A New Analytical SN Solution in Slab GeometryA New Analytical SN Solution in Slab Geometry Dean...

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.

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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

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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

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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

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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

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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

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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

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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/

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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/

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