EJ2210 - Analysis of Electrical Machines

FEM-Based Analysis of Induction Machines

Oskar WallmarkAssociate Professor, PhD

Laboratory of Electrical Energy Conversion

Royal Institute of Technology

Stockholm, Sweden

May 4, 2012

◮ Many of you have previous experience with FEM.

◮ Which of Maxwell’s equations are solved for in thisapplication?

◮ How are magnetic field lines determined?

◮ How is the flux linkage (and voltage) in the coils computed?

◮ How can torque be computed?

◮ Why is the mesh density so high in the air gap?

◮ Isn’t the global accuracy of the obtained solution disturbed ifthe mesh density is high only in the air gap?

Review of Maxwell’s Equations

Basic quantities:

◮ Electric field strength E=Ex x + Ey y + Ez z [V/m].

◮ Magnetic field strength H [A/m].

◮ Electric flux density D [C/m2].

◮ Magnetic flux density B [Vs/m2], [T].

◮ Current density J [A/m2].

◮ Electric charge density ρ [C/m3].

◮ Note that E, H, D, B and J are vector fields and ρ is a scalarfield.

Review of Maxwell’s Equations

And X 1 said:

∇× E = −∂B

∂t(Faraday’s law)

∇×H = J+∂D

∂t≈ J (Ampere’s law, quasi static form)

∇ ·D = ρ (Gauss’ law for electric fields)

∇ ·B = 0 (Gauss’ law for magnetic fields).

◮ The current density represented by ∂D/∂t can be neglectedbelow radio frequencies (D=εE and ε is “very small,”ε0=8.85 · 10−12 As/Vm).

◮ Four coupled partial differential equations (PDEs).

◮ For our application, is it possible to reduce the above to onePDE?

1Insert suitable object of worship.

Magnetic Vector Potential

◮ Introduce the magnetic vector potential defined as B=∇×A.

◮ A is not uniquely defined from B=∇× A, but is if we alsodefine ∇ ·A. We choose ∇ ·A=0 for simplicity.

◮ From Faraday’s law, we have (changed order of ∇× and ∂/∂t)

∇× E = −∇×∂A

∂t. (1)

◮ From (1), we have (since ∇×∇V ≡0)

E = −∂A

∂t−∇V (2)

where V is the reduced electric scalar potential.

Magnetic Vector Potential

◮ Ohm’s law tells us that J=σE, where σ is the electrical

conductivity [S/m].

◮ B and H are related as B=µH, where µ is the permeability

[Vs/Am].

◮ Express Ampere’s law (quasi static) using A yields

∇×

(1

µ∇× A

)

= J (3)

∇×

(1

µ∇× A

)

+ σ∂A

∂t+ σ∇V = 0 (4)

where (3) is solved in regions where J is defined (stator slots)and areas “without” current densities (laminations) and (4) issolved in regions where eddy currents may be induced (rotorbars).

Magnetic Vector Potential

◮ In an electric machine, axial effects can often be neglected,i.e, B=Bx x + By y and J=Jz z is now assumed (a twodimensional problem).

◮ Since ∇×

(1

µ∇× A

)

=J and J=Jz z ⇒ A=Az z.

◮ To summarize: B has only x- and y -components and J and A

have only a z-component.

Magnetic Vector Potential

◮ How is the term σ∇V in (4) determined?

◮ The two first components on the right-hand side of (4)contain only z-components. Hence, ∇V contains only az-component, i.e.,

∇V =∂V

∂xx +

∂V

∂yy +

∂V

∂zz =

∂V

∂zz.

V could be a linear function of z (V =C0 + C1z), but sincethis only yields an additional constant ∇V =C1z, V ≡0 ischosen in our application.

◮ A non-zero V can be used to force constraints on the currentin conductive regions. Thereby, a connection to a circuit withvoltage sources as input can be realized.

Problem Formulation

◮ Let’s review the equations to solve again (two dimensionalcase)

∇×

(1

µ∇× Az z

)

= Jz z (5)

∇×

(1

µ∇× Az z

)

+ σ∂Az z

∂t= 0. (6)

◮ Eqs. (5) and (6) are solved in FEM-based softwares such asFlux, JMAG, Opera, Comsol (including three-dimensionalformulations).

