T RANSIENTS AND EIGENVALUE ANALYSIS

PC-IMD can calculate starting transients. To model the capability of soft-starters with point-on-wave

switching, the AC supply voltage can be switched in two stages, so that the DC offset can be suppressed. This practically eliminates the oscillatory torque which occurs when the three lines close simultaneously, and considerably lightens the duty on the shaft coupling, [Wood, 1965].

PC-IMD can also calculate the eigenvalues of the motor operating from a sinusoidal voltage source at

zero slip over a range of frequencies from the normal line frequency down to zero. The resulting root- locus diagram is indicative of the relative stability of the motor when fed from a variable-frequency inverter without shaft position or velocity feedback.

The theory of these dynamic calculations is presented here.

Reference Frame Transformations

PC-IMD’s dynamic calculations use two sets of d,q axes:

1. synchronously rotating (Kron), for eigenvalue analysis; and

2. d,q axes fixed on the rotor (Park) for transient analysis (starting calculations).

The conventions are the same as in Fitzgerald and Kingsley [1961], with only one or two minor changes of notation: see Fig. 3.15.4 _{Although this approach is rather old-fashioned it is practical and clear. PC-} IMD does not include zero-sequence voltages or currents, so the theory works only for three-wire

connections and the transformation matrices are not square.

The transformation from (a,b,c) variables (i.e., direct phase variables) to dq axes is eqn. (3.29), where 2 is the angle between the d-axis and the axis of phase a in elec. rad. The same transformation (and its inverse, eqn. (3.30)) is used for voltages, currents, and flux-linkages.

v_d ' 2

3 [vacos 2 % vbcos (2 ! 2B/3) % vccos (2 % 2B/3)];

v_q ' ! 2

3 [vasin 2 % vbsin (2 ! 2B/3) % vcsin (2 % 2B/3)].

v_ai_a % v_bi_b % v_ci_c ' 3 2[vdid % vqiq] (3.31) R_a ' L_aai_a % L_abi_b % L_aci_c % L_aAi_A % L_aBi_B % L_aCi_C R_b ' L_bai_a % L_bbi_b % L_bci_c % L_bAi_A % L_bBi_B % L_bCi_C R_c ' L_cai_c % L_cbi_b % L_cci_c % L_cAi_A % L_cBi_B % L_cCi_C (3.32) L_aA(2) ' L_aA cos 2_r L_aB(2) ' L_aA cos (2_r % 2B/3) L_aC(2) ' L_aA cos (2_r ! 2B/3) (3.33) R_q ' L₁₁i_q % L₁₂i_Q. _(3.35) R_d ' 2

3[Ra cos 2 % Rb cos (2!2B/3) % Rc cos (2%2B/3)] ' L11id % L12iD (3.34)

R_D ' L₂₂i_D % L₁₂i_d;

R_Q ' L₂₂i_Q % L₁₂i_q. (3.36)

Inverse:

At this stage it is not specified whether the dq axes are fixed to the rotor or to the stator or to the flux, or what. All that is necessary to specialize the transformation to one of the classic frames of reference is to constrain 2 appropriately.

The total electrical instantaneous power at the terminals is

Flux-linkages and inductances

There are three phases a, b, c on the stator and three on the rotor, A,B,C. For the stator,

The self-inductances are constant and symmetrical: Laa

= L

bb

= L

cc

;

LAA

= L

BB

= L

CC

. The stator-stator

mutual inductances are constant: Lab

= L

bc

= L

ca

= L

ba

= L

cb

= L

ac

. The rotor-rotor mutual inductances

are constant: LAB

= L

BC

= L

CA

= L

BA

= L

CB

= L

AC

.

The stator-rotor mutual inductances vary sinusoidally with rotor position 2r

: for example,

A similar set of equations applies for L_bA(2), L_bB(2), L_bC(2),L_cA(2), L_cB(2), and L_cC(2), with appropriate angles taken from Fig. 3.15. Also an equation similar to eqn. (3.32) is written for the rotor flux-linkages R_A

,R

_B

, and R

_C

.

