Synchronous machine with PM excitation Two-axis model

Transcription

1 Synchronous machine with PM excitation q Two-axis model q i q u q d i Q d Q D i d N S i D u d

2 Voltage, flux-linkage and motion equations for a PM synchronous machine dd ud Ri s d q dt dq uq Ri s q d dt d R D DD i dt dq RQQ i dt L i L i d dd ddd L i L i D ddd DD L i L i q qq qqq L i L i Q qqq QQ pm pmd 3 J d p di id T 2 p dt d dt q q m

3 Load angle For synchronous machines, the position angle of the rotor is defined differently from the rotor angle used, for instance, for induction machines. d dt is replaced by s d dt where the load angle is the angle between the excitation voltage vector u p (on q-axis) and stator voltage q u p r u s d vector u r s u u d q uˆ sin s uˆ cos s N S

4 PM synchronous machine in a steady state Space vectors are constants in rotor frame of reference r s d q u u ju constant r s d q i i ji constant The time-derivatives of flux linkages vanish d d d q dt dt The currents of the damper windings are zero i D i Q

5 Space-vector diagram for a PM synchronous machine u u Ri d s d q Ri q s q d d q Li dd Li qq pm pm q Li u p Ri s r u s qq r s jli r j s dd r r s s s d d q q p u R i jl i L i u u p juˆ p j pm i d pm r d i ˆ d is sin i ˆ q is cos i q i r s r s

6 Electromagnetic torque pm 3 3 T p i i p L L i i i 2 2 e dq qd d q dq q Neglecting the stator resistance u dussin Li qq u u cos Ldd i uˆ q s p => i i d q u s cos uˆ L d us sin L q p T e 3 L d Lq uˆscos uˆp uˆssin uu ˆpˆssin p L dlq Lq 2 3 p uu ˆsˆp uˆ s 1 1 sin sin 2 2 X 2 d Xq Xd

7 Synchronous reluctance machine Current supply 3 T pl L ii 2 e d q dq Voltage supply T e 3puˆ2 s Xq X d sin2 The effect of stator resistance has been neglected.

8 Multiply-excited magnetic systems Electric system(s) Coupling field Mechanical system Energy balance expressed for a magnetic field system with a rotor N N j f m i e dt i d dw dw dw Td j j j1 j1 where e j, i j and j are the electromotive forces, currents and flux linkages of the exciting circuits. W f is the energy of the electromagnetic field and W m is the mechanical work done on the system. d is an angular displacement and T the electromagnetic torque. f

9 Principle of virtual work I Taking the currents and rotation angle as free variables, we can write i Wf Wf i, j j, where i is a vector containing all the N currents of the system. The total differences are N i, W i, W f f dwf dij d j1 ij N,, j i j i d j dij d j 1 ij Substituting the differences in the energy balance equation

10 N Principle of virtual work II i, i, N j j ik dij d j ij k1 1 N i, W i, W f f dik d Td i k1 k Taking the virtual displacements as common multipliers i, W i, j f i di di i j i k N N N k j k k1 j1 k1 i, W i, N j f ik Td k1

11 Principle of virtual work III and noticing that i and are independent from each other, we get By defining electromagnetic coenergy W c a simple equation is obtained for the electromagnetic torque f 1,, (, ) N j j j W T i i i i c f 1, N j j j W i W i c, (, ) W T i i

12 Principle of virtual work IV Another alternative is to use flux linkages as free variables. This leads to torque equation T(, ) W f, For a detailed derivation of this equations, see for instance [Krause & Wasynczuk]. If the flux linkages are the free variables (state variables) that define the magnetic state of the system, the magnetic energy should be integrated from equation,, dw i d Td f N j1 j This integration is considered in the case of two electric circuits. j

13 System with two electrical inputs Electric system(s) Coupling field Mechanical system,,,,,, T,, d dw i d i d f As the coupling field is assumed to be conservative, we can freely choose the path for integrating the energy from point (,,) to point ( ) in the state space. The easiest way to integrate is to allow only one state variable to change at a time and to divide the integration into three consecutive parts

14 System with two electrical inputs II Electric system(s) Coupling field Mechanical system Starting from the displacement angle, the energy becomes f,,,, W T d f 1f 2f f 2f i,, d 1f f 2 i,, d f f 1 1f f 1 2f 2

15 System with two electrical inputs III The first integral vanishes as the integration path sets the magnetic fluxes to zero (no magnetic forces). The two remaining terms are 2f,,,, W i d f 1f 2f f f 2 1f i,, d f f 1 If the system is magnetically linear, we can use inductances that depend on the rotation angle, only L i M i L i M i

16 System with two electrical inputs IV For integrating the energy expression, the currents have to be solved from the flux-linkage equations i L M ; i M L L L M L L M W L1f2,, d L L M 2f f 1f 2f f 2 1 f 2 f f 1f L 2f1 Mff L L M 1 f 2 f f 2 d 2 1

17 System with two electrical inputs V The energy becomes W L f Mf 2 L f L L M 2 L L M 2 2 f 1f 2 2 2f 1 f 2 f f 1 f 2 f f L L M 1 f 2 f f 1f 2f From which we can calculate the torque T(, ) W f, only the derivatives may become quite cumbersome.

