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;
DC = 1 ê H3 StrCL; LC =
DC
SCa
;
Flux_Profile_2group.nb
Print@" Core parameters "D
Print@"
Fast group:
Str1 = ",
StrC ê. params1, " cm^-1 DC1 = ", DC ê. params1,
" cm LC1 = ", LC ê. params1, " cm"D
Print@"
Thermal group: Str2 = ",
StrC ê. params2, " cm^-1 DC2 = ", DC ê. params2,
" cm LC1 = ", LC ê. params2, " cm"D
H* Diffusion constants for reflector *L
StrRef = SRefa + H1 - mRefL SRefs;
lex = 1 ê H3 StrRefL;
DR = 1 ê H3 StrRefL; LR =
DR
;
SRefa
Print@" Reflector parameters "D
Print@"
Fast group:
Str1 = ",
StrRef ê. params1, " cm^-1 DR1 = ", DR ê. params1,
" cm LR1 = ", LR ê. params1, " cm"D
Print@"
Thermal group: Str2 = ",
StrRef ê. params2, " cm^-1 DR2 = ", DR ê. params2,
" cm LR1 = ", LR 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
DC1 = 0.100939 cm
LC1 = 1.04757
Thermal group:
Str2 = 3.43955 cm^-1
DC2 = 0.0969119 cm
Fast group:
Str1 = 0.360201 cm^-1
DR1 = 0.92541 cm
LR1 = 58.2432
cm
Thermal group:
Str2 = 0.360201 cm^-1
DR2 = 0.92541 cm
LR1 = 58.2432
cm
cm
LC1 = 0.776812
cm
Reflector parameters
3
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
2
D1 “2 F1 == H-Sa1 - Ss1->2 L F1 + n Sf
F2
k
The diffusion equations for the two velocity group fluxes are:
D2 “2 F2 == -Sa2 F2 + Ss1->2 F1
For cylindrical geometry these factor into r and z components: F = R[r] Z[z].
Reqn =
1
D@Hr R '@rDL, rD == -Br2 R@rD;
r
Zeqn = Z ''@zD ã -Bz2 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
R1soln@r_D
Z2soln@z_D
R2soln@r_D
=
=
=
=
Az1 Cos@Bz zD
Ar1 BesselJ@0, Br rD
Az2 Cos@Bz zD
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
z0soln =
z0 ê. FindRoot@BesselJ@0, z0D ã 0, 8z0, 2.5<D;
Bconstants = 8Br -> z0soln ê Rex, Bz -> p ê Hex<;
Bnet = Br2 + Bz2 ê. Bconstants ê. geom;
PNL = 1 ê H1 + Bnet ^ 2 LC ^ 2L ê. params1 ê. geom;
r21 =
SCs ê HSCa + DC 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
5
6
Flux_Profile_2group.nb
20
15
10
5
10
20
30
40
Flux_Profile_2group.nb
20
15
10
5
5
10
15
20
7
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:
DC1 “2 FC1 ã HSaC1 + S12 C L FC1 -
[1a]
DR1 “2 FR1 == HSaR1 + S12 R L FR1
1
k
Hn1 Sf1 FC1 + n2 Sf2 FC2 L
[1b]
DC2 “2 FC2 == SaC2 FC2
-
SsC1 FC1
[1c]
DR2 “2 FR2 == SaR2 FR2 - SsR1 FR1
[1d]
In the following we will solve for the conditions that give k= 1. For slab symmetry “2 F ->
d2
dx2
