PHGN590. Introduction to Nuclear Reactor Physics. Two Group Flux Profile

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PHGN590 Introduction to Nuclear Reactor Physics Two Group Flux Profile J. A. McNeil Physics Department Colorado School of Mines 4/2009

2 Flux_Profile_2group.nb Two velocity groups ü Parameters based on USGS Triga (group 1-> fast, group 2 -> thermal) H* Geometry is given in inches and converted to cm *L geom = 8RC Ø 2.54 H10 + 5 ê 16L, HC Ø 2.54 µ 15, RRef Ø 2.54 H21 + 5 ê 8L, HRef Ø 2.54 H15 + 2 µ 3.47L<; Print@" Reactor geometry: Core Radius = ", RC ê. geom, " cm Core Height = ", HC ê. geom, " cm"d Print@" Reflector radius = ", RRef ê. geom, " cm Reflector Height = ", HRef ê. geom, " cm"d params1 = 8n Ø 2.07, Sf Ø.00258, SCa Ø H0.0894 + 0.00258L, SRefa Ø 0.0002728, SCs Ø 3.29, SRefs Ø.3811, mc Ø.025, mref Ø 1 ê 18.<; params2 = 8n Ø 2.07, Sf Ø.0712, SCa Ø H0.0894 + 0.0712L, SRefa Ø 0.0002728, SCs Ø 3.29, SRefs Ø.3811, mc Ø.025, mref Ø 1 ê 18.<; H* Diffusion constants for core *L StrC = SCa + Sf + H1 - mcl SCs; D C = 1 ê H3 StrCL; L C = D C SCa ;

Flux_Profile_2group.nb 3 Print@" Core parameters "D Print@" Fast group: Str1 = ", StrC ê. params1, " cm^-1 D C1 = ", D C ê. params1, " cm L C1 = ", L C ê. params1, " cm"d Print@" Thermal group: Str2 = ", StrC ê. params2, " cm^-1 D C2 = ", D C ê. params2, " cm L C1 = ", L C ê. params2, " cm"d H* Diffusion constants for reflector *L StrRef = SRefa + H1 - mrefl SRefs; lex = 1 ê H3 StrRefL; D R = 1 ê H3 StrRefL; L R = D R SRefa ; Print@" Reflector parameters "D Print@" Fast group: Str1 = ", StrRef ê. params1, " cm^-1 D R1 = ", D R ê. params1, " cm L R1 = ", L R ê. params1, " cm"d Print@" Thermal group: Str2 = ", StrRef ê. params2, " cm^-1 D R2 = ", D R ê. params2, " cm L R1 = ", L R ê. params2, " cm"d Reactor geometry: Core Radius = 26.1938 cm Core Height = 38.1 cm Reflector radius = 54.9275 cm Reflector Height = 55.7276 cm Core parameters Fast group: Str1 = 3.30231 cm^-1 D C1 = 0.100939 cm L C1 = 1.04757 cm Thermal group: Str2 = 3.43955 cm^-1 D C2 = 0.0969119 cm L C1 = 0.776812 cm Reflector parameters Fast group: Str1 = 0.360201 cm^-1 D R1 = 0.92541 cm L R1 = 58.2432 cm Thermal group: Str2 = 0.360201 cm^-1 D R2 = 0.92541 cm L R1 = 58.2432 cm

4 Flux_Profile_2group.nb Rex = HRRef + lexl ê. params1 ê. geom; Hex = HHRef + 2 lexl ê. params1 ê. geom; Print@" Extrapolation distance: lex = ", lex ê. params1, " cm"d Extrapolation distance: lex = 0.92541 cm ü Bare Cylindrical Reactor The diffusion equations for the two velocity group fluxes are: D 1 2 F 1 == H-Sa 1 - Ss 1->2 L F 1 + n Sf 2 F k 2 D 2 2 F 2 == -Sa 2 F 2 + Ss 1->2 F 1 For cylindrical geometry these factor into r and z components: F = R[r] Z[z]. Reqn = 1 r D@Hr R'@rDL, rd == -Br2 R@rD; Zeqn = Z''@zD ã -Bz 2 Z@zD; For both groups the boundary conditions are R'[0]=Z'[0]==0 and R[Rex]=Z[- Hex/2]==0 which gives: Z1soln@z_D = Az1 Cos@Bz zd R1soln@r_D = Ar1 BesselJ@0, Br rd Z2soln@z_D = Az2 Cos@Bz zd R2soln@r_D = Ar2 BesselJ@0, Br rd Az1 Cos@Bz zd Ar1 BesselJ@0, Br rd Az2 Cos@Bz zd Ar2 BesselJ@0, Br rd where Bz = p / Hex and Br = z0 / Rex, where z0 is the first zero of J0:

