Lecture 2. Introduction to FEM. What it is? What we are solving? Potential formulation Why? Boundary conditions

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1 Introduction to FEM What it is? What we are solving? Potential formulation Why? Boundary conditions Lecture 2

2 Notation Typical notation on the course: Bolded quantities = matrices (A) and vectors (a) Unit vector = e z x = position = (x,y) Operators (click for explanatory material) Gradient of scalar function f = x fe x + y fe y + z fe z Divergence of vector function f = f x e x + f y e y + f y e y f = x f x + y f y + z f z Curl of vector function f = e x e y e z x y z f x f y f z

3 Notation Special cases: Curl of z-directional vector f = fe z f = f y e x f x e y Curl of a 2D vector (xy-plane) f(x, y) = f x e x + f y e y f = f y x f x y e z Matrix transpose A T and inverse A 1 Dot product of two vectors a b Good to know: Curl of any gradient is always zero: φ 0 Divergence of any curl is always zero: w 0 Wikipedia: Vector calculus identities

4 Idea of FEM We want to find a function A(x) that satisfies some partial differential equation on some domain (Ω) + some boundary conditions For (a convenient) example ν x A x = J x Known functions of the position x

5 Idea of FEM ν x A x = J x IDEA: write A(x) as a weighted sum of some known functions N(x) and some unknown coefficients a: n A x a j N j x = A(x) j=1 The functions N are typically called shape functions (or trial functions) Compare to Fourier series The problem of finding an unknown function now reduced to finding unknown scalar coefficients

6 Idea of FEM In other words: we want to find the coefficients a n A x = a j N j x j=1 in such a way that the equation ν A J is satisfied as well as possible Please note that the end result still is a function It s simply defined by a set of scalar coefficients Takeaway = we have an approximation for A(x) at each and any x useful later

7 What we are solving Next: where does the div-grad equation come from? ν A = J

8 What we are solving? Five field variables (E, D, H, B, J) are needed to present a complete electromagnetic field. D B 0 B E t D H J t Maxwell s equations D E J E B H Material equations + Boundary conditions are needed

9 Boundary conditions = some info about the solution on the outer boundary E.q. values, derivatives, some combination Have to know what happens here Why we need these? Because of (rather complex) maths = if we leave part of the boundary free, we get infinitely many solutions satistying the fixed part + Maxwell s eqs inside. Maxwell s equations satisfied in here

10 Interface conditions for magnetics Boundaries inside problem domain 1 The tangential component of field strength is continuous H H t1 H t2 Comes from H = 0 (on the thin boundary) B The normal component of flux density is continuous B = μh B n1 B n2 Comes from B = 0 Takeaway: the normal component of H and tangential component of B often have jumps whenever μ does.

11 Reluctivity In electromechanics, the reluctivity function ν is often used instead of the permeability H = νb ν = 1 μ Easier notation

12 Simplifications related to timedependence Dynamic problems: Complete Maxwell s equations needed (wave equation) Quasi-static problems: Displacement current neglected D 0 t Static problems: Induced electric field is also neglected B 0 t

13 Next Introducing the vector potential Definition Properties

14 Vector potential Problem: Even in static case, we have two equations for one unknown υb = J B = 0 Ampere s law Gauss s law Let s define B by a vector potential A(x) B = A Now Gauss s law is satisfied automatically since any vector A satisfies ( cuz maths) A = 0

15 Vector potential By definition, the flux density B is B Bxex Byey A A x, y ez A Bx By y A x

16 Vector potential Flux lines In 2D, the vector potential is constant in the direction of the flux density B This can be seen by computing the directed derivative (how much A changes in direction of B): A A A B = A A = x A y y A x = 0

17 Vector potential Flux lines In other words: the flux lines correspond to the equipotential lines of A = isolines = contour lines = lines at which A is constant Compare to the contours ( height lines ) in a map Values of A Flux lines

18 Vector potential Note: vector potential is not unique in 3D * * Several A give the same B A Af => A Af Af A A gauge condition is often included to ensure uniqueness A = 0 In 2D, it is satisfied automatically no need for concern

19 Vector potential formulation Let s combine the potential with Maxwell s equations: Partial differential equation for the vector potential H J H B => B J => A J B A Nice to know: called curl-curl equation

20 Formulation in 2D Current density and vector potential point in the same direction J J x, y ez A Ax, ye z Unit vector of z-direction

21 By writing open Formulation in 2D υ A e z = Je z we get or simply A A e x z Je x y y z ( A) J There s our original example, wooo!

22 Formulation in 2D ( A) J This is the so-called div-grad equation Poisson s equation if ν is constant Appears in wildly diverse problems Thermal problem Electric potential Magnetic scalar potential Will take a look on these later Boundary conditions have different meanings in different problems

23 Boundary conditions = have to know something about the solution on the outer boundary Three common conditions for the vector potential formulation Dirichlet condition = A is constant = field is parallel to boundary Homogeneous Neumann condition = ν A perpendicular to boundary (alt. notation A n n = 0 = field is = A n) (Anti)periodic boundary = A 1 = ±A 2 on some parts of the boundary = used for modelling symmetry sectors

24 Interface conditions Interfaces (= boundaries wholly inside problem domain) are automatically included 1 H The conditions H t1 = H t2 and B n1 = B n2 are consequence of the Maxwells s equations Since we are approx. solving the equations, these conditions are also approx. satisfied B Will take a closer look later during the course, in the exercises B = μh

25 Example problem Magnetic field in the air gap and slots of a machine; = 0 (constant) J J J Nice to know = since ν is constant, the div-grad equation is reduced to the Poisson s equation ν 0 A = J ν 0 A = J 2 A = μ 0 J

26 Boundary conditions in the example The permeability of iron is much larger than those of the air and conductors On the air-iron interface Air Fe 1 Air 1 Fe Air Fe H 0 t Ht => Bt Bt => Bt Bt 0 0 Fe Fe Flux is almost perpendicular to the iron surface (Homogeneous) Neumann boundary condition ν A n = 0

27 Boundary conditions in the example Solution region can be reduced because of symmetry A 0 2 A 0 n A 0 Neumann condition on the right boundary No flux goes through left boundary. Dirichlet boundary condition A is constant: often A = 0 A n 0 2 A J 0 A n 0

28 Solution by finite element method 1. The problem domain is divided into small triangles called elements shape functions are generated based on these 2. Boundary conditions are applied 3. Computer solves the problem and draws nice pics for ya 4. Profit!

29 Assignment 1 Levitation melting b.gif Simulate this in FEMM Axisymmetric geometry Harmonic = sinusoidal current supply

30 Today we Conclusion 1. Understood the basic idea of FEM 2. Introduced the vector potential 3. and used it to derive the div-grad equation from the Maxwell s equations 4. Learned the most typical boundary conditions