Sliding Interfaces for Eddy Current Simulations Raffael Casagrande - PowerPoint PPT Presentation

Sliding Interfaces for Eddy Current Simulations Raffael Casagrande Supervisor: Prof. Dr. Ralf Hiptmair Seminar of Applied Mathematics ETH Zürich April 17th, 2013 Raffael Casagrande (ETH Zürich) Sliding Interfaces for Eddy Current April 17th, 2013 1 / 25

Outline Introduction 1 Motivation Deriving the eddy current model 2 Maxwell’s Equations in a moving frame The eddy current model in a moving frame Discontinuous Galerkin Formulation 3 DG Theory Aspects of the implementation Results and Conclusion 4 Raffael Casagrande (ETH Zürich) Sliding Interfaces for Eddy Current April 17th, 2013 2 / 25

Motivation Generator circuit breakers ◮ translational motion Electric engines ◮ rotation Raffael Casagrande (ETH Zürich) Sliding Interfaces for Eddy Current April 17th, 2013 3 / 25

Motivation Generator circuit breakers ◮ translational motion Electric engines ◮ rotation Raffael Casagrande (ETH Zürich) Sliding Interfaces for Eddy Current April 17th, 2013 3 / 25

Outline Introduction 1 Motivation Deriving the eddy current model 2 Maxwell’s Equations in a moving frame The eddy current model in a moving frame Discontinuous Galerkin Formulation 3 DG Theory Aspects of the implementation Results and Conclusion 4 Raffael Casagrande (ETH Zürich) Sliding Interfaces for Eddy Current April 17th, 2013 4 / 25

Outline Introduction 1 Motivation Deriving the eddy current model 2 Maxwell’s Equations in a moving frame The eddy current model in a moving frame Discontinuous Galerkin Formulation 3 DG Theory Aspects of the implementation Results and Conclusion 4 Raffael Casagrande (ETH Zürich) Sliding Interfaces for Eddy Current April 17th, 2013 5 / 25

Maxwell’s Equations curl E + ∂ B div B = 0 ∂ t = 0 div E = ρ ∂ E curl B − 1 ∂ t = µ 0 ( j f + j i ) . c 2 ε 0  Quasistatic model for ⇓ j f = σ E , c → ∞  slowly varying Electric fields (High conductivities)  Eddy Current Model curl E + ∂ B div B = 0 ∂ t = 0 div E = ρ � σ E + j i � curl B = µ 0 ε 0 Raffael Casagrande (ETH Zürich) Sliding Interfaces for Eddy Current April 17th, 2013 6 / 25

Maxwell’s Equations curl E + ∂ B div B = 0 ∂ t = 0 div E = ρ ∂ E curl B − 1 ∂ t = µ 0 ( j f + j i ) . c 2 ε 0  Quasistatic model for ⇓ j f = σ E , c → ∞  slowly varying Electric fields (High conductivities)  Eddy Current Model curl E + ∂ B div B = 0 ∂ t = 0 div E = ρ � σ E + j i � curl B = µ 0 ε 0 Raffael Casagrande (ETH Zürich) Sliding Interfaces for Eddy Current April 17th, 2013 6 / 25

Maxwells equations are invariant under Lorentz transformation if E and B transform as E = γ ( E + V × B ) − ( γ − 1 )( E · ˆ ˜ V )ˆ V � B − V × E � ˜ − ( γ − 1 )( B · ˆ V )ˆ B = γ V c 2 1 ˆ V = ˆ γ := V / | ˆ V | � 1 − v 2 / c 2 ⇓ c → ∞ ˜ E = E + V × B ˜ B = B It can be shown that the eddy current model is also invariant under Rotation !!! Raffael Casagrande (ETH Zürich) Sliding Interfaces for Eddy Current April 17th, 2013 7 / 25

Maxwells equations are invariant under Lorentz transformation if E and B transform as E = γ ( E + V × B ) − ( γ − 1 )( E · ˆ ˜ V )ˆ V � B − V × E � ˜ − ( γ − 1 )( B · ˆ V )ˆ B = γ V c 2 1 ˆ V = ˆ γ := V / | ˆ V | � 1 − v 2 / c 2 ⇓ c → ∞ ˜ E = E + V × B ˜ B = B It can be shown that the eddy current model is also invariant under Rotation !!! Raffael Casagrande (ETH Zürich) Sliding Interfaces for Eddy Current April 17th, 2013 7 / 25

Two eddy current formulations Temporal gauged Potential formulation : µ curl A + σ∂ A curl 1 ∂ t = j i A ( t = 0 ) = 0 curl A × n = 0 on ∂ Ω H -formulation : σ curl H + µ∂ H curl 1 ∂ t = curl 1 σ j i H ( t = 0 ) = 0 H = 0 on ∂ Ω Raffael Casagrande (ETH Zürich) Sliding Interfaces for Eddy Current April 17th, 2013 8 / 25

Two eddy current formulations Temporal gauged Potential formulation (Rest frame): µ curl A + σ∂ A curl 1 ∂ t = j i + σ V × curl A A ( t = 0 ) = 0 curl A × n = 0 on ∂ Ω H -formulation (Rest frame): σ curl H + µ∂ H curl 1 ∂ t = curl 1 σ j i + curl ( µ V × H ) H ( t = 0 ) = 0 H = 0 on ∂ Ω Raffael Casagrande (ETH Zürich) Sliding Interfaces for Eddy Current April 17th, 2013 8 / 25