Flux_Profile_2group.nb 5 z0soln = z0 ê. FindRoot@BesselJ@0, z0d ã 0, 8z0, 2.5<D; Bconstants = 8Br -> z0soln ê Rex, Bz -> p ê Hex<; Bnet = Br 2 + Bz 2 ê. Bconstants ê. geom; PNL = 1 ê H1 + Bnet^2 L C ^2L ê. params1 ê. geom; r21 = SCs ê HSCa + D C Bnet^2L ê. params2 ê. Bconstants; kmax = HHHn SfL ê. params1l + r21 HHn SfL ê. params2ll ê HBnet ^ 2 + HHSCa + SCsL ê. params2ll ê. Bconstants; Print@" Buckling constants Hcm^-1L: Br = ", Br ê. Bconstants, " Bz = ", Bz ê. Bconstants, " BHtotalL = ", BnetD Print@" Nonleakage probability = ", PNLD Print@" Ratio of thermal to fast flux = ", r21d Print@ " Maximum neutron multiplication: k = ", kmaxd Buckling constants Hcm^-1L: Br = 0.0430564 Bz = 0.054562 BHtotalL = 0.0695044 Nonleakage probability = 0.994727 Ratio of thermal to fast flux = 20.4261 Maximum neutron multiplication: k = 0.872779 soln = 8Br -> z0soln ê Rex, Bz -> p ê Hex, Ar1 Ø 1, Az1 Ø 1, Ar2 Ø r21, Az2 Ø r21<; Plot@8 R1soln@rD ê. soln, R2soln@rD ê. soln<, 8r, 0, Rex<D Plot@8 Z1soln@zD ê. soln, Z2soln@zD ê. soln<, 8z, 0, Hex ê 2<D

6 Flux_Profile_2group.nb 20 15 10 5 10 20 30 40

Flux_Profile_2group.nb 7 20 15 10 5 5 10 15 20

8 Flux_Profile_2group.nb ü Reflected Slab Reactor Consider a slsb reactor having a core of width, a, which is surrounded by an infinite reflector. There are two velocity groups and two regions (core and reflector) giving the following four diffusion equations for the neutron flux: [1a] D C1 2 F C1 ã HS ac1 + S 12 C L F C1-1 k Hn 1 S f1 F C1 + n 2 S f2 F C2 L [1b] D R1 2 F R1 == HS ar1 + S 12 R L F R1 [1c] D C2 2 F C2 == S ac2 F C2 - S sc1 F C1 D R2 2 F R2 == S ar2 F R2 - S sr1 F R1 [1d] In the following we will solve for the conditions that give k= 1. For slab symmetry 2 F -> d2 F. The (8) boundary conditions are: dx 2 F C1 '[0] = F C2 '[0] = 0 F C1 A a 2 E = F R1A a 2 E J C1 A a 2 E = J R1A a 2 E F C2 A a 2 E = F R2A a 2 E J C2 A a 2 E = J R2A a 2 E F R1 @ D = F R2 @ D = 0 [2a,b] [2c] [2d] [2e] [2f] [2g,h] The general solutions, satisfying [2a,b] and [2g,h] and normalized by the first sineterm, are: F C1 @x_d = Cos@m xd + r C Cosh@n xd; F C2 @x_d = S 1 Cos@m xd + S 2 r C Cosh@n xd; F R1 @x_d = r F Exp@-k 1 xd; F R2 @x_d = r G Exp@-k 2 xd + S 3 r F Exp@-k 1 xd;

Flux_Profile_2group.nb 9 We will work only in the positive x region and use reflection symmetry to obtain the solution in the negative x region. There remain 10 parameters, {m, n, k 1, k 2, r C, r F, r G, S 1, S 2, S 3 <, to be determined. Substitution in to Eq.[1c,d] gives: k 1 soln = S ar1 + S 12 R D R1 ; k 2 soln = S ar2 D R2 ; Substitution into [1a,b] and equating sine and sinh-terms gives: S 1 soln = S 2 soln = S 3 soln = S 12 C S ac2 + D C2 m 2 ; S 12 C S ac2 - D C2 n ; 2 D R1 S 12 R ; D R1 S ar2 - D R2 S ar1 - D R2 S 12 R The joining conditions for the currents at r = R gives the flux ratios, { r C, r F, r G <:

