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A New Analytical SNSolution in Slab Geometry
Dean Wang, Tseelmaa ByambaakhuuUniversity of Massachusetts Lowell November 1, 2017
2017 ANS Winter Meeting, Washington DC
Why another solution?
• Previous work: • Chandrasekhar 1960; Vargas 1997; Warsa 2002; Ganapol
2008; Goncalez 2011, …• Solution methods: Separation of variables, Green’s function,
Laplace transfer, and decomposition method.• Our solution techniques:
- Eigen decomposition: a system of coupled SN PDEs is decoupled into a system of separate ODEs.
- Boundary treatment: the left and right incoming angular flux vectors are combined into one single vector.
- Derivation: the whole derivation process is based on linear algebra.
- Solution: a truly closed-form analytical expression.
2
Problem StatementFind the solution of the monoenergetic SN equation in slab geometry:
𝛍𝑑𝑑𝑥𝚿 + Σ'𝚿 =
Σ)2 𝐖𝚿+
𝑄2 1
𝚿 = 𝜓/ 𝜓0 … 𝜓2 𝑻, angular flux vector;
𝛍 =𝝁
−𝝁 , 𝑁×𝑁 matrix consisting of Gauss-Legendre quadrature direction cosine values, and
𝝁 = diag(𝜇>) > 0, 𝑛 = 1,… , 20
𝐖 = 𝒘 𝒘𝒘 𝒘 , 𝑁×𝑁 matrix consisting of Gauss-Legendre quadrature weights, and in which
𝒘 =
𝑤/ 𝑤0 … 𝑤FG
𝑤/ 𝑤0 … 𝑤FG
⋮ ⋮ ⋱ 𝑤FG
𝑤/ 𝑤0 … 𝑤FG
, 20×2
0matrix, and ∑>K/
FG 𝑤> = 1; 𝟏 = 1 1 … 1 𝑻;
Σ', total macroscopic cross section; ΣM, macroscopic scattering cross section;𝑄, constant neutron source.
where
L
3
Solution
𝑑𝑑𝑥𝚿 + Σ'𝛍N/ 𝐈 −
c2𝐖 𝚿 = 𝐪
𝑐 = STSU
, scattering ratio
𝐪 = V0 𝛍
N𝟏𝟏
where
Σ'𝛍N/ 𝐈 −c2𝐖 = 𝐑𝚲𝐑N/
Matrix eigen decomposition:
𝚲 = 𝚲Y𝚲N
, and in which
𝚲Y = diag(𝜆>), 𝑛 = 1,…20; and
𝚲N = diag(𝜆>), 𝑛 = 20,…𝑁
where
4
Solution𝑑𝑑𝑥 𝐑
N/𝚿 + 𝚲𝐑N/𝚿 = 𝐑N/𝐪
Let 𝕐 =
𝑦/𝑦0⋮𝑦2
= 𝐑N𝟏𝚿, and 𝐛 = 𝐑N𝟏𝐪, we have
𝑑𝑑𝑥𝕐 + 𝚲𝕐 = 𝐛
Integrating gives the analytical solution:
𝕐 = 𝚲N/𝐛 − eN_𝚲𝒂
where
𝒂 = 𝑎/ 𝑎0 … 𝑎2 𝑻
5
Solution
where 𝒂Y𝒂N can be determined by the boundary conditions at 𝑥 = 0 and 𝐿:
𝕐Y𝕐N
= 𝚲YN/𝐛Y − eN_𝚲d𝒂Y𝚲NN/𝐛N − eN_𝚲e𝒂N
𝒂Y = 𝚲YN/𝐛Y − 𝕐Yf , 𝑥 = 0
𝒂N = eg𝚲e𝚲NN/𝐛N − eg𝚲e𝕐Nh , 𝑥 = 𝐿,
𝚿Y𝟎
𝚿N𝐋 = 𝐈
𝟎 𝐑 𝕐Yf
𝕐Nf+ 𝟎
𝐈 𝐑𝕐Yh
𝕐Nh
After some algebra:
where 𝕐Yf