F. The (8) boundary conditions are:
FC1 A a2 E = FR1 A a2 E
FC1 '[0] = FC2 '[0] = 0
[2a,b]
JC1 A a2 E = JR1 A a2 E
[2c]
JC2 A a2 E = JR2 A a2 E
FR1 @¶D = FR2 @¶D = 0
[2e]
FC2 A a2 E = FR2 A a2 E
[2d]
[2f]
[2g,h]
The general solutions, satisfying [2a,b] and [2g,h] and normalized by the first sineterm, are:
FC1 @x_D
FC2 @x_D
FR1 @x_D
FR2 @x_D
=
=
=
=
Cos@m xD + rC Cosh@n xD;
S1 Cos@m xD + S2 rC Cosh@n xD;
rF Exp@-k1 xD;
rG Exp@-k2 xD + S3 rF Exp@-k1 xD;
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, k1 , k2 , rC , rF , rG , S1 , S2 , S3 <, to be determined. Substitution in to Eq.[1c,d]
gives:
Flux_Profile_2group.nb
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, k1 , k2 , rC , rF , rG , S1 , S2 , S3 <, to be determined. Substitution in to Eq.[1c,d]
gives:
k1 soln =
k2 soln =
SaR1 + S12 R
DR1
SaR2
DR2
;
;
Substitution into [1a,b] and equating sine and sinh-terms gives:
S1 soln =
S2 soln =
S3 soln =
S12 C
SaC2 + DC2 m2
S12 C
2
;
;
SaC2 - DC2 n
DR1 S12 R
DR1 SaR2 - DR2 SaR1 - DR2 S12 R
;
The joining conditions for the currents at r = R gives the flux ratios,
{ rC , rF , rG <:
9
10
Flux_Profile_2group.nb
a
a
eq1 = FC1 B F ã FR1 B F;
2
2
eq2 = DC1 D@FC1 @xD, xD == DR1 D@FR1 @xD, xD ê. x Ø
a
;
2
rCFsoln = Flatten@Solve@8eq1, eq2<, 8rC , rF <DD;
rCsoln = rC ê. rCFsoln;
rFsoln = rF ê. rCFsoln;
a
a
eq3 = FC2 B F ã FR2 B F;
2
2
eq4 = DC2 D@FC2 @xD, xD == DR2 D@FR2 @xD, xD ê. x Ø
a
;
2
rFGsoln = Flatten@Solve@8eq3, eq4<, 8rG , rF <DD;
rGsoln = Simplify@rG ê. rFGsoln ê. rC -> rCsoln D;
rFsoln2 = Simplify@rF ê. rFGsoln ê. rC -> rCsoln D;
Print@" rC = ", rCsoln D
Print@" rF = ", rFsoln D
Print@" rG = ", rGsoln D
-m SinA
rC = -
‰
rF =
rG =
n
an
SinhA 2 E
a k1
2
am
E
2
Im CoshA
n
1
an
E
2
n
DC1 +
an
E
2
‰
a k2
2
am
E
2
DR1 k1
an
CoshA 2 E
SinA
an
SinhA 2 E
DR2 Hk1 - k2 L
n SinhA
DC1 + CosA
am
E
2
DC2 S2 Im SinA
an
SinhA 2 E
+ n CosA
DC1 +
-m SinB
DC1 +
DR1 k1
am
E
2
an
CoshA 2 E
am
2
am
E
2
SinhA
DC1
DR1 k1
F DC2 S1 + CosB
DC1 - CosA
an
CoshA 2 E
an
EM
2
am
E
2
DR1 k1
am
2
F DR2 S1 k1 +
DR1 k1 M
CoshA
-
an
E
2
DR2 S2 k1 I-m SinA
n
an
SinhA 2 E
DC1 +
am
E
2
DC1 + CosA
an
CoshA 2 E
am
E
2
DR1 k1 M
DR1 k1
The second solution for rF 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 = 8n1 Ø 2.47, Sf1 Ø .002524, SaC1 Ø 0.003959,
SaR1 Ø 0.0, SsC1 Ø 0.4930, SsR1 Ø .6871,