10 Flux_Profile_2group.nb eq1 = F C1 B a 2 F ã F R1B a 2 F; eq2 = D C1 D@F C1 @xd, xd == D R1 D@F R1 @xd, xd ê. x Ø a 2 ; rcfsoln = Flatten@Solve@8eq1, eq2<, 8r C, r F <DD; r Csoln = r C ê. rcfsoln; r Fsoln = r F ê. rcfsoln; eq3 = F C2 B a 2 F ã F R2B a 2 F; eq4 = D C2 D@F C2 @xd, xd == D R2 D@F R2 @xd, xd ê. x Ø a 2 ; rfgsoln = Flatten@Solve@8eq3, eq4<, 8r G, r F <DD; r Gsoln = Simplify@r G ê. rfgsoln ê. r C -> r Csoln D; r Fsoln2 = Simplify@r F ê. rfgsoln ê. r C -> r Csoln D; Print@" r C Print@" r F Print@" r G = ", r Csoln D = ", r Fsoln D = ", r Gsoln D r C = - -m SinA a m 2 E D C1 + CosA a m 2 E D R1 k 1 n SinhA a n 2 E D C1 + CoshA a n 2 E D R1 k 1 r F = r G = a k 1 2 Im CoshA a n 2 E SinA a m 2 E + n CosA a m 2 E SinhA a n 2 EM D C1 n SinhA a n 2 E D C1 + CoshA a n 2 E D R1 k 1 1 D R2 Hk 1 - k 2 L a k2 2 -m SinB a m 2 F D C2 S 1 + CosB a m 2 F D R2 S 1 k 1 + n SinhA a n 2 E D C2 S 2 Im SinA a m 2 E D C1 - CosA a m 2 E D R1 k 1 M n SinhA a n 2 E D C1 + CoshA a n 2 E D R1 k 1 - CoshA a n 2 E D R2 S 2 k 1 I-m SinA a m 2 E D C1 + CosA a m 2 E D R1 k 1 M n SinhA a n 2 E D C1 + CoshA a n 2 E D R1 k 1 The second solution for r F will be used to obtain the reactor width, a, self-consistently. We start with the value for a bare reactor based on the reactor parameters are taken from Meem, Two Group Reactor Theory,

Flux_Profile_2group.nb 11 H* Meems parameters *L params1 = 8n 1 Ø 2.47, S f1 Ø.002524, S ac1 Ø 0.003959, S ar1 Ø 0.0, S sc1 Ø 0.4930, S sr1 Ø.6871, z C Ø 0.7751, z R Ø.8203, E 0 Ø 1.62 µ 10^6, E th Ø 0.1, mc Ø.5078, mref Ø.5391<; params2 = 8n 2 Ø 2.47, S f2 Ø.06147, S ac2 Ø 0.09089, S ar2 Ø 0.01966, S sc2 Ø 4.715, S sr2 Ø 6.529, mc Ø.650086, mref Ø.6533<; Calculate the various parameters appropriate to this reactor: S 12 Cvalue = z C S sc1 ê Log@E 0 ê E th D ê. params1; S 12 Rvalue = z R S sr1 ê Log@E 0 ê E th D ê. params1; PrintA" S 12 C = ", S 12 Cvalue, " cm -1 S 12 R = ", S 12 Rvalue, " cm -1 "E StrC1 = H1 - mcl S sc1 ê. params1; StrC2 = H1 - mcl S sc2 ê. params2; StrR1 = H1 - mrefl S sr1 ê. params1; StrR2 = H1 - mrefl S sr2 ê. params2; D C1value = 1 ê H3 HS ac1 + StrC1LL ê. params1; D C2value = 1 ê H3 HS ac2 + StrC2LL ê. params2; D R1value = 1 ê H3 HS ar1 + StrR1LL ê. params1; D R2value = 1 ê H3 HS ar2 + StrR2LL ê. params2; Print@" D C1 = ", D C1value, " cm D C2 = ", D C2value, " cm D R1 = ", D R1value, " cm D R2 = ", D R2value, " cm"d Dparams = 8D C1 -> D C1value, D C2 -> D C2value, D R1 -> D R1value, D R2 -> D R2value, S 12 C -> S 12 Cvalue, S 12 R -> S 12 Rvalue <; k 1 value = k 1 soln ê. Dparams ê. params1; k 2 value = k 2 soln ê. Dparams ê. params2;