◮ Note that only (5) has to be solved if eddy currents areneglected ⇒ A PM motor can be analyzed using only (5).

Drawing Flux Lines

◮ The FEM solver solves for A and B is computed fromB=∇×A. In the two-dimensional case

B = Bx x + By y = ∇× Az z =∂Az

∂yx −

∂Az

∂xy .

◮ Flux lines are lines to which the flux density is parallel.

◮ If Az is constant along the x-direction ⇒ Bx=∂Az/∂y andBy =−∂Az/∂x = 0.

◮ If Az is constant along the y -direction ⇒ Bx =∂Az/∂y=0and By =−∂Az/∂x .

◮ Same argument holds for any direction in which Az is constant⇒ Flux lines can be defined as lines on which Az is constant.

Boundary Conditions

◮ Proper boundary conditions need to be added to solve (5) and(6).

◮ Setting Az=0 on the outer stator periphery will force the fluxlines to be parallel along this boundary (i.e., the flux density isconfined within the machine).

◮ This boundary condition is called a homogenous Dirichletboundary condition.

◮ If a complete machine is simulated, the boundary conditionabove is sufficient.

Boundary Conditions

◮ If only a portion of the machine is simulated, proper boundaryconditions need to be added.

◮ The (homogenous) Neumann boundary condition

∂Az

∂n= 0

forces the flux lines to cross the boundary in parallel to thenormal direction.

◮ Antiperiodic boundary conditions such as

Az(r , ϕ) = −Az

(

r , ϕ+(2k − 1)π

p/2

)

, k = 1, 2, 3, . . . , p/2

enable the study of an odd number of poles (including a singlepole).

Computing Flux and Flux Linkage

◮ For an electric machine with active length La, the flux φbetween points (x1, y1) and (x2, y2) can be computed as

φ =

∫

S

B · ndS =

∫

S

(∇× A) · ndS

=

∮

C

A · d r = La (Az(x1, y1)− Az(x2, y2)) .

◮ For massive conductors with cross sections SCu, the flux canbe computed using the average value Az on the conductorsurface

φ =La

SCu

(∫

S+Cu

AzdS −

∫

S−

Cu

AzdS

)

.

◮ Flux linkage ψ simply obtained by multiplying φ by thenumber of turns.

◮ Induced voltage v=dψ/dt.

Computing Torque

◮ How can torque be computed?

◮ Let’s start with the well known Lorenz force law (force oncharge q traveling with the velocity v

F = q (E+ (v × B)) .

◮ Lorenz force law is an empirical statement and can be used todefine E and B.

◮ Lorenz force law in differential form, assuming vacuum µ=µ0and neglecting the contribution from E

dF = fdV = J× BdV =

(1

µ0∇× B

)

× B

︸ ︷︷ ︸

= f

dV .

Computing Torque

◮ Hence, f=

(1

µ0∇× B

)

× B. The x-component of f can be

expressed as (assuming Bz=0)

fx =1

µ0

(

−By∂By

∂x+ By

∂Bx

∂y

)

.

◮ Add and subtract the term (1/µ0)Bx∂Bx/∂x and use theidentity ∂(B2

x )/∂x=2Bx∂Bx/∂x yields

fx =1

µ0

1

2

∂

∂x

(B2x

)+ By

∂Bx

∂y−

1

2

∂

∂x(B2

x + B2y

︸ ︷︷ ︸

= B2

)

.

Computing Torque

◮ Further manipulation yields

fx =1

µ0

∂

∂x

(

B2x −

B2

2

)

+∂(BxBy )

∂y− Bx ∇ ·B

︸ ︷︷ ︸

= 0

=1

µ0

(∂

∂x

(

B2x −

B2

2

)

+∂(BxBy)

∂y

)

.

◮ Now, fx can be expressed as fx =∇ · Sx where

Sx =1

µ0

((B2x − B2/2

)x + BxBy y

).

◮ Similarly, fy can be expressed as fy =∇ · Sy where

Sy =1

µ0

(BxBy x +

(B2y − B2/2

)y).