zC Ø 0.7751, zR Ø .8203, E0 Ø 1.62 µ 10 ^ 6,
Eth Ø 0.1, mC Ø .5078, mRef Ø .5391<;
params2 = 8n2 Ø 2.47, Sf2 Ø .06147,
SaC2 Ø 0.09089, SaR2 Ø 0.01966, SsC2 Ø 4.715,
SsR2 Ø 6.529, mC Ø .650086, mRef Ø .6533<;
Calculate the various parameters appropriate to this reactor:
S12 Cvalue = zC SsC1 ê Log@E0 ê Eth D ê. params1;
S12 Rvalue = zR SsR1 ê Log@E0 ê Eth D ê. params1;
PrintA" S12 C = ", S12 Cvalue ,
" cm-1
StrC1
StrC2
StrR1
StrR2
=
=
=
=
S12 R = ", S12 Rvalue , " cm-1 "E
H1 - mCL SsC1 ê. params1;
H1 - mCL SsC2 ê. params2;
H1 - mRefL SsR1 ê. params1;
H1 - mRefL SsR2 ê. params2;
DC1value = 1 ê H3 HSaC1 + StrC1LL ê. params1;
DC2value = 1 ê H3 HSaC2 + StrC2LL ê. params2;
DR1value = 1 ê H3 HSaR1 + StrR1LL ê. params1;
DR2value = 1 ê H3 HSaR2 + StrR2LL ê. params2;
Print@" DC1 = ", DC1value ,
" cm DC2 = ", DC2value , " cm DR1 = ",
DR1value , " cm DR2 = ", DR2value , " cm"D
Dparams = 8DC1 -> DC1value , DC2 -> DC2value , DR1 -> DR1value ,
DR2 -> DR2value , S12 C -> S12 Cvalue , S12 R -> S12 Rvalue <;
k1 value = k1 soln ê. Dparams ê. params1;
k2 value = k2 soln ê. Dparams ê. params2;
12
Flux_Profile_2group.nb
PrintA" k1 = ", k1 value ,
" cm-1
Lnsol =
k2 = ", k2 value , " cm-1 "E
DC1
S12 C + SaC1 - n1 Sf1
;
Lnvalue = Lnsol ê. Dparams ê. params1;
Print@" Ln = ", Lnvalue , " cm"D
slabparams = 8DC1 -> DC1value , DC2 -> DC2value ,
DR1 -> DR1value , DR2 -> DR2value , S12 C -> S12 Cvalue ,
S12 R -> S12 Rvalue , Ln -> Lnvalue , R Ø Rvalue <;
S12 C = 0.0230188 cm-1
S12 R = 0.0339524 cm-1
DC1 = 1.35164 cm DC2 = 0.19149
cm DR1 = 1.05257 cm DR2 = 0.14599 cm
k1 = 0.179601 cm-1
k2 = 0.36697 cm-1
Ln = 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
anew = 0;
k¶ = n1 Sf1 ê SaC1 ê. params1;
DC1value
L1 =
SaC1
ê. params1; B1 =
Hk¶ - 1L ë IL12 M ;
ê. params2; B2 =
Hk¶ - 1L ë L22 ;
k¶ = n2 Sf2 ê SaC2 ê. params2;
DC2value
L2 =
a0 = p í
SaC2
B1 B2 ;
Print@" a HguessL = ", a0 , " 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 a0 .
RUN SOLUTION HERE -->
avalue = If@anew < 1, a0 , anew D;
slabparams = 8DC1 -> DC1value , DC2 -> DC2value ,
DR1 -> DR1value , DR2 -> DR2value , S12 C -> S12 Cvalue ,
S12 R -> S12 Rvalue , Ln -> Lnvalue , a Ø avalue <
8DC1 Ø 1.35164, DC2 Ø 0.19149,
DR1 Ø 1.05257, DR2 Ø 0.14599, S12 C Ø 0.0230188,
S12 R Ø 0.0339524, Ln Ø 8.07216, a Ø 12.1113<
Find the remaining parameters needed to specify the solutions.