12 Flux_Profile_2group.nb PrintA" k 1 = ", k 1 value, " cm -1 k 2 = ", k 2 value, " cm -1 "E L nsol = D C1 S 12 C + S ac1 - n 1 S f1 ; L nvalue = L nsol ê. Dparams ê. params1; Print@" L n = ", L nvalue, " cm"d slabparams = 8D C1 -> D C1value, D C2 -> D C2value, D R1 -> D R1value, D R2 -> D R2value, S 12 C -> S 12 Cvalue, S 12 R -> S 12 Rvalue, L n -> L nvalue, R Ø R value <; S 12 C = 0.0230188 cm -1 S 12 R = 0.0339524 cm -1 D C1 = 1.35164 cm D C2 = 0.19149 cm D R1 = 1.05257 cm D R2 = 0.14599 cm k 1 = 0.179601 cm -1 k 2 = 0.36697 cm -1 L n = 8.07216 cm The initial guess for the slab width is estimated from the buckling constant using the geometric mean of the fast and thermal buckling:

Flux_Profile_2group.nb 13 a new = 0; k = n 1 S f1 ê S ac1 ê. params1; L1 = D C1value S ac1 ê. params1; B1 = Hk - 1L ë IL1 2 M ; k = n 2 S f2 ê S ac2 ê. params2; L2 = D C2value S ac2 ê. params2; B2 = Hk - 1L ë L2 2 ; a 0 = p í B1 B2 ; Print@" a HguessL = ", a 0, " cm"d a HguessL = 20.6498 cm The parameters list needs to be re-run once the self consistent radius is determined. Initially, however, set it to a 0. RUN SOLUTION HERE --> a value = If@a new < 1, a 0, a new D; slabparams = 8D C1 -> D C1value, D C2 -> D C2value, D R1 -> D R1value, D R2 -> D R2value, S 12 C -> S 12 Cvalue, S 12 R -> S 12 Rvalue, L n -> L nvalue, a Ø a value < 8D C1 Ø 1.35164, D C2 Ø 0.19149, D R1 Ø 1.05257, D R2 Ø 0.14599, S 12 C Ø 0.0230188, S 12 R Ø 0.0339524, L n Ø 8.07216, a Ø 12.1113< Find the remaining parameters needed to specify the solutions. b m = b n = 1 L + S ac2 2 n D C2 ê. slabparams ê. params1 ê. params2;

14 Flux_Profile_2group.nb - 1 L n 2 + S ac2 D C2 ê. slabparams ê. params1 ê. params2; c = - n 2 S f2 S 12 C D C1 D C2 + S ac2 L n 2 D C2 ê. slabparams ê. params1 ê. params2; m value = I-b m + SqrtAb m 2-4 cem ë 2 ; n value = I-b n + SqrtAb n 2-4 cem ë 2 ; S 1 value = S 1 soln ê. 8m -> m value < ê. slabparams ê. params2; S 2 value = S 2 soln ê. 8n Ø n value < ê. slabparams ê. params2; S 3 value = S 3 soln ê. slabparams ê. params2 ê. params1; musparams = 8m -> m value, n -> n value, S 1 -> S 1 value, S 2 -> S 2 value, S 3 -> S 3 value, k 1 -> k 1 value, k 2 -> k 2 value <; r Cvalue = r Csoln ê. slabparams ê. params2 ê. params1 ê. musparams; r Fvalue = r Fsoln ê. slabparams ê. params2 ê. params1 ê. musparams; r Gvalue = r Gsoln ê. slabparams ê. params2 ê. params1 ê. musparams; solnparams = 8m -> m value, n -> n value, S 1 -> S 1 value, S 2 -> S 2 value, S 3 -> S 3 value, k 1 -> k 1 value, k 2 -> k 2 value <; rparams = 8r C -> r Cvalue, r F -> r Fvalue, r G -> r Gvalue < ê. solnparams; PrintA" m = ", m value, " cm -1 n = ", n value, " cm -1 "E Print@" S 1 = ", S 1 value, " S 2 = ", S 2 value, " S 3 = ", S 3 value D