Computing Torque

◮ We have

f = fx x + fy y = ∇ · Sx x +∇ · Sy y =

[∇ · Sx

∇ · Sy

]

. (7)

◮ Eq. (7) can be expressed as

f = ∇ · S where S =1

µ0

[B2x − B2/2 BxBy

ByBx B2y − B2/2

]

. (8)

◮ The matrix S is called Maxwell’s stress tensor (assuming atwo-dimensional geometry and without electrical fields).

◮ Note that ∇ · S is a vector.

Computing Torque

◮ What is the point of deriving the expression f = ∇ · S?

◮ Answer: Because we can now use the divergence theorem

F =

∫

df =

∫

V

fdV =

∫

V

∇ · SdV =

∮

S

S · ds. (9)

◮ From (9), we have

df=S · ds. (10)

◮ From (10), the differential torque dT can be expressed as

dT = rdfϕ (11)

where dfϕ is the tangential component of df, i.e., dfϕ=df · ϕ.

Computing Torque

◮ Let’s find dT expressed using cylindrical coordinatesB=Br r + Bϕϕ

dT = rdfϕ = rdf · ϕ = r (S · ds) · ϕ. (12)

◮ We have ds=Lardϕr=Lardϕ(sinϕx + cosϕy). Also,ϕ=− sinϕx + cosϕy .

◮ We can also express Bx and By (in the matrix S) in terms ofBr and Bϕ according to

Bx = Br cosϕ− Bϕ sinϕ

By = Br sinϕ+ Bϕ cosϕ.

◮ Insert the expressions above in (12) yields

T =

∫

dT = Phew! =Lar

2

µ0

∫ 2π

0

BrBϕdϕ.

Computing Torque

◮ Let’s recapitulate:

T =Lar

2

µ0

∫ 2π

0

BrBϕdϕ.

◮ The integral should be evaluated in the air gap.

◮ Both Br and Bϕ are required in the computation.

◮ Br and Bϕ are obtained from B=∇× A.

◮ A is solved for in each node. Between the nodes, linearapproximations are assumed.

◮ Hence, to calculate derivatives of A (to obtain components ofB), a suitably high mesh density is needed.

◮ Since the torque contains a product of two components of B,the mesh density in the air gap needs to be even higher.

◮ Other methods for torque computation are also available(virtual work and the flux times current relationship).

Can we now answer the initial questions?

◮ Which of Maxwell’s equations are solved for in thisapplication?-Answer: We combined them and derived a single equation tobe solved involving only Az (the magnetic vector potential).

◮ How are magnetic field lines determined?-Answer: Magnetic field lines are simply contours where Az isconstant.

◮ How is the flux linkage (and voltage) in the coils computed?-Answer: The flux can be obtained simply from the differenceof Az between the two points.

◮ How can torque be computed?-Answer: A common method was presented where the torquewas found by evaluating a derived line integral (obtained fromMaxwell’s stress tensor) in the air gap.

Can we now answer the initial questions?

◮ Why is the mesh density so high in the air gap?-Answer: A product of two components of B is required in thetorque computation.

◮ Isn’t the global accuracy of the obtained solution disturbed ifthe mesh density is high only in the air gap?-Answer: Globally, the mesh density only needs to be so highso that Az can be resolved with sufficient accuracy.

Introduction to FEMM

◮ FEMM is a free finite element based software.

◮ FEMM can be downloaded from www.femm.info

◮ FEMM solves two-dimensional (planar and axial symmetry)magnetostatic, time-harmonic magnetic, electrostatic andstatic heat flow problems.

Introduction to FEMM

◮ We need to solve

∇×

(1

µ∇× Az z

)

= Jz z (13)

∇×

(1

µ∇× Az z

)

+ σ∂Az z

∂t= 0. (14)

◮ FEMM can solve (13) but not (14) in its present form.

Introduction to FEMM

◮ Assume that all fields are varying sinusoidally with a single,fixed frequency ω.

◮ Introduce the jω-phasor notation

A = Az(x , y)z ejωt .

◮ Insert A into (14) yields

∇×

(1

µ∇× Az z

)

+ jωσAz z = 0. (15)

◮ FEMM can solve (15). To obtain Az =Az(x , y , t), we compute

Az = ℜ(Azejωt) = ℜ(Az) cos ωt −ℑ(Az) sinωt. (16)

◮ Note that the output of FEMM (Az , Bx , By , ...) are hencecomplex quantities and time dependence is obtained byapplying (16).