18 Using the currents as the state variables and working with the coenergy would have led to much simpler equations An integration similar to the case of energy, gives coenergy and torque System with two electrical inputs VI,,,, W i i i i W c f 1 2 dw i, i, di di Td c W i, i, L i L i M i i 2 2 c 1f 2f f 1 f 1f 2 f 2f f 1f 2f T,, 2 2 i i Wc i1 i2 1 dl1 2 dl2 dm ii 12 2 d 2 d d i, i 1 2

19 Example: Two-phase reluctance motor The rotor of a fourpole, two-phase cage induction motor has been forced to a somewhat quadratic form. Question: Does this construction work as a synchronous reluctance motor?

20 Measured inductances Inductance [H] Inductance [H] Rotation angle [deg] Rotation angle [deg] Self inductance Mutual inductance Self inductance Mutual inductance Self and mutual inductances as a function of rotor position; excitation current in phase A. Self and mutual inductances as a function of rotor position; excitation current in phase B.

21 Fitting a simple analytical model Inductance [H] Inductance [H] Rotation angle [deg] Rotation angle [deg] Self inductance Mutual inductance Self inductance Mutual inductance Appr. of L Appr. of M Appr. of L Appr. of M sin2 sin 2 L A B p M C p M sin2 sin 2 L A B p M C p M A mh; B mh; 1 4 ; C 25.1 mh; M 8 A mh; B mh; 2 ; C 25.1 mh; M 8

22 Calculating the torque of the machine L 1 A B p 4 L2 ABsin2p M Csin 2p 8 sin 2 A43.3 mh B 15.1 mh C 25.1 mh T i i dl dl dm ii 12 2 d 2 d d dl1 2pBcos 2p 4 d dl2 2pBcos2p d dm 2pCcos 2p 8 d t p i1 I t i2 I t cos sin

23 2 2 Torque of the machine i 1 i T 2pBcos 2 2p 4 2pBcos 2p i12 i 2pC cos 2p pbi cos t cos 2t p pbi sin t cos2t 2 2pCI sin t cos t cos 2 tp pbi cos t cos2tpbi sin t cos2t 2 2pCI sin t cos t 2 sin2t pi Bcos2 t cos2tcsin 2 t sin 2t 4

24 1 2 p 2 f 5 Hz I 18 A 2 3 B 15.1 mh C 25.1 mh 2 Torque [Nm] Torque of the machine II T pi Bcos2 t cos2tcsin2 t sin2t CBsin2tsin2t 2Bcos2 t cos2t Bsin2 t sin2t pi 2 cos 2 cos4 2 pi BC BC t Black curve: Analytical model; Red curve: FEA Time [s]

25 Results from FEA Steady state operation at the synchronous speed. The load torque is 15 Nm. The simple analytical model assumed currents varying sinusoidally in time. In practice, the currents have strong higher harmonics, which result from the shape of the pole shoes and saturation of the iron core. The high-frequency oscillations in the torque result from the slotting of the stator and rotor surfaces. Phase voltages [V] Phase currents [A] Time [ms] 2 Torque [Nm] Time [ms] Time [s]

26 Example II: Torque of a synchronous RM The torque of a synchronous reluctance motor should be computed based on the basic energy or coenergy formulations. Rated values? P = 4 kw U s = 4 V f = 5 Hz The machine has an ordinary three-phase stator winding. No windings or permanent magnets in the rotor.

27 Energy [J] Torque of a synchronous reluctance motor II The energy and coenergy are integrated from the currents and fluxlinkages by first turning the rotor to the expected load angle and then exiting the windings gradually one by one each in about 3 steps. At each step, the field and flux linkages are solved by FEA. The process has to be done for two position angles to compute the numerical derivative of coenergy in the equation of the torque Rotor position 6. deg. Phase A Phase B Phase C All the currents at rated values Number of field solution Energy Coenergy Energy and coenergy have different values <=> non-linear case.

28 Torque of a synchronous reluctance motor III 6 5 Coenergy at two rotor angles Coenergy [J] W c i k k k di k alfa = 6. deg alfa = 6.1 deg Number of solution interval Torque from two methods Torque at the rated load Torque [Nm] c,1 c,2 2 Max r 1 2 W W l Tcoe ; T BBrd Coenergy Mawell's stress Number of solution interval

29 Conclusions from the basic coenergy formulation This method, in which the energies are integrated numerically from the currents and flux-linkages, requires a lot of computation. Because of nonlinearity, 18 field solutions were done to get one torque value. The amount of computation can be reduced significantly when the energy is integrated directly from the fields B Wc HB d dv H B V The integral can be computed directly from the magnetisation characteristics of the materials and fields at the operation point. Only two field solutions are needed. However, the numerical derivative of coenergy always requires the two field solutions and this would be frustrating, for instance, within time-stepping simulations of rotating electrical machines.