Flux_Profile_2group.nb 15 PrintA" k 1 = ", k 1 value, " cm -1 k 2 = ", k 2 value, " cm -1 "E Print@" r C = ", r Cvalue, " r F = ", r Fvalue, " r G = ", r Gvalue D m = 0.11126 cm -1 n = 0.708782 cm -1 S 1 = 0.246823 S 2 = -4.33564 S 3 = 2.27093 k 1 = 0.179601 cm -1 k 2 = 0.36697 cm -1 r C = -0.00128555 r F = 2.1793 r G = -11.731 Test the solution by examining the matching conditions at the reactor boundary, a/2: FC1 = F C1 B a F ê. params1 ê. slabparams ê. Dparams ê. 2 solnparams ê. rparams; FR1 = F R1 B a F ê. params1 ê. slabparams ê. Dparams ê. 2 solnparams ê. rparams; PrintB" F C1 @ a D = ", FC1, " 2 F R1@ a D = ", FR1F 2 FC2 = F C2 B a F ê. params1 ê. params2 ê. slabparams ê. 2 Dparams ê. solnparams ê. rparams; FR2 = F R2 B a F ê. params1 ê. params2 ê. slabparams ê. 2 Dparams ê. solnparams ê. rparams; PrintB" F C2 @ a D = ", FC2, " 2 F R2@ a D = ", FR2F 2

16 Flux_Profile_2group.nb jc1 = D C1 D@F C1 @xd, xd ê. x Ø a 2 ê. params1 ê. slabparams ê. Dparams ê. solnparams ê. rparams; jr1 = D R1 D@F R1 @xd, xd ê. x Ø a 2 ê. params1 ê. slabparams ê. Dparams ê. solnparams ê. rparams; PrintB" J C1 @ a D = ", jc1, " 2 J R1@ a D = ", jr1f 2 jc2 = D C2 D@F C2 @xd, xd ê. x Ø a 2 ê. params1 ê. params2 ê. slabparams ê. Dparams ê. solnparams ê. rparams; jr2 = D R2 D@F R2 @xd, xd ê. x Ø a 2 ê. params1 ê. params2 ê. slabparams ê. Dparams ê. solnparams ê. rparams; PrintB" J C2 @ a D = ", jc2, " 2 J R2@ a D = ", jr2f 2 F C1 @ a 2 D = 0.734476 F R1@ a 2 D = 0.734476 F C2 @ a 2 D = 0.396706 F R2@ a 2 D = 0.396706 J C1 @ a 2 D = -0.138848 J R1@ a 2 D = -0.138848 J C2 @ a 2 D = 0.0243718 J R2@ a 2 D = 0.0243718 The first attempt using the initial guess for a does not quite satisfy the matching conditions. To find the correcd slab width, find the value of a which gives the same flux ratio, r F, for the two cases:

Flux_Profile_2group.nb 17 a new = a ê. FindRoot@Hr Fsoln - r Fsoln2 L ê. params1 ê. params2 ê. Dparams ê. solnparams, 8a, a value <D; Print@" Self-consistent a = ", a new, " cm"d Self-consistent a = 12.1113 cm --> Now re-run the solutions to show that the matching conditions are now exactly satisfied (go to "RUN SOLUTION HERE" and re-run parameter list ). F1plot@x_D = If@x < a value ê 2, HF C1 @xdl ê. solnparams ê. rparams, HF R1 @xdl ê. solnparams ê. rparamsd; F2plot@x_D = If@x < a value ê 2, HF C2 @xdl ê. solnparams ê. rparams, HF R2 @xdl ê. solnparams ê. rparamsd; Plot@8F1plot@xD, F2plot@xD<, 8x,.01, 2 a value <, PlotRange Ø AllD 1.0 0.8 0.6 0.4 5 10 15 20 0.0

18 Flux_Profile_2group.nb Reflected Spherical Reactor Consider a spherical reactor having a core of radius, R, which is surrounded by an infinite reflector. There are two velocity groups and two regions (core and reflector) giving the following four diffusion equations for the neutron flux: [1a] D C1 2 F C1 ã HS ac1 + S 12 C L F C1-1 k Hn 1 S f1 F C1 + n 2 S f2 F C2 L [1b] D R1 2 F R1 == HS ar1 + S 12 R L F R1 [1c] D C2 2 F C2 == S ac2 F C2 - S sc1 F C1 [1d] D R2 2 F R2 == S ar2 F R2 - S sr1 F R1 For spherical symmetry we define F = u/r and, 2 F -> 1 d2 r dr 2 u. The (8) boundary conditions are: u C1 [0] = u C2 [0] = 0 [2a,b] u C1 @RD = u R1 @RD [2c] J C1 @RD = J R1 @RD [2d] u C2 @RD = u R2 @RD [2e] J C2 @RD = J R2 @RD [2f] u R1 @ D = u R2 @ D = 0 [2g,h] The general solutions, satisfying [2a,b] and [2g,h] and normalized by the first sineterm, are: u C1 @r_d = Sin@m rd + r C Sinh@n rd; u C2 @r_d = S 1 Sin@m rd + S 2 r C Sinh@n rd; u R1 @r_d = r F Exp@-k 1 rd; u R2 @r_d = r G Exp@-k 2 rd + S 3 r F Exp@-k 1 rd; There remain 10 parameters, {m, n, k 1, k 2, r C, r F, r G, S 1, S 2, S 3 <, to be deter-