Introduction to FEMM

◮ Caution: By considering only a single stator and rotorfrequency, effects due to harmonics are neglected. This cancause large errors in the predicted torque for large slip values.

◮ For the phasor notation to be mathematically correct, alloperators need to be linear. Therefore, the permeability µshould be constant (magnetic saturation cannot be modeledexactly).

◮ Saturation can be modeled approximately by inserting anequivalent permeability as outlined in, e.g.,N. Bianchi, Electrical Machine Analysis Using Finite Elements,CRC Taylor & Francis, 2005.

◮ Note that if saturation is neglected, care must be taken whenmodeling machines with closed rotor slots.

A Very Brief Introduction to FEM

◮ The equations to solve can be expressed as

Lφ = f .

◮ φ is the unknown function (Az in our case). Fortwo-dimensional problems φ=φ(x , y).

◮ L is the differential operator (=∇×1

µ∇× in our case).

◮ f is the forcing function (=Jz in our case).

Galerkin’s Method

◮ Introduce φ∗ that approximates the exact solution φ.

◮ Now, φ∗ is expressed as

φ∗ =N∑

j=1

Φjνj

where Φj are unknown coefficients (constants) that must bedetermined and νj is a defined set of base functions.

◮ Introduce the residual r=Lφ∗ − f which ideally should bezero.

Galerkin’s Method

◮ Instead of demanding r=0, let’s relax that condition and letthe integral (volume integral over the problem domain V ) ofthe residuals ri , weighted with some weight function wi , bezero instead. Hence,

ri =

∫

V

wi (Lφ∗− f ) dV = 0.

◮ Now, select wi =vi (known as Galerkin’s method) yields

ri =

∫

V

νiL

N∑

j=1

Φjνj

− νi fdV = 0, i = 1, 2, 3, . . . ,N

(17)

Galerkin’s Method

◮ Eq. (17) can be expressed on matrix form as

SΦ = T (18)

where

Sij =

∫

V

νiLνjdV

Φ = [Φ1 Φ2 Φ3 · · · ΦN ]T

Tij =

∫

V

νi fdV .

Galerkin’s Method

◮ We have transformed the original problem (the PDE) to (18)which is a large matrix equation with the unknowns Φi .

◮ The resulting large matrix S is often sparse and symmetrical.

◮ Much research has been focused on developing methods tosolve (18) with high computational efficiency given that S issparse and symmetrical.

◮ Other approaches also exist that transforms the originalproblem into a large matrix equation (e.g. the Rayleigh-Ritz’svariational approach).

The Finite Element Method

◮ Note that we have not defined the base functions νj yet.

◮ In the finite element approach, the problem domain is typicallydivided into a number of triangles (two-dimensional case).

◮ The (unkown) solution in element (triangle) m can beexpressed as

φ∗m(x , y) =

3∑

j=1

Φmjνmj

where Φmj is the value of φ in the jth node of the mthelement.

The Finite Element Method

◮ In the finite element method, the base functions νmj arechosen so that

νmj(x , y) =

{1, in node j

0, in all other nodes.

◮ Between all nodes, linear interpolation is used.

◮ The resulting matrix S becomes sparse and the non-zeroelements are simple to compute ⇒ computational efficiency.

Modeling Non-Linear Materials

◮ In electrical steels, the permeability µ is strongly nonlinear.Hence, µ=µ(B).

◮ We have seen how the finite element approach results in alarge matrix equation.

◮ The nonlinearity is typically solved using the Newton-Raphsonapproach.

◮ As an example, for the nonlinear system F(x)=0, the solutionx is found by the iteration

xk+1 = xk −

(∂F

∂x

)−1

F(xk)

where the partial derivatives are evaluated at x=xk .

◮ The approach above is implemented and the iterationcontinues until the change is smaller than a certain threshold.

What have we learnt?

◮ We have seen how the original PDE can be transformed into alarge matrix equation.

◮ The resulting solution is an approximation of the exactsolution.

◮ We have seen how the finite element approach results in amatrix S that is sparse and where the non-zero elements aresimple to compute.

◮ Much more can be said about the finite element method butwhat mentioned is sufficient for our purposes.