The three stator phases will be replaced by two fictitious, orthogonal coils d and q fixed to (and rotating with) the d- and q- axes respectively. The flux-linkage of the d-coil is obtained by applying the voltage transformation to Ra

, R

b

and R

c

:

where L₁₁ = L_aa ! L_ab and L₁₂ = 3/2 L_aA and much algebra has been omitted. Similarly in the q-axis,

For the rotor D and Q coils, the transformation is the same but with 2_r

instead of 2; thus

where L₂₂ = L_AA ! L_AB, and again L₁₂ = 3/2 L_aA.

v_a ' v_d cos 2 ! v_q sin 2;

v_b ' v_d cos (2 ! 2B/3) ! v_q sin (2 ! 2B/3);

v_c ' v_d cos (2 % 2B/3) ! v_q sin (2 % 2B/3).

v_a ' R_ai_a % pR_a; v_b ' R_ai_b % pR_b; v_c ' R_ai_c % pR_{c (3.37)}

v_d ' R_di_d % pR_d ! p2.R_q;

v_q ' R_qi_q % pR_q % p2.R_d. (3.38)

Fig. 3.16 d-axis aligned with phase a

Physical attributes of the fictitious d and q coils

Consider the instant when the d-axis is aligned with the axis of phase a, and ia

= I, i

b

= i

c

=!½I in the 3-phase winding,

Fig. 3.16. According to the reference frame transformation

id

= I. If the flux-linkage in phase a is Q, the flux-linkages of

phases b and c must be !½Q each, so that by the reference eframe transformation Rd

=

2/3 × Q × [1 !½(!½)!½(!½)] =

Q. The flux-linkage per ampere in the d-coil, is Rd

/

id

= Q/I.

The d-coil evidently has the same number of turns as the a- coil. If we consider I to be a DC current then the voltage drop in phase a is va

= R

aI

and the voltage-drops in phases

b and c are vb

= v

c

= !R

aI/2

. According to the reference frame transformation, v

d

= 2/3 × R

aI

× [1

!½(!½)!½(!½)] = R_aI

. The resistance of the d-coil is evidently v

_d

/

i_d

= R

_a

. This is consistent with

having the same number of turns, provided that the total cross-section of copper in the conductors of the d-coil is the same as that of the a-coil.

Consider the stator flux-linkage produced by the stator currents alone at the position shown in Fig. 3.16, 2_r

= 0. R

_a

is made up of L

_aaI

due to self-flux-linkage, plus (!½I) × L

ab

mutually coupled from each of

phases b and c, for a total of (L_aa

!

L_ab

)

I

= L

11I

. The flux-linkage of phase b is L

aa

× (!½I) due to self

flux-linkage, with Lab

× (!½I) mutually coupled from phase c, and L

abI

mutually coupled from phase

a, for a total of !½L11I

. The flux-linkage of phase c is the same. So R

d = 2/3 × L11I ×

[1+(!½)×(!½)+(!½)×(!½)] = L₁₁I

. According to the transformation equation the d-coil current is I, so

the apparent inductance is L11

. The d-coil represents all three stator phase coils along the d-axis,

including their mutual coupling.

Now consider the stator flux-linkage produced by the rotor currents alone, with current I flowing in coil

A and !½I in coils B and C. The flux-linkage of stator coil a is Ra

= L

aAI

from coil A and !½L

aA× (!½I)

from each of coils B and C, for a total of LaAI

× [1 + (!½) × (!½) + (!½) × (!½)] = 3/2 × L

aAI. The flux- linkage of stator coil b is R_b

comprising !½L

aA

× I from coil A, L

aA

× (!½I) from coil B, and !½L

aA

×

(!½I) from coil C for a total of !¾LaAI. The flux-linkage of coil c is the same, so that according to the

reference frame transformation Rd = 2/3 × [3/2 + (!½) × (!¾) + (!½) × (!¾)] × LaAI = 3/2 × LaAI

.

According to the transformation the current iD

is I × 2/3 × [1 + (!½) × (!½)+(!½) × (!½)] = I, so that

the apparent mutual inductance is 3/2 × LaA

.

Voltage equations

For the stator

where p is the operator d/dt, and we have assumed that Ra

= R

b

= R

c

. For the stator d- and q-coils the

voltage equations are derived by incorporating eqns. (3.37) into the reference-frame transformation equations for vd

and v

q

. When p operates on the trigonometric functions of 2, we get the speed voltages.