Flux_Profile_2group.nb 19 1 2 C F G 1 2 3< mined. Substitution in to Eq.[1c,d] gives: k 1 soln = S ar1 + S 12 R D R1 ; k 2 soln = S ar2 D R2 ; Substitution into [1a,b] and equating sine and sinh-terms gives: S 1 soln = S 2 soln = S 3 soln = S 12 C S ac2 + D C2 m 2 ; S 12 C S ac2 - D C2 n ; 2 D R1 S 12 R ; D R1 S ar2 - D R2 S ar1 - D R2 S 12 R The joining conditions for the currents at r = R gives the flux ratios, { r C, r F, r G <:

20 Flux_Profile_2group.nb eq1 = u C1 @RD == u R1 @RD; eq2 = D C1 D@u C1 @rd ê r, rd == D R1 D@u R1 @rd ê r, rd ê. r Ø R; rcfsoln = Flatten@Solve@8eq1, eq2<, 8r C, r F <DD; r Csoln = r C ê. rcfsoln; r Fsoln = r F ê. rcfsoln; eq3 = u C2 @RD == u R2 @RD; eq4 = D C2 D@u C2 @rd ê r, rd == D R2 D@u R2 @rd ê r, rd ê. r Ø R; rfgsoln = Flatten@Solve@8eq3, eq4<, 8r G, r F <DD; r Gsoln = Simplify@r G ê. rfgsoln ê. r C -> r Csoln D; r Fsoln2 = Simplify@r F ê. rfgsoln ê. r C -> r Csoln D; Print@" r C Print@" r F Print@" r G = ", r Csoln D = ", r Fsoln D = ", r Gsoln D R m Cos@R md D C1 - Sin@R md D C1 + Sin@R md D R1 + R Sin@R md D R1 k 1 r C = - R n Cosh@R nd D C1 - Sinh@R nd D C1 + Sinh@R nd D R1 + R Sinh@R nd D R1 k 1 r F = r G = R k 1 R Hn Cosh@R nd Sin@R md - m Cos@R md Sinh@R ndl DC1 R n Cosh@R nd D C1 - Sinh@R nd D C1 + Sinh@R nd D R1 + R Sinh@R nd D R1 k 1 -I R k 2 HD R1 H1 + R k 1 L HD C2 HHR m Cos@R md - Sin@R mdl Sinh@R nd S 1 + Sin@R md H-R n Cosh@R nd + Sinh@R ndl S 2 L + Sin@R md Sinh@R nd D R2 HS 1 - S 2 L H1 + R k 1 LL + D C1 HHR m Cos@R md - Sin@R mdl HR n Cosh@R nd - Sinh@R ndl D C2 HS 1 - S 2 L + D R2 HSin@R md HR n Cosh@R nd - Sinh@R ndl S 1 + H-R m Cos@R md + Sin@R mdl Sinh@R nd S 2 L H1 + R k 1 LLLM ë HR D R2 HHR n Cosh@R nd - Sinh@R ndl D C1 + Sinh@R nd D R1 H1 + R k 1 LL H-k 1 + k 2 LL The second solution for r F will be used to obtain the reactor radius, R, self-consistently. We start with the value for a bare reactor based on the reactor parameters are taken from Meem, Two Group Reactor Theory,