bm =
1
L2n
+
SaC2
DC2
ê. slabparams ê. params1 ê. params2;
bn =
;
13
14
Flux_Profile_2group.nb
1
-
L2n
c= -
+
SaC2
DC2
ê. slabparams ê. params1 ê. params2;
n2 Sf2 S12 C
DC1 DC2
+
SaC2
L2n
DC2
ê. slabparams ê. params1 ê.
params2;
mvalue =
nvalue =
I-bm + SqrtAbm 2 - 4 cEM ë 2 ;
I-bn + SqrtAbn 2 - 4 cEM ë 2 ;
S1 value =
S1 soln ê. 8m -> mvalue < ê. slabparams ê. params2;
S2 value = S2 soln ê. 8n Ø nvalue < ê. slabparams ê.
params2;
S3 value = S3 soln ê. slabparams ê. params2 ê. params1;
muSparams =
8m -> mvalue , n -> nvalue , S1 -> S1 value , S2 -> S2 value ,
S3 -> S3 value , k1 -> k1 value , k2 -> k2 value <;
rCvalue = rCsoln ê. slabparams ê. params2 ê. params1 ê.
muSparams;
rFvalue = rFsoln ê. slabparams ê. params2 ê. params1 ê.
muSparams;
rGvalue = rGsoln ê. slabparams ê. params2 ê. params1 ê.
muSparams;
solnparams = 8m -> mvalue , n -> nvalue ,
S1 -> S1 value , S2 -> S2 value , S3 -> S3 value ,
k1 -> k1 value , k2 -> k2 value <;
rparams = 8rC -> rCvalue , rF -> rFvalue , rG -> rGvalue < ê.
solnparams;
PrintA" m = ", mvalue ,
" cm-1
n = ", nvalue , " cm-1 "E
Print@" S1 = ", S1 value , " S2 = ",
S2 value , " S3 = ", S3 value D
Flux_Profile_2group.nb
PrintA" k1 = ", k1 value ,
" cm-1 k2 = ", k2 value , " cm-1 "E
Print@" rC = ", rCvalue , " rF = ",
rFvalue , " rG = ", rGvalue D
m = 0.11126
S1 = 0.246823
cm-1
n = 0.708782
cm-1
S2 = -4.33564 S3 = 2.27093
k1 = 0.179601 cm-1
k2 = 0.36697 cm-1
rC = -0.00128555 rF = 2.1793 rG = -11.731
Test the solution by examining the matching conditions at the reactor boundary,
a/2:
a
FC1 = FC1 B F ê. params1 ê. slabparams ê. Dparams ê.
2
solnparams ê. rparams;
a
FR1 = FR1 B F ê. params1 ê. slabparams ê. Dparams ê.
2
solnparams ê. rparams;
a
a
PrintB" FC1 @ D = ", FC1, " FR1 @ D = ", FR1F
2
2
a
FC2 = FC2 B F ê. params1 ê. params2 ê. slabparams ê.
2
Dparams ê. solnparams ê. rparams;
a
FR2 = FR2 B F ê. params1 ê. params2 ê. slabparams ê.
2
Dparams ê. solnparams ê. rparams;
a
a
PrintB" FC2 @ D = ", FC2, " FR2 @ D = ", FR2F
2
2
15
16
Flux_Profile_2group.nb
jC1 = DC1 D@FC1 @xD, xD ê. x Ø
ê. params1 ê.
2
slabparams ê. Dparams ê.
solnparams ê. rparams;
a
jR1 = DR1 D@FR1 @xD, xD ê. x Ø
ê. params1 ê.
2
slabparams ê. Dparams ê.
solnparams ê. rparams;
a
a
PrintB" JC1 @ D = ", jC1, " JR1 @ D = ", jR1F
2
2
a
jC2 =
DC2 D@FC2 @xD, xD ê. x Ø
ê. params1 ê. params2 ê.
2
slabparams ê. Dparams ê.
solnparams ê. rparams;
a
jR2 = DR2 D@FR2 @xD, xD ê. x Ø
ê. params1 ê.
2
params2 ê. slabparams ê.