Missing out a lot of algebra,

In synchronously rotating axes p2 = T = 2Bf. In d,q axes fixed to the rotor, p2 = T_r

,

i.e., the rotor angular velocity in elec rad/s. Note that Rd

= R

q

= R

a

.

A similar procedure applies to the D- and Q-coils on the rotor, except that they are themselves rotating at angular velocity Tr

, the physical angular velocity of the rotor in elec rad/s. The result is

v_D ' 0 ' R_Di_D % pR_D ! p2_s.R_Q v_Q ' 0 ' R_Qi_Q % pR_Q % p2_s.R_D. (3.39) T_e ' 3 2 P 2 [Rdiq ! Rqid] [Nm]. (3.40) T_e ' J dTr dt × 2 P % DTm (3.41) pR_d ' v_d ! R_di_d % TR_q pR_q ' v_q ! R_qi_q ! TR_d pR_D ' ! R_Di_D % (T ! T_r)R_Q pR_Q ' ! R_Qi_Q ! (T ! T_r)R_D pT_r ' P 2J [Te ! DTm] (3.42) v_a ' v_pk cos (Tt % *); v_b ' v_pk cos (Tt % * ! 2B/3); v_c ' v_pk cos (Tt % * % 2B/3). (3.43) v_d ' v_pk cos (Tt ! 2 % *); v_q ' v_pk sin (Tt ! 2 % *). (3.44) v_d ' v_pkcos *; v_q ' v_pksin *. (3.45)

where p2_s

is the slip velocity in elec rad/s, i.e. the angular velocity of the d,q axes relative to a point

fixed on the rotor. Thus in synchronously rotating axes p2s

= T!T

r

, with T = 2B f, but in d,q axes fixed

to the rotor, p2s

= 0. Note that R

D

= R

Q

= R

A.

The electromagnetic torque is determined from the power associated with the stator current and the speed voltages, divided by the angular velocity in mechanical rad/s. If P is the number of poles,

The dynamical equation of motion is

where Tr

is in elec rad/s and T

m = Tr × 2/P is the rotor speed in rad/s. D is a viscous damping term.

Transient simulation

Transients are calculated by integrating the differential eqns. (3.38),(3.39) and (3.40) in the form.

where T is the velocity of the rotating reference frame in elec rad/s. At the end of each timestep, new values of R_d

, R

q

, R

D

, R

Q

, and T

r

become available and the currents must be updated by solving eqns.

(3.34!3.36). Likewise the torque must be updated by means of eqn. (3.40). A typical timestep would be 0.002 s, but it depends on the time-constants of the particular motor. The system of equations is non- linear because of products like Rdiq

in the torque equation and T R

d

in the voltage equation.

If the d,q axes are fixed to the rotor, the speed voltages in eqns. (3.42) vanish, since T = Tr

. The

integration is "driven" by the applied voltages vd

and v

q

, which are defined in terms of v

a

,

vb

, and v

c

.

while the flux-linkages and currents are "outputs". For example, suppose the supply is defined by

where T = 2B f and * is an arbitrary phase angle. According to the transformation equations

If 2 = 2r the d,q axes are fixed to the rotor. If 2 = Tt they rotate at synchronous speed, with T = 2B f.

v_ab ' v_pk 3 cos (Tt % B/6); v_bc ' v_pk 3 cos (Tt ! B/2); v_ca ' v_pk 3 cos (Tt % 5B/6). (3.46) L_LL ' L_aa % L_bb ! 2L_ab ' 2L₁₁. _(3.47) i_d ' 2

3ia [cos 2 ! cos (2 ! 2B/3)] ' 2ia/ 3 cos (2 % B/6);

i_q ' !2

3ia [sin 2 ! sin (2 ! 2B/3)] ' !2ia/ 3 sin (2 % B/6)

(3.48)

pR_D ' !R_Di_D. _(3.49)

Fig. 3.17 Non-simultaneous switching of the supply phases

Non-simultaneous switching of the phases

Fig. 3.17 shows the connection of the motor to the supply. The contacts in line a are assumed to be already closed, with no current flowing. Those in line b close at the "point on wave" Tt = 21, and line

c closes at the angle Tt = 2₁ + 2₂; that is, after a delay 2₂/T measured from the closure of line b.