Two eddy current formulations Temporal gauged Potential formulation (Moving frame) : A + σ∂ ˜ curl 1 A ˜ curl ˜ ˜ ∂ t = ˜ j i µ ˜ A ( t = 0 ) = 0 curl ˜ ˜ A × n = 0 on ∂ Ω H -formulation (Moving frame): H + µ∂ ˜ H curl 1 curl 1 ˜ curl ˜ ˜ ˜ ˜ j i ∂ t = σ σ ˜ H ( t = 0 ) = 0 ˜ H = 0 on ∂ Ω Note: If j i is smooth enough, 1 µ curl A = H ⇒ Do the same simulation and compare the two models (Primal & Dual formulation). Raffael Casagrande (ETH Zürich) Sliding Interfaces for Eddy Current April 17th, 2013 8 / 25

Transformation laws The coordinates of the moving frame ( ˜ x ) are related to the rest frame ( x ) by x = T ( t )˜ x + r ( t ) . T : Rotation matrix. Transformation laws T ˜ T ˜ E = E + V × B B = B T ˜ j i = j i T ˜ H = H T ˜ j f = j f T ˜ V = − V � t T T grad ( V · A ) T ˜ A = A − T 0 ⇒ Use transformation laws to derive transmission conditions at sliding interface. Raffael Casagrande (ETH Zürich) Sliding Interfaces for Eddy Current April 17th, 2013 9 / 25

Transformation laws The coordinates of the moving frame ( ˜ x ) are related to the rest frame ( x ) by x = T ( t )˜ x + r ( t ) . T : Rotation matrix. Transformation laws T ˜ T ˜ E = E + V × B B = B T ˜ j i = j i T ˜ H = H T ˜ j f = j f T ˜ V = − V � t T T grad ( V · A ) T ˜ A = A − T 0 ⇒ Use transformation laws to derive transmission conditions at sliding interface. Raffael Casagrande (ETH Zürich) Sliding Interfaces for Eddy Current April 17th, 2013 9 / 25

Transformation laws The coordinates of the moving frame ( ˜ x ) are related to the rest frame ( x ) by x = T ( t )˜ x + r ( t ) . T : Rotation matrix. Transformation laws T ˜ T ˜ E = E + V × B B = B T ˜ j i = j i T ˜ H = H T ˜ j f = j f T ˜ V = − V � t T T grad ( V · A ) T ˜ A = A − T 0 ⇒ Use transformation laws to derive transmission conditions at sliding interface. Raffael Casagrande (ETH Zürich) Sliding Interfaces for Eddy Current April 17th, 2013 9 / 25

Outline Introduction 1 Motivation Deriving the eddy current model 2 Maxwell’s Equations in a moving frame The eddy current model in a moving frame Discontinuous Galerkin Formulation 3 DG Theory Aspects of the implementation Results and Conclusion 4 Raffael Casagrande (ETH Zürich) Sliding Interfaces for Eddy Current April 17th, 2013 10 / 25

DG Formulation of the Eddy Current Model σ∂ A ∂ t + curl 1 µ curl A = j i curl A × n = 0 on ∂ Ω DG Variational formulation Find A ( i ) h ∈ V h , i = 1 , . . . , N such that for all A ′ h ∈ V h , we have � σ A ( i + 1 ) − A ( i ) � � � ( A ( i + 1 ) h h , A ′ + a SWIP , A ′ j i , ( i + 1 ) , A ′ h ) = h h h h δ t � 3 , P k � ∀ T ∈ T h , v | t ∈ P k � � P k � v ∈ L 2 (Ω) � Where V h := 3 ( T h ) d ( T h ) := d ( T ) . Raffael Casagrande (ETH Zürich) Sliding Interfaces for Eddy Current April 17th, 2013 11 / 25

Symmetric-Weighted-Interior-Penalty Bilinear form a SWIP h � 1 a SWIP ( A h , A ′ µ curl h A h · curl h A ′ h ) = h h Ω � 1 � � � A ′ � � − · µ curl h A h h T F ω F ∈F i h � 1 � � � µ curl h A ′ − · [ A h ] T h F ω F ∈F i h ηγ µ, F � � � A ′ � + [ A h ] T · h T h F F F ∈F i h { A h } ω = ω 1 A h , 1 + ω 2 A h , 2 , [ A h ] T = n F × ( A h , 1 − A h , 2 ) (1) µ 1 µ 2 2 ω 1 = , ω 2 = , γ µ, F = (2) µ 1 + µ 2 µ 1 + µ 2 µ 1 + µ 2 Raffael Casagrande (ETH Zürich) Sliding Interfaces for Eddy Current April 17th, 2013 12 / 25

Convergence Under regularity conditions on the mesh sequence (matching) and assuming the exact solution A is smooth enough we can prove � 1 / 2 � N √ σ ( A ( N ) − A ( N ) 2 � � � � A ( i ) − A ( i ) � � ) L 2 (Ω) + C stab δ t ≤ � � � � h h � � � SWIP i = 1 C 1 h k + C 2 δ t 1 / 2 � � Ct F � � � ∂ 2 A ( t ) where C 1 = max t ∈ [ 0 , t F ] | A ( t ) | H k + 1 (Ω) and C 2 = max t ∈ [ 0 , t F ] L 2 (Ω) The � � ∂ t 2 � constants C 1 , C 2 and C are independent of h and δ t . 1 / 2   2 � � γ mu , F 1 � � [ A ] T � 2 � � | A | SWIP := + √ µ curl h A � �  L 2 ( F )  h F � � L 2 (Ω) F ∈F h Raffael Casagrande (ETH Zürich) Sliding Interfaces for Eddy Current April 17th, 2013 13 / 25