Dparams ê. solnparams ê. rparams;
a
a
PrintB" JC2 @ D = ", jC2, " JR2 @ D = ", jR2F
2
2
a
FC1 @ D = 0.734476
2
a
FC2 @ D = 0.396706
2
a
JC1 @ D = -0.138848
2
a
JC2 @ D = 0.0243718
2
a
a
FR1 @ D = 0.734476
2
a
FR2 @ D = 0.396706
2
a
JR1 @ D = -0.138848
2
a
JR2 @ D = 0.0243718
2
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, rF , for the two cases:
Flux_Profile_2group.nb
17
anew = a ê.
FindRoot@HrFsoln - rFsoln2 L ê. params1 ê. params2 ê.
Dparams ê. solnparams, 8a, avalue <D;
Print@" Self-consistent a = ", anew , " 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 < avalue ê 2, HFC1 @xDL ê. solnparams ê. rparams,
HFR1 @xDL ê. solnparams ê. rparamsD;
F2plot@x_D = If@x < avalue ê 2,
HFC2 @xDL ê. solnparams ê. rparams,
HFR2 @xDL ê. solnparams ê. rparamsD;
Plot@8F1plot@xD, F2plot@xD<,
8x, .01, 2 avalue <, PlotRange Ø AllD
1.0
0.8
0.6
0.4
5
0.0
ü Reflected Spherical Reactor
10
15
20
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:
DC1 “2 FC1 ã HSaC1 + S12 C L FC1 -
[1a]
DR1 “2 FR1 == HSaR1 + S12 R L FR1
1
k
Hn1 Sf1 FC1 + n2 Sf2 FC2 L
[1b]
DC2 “2 FC2 == SaC2 FC2
-
SsC1 FC1
[1c]
DR2 “2 FR2 == SaR2 FR2 - SsR1 FR1
[1d]
For spherical symmetry we define F = u/r and, “2 F ->
ary conditions are:
uC1 [0] = uC2 [0] = 0
uC1 @RD = uR1 @RD
JC1 @RD = JR1 @RD
uC2 @RD = uR2 @RD
JC2 @RD = JR2 @RD
uR1 @¶D = uR2 @¶D = 0
1 d2
r dr2
u. The (8) bound[2a,b]
[2c]
[2d]
[2e]
[2f]
[2g,h]
The general solutions, satisfying [2a,b] and [2g,h] and normalized by the first sineterm, are:
uC1 @r_D
uC2 @r_D
uR1 @r_D
uR2 @r_D
=
=
=
=
Sin@m rD + rC Sinh@n rD;
S1 Sin@m rD + S2 rC Sinh@n rD;
rF Exp@-k1 rD;
rG Exp@-k2 rD + S3 rF Exp@-k1 rD;
There remain 10 parameters, {m, n, k1 , k2 , rC , rF , rG , S1 , S2 , S3 <, to be determined. Substitution in to Eq.[1c,d] gives:
ü
There remain 10 parameters, {m, n, k1 , k2 , rC , rF , rG , S1 , S2 , S3 <, to be determined. Substitution in to Eq.[1c,d] gives:
Flux_Profile_2group.nb
k1 soln =
k2 soln =
SaR1 + S12 R
DR1
SaR2
DR2
;
;
Substitution into [1a,b] and equating sine and sinh-terms gives:
S1 soln =
S2 soln =
S3 soln =
S12 C
SaC2 + DC2 m2
S12 C
2
;
;
SaC2 - DC2 n
DR1 S12 R
DR1 SaR2 - DR2 SaR1 - DR2 S12 R
;
The joining conditions for the currents at r = R gives the flux ratios,
{ rC , rF , rG <:
19
20
Flux_Profile_2group.nb