With line-neutral voltages given by eqns. (3.43), then with * = 0 the line-line voltages are

During the interval 2₁ # Tt # 2₂ v_c

is undefined, but i

_c

= 0 and i

_a

= !i

_b

. Current flows in the loop ab. No

torque is produced, and the rotor remains stationary. A particular case of interest is to close line b at a peak of v_ab

, that is, when Tt = !B/6, and then close line c after a further delay of 90E (2

1 = 0; 22 = 90E).

This strategy eliminates most of the oscillatory component in the transient torque, [Wood, 1965]. During the delay, stator current flows only in the loop through phases a and b. The line-line resistance is R_LL

= 2R

a

and the line-line inductance L

LL

is

Obviously current is induced in the rotor. The rotor circuit can be reduced to a single circuit by fixing the d-axis to the rotor and aligning it with the axis of the effective stator coil ab that results from the loop current flowing in lines a and b. Then the whole circuit is reduced to that of a single-phase transformer. With i_c

= 0 we have

and the required alignment is achieved by setting 2 = 2r = !B/6, with id

=2 i

a/%3 and iq

= 0: that is, all

the stator current is in the d-axis and none in the q-axis.

On the rotor, the Q coil is orthogonal to the axis of the effective stator coil ab and so R_Q

= i

Q

= 0, while

eqn. (3.42) holds for RD

with T = T

r

= 0:

The mutual inductance between the rotor D-coil and the stator d-coil is L₁₂ = 3/2 × L_aA

. Since the d- and

a- coils have the same number of turns, the mutual inductance between the D-coil and phase a is also

R_LL ' L_LLi_a % 3L₁₂i_D R_D ' L₁₂i_d % L_Di_D (3.50) R_d ' L₁₁i_d % L₁₂i_D; R_D ' L₁₂i_d % L_Di_D. (3.52) pR_LL ' v_LL ! 3 2 RLLid. (3.53)

Fig. 3.18 Alignment of dq axes with the conducting loop in phases a and b.

R_LL ' R_ab ' R_a ! R_b ' 3 R_d (3.51)

L12 cos (!B/6) = %3/2 × LaA

. The axis of phase b is at an angle 5B/6 relative to the d-axis, so that with

current in the stator loop ab and ib = !ia phase b makes an equal contribution to Rd

. Therefore the total

mutual inductance M_D-LL

is 2 × L

12

× %3/2 = %3 L

12

, and

where R_LL is the flux-linkage of the series connection of phases ab. Using the inverse reference-frame transformation it can be shown that

and since ia = /3/2 id and LLL = 2 L11, we can write eqns. (3.50) as

This agrees with eqns. (3.34) and (3.36), since L_D = L₂₂ . The voltage equation for the stator loop is

The solution during the delay interval can therefore proceed by integrating eqns. (3.49) and (3.53), updating the currents at each timestep using eqns. (3.50); there is no torque and no rotation. After line

c closes, all three motor terminal voltages are known: eqn. (3.45) can be used for vd

and v

q

, and the

solution can proceed in d,q axes as before, provided that the final currents at the end of the delay are transformed into initial values for id

and i

q.

v_d ' R_di_d ! TR_q ' R_di_d ! T(L₁₁i_q % L₁₂i_Q);

v_q ' R_qi_q % TR_d ' R_qi_q % T(L₁₁i_d % L₁₂i_D). (3.54)

i_a ' [i_d cos 2 ! i_q sin 2] ' i_d cos Tt % i_q cos (Tt % B/2). _(3.55)

V_d % j V_q ' [R₁ % j TL₁₁] (I_d % j I_q) % j TL₁₂(I_D % j I_Q) _(3.56) 0 ' [R₂ % j sTL₂₂] (I_D % j I_Q) % j sTL₁₂(I_d % j I_q). _(3.57) a₅₁ ' 3 2 (P/2)2 J [ iq0 ! R_q0 L_sN] (3.59) a₅₂ ' ! 3 2 (P/2)2 J [ id0 ! R_d0 L_sN] (3.60) Steady-state operation

