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: € σtr= 8.246b € σa= 2.844b € ρ= 18.75gcm3 € νσf= 5.297neutron ⋅ b Compare your result to the actual critical mass of the Godiva Experiment composed of 93.9 % enriched U235 where € Mcritical= 48.8kg c. Calculate these values for a Pu239 spherical reactor with the following parameters: € σtr= 6.8b € σc= 0.26b € ρ= 19.74gcm3 € ν= 2.98neutrons € σf= 1.85b Compare your result to the actual critical mass of the Jezebel Experiment composed of pure Pu239 where € Mcritical= 20.53kg d. Discuss the criticality situation for a sphere made out of U238 with the following data: € σtr= 6.9b € σc= 0.16b € ρ= 19.05gcm3 € ν= 2.6neutrons € σf= 0.095b Discuss the results of your calculations for the attainable critical masses for these materials. a. Start by equating expressions for geometric and material buckling € Bg2= Bm2 € πRc      2=k∞−1L2 € Rcπ=Lk∞−1 € Rc=πLk∞−1 € Vc=4π3Rc3=4π3πLk∞−1        3=4π4L33 k∞−1( )32 € Vc=4π4L33 k∞−1( )32 € Mc=ρVc=4π4L3ρ3 k∞−1( )32 € Mc=4π4L3ρ3 k∞−1( )32 For these equations, the following relationships hold: € L2=DΣa and € k∞=ηεpf =νσfσa      1( )1( )σaFσaT      But, all three reactors are in question are bare reactors, so € k∞=νσfσa      !since!€ σaT=σaF.!Given!these!expressions,!we!can!now!calculate!critical!radius,!volume,!and!mass.!!!b. For the bare U235 spherical reactor, calculations are as follows: € k∞=ηεpf =νσfσa      1( )1( )1( )=5.2972.844      = 1.86252neutronsN =ρAvM=18.75gcm3( )6.02 ×1023neutronsmol( )235gmol= 4.803 ×1022neutronscm3Σa= Nσa= 4.803 ×1022neutronscm3( )2.844 ×10−24cm2( )= 0.13661cmΣtr= Nσtr= 4.803 ×1022neutronscm3( )8.246 ×10−24cm2( )= 0.3961cmL =DΣa=13ΣtrΣa=13 0.396( )0.1366( )= 2.48cmRc=π2.48cm( )1.86252 −1( )= 8.396cmVc=43πRc3=43π8.396cm( )3= 2479.44cm3Mc= 18.75gcm3( )2479.44cm3( )= 46.49kg € Error =48.8 − 46.49( )48.8= 4.8% 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: € k∞=2.98( )1.85( )1.85 + 0.26( )      = 2.6128neutronsN =ρAvM=19.74gcm3( )6.02 ×1023neutronsmol( )239gmol= 4.972 ×1022neutronscm3Σa= Nσa= 4.972 ×1022neutronscm3( )2.11×10−24cm2( )= 0.10491cmΣtr= Nσtr= 4.972 ×1022neutronscm3( )6.8 ×10−24cm2( )= 0.3381cmL =DΣa=13ΣtrΣa=13 0.338( )0.1049( )= 3.065cmRc=π3.065cm( )2.6128 −1( )= 7.58cmVc=43πRc3=43π7.58cm( )3= 1826.68cm3Mc= 19.74gcm3( )1826.68cm3( )= 36.06kg € Error =20.53 − 36.0620.53= 75.65% The reason for the large error here is the approximation often made for calculating the diffusion coefficient € D =13 Σa+ Σtr( )≈13 Σtr( )      is not a good one to make since the 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 =DΣa= 3 Σtr+ Σa( )Σa[ ]−1= 3 0.338 + 0.1049( )0.1049( )[ ]−1= 2.679cmRc=π2.679cm( )2.6128 −1( )= 6.626cmVc=43πRc3=43π6.626cm( )3= 1218.61cm3Mc= 19.74gcm3( )1218.61cm3( )= 24.06kg The error, while still large, can now be explained by the assumptions inherent in neutron diffusion theory. € Error =20.53 − 24.0620.53= 17.19%d. For a bare U238 spherical reactor, the calculations are as follows: € k∞=2.6( )0.095( )0.16 + 0.095( )      = 0.968neutrons € Rc=πL0.968 −1( )=not real 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. € σtr= 8.246b € σa= 2.844b € ρ= 18.75gcm3 € νσf= 5.297neutron ⋅ b 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 the geometry. Thus, it will be the same for each reactor shape and is given below using the same equations and calculations as in HW 24 above: € k∞= 1.86252neutrons € Σa= 0.13661cm € Σtr= 0.3961cm € L2= 2.48cm( )2= 6.161cm2 € Bm2=k∞−1L2=1.86252 −1( )6.161= 0.139998 The geometric buckling is different depending on the reactor shape and is given below: Spherical: € Bg=πRc      2= 0.139998 € Rc=π0.139998= 8.396cm € Vc=43πRc3=43π8.396cm( )3= 2479.45cm3Mc=ρVc= 18.75gcm3( )2479.45cm3( )= 46.49kg Cylindrical: € Bg=πHc      2+2.405Rc      2=π2Rc      2+2.405Rc      2=π2( )2+ 2.405( )2Rc( )2= 0.139998 € € Rc=π2( )2+ 2.405( )20.139998( )= 7.678cmVc=πRc2Hc=π7.678cm( )22 ⋅ 7.678cm( )= 2843.63cm3Mc=ρVc= 18.75gcm3( )= 53.32kg Cubic: € Bg= 3πac      2= 0.139998 € Rc=3π0.139998= 14.544cm € Vc= ac3= 14.544cm( )3= 3073.92cm3Mc=ρVc= 18.75gcm3( )3073.92cm3( )= 57.64kg The net result should not be surprising considering the surface area to volume relationship of these shapes: € Mcsphere( )< Mccylinder( )< Mccube(