eq1 = uC1 @RD == uR1 @RD;
eq2 =
DC1 D@uC1 @rD ê r, rD == DR1 D@uR1 @rD ê r, rD ê. r Ø R;
rCFsoln = Flatten@Solve@8eq1, eq2<, 8rC , rF <DD;
rCsoln = rC ê. rCFsoln;
rFsoln = rF ê. rCFsoln;
eq3 = uC2 @RD == uR2 @RD;
eq4 =
DC2 D@uC2 @rD ê r, rD == DR2 D@uR2 @rD ê r, rD ê. r Ø R;
rFGsoln = Flatten@Solve@8eq3, eq4<, 8rG , rF <DD;
rGsoln = Simplify@rG ê. rFGsoln ê. rC -> rCsoln D;
rFsoln2 = Simplify@rF ê. rFGsoln ê. rC -> rCsoln D;
Print@" rC = ", rCsoln D
Print@" rF = ", rFsoln D
Print@" rG = ", rGsoln D
rC = -
rF =
R m Cos@R mD DC1 - Sin@R mD DC1 + Sin@R mD DR1 + R Sin@R mD DR1 k1
R n Cosh@R nD DC1 - Sinh@R nD DC1 + Sinh@R nD DR1 + R Sinh@R nD DR1 k1
‰R k1 R Hn Cosh@R nD Sin@R mD - m Cos@R mD Sinh@R nDL DC1
R n Cosh@R nD DC1 - Sinh@R nD DC1 + Sinh@R nD DR1 + R Sinh@R nD DR1 k1
rG =
-I‰R k2 HDR1 H1 + R k1 L HDC2 HHR m Cos@R mD - Sin@R mDL Sinh@R nD S1 + Sin@R mD H-R n Cosh@R nD + Sinh@R nDL S2 L + Sin@R mD
Sinh@R nD DR2 HS1 - S2 L H1 + R k1 LL + DC1 HHR m Cos@R mD - Sin@R mDL HR n Cosh@R nD - Sinh@R nDL DC2 HS1 - S2 L +
DR2 HSin@R mD HR n Cosh@R nD - Sinh@R nDL S1 + H-R m Cos@R mD + Sin@R mDL Sinh@R nD S2 L H1 + R k1 LLLM ë
HR DR2 HHR n Cosh@R nD - Sinh@R nDL DC1 + Sinh@R nD DR1 H1 + R k1 LL H-k1 + k2 LL
The second solution for rF 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 = 8n1 Ø 2.47, Sf1 Ø .002524, SaC1 Ø 0.003959,
SaR1 Ø 0.0, SsC1 Ø 0.4930, SsR1 Ø .6871,
zC Ø 0.7751, zR Ø .8203, E0 Ø 1.62 µ 10 ^ 6,
Eth Ø 0.1, mC Ø .5078, mRef Ø .5391<;
params2 = 8n2 Ø 2.47, Sf2 Ø .06147,
SaC2 Ø 0.09089, SaR2 Ø 0.01966, SsC2 Ø 4.715,
SsR2 Ø 6.529, mC Ø .650086, mRef Ø .6533<;
Calculate the various parameters appropriate to this reactor:
S12 Cvalue = zC SsC1 ê Log@E0 ê Eth D ê. params1;
S12 Rvalue = zR SsR1 ê Log@E0 ê Eth D ê. params1;
PrintA" S12 C = ", S12 Cvalue ,
" cm-1
StrC1
StrC2
StrR1
StrR2
=
=
=
=
S12 R = ", S12 Rvalue , " cm-1 "E
H1 - mCL SsC1 ê. params1;
H1 - mCL SsC2 ê. params2;
H1 - mRefL SsR1 ê. params1;
H1 - mRefL SsR2 ê. params2;
DC1value = 1 ê H3 HSaC1 + StrC1LL ê. params1;
DC2value = 1 ê H3 HSaC2 + StrC2LL ê. params2;
DR1value = 1 ê H3 HSaR1 + StrR1LL ê. params1;
DR2value = 1 ê H3 HSaR2 + StrR2LL ê. params2;
Print@" DC1 = ", DC1value ,
" cm DC2 = ", DC2value , " cm DR1 = ",
DR1value , " cm DR2 = ", DR2value , " cm"D
