A A New An Analytical S N So Solut ution i n in Sl n Slab Ge b - PowerPoint PPT Presentation

A A New An Analytical S N So Solut ution i n in Sl n Slab Ge b Geome metry y Dean Wang, Tseelmaa Byambaakhuu University 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 S N 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 Statement Find the solution of the monoenergetic S N equation in slab geometry: L 𝛎 𝑒 𝑒𝑦 𝛀 + Σ ' 𝛀 = Σ ) 2 𝐗𝛀 + 𝑅 2 1 where 𝛀 = 𝜔 / 𝜔 0 … 𝜔 2 𝑼 , angular flux vector; 𝛎 = 𝝂 −𝝂 , 𝑂×𝑂 matrix consisting of Gauss-Legendre quadrature direction cosine values, and 2 𝝂 = diag(𝜈 > ) > 0 , 𝑜 = 1, … , 0 𝐗 = 𝒙 𝒙 𝒙 , 𝑂×𝑂 matrix consisting of Gauss-Legendre quadrature weights, and in which 𝒙 𝑥 / 𝑥 0 … 𝑥 F G 𝑥 / 𝑥 0 … 𝑥 F F 2 2 G 0 matrix, and ∑ >K/ 1 𝑼 ; 𝒙 = 0 × G 𝑥 > = 1 ; 𝟐 = 1 , 1 … ⋮ ⋮ ⋱ 𝑥 F G 𝑥 / 𝑥 0 … 𝑥 F G Σ ' , total macroscopic cross section; Σ M , macroscopic scattering cross section; 3 𝑅 , constant neutron source.

Solution 𝑒𝑦 𝛀 + Σ ' 𝛎 N/ 𝐉 − c 𝑒 2 𝐗 𝛀 = 𝐫 where 𝑑 = S T S U , scattering ratio 𝐫 = V 0 𝛎 N𝟐 𝟐 Matrix eigen decomposition: Σ ' 𝛎 N/ 𝐉 − c 2 𝐗 = 𝐒𝚳𝐒 N/ where 𝚳 = 𝚳 Y 𝚳 N , and in which 2 𝚳 Y = diag(𝜇 > ) , 𝑜 = 1, … 0 ; and 𝑜 = 2 𝚳 N = diag(𝜇 > ) , 0 , … 𝑂 4

Solution 𝑒 𝑒𝑦 𝐒 N/ 𝛀 + 𝚳𝐒 N/ 𝛀 = 𝐒 N/ 𝐫 𝑧 / 𝑧 0 = 𝐒 N𝟐 𝛀 , and 𝐜 = 𝐒 N𝟐 𝐫 , we have Let 𝕑 = ⋮ 𝑧 2 𝑒 𝑒𝑦 𝕑 + 𝚳𝕑 = 𝐜 Integrating gives the analytical solution: 𝕑 = 𝚳 N/ 𝐜 − e N_𝚳 𝒃 where 𝒃 = 𝑏 / 𝑏 0 … 𝑏 2 𝑼 5

Solution N/ 𝐜 Y − e N_𝚳 d 𝒃 Y 𝕑 Y 𝕑 N = 𝚳 Y N/ 𝐜 N − e N_𝚳 e 𝒃 N 𝚳 N where 𝒃 Y 𝒃 N can be determined by the boundary conditions at 𝑦 = 0 and 𝑀 : f , N/ 𝐜 Y − 𝕑 Y 𝒃 Y = 𝚳 Y 𝑦 = 0 h , 𝒃 N = e g𝚳 e 𝚳 N N/ 𝐜 N − e g𝚳 e 𝕑 N 𝑦 = 𝑀 , f where 𝕑 Y h can be determined by the following equation: 𝕑 N 𝟏 f h 𝛀 Y 𝟏 𝐒 𝕑 Y 𝐉 𝐒 𝕑 Y 𝐌 = 𝐉 f + 𝟏 h 𝕑 N 𝕑 N 𝛀 N After some algebra: N/ 𝛀 Y N/ 𝐒 𝟐𝟑 𝐉 − e g𝚳 e f 𝐒 𝟐𝟑 e g𝚳 e 𝟏 𝐒 𝟐𝟑 e g𝚳 e N/ 𝐜 Y 𝕑 Y 𝐒 𝟐𝟐 𝐒 𝟐𝟐 𝚳 Y = 𝐌 − × h 𝐒 𝟑𝟐 e Ng𝚳 d 𝐒 𝟑𝟐 e Ng𝚳 d 𝐒 𝟑𝟐 𝐉 − e Ng𝚳 d N/ 𝐜 N 𝕑 N 𝐒 𝟑𝟑 𝛀 N 𝐒 𝟑𝟑 𝚳 N 6