𝕐Nhcan be determined by the following equation:
𝕐Yf
𝕐Nh= 𝐑𝟏𝟏 𝐑𝟏𝟐eg𝚲e
𝐑𝟐𝟏eNg𝚲d 𝐑𝟐𝟐
N/𝚿Y𝟎
𝚿N𝐋 − 𝐑𝟏𝟏 𝐑𝟏𝟐eg𝚲e
𝐑𝟐𝟏eNg𝚲d 𝐑𝟐𝟐
N/
×𝐑𝟏𝟐 𝐈 − eg𝚲e
𝐑𝟐𝟏 𝐈 − eNg𝚲d𝚲YN/𝐛Y𝚲NN/𝐛N
6
Solution
𝚿 = 𝚿Y𝚿N
= 𝐑 𝕐Y𝕐N
= 𝐑 l 𝐈 − eN_𝚲de gN_ 𝚲e
𝚲N𝟏𝐑N𝟏 V0𝛍N𝟏𝟏 m+ eN_𝚲d
e gN_ 𝚲e𝕐Yf
𝕐Nh
𝛍N𝟏𝟏 = 𝐑𝚲𝐑N𝟏 𝟏SU
𝐈 − n0𝐖
N/𝟏 = 𝐑𝚲𝐑N𝟏 /
SU /No𝟏
7
Final Solution
Φ = 𝕎𝑻𝚿 = VSU /No
−𝕎r𝐑 eN_𝚲de gN_ 𝚲e
𝐑N𝟏× V0
/SU /No
𝟏 − 𝐑 𝕐Yf
𝕐Nh
Remark:
• Diffusion limit: 𝚿 ≈ V0
/SU /No
𝟏 = t0𝟏, as Σ' → ∞
• Thin limit: 𝚿 ≈ 𝐑 𝕐Yf
𝕐Nh= 𝚿Y
𝟎
𝚿N𝑳 , as Σ' → 0
𝚿 = V0
/SU /No
𝟏 − 𝐑 eN_𝚲de gN_ 𝚲e
𝐑N𝟏× V0
/SU /No
𝟏 − 𝐑 𝕐Yf
𝕐Nh
Particular Solution Homogenous Solution
8
Eigen Decomposition
Conditioning of Eigenvalues:
Cond 𝜆 =𝑢 0 𝑤 0
𝑢, 𝑤 ~1
where 𝑢 and 𝑤 are the right and left eigenvectors associated with 𝜆.
𝐴 ≡ Σ'𝛍N/ 𝐈 −c2𝐖 = 𝐑𝚲𝐑N/
Conditioning of Eigenvectors:
Cond 𝑢 = 𝑆 𝜆 𝐼 − 𝑃 0
where 𝑆 𝜆 is the reduced resolvent of 𝐴 at 𝜆, and 𝑃 is the spectral projector associated with 𝜆.
for Matlab “eig” function
9Saad 2011
SN Angular Convergence
1.0E-15
1.0E-13
1.0E-11
1.0E-09
1.0E-07
1.0E-05
1.0E-03
1.0E-01
1 10 100 1000 10000 100000
L1 E
rror
N
c = 0
c = 0.4
c = 0.8
c = 0.99
Σ� = 1 cmN/ and L = 1 cm
1.0E-111.0E-101.0E-091.0E-081.0E-071.0E-061.0E-051.0E-041.0E-031.0E-021.0E-01
1 10 100 1000 10000 100000
L1 E
rror
N
sigma_t = 1 cm^-1
sigma_t = 5 cm^-1
sigma_t = 10 cm^-1
𝑐 = 0.8 and L = 1 cm
1.0E-111.0E-101.0E-091.0E-081.0E-071.0E-061.0E-051.0E-041.0E-031.0E-021.0E-01
1 10 100 1000 10000 100000
L1 E
rror
N
L = 1 cmL = 5 cmL = 10 cm
Σ� = 1 cmN/ and c = 0.8
𝜀 = 0.98𝑁N0./�
Gauss-Legendre Quadrature
10Reference: 𝑁 = 2/� = 16394
SC Spatial Error – S10 1D Slab
1.0E-16
1.0E-14
1.0E-12
1.0E-10
1.0E-08
1.0E-06
1.0E-04
1.0E-02
0.000001 0.00001 0.0001 0.001 0.01 0.1
L1 E
rror
Mesh Size (cm)
c = 0.8c = 0.2c = 0.001c = 0
~𝑂 ℎN/
~𝑂 ℎ0
10 cm
Σ' = 2 cmN/, 𝑄 = 1 cmN�sN/
11
Inhomogeneous Case
-0.4
0
0.4
0.8
1.2
0 10 20 30 40 50 60 70 80
Scal
ar F
lux
Mesh Points
AnalyticalDDSC
Σ' = 50 cmN/
𝑐 = 0.6𝐿2 = 4 cmℎ = 0.1 cm𝑄 = 1 cmN�sN/
Σ' = 2 cmN/
𝑐 = 0.6𝐿1 = 2 cmℎ = 0.1 cm𝑄 = 1 cmN�sN/
Σ' = 2 cmN/
𝑐 = 0.6𝐿3 = 2 cmℎ = 0.1 cm𝑄 = 1 cmN�sN/
12