Dparams = 8DC1 -> DC1value , DC2 -> DC2value , DR1 -> DR1value ,
DR2 -> DR2value , S12 C -> S12 Cvalue , S12 R -> S12 Rvalue <;
k1 value = k1 soln ê. Dparams ê. params1;
k2 value = k2 soln ê. Dparams ê. params2;
22
Flux_Profile_2group.nb
PrintA" k1 = ", k1 value ,
" cm-1
Lnsol =
k2 = ", k2 value , " cm-1 "E
DC1
S12 C + SaC1 - n1 Sf1
;
Lnvalue = Lnsol ê. Dparams ê. params1;
Print@" Ln = ", Lnvalue , " cm"D
sphereparams = 8DC1 -> DC1value , DC2 -> DC2value ,
DR1 -> DR1value , DR2 -> DR2value , S12 C -> S12 Cvalue ,
S12 R -> S12 Rvalue , Ln -> Lnvalue , R Ø Rvalue <;
S12 C = 0.0230188 cm-1
S12 R = 0.0339524 cm-1
DC1 = 1.35164 cm DC2 = 0.19149
cm DR1 = 1.05257 cm DR2 = 0.14599 cm
k1 = 0.179601 cm-1
k2 = 0.36697 cm-1
Ln = 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
Rnew = 0;
k¶ = n1 Sf1 ê SaC1 ê. params1;
DC1value
L1 =
SaC1
ê. params1; B1 =
Hk¶ - 1L ë IL12 M ;
ê. params2; B2 =
Hk¶ - 1L ë L22 ;
k¶ = n2 Sf2 ê SaC2 ê. params2;
DC2value
L2 =
R0 = p í
SaC2
B1 B2 ;
Print@" R HguessL = ", R0 , " 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 R0 .
RUN SOLUTION HERE -->
Rvalue = If@Rnew < 1, R0 , Rnew D;
sphereparams = 8DC1 -> DC1value , DC2 -> DC2value ,
DR1 -> DR1value , DR2 -> DR2value , S12 C -> S12 Cvalue ,
S12 R -> S12 Rvalue , Ln -> Lnvalue , R Ø Rvalue <
8DC1 Ø 1.35164, DC2 Ø 0.19149,
DR1 Ø 1.05257, DR2 Ø 0.14599, S12 C Ø 0.0230188,
S12 R Ø 0.0339524, Ln Ø 8.07216, R Ø 19.5914<
Find the remaining parameters needed to specify the solutions.
bm =
1
L2n
+
SaC2
DC2
ê. sphereparams ê. params1 ê. params2;
23
24
Flux_Profile_2group.nb
bn = -
1
L2n
+
SaC2
DC2
ê. sphereparams ê. params1 ê.
params2;
n2 Sf2 S12 C
SaC2
c= +
ê. sphereparams ê.
2
DC1 DC2
Ln DC2
params1 ê. params2;
mvalue =
nvalue =
I-bm + SqrtAbm 2 - 4 cEM ë 2 ;
I-bn + SqrtAbn 2 - 4 cEM ë 2 ;
S1 value =
S1 soln ê. 8m -> mvalue < ê. sphereparams ê. params2;
S2 value = S2 soln ê. 8n Ø nvalue < ê. sphereparams ê.
params2;
S3 value = S3 soln ê. sphereparams ê. params2 ê. params1;
muSparams =
8m -> mvalue , n -> nvalue , S1 -> S1 value , S2 -> S2 value ,
S3 -> S3 value , k1 -> k1 value , k2 -> k2 value <;
rCvalue = rCsoln ê. sphereparams ê. params2 ê.
params1 ê. muSparams;
rFvalue = rFsoln ê. sphereparams ê. params2 ê.
params1 ê. muSparams;
rGvalue = rGsoln ê. sphereparams ê. params2 ê.