Solution 𝛀 = 𝛀 Y 𝛀 N = 𝐒 𝕑 Y 𝕑 N f 𝕑 Y = 𝐒 l 𝐉 − e N_𝚳 d + e N_𝚳 d V 𝚳 N𝟐 𝐒 N𝟐 0 𝛎 N𝟐 𝟐 m h e gN_ 𝚳 e e gN_ 𝚳 e 𝕑 N N/ 𝛎 N𝟐 𝟐 = 𝐒𝚳𝐒 N𝟐 𝟐 n / 𝟐 = 𝐒𝚳𝐒 N𝟐 S U 𝐉 − 0 𝐗 S U /No 𝟐 7

Final Solution f S U /No 𝟐 − 𝐒 𝕑 Y S U /No 𝟐 − 𝐒 e N_𝚳 d V / V / e gN_ 𝚳 e 𝐒 N𝟐 × 𝛀 = h 𝕑 N 0 0 Particular Solution Homogenous Solution f S U /No 𝟐 − 𝐒 𝕑 Y S U /No − 𝕏 r 𝐒 e N_𝚳 d V V / Φ = 𝕏 𝑼 𝛀 = e gN_ 𝚳 e 𝐒 N𝟐 × h 𝕑 N 0 Remark: V / t Diffusion limit: 𝛀 ≈ S U /No 𝟐 = 0 𝟐 , as Σ ' → ∞ • 0 f 𝟏 𝛀 ≈ 𝐒 𝕑 Y = 𝛀 Y Thin limit: 𝑴 , as Σ ' → 0 • h 𝕑 N 𝛀 N 8

Eigen Decomposition 𝐵 ≡ Σ ' 𝛎 N/ 𝐉 − c 2 𝐗 = 𝐒𝚳𝐒 N/ Conditioning of Eigenvalues : 𝑣 0 𝑥 0 for Matlab “eig” function Cond 𝜇 = ~1 𝑣, 𝑥 where 𝑣 and 𝑥 are the right and left eigenvectors associated with 𝜇 . Conditioning of Eigenvectors: Cond 𝑣 = 𝑇 𝜇 𝐽 − 𝑄 0 where 𝑇 𝜇 is the reduced resolvent of 𝐵 at 𝜇 , and 𝑄 is the spectral projector associated with 𝜇 . Saad 2011 9

S N Angular Convergence Gauss-Legendre Quadrature Σ • = 1 cm N/ and L = 1 cm 1.0E-01 1.0E-03 1.0E-05 L1 Error 1.0E-07 c = 0 1.0E-09 c = 0.4 𝜁 = 0.98𝑂 N0./ˆ 1.0E-11 c = 0.8 1.0E-13 c = 0.99 1.0E-15 1 10 100 1000 10000 100000 N Σ • = 1 cm N/ and c = 0.8 𝑑 = 0.8 and L = 1 cm 1.0E-01 1.0E-01 sigma_t = 1 cm^-1 1.0E-02 1.0E-02 L = 1 cm sigma_t = 5 cm^-1 1.0E-03 1.0E-03 L = 5 cm 1.0E-04 1.0E-04 sigma_t = 10 cm^-1 1.0E-05 1.0E-05 L = 10 cm L1 Error L1 Error 1.0E-06 1.0E-06 1.0E-07 1.0E-07 1.0E-08 1.0E-08 1.0E-09 1.0E-09 1.0E-10 1.0E-10 1.0E-11 1.0E-11 1 10 100 1000 10000 100000 1 10 100 1000 10000 100000 N N Reference: 𝑂 = 2 /‰ = 16394 10

SC Spatial Error – S10 1D Slab 10 cm Σ ' = 2 cm N/ , 𝑅 = 1 cm N• s N/ c = 0.8 1.0E-02 c = 0.2 1.0E-04 c = 0.001 ~𝑃 ℎ 0 1.0E-06 c = 0 1.0E-08 L1 Error 1.0E-10 1.0E-12 ~𝑃 ℎ N/ 1.0E-14 1.0E-16 0.000001 0.00001 0.0001 0.001 0.01 0.1 Mesh Size (cm) 11

Inhomogeneous Case 1.2 Analytical DD 0.8 SC Scalar Flux Σ ' = 50 cm N/ 𝑑 = 0.6 0.4 𝑀2 = 4 cm ℎ = 0.1 cm 𝑅 = 1 cm N• s N/ Σ ' = 2 cm N/ Σ ' = 2 cm N/ 0 𝑑 = 0.6 𝑑 = 0.6 𝑀3 = 2 cm 𝑀1 = 2 cm ℎ = 0.1 cm ℎ = 0.1 cm 𝑅 = 1 cm N• s N/ 𝑅 = 1 cm N• s N/ -0.4 0 10 20 30 40 50 60 70 80 Mesh Points 12