The theory so far assumes that we know the inductances L11

,

L12

etc., but we need to relate these to the

familiar "equivalent circuit" parameters of the induction motor: R1

,

R2

,

X1

,

X2

, and X

m. In the steady

state the terms in eqns. (3.42) containing the p operator are zero so that

We can combine these equations by writing V = Vd

+ jV

q

where V

d

= v

d/%2 and Vq

= v

q/%2 and similarly

with the currents. This can be seen as follows: if 2 = Tt then From eqn. (3.54) it follows that

where R₁

= R

_a

and L

₁₂

= 3/2 × L

_aA

. We can write I

₂

= I

_D

+ j I

_Q

directly for the "rotor current", while

for the "stator current" I1

= I

d + jIq

. The construction of the equivalent circuit requires also the

equivalent of eqn. (3.56) for the rotor, i.e.

where R2

= R

A

= R

D

. Eqns. (3.56) and (3.57) represent the conventional equivalent circuit and so we

have done enough to show that R_d

= R

q

= R

a

;

L11

= [X

1

+ X

m

]/T,

L22

= [X

2

+ X

m

]/T, and L

12

= 3/2 × L

aA

= Xm

/T. Note the appearance of slip s in eqn. (3.57): by dividing this equation throughout by s, the

rotor circuit is "referred" to the stator with the same frequency T and the familiar R2/s appears as the

only manifestation of rotation.

Small-signal analysis

For this we assume small perturbations (denoted by )) about a steady operating point (denoted by subscript 0): vd

= v

d0

+ )v

d; vq

= v

q0

+ )v

q; id

= i

d0

+ )i

d; iq

= i

q0

+ )i

q; iD

= i

D0

+ )i

D; iQ

= i

Q0

+ )i

Q; Rd =

R_d0 + )Rd; Rq = Rq0 + )Rq; RD = RD0

+ )R

D; RQ

= R

Q0 + )RQ; Tr

= T

r0 + )Tr; and Te = Te0 + )Te. T is missing

because it is considered fixed and equal to 2Bf, but T_r

can vary. If we substitute these into the voltage

eqn. (3.42) and the torque eqn. (3.40) and simplify by subtracting the "steady-state" components, while ignoring products of the form )x)y, then after some grinding we get

p)Rd !1/JsN T kr/JsN 0 0 )Rd )vd p)Rq !Tr0 !1/JsN 0 kr/JsN 0 )Rq )vq p)RD = ks/JrN 0 !1/JrN T ! Tr0 !Rq0 )RD + 0 (73) p)R_Q 0 k_s/J_rN 0 !1/J_rN R_d0 )R_Q 0 p)T_r a₅₁ a₅₂ a₅₃ a₅₄ !D/J )T_r 0 where

a₅₃ ' 3 2 (P/2)2 J k_r L_sNRq0 (3.61) a₅₄ ' ! 3 2 (P/2)2 J k_r L_sNRd0 (3.62) v_d0 ' R_di_d0 ! TR_q0; R_q0 ' L₁₁i_q0 v_q0 ' R_qi_q0 % TR_d0; R_d0 ' L₁₁i_d0 (3.63) i_d0 ' TL11vq0 % Ravd0 R_a2 _{% T}2_L 112 ; i_q0 ' ! TL11vd0 ! Ravq0 R_a2 _{% T}2_L 112 (3.64)

Fig. 3.19 Equivalent circuit of split-phase motor. During start-up, the cut-out switch switches from "start" to "run" at a certain percentage of the synchronous speed. The OC (open-circuit) position of the cut-out switch simply represents here the electrical connections of a single-phase motor.

and JsN = LsN/R1, Jr

N = L

rN/R2, LsN = (LsLr ! M2)/Lr

,

LrN = (LsLr ! M2)/Ls, ks = M/Ls, kr = M/Lr

, and P is

the number of poles. The notation has been changed using L_s = L₁₁; L_r = L₂₂, and M = L₁₂ so that the matrix eqn. (3.58) can be more easily compared with Vas' eqn. 2.10-36 [9] which is derived using space- vector theory.

In the calculation of the coefficients a51..a54, the determination of Rd0 and Rq0 generally requires the

inversion of a 4 × 4 matrix but this can be avoided if the eigenvalues are calculated for the zero-slip condition. This makes iD0 = iQ0

= 0 and T!T

r0 = 0. From eqn. (3.57) the steady-state conditions are

for which the solution is

In document Miller, T.J.E. - SPEED's Electric Motors (Page 143-150)