Flux_Profile_2group.nb 21 H* Meems parameters *L params1 = 8n 1 Ø 2.47, S f1 Ø.002524, S ac1 Ø 0.003959, S ar1 Ø 0.0, S sc1 Ø 0.4930, S sr1 Ø.6871, z C Ø 0.7751, z R Ø.8203, E 0 Ø 1.62 µ 10^6, E th Ø 0.1, mc Ø.5078, mref Ø.5391<; params2 = 8n 2 Ø 2.47, S f2 Ø.06147, S ac2 Ø 0.09089, S ar2 Ø 0.01966, S sc2 Ø 4.715, S sr2 Ø 6.529, mc Ø.650086, mref Ø.6533<; Calculate the various parameters appropriate to this reactor: S 12 Cvalue = z C S sc1 ê Log@E 0 ê E th D ê. params1; S 12 Rvalue = z R S sr1 ê Log@E 0 ê E th D ê. params1; PrintA" S 12 C = ", S 12 Cvalue, " cm -1 S 12 R = ", S 12 Rvalue, " cm -1 "E StrC1 = H1 - mcl S sc1 ê. params1; StrC2 = H1 - mcl S sc2 ê. params2; StrR1 = H1 - mrefl S sr1 ê. params1; StrR2 = H1 - mrefl S sr2 ê. params2; D C1value = 1 ê H3 HS ac1 + StrC1LL ê. params1; D C2value = 1 ê H3 HS ac2 + StrC2LL ê. params2; D R1value = 1 ê H3 HS ar1 + StrR1LL ê. params1; D R2value = 1 ê H3 HS ar2 + StrR2LL ê. params2; Print@" D C1 = ", D C1value, " cm D C2 = ", D C2value, " cm D R1 = ", D R1value, " cm D R2 = ", D R2value, " cm"d Dparams = 8D C1 -> D C1value, D C2 -> D C2value, D R1 -> D R1value, D R2 -> D R2value, S 12 C -> S 12 Cvalue, S 12 R -> S 12 Rvalue <; k 1 value = k 1 soln ê. Dparams ê. params1; k 2 value = k 2 soln ê. Dparams ê. params2;

22 Flux_Profile_2group.nb PrintA" k 1 = ", k 1 value, " cm -1 k 2 = ", k 2 value, " cm -1 "E L nsol = D C1 S 12 C + S ac1 - n 1 S f1 ; L nvalue = L nsol ê. Dparams ê. params1; Print@" L n = ", L nvalue, " cm"d sphereparams = 8D C1 -> D C1value, D C2 -> D C2value, D R1 -> D R1value, D R2 -> D R2value, S 12 C -> S 12 Cvalue, S 12 R -> S 12 Rvalue, L n -> L nvalue, R Ø R value <; S 12 C = 0.0230188 cm -1 S 12 R = 0.0339524 cm -1 D C1 = 1.35164 cm D C2 = 0.19149 cm D R1 = 1.05257 cm D R2 = 0.14599 cm k 1 = 0.179601 cm -1 k 2 = 0.36697 cm -1 L n = 8.07216 cm The initial guess for the radius is estimated from the buckling constant using the geometric mean of the fast and thermal buckling:

Flux_Profile_2group.nb 23 R new = 0; k = n 1 S f1 ê S ac1 ê. params1; L1 = D C1value S ac1 ê. params1; B1 = Hk - 1L ë IL1 2 M ; k = n 2 S f2 ê S ac2 ê. params2; L2 = D C2value S ac2 ê. params2; B2 = Hk - 1L ë L2 2 ; R 0 = p í B1 B2 ; Print@" R HguessL = ", R 0, " cm"d R HguessL = 20.6498 cm The parameters list needs to be re-run once the self consistent radius is determined. Initially, however, set it to R 0. RUN SOLUTION HERE --> R value = If@R new < 1, R 0, R new D; sphereparams = 8D C1 -> D C1value, D C2 -> D C2value, D R1 -> D R1value, D R2 -> D R2value, S 12 C -> S 12 Cvalue, S 12 R -> S 12 Rvalue, L n -> L nvalue, R Ø R value < 8D C1 Ø 1.35164, D C2 Ø 0.19149, D R1 Ø 1.05257, D R2 Ø 0.14599, S 12 C Ø 0.0230188, S 12 R Ø 0.0339524, L n Ø 8.07216, R Ø 19.5914< Find the remaining parameters needed to specify the solutions. b m = 1 L + S ac2 2 n D C2 ê. sphereparams ê. params1 ê. params2;