params1 ê. muSparams;
solnparams = 8m -> mvalue , n -> nvalue ,
S1 -> S1 value , S2 -> S2 value , S3 -> S3 value ,
k1 -> k1 value , k2 -> k2 value <;
rparams = 8rC -> rCvalue , rF -> rFvalue , rG -> rGvalue < ê.
solnparams;
Print@" m = ", mvalue , " n = ", nvalue D
Print@" S1 = ", S1 value ,
" S2 = ", S2 value , " S3 = ", S3 value D
Flux_Profile_2group.nb
PrintA" k1 = ", k1 value ,
" cm-1 k2 = ", k2 value , " cm-1 "E
Print@" rC = ", rCvalue , " rF = ",
rFvalue , " rG = ", rGvalue D
m = 0.11126 n = 0.708782
S1 = 0.246823
S2 = -4.33564 S3 = 2.27093
k1 = 0.179601 cm-1
k2 = 0.36697 cm-1
rC = -9.30729 µ 10-8 rF = 25.9892 rG = -1763.24
Test the solution by examining the matching conditions at the reactor radius, R:
25
26
Flux_Profile_2group.nb
uC1 = uC1 @RD ê. params1 ê. sphereparams ê. Dparams ê.
solnparams ê. rparams;
uR1 = uR1 @RD ê. params1 ê. sphereparams ê. Dparams ê.
solnparams ê. rparams;
Print@" uC1 @RD = ", uC1, " uR1 @RD = ", uR1D
uC2 = uC2 @RD ê. params1 ê. params2 ê. sphereparams ê.
Dparams ê. solnparams ê. rparams;
uR2 = uR2 @RD ê. params1 ê. params2 ê. sphereparams ê.
Dparams ê. solnparams ê. rparams;
Print@" uC2 @RD = ", uC2, " uR2 @RD = ", uR2D
jC1 = DC1 D@uC1 @rD ê r, rD ê. r Ø R ê. params1 ê.
sphereparams ê. Dparams ê.
solnparams ê. rparams;
jR1 = DR1 D@uR1 @rD ê r, rD ê. r Ø R ê. params1 ê.
sphereparams ê. Dparams ê.
solnparams ê. rparams;
Print@" JC1 @RD = ", jC1, " JR1 @RD = ", jR1D
jC2 =
DC2 D@uC2 @rD ê r, rD ê. r Ø R ê. params1 ê. params2 ê.
sphereparams ê. Dparams ê.
solnparams ê. rparams;
jR2 = DR2 D@uR2 @rD ê r, rD ê. r Ø R ê. params1 ê.
params2 ê. sphereparams ê.
Dparams ê. solnparams ê. rparams;
Print@" JC2 @RD = ", jC2, " JR2 @RD = ", jR2D
uC1 @RD = 0.770309
uC2 @RD = 0.418968
uR1 @RD = 0.770309
uR2 @RD = 0.418968
JC1 @RD = -0.00954537
JC2 @RD = 0.00113739
JR1 @RD = -0.00954537
JR2 @RD = 0.00113739
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, rF , for the two cases:
Flux_Profile_2group.nb
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, rF , for the two cases:
Rnew = R ê.
FindRoot@HrFsoln - rFsoln2 L ê. params1 ê. params2 ê.
Dparams ê. solnparams, 8R, Rvalue <D;
Print@" Self-consistent R = ", Rnew , " 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").
27
28
Flux_Profile_2group.nb
F1plot@r_D =
If@r < Rvalue , HuC1 @rD ê rL ê. solnparams ê. rparams,
HuR1 @rD ê rL ê. solnparams ê. rparamsD;
F2plot@r_D = If@r < Rvalue ,
HuC2 @rD ê rL ê. solnparams ê. rparams,
HuR2 @rD ê rL ê. solnparams ê. rparamsD;
Plot@8F1plot@rD, F2plot@rD<, 8r, .01, 1.5 Rvalue <D
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