4 9 Compare the critical volumes and masses of fast reactors composed of U235 in the following geometrical shapes a A spherical reactor core same as last problem b A cubical reactor core c A cylindrical reactor core with H 2R f 5 297neutron b 18 75 cmg 3 tr 8 246b a 2 844b The critical dimensions of the reactors in question can be determined by equating the material buckling with the geometric buckling The material buckling by definition is not dependent on reactor shape and is given below using the same the geometry Thus it will be the same for each equations and calculations as in HW 24 above 2 tr 0 396 cm1 L2 2 48cm 6 161cm 2 k 1 86252neutrons a 0 1366 cm1 k 1 1 86252 1 Bm2 2 0 139998 L 6 161 shape and is given below The geometric buckling is different depending on the reactor 2 Spherical Rc 8 396cm B 0 139998 g 0 139998 Rc 4 4 3 Vc Rc3 8 396cm 2479 45cm 3 3 3 Cylindrical M c Vc 18 75 cmg 3 2479 45cm 3 46 49kg 2 2 405 2 2 2 405 2 2 2 2 405 2 Bg 0 139998 2 H c Rc 2Rc Rc Rc 2 Rc 2 2 2 405 7 678cm 0 139998 2 Vc Rc2 H c 7 678cm 2 7 678cm 2843 63cm 3 M c Vc 18 75 cmg 3 53 32kg Cubic 2 Bg 3 0 139998 ac Rc 3 14 544cm 0 139998 3 Vc ac3 14 544cm 3073 92cm 3 M c Vc 18 75 cmg 3 3073 92cm 3 57 64kg The net result should not be surprising considering the surface area to volume relationship of these shapes M c sphere M c cylinder M c cube