24 Flux_Profile_2group.nb b n = - 1 L n 2 + S ac2 D C2 ê. sphereparams ê. params1 ê. params2; c = - n 2 S f2 S 12 C D C1 D C2 + S ac2 L n 2 D C2 ê. sphereparams ê. params1 ê. params2; m value = I-b m + SqrtAb m 2-4 cem ë 2 ; n value = I-b n + SqrtAb n 2-4 cem ë 2 ; S 1 value = S 1 soln ê. 8m -> m value < ê. sphereparams ê. params2; S 2 value = S 2 soln ê. 8n Ø n value < ê. sphereparams ê. params2; S 3 value = S 3 soln ê. sphereparams ê. params2 ê. params1; musparams = 8m -> m value, n -> n value, S 1 -> S 1 value, S 2 -> S 2 value, S 3 -> S 3 value, k 1 -> k 1 value, k 2 -> k 2 value <; r Cvalue = r Csoln ê. sphereparams ê. params2 ê. params1 ê. musparams; r Fvalue = r Fsoln ê. sphereparams ê. params2 ê. params1 ê. musparams; r Gvalue = r Gsoln ê. sphereparams ê. params2 ê. params1 ê. musparams; solnparams = 8m -> m value, n -> n value, S 1 -> S 1 value, S 2 -> S 2 value, S 3 -> S 3 value, k 1 -> k 1 value, k 2 -> k 2 value <; rparams = 8r C -> r Cvalue, r F -> r Fvalue, r G -> r Gvalue < ê. solnparams; Print@" m = ", m value, " n = ", n value D Print@" S 1 = ", S 1 value, " S 2 = ", S 2 value, " S 3 = ", S 3 value D

Flux_Profile_2group.nb 25 PrintA" k 1 = ", k 1 value, " cm -1 k 2 = ", k 2 value, " cm -1 "E Print@" r C = ", r Cvalue, " r F = ", r Fvalue, " r G = ", r Gvalue D m = 0.11126 n = 0.708782 S 1 = 0.246823 S 2 = -4.33564 S 3 = 2.27093 k 1 = 0.179601 cm -1 k 2 = 0.36697 cm -1 r C = -9.30729 µ 10-8 r F = 25.9892 r G = -1763.24 Test the solution by examining the matching conditions at the reactor radius, R:

26 Flux_Profile_2group.nb uc1 = u C1 @RD ê. params1 ê. sphereparams ê. Dparams ê. solnparams ê. rparams; ur1 = u R1 @RD ê. params1 ê. sphereparams ê. Dparams ê. solnparams ê. rparams; Print@" u C1 @RD = ", uc1, " u R1 @RD = ", ur1d uc2 = u C2 @RD ê. params1 ê. params2 ê. sphereparams ê. Dparams ê. solnparams ê. rparams; ur2 = u R2 @RD ê. params1 ê. params2 ê. sphereparams ê. Dparams ê. solnparams ê. rparams; Print@" u C2 @RD = ", uc2, " u R2 @RD = ", ur2d jc1 = D C1 D@u C1 @rd ê r, rd ê. r Ø R ê. params1 ê. sphereparams ê. Dparams ê. solnparams ê. rparams; jr1 = D R1 D@u R1 @rd ê r, rd ê. r Ø R ê. params1 ê. sphereparams ê. Dparams ê. solnparams ê. rparams; Print@" J C1 @RD = ", jc1, " J R1 @RD = ", jr1d jc2 = D C2 D@u C2 @rd ê r, rd ê. r Ø R ê. params1 ê. params2 ê. sphereparams ê. Dparams ê. solnparams ê. rparams; jr2 = D R2 D@u R2 @rd ê r, rd ê. r Ø R ê. params1 ê. params2 ê. sphereparams ê. Dparams ê. solnparams ê. rparams; Print@" J C2 @RD = ", jc2, " J R2 @RD = ", jr2d u C1 @RD = 0.770309 u R1 @RD = 0.770309 u C2 @RD = 0.418968 u R2 @RD = 0.418968 J C1 @RD = -0.00954537 J R1 @RD = -0.00954537 J C2 @RD = 0.00113739 J R2 @RD = 0.00113739

Flux_Profile_2group.nb 27 The first attempt using the initial guess for R does not quite satisfy the matching conditions. To find the correcd radius, find the value of R which gives the same flux ratio, r F, for the two cases: R new = R ê. FindRoot@Hr Fsoln - r Fsoln2 L ê. params1 ê. params2 ê. Dparams ê. solnparams, 8R, R value <D; Print@" Self-consistent R = ", R new, " cm"d Self-consistent R = 19.5914 cm Now re-run the solutions to show that the matching conditions are now exactly satisfied (go to "RUN SOLUTION HERE").

28 Flux_Profile_2group.nb F1plot@r_D = If@r < R value, Hu C1 @rd ê rl ê. solnparams ê. rparams, Hu R1 @rd ê rl ê. solnparams ê. rparamsd; F2plot@r_D = If@r < R value, Hu C2 @rd ê rl ê. solnparams ê. rparams, Hu R2 @rd ê rl ê. solnparams ê. rparamsd; Plot@8F1plot@rD, F2plot@rD<, 8r,.01, 1.5 R value <D 0.10 0.08 0.06 0.04 0.02 5 10 15 20 25 30

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