Two Reaction Theory of Synchronous Machines Generalized Method of Analysis I

Two Reaction Theory of Synchronous MachinesGeneralized Method of Analysis-Part I

BY R. H. PARK*Associate, A. I. E. E.

Synopsis.-Starting with the basic assumption of no saturation In addition, new and more accurate equivalent circuits areor hysteresis, and with distribution of armature phase m. m. f. developed for synchronous and asynchronous machines operatingeffectively sinusoidal as far as regards phenomena dependent upon in parallel, and the domain of validity of such circuits is established.rotor position, general formulas are developed for current, voltage, Throughout, the treatment has been generalized to include salientpower, and torque under steady and transient load conditions. poles and an arbitrary number of rotor circuits. The analy-Special detailed formulas are also developed which permit the sis is thus adapted to machines equipped with field pole collars,determination of current and torque on three-phase short circuit, or with amortisseur windings of any arbitrary construction.during starting, and when only small deviations from an average It is proposed to continue the analysis in a subsequent paper.operating angle are involved. * * * * *

T HIS paper presents a generalization and extension ia, ib, i, = per unit instantaneous phase currentsof the work of Blondel, Dreyfus, and Doherty eay eby e, = per unit instantaneous phase voltagesand Nickle, and establishes new and general *a, 'rb, V1 = per unit instantaneous phase linkages

methods of calculating current power and torque insalient and non-salient pole synchronous mac hines,under both transient and steady load conditions. d

Attention is restricted to symmetrical three-phaset P dtmachines with field structure symmetrical about Then there isthe axes of the field winding and interpolar space, ea=nthereribut salient poles and an arbitrary number of rotorl ea P 1a-rcircuits is considered. eb = P 4'b-r 'ib

Idealization is resorted to, to the extent that satura- e, = p Vlc - r i(1)tion and hysteresis in every magnetic circuit and eddy It has been shown previously' that

Axis of Phase a 21a Id COS 6- Iq sin 0X0~__ Xd +Xq [ ib___ 1

i

D \ \ - 3 ~~~~~~~~~~~~~~~[ia+ ib + icl - a-3 [3+b+c 2 J

Direction ofRotatiion Xf - Xq [ia cos 2 0+ ib Cos (2 6 - 120)

3

+ i, cos (2 6 + 120)]Quadrature Axis \1b Id cos (6- 120)- Iq sin (6- 120)

Axis of Phase b xis of Phase c ia + ib +ib 2Xd + xq ]i ic + ia_

FIG. 1 Xd- Xqd q[a COS (2 0 - 120) + ib COS (2 0 + 120)

currents in the armature iron are neglected, and in 3the assumption that, as far as concerns effects depend- + i, cos 2 0] (2)ing on the position of the rotor, each armature winding I = Id cos (6 + 120)may be regarded as, in effect, sinusoidally distributed.3 ia + ib + icA. Fundamental Circuit Equations -Iq sin (6 + 120) -xo

Consider the ideal synchronous machine of Fig. 1,and let

Xd + X, - a+i

*General Engg. Dept., General Electric Company, Schenec- - 3 I C - 9 J

tady, N. Y.LtSingle-phase machines may be regarded as three-phase xd - Xz o 26-- 2)±1 o

machines with one phase open circuited. _ 3 [aCs( 2)+i OtStator for a machine with stationary field structure.33For numbered references see Bibliography.Presented at the Winter Convention of the A. I. E. E., New York, + i.c os (2 6 - 120)]

N. Y., Jan. 28-Feb. i, 1929. where,716

29-33

July 1929 PARK: SYNCHRONOUS MACHINES 717

Id = per-unit excitation in direct axis If there is one additional rotor circuit in the directI, = per-unit excitation in quadrature axis axis there is,Xd = direct synchronous reactance E-Ix, = quadraturesynchronousreactance I = I + XflId - (Xd-Xd) id To px0 = zero phase-sequence reactance

As shown in the Appendix, if normal linkages in -Ildthe field circuit are defined as those obtaining at no Fld = Xlld Ild + Xfld I -Xmi d id = Toload* there is in the case of no rotor circuits in thedirect axis in addition to the field, which gives,

4) = per-unit instantaneous field linkages [Xlld - Xfld] Told P + 1= I- (Xd - Xd') id G (p) A (p)

where,I = per-unit instantaneous field current To Told [Xlld (Xd - Xd') - Xfld Xmldl p2

2 + [(Xd X'd) Told + Xmld To] P

id = 3 Iia cos 0 + ib cos (O - 120) + i cos (f9 + 120)1 Xd (p) XXd A (p)

(3) where,On the other hand, if n additional rotor circuits A (p)=[XI1d-X.fXd21 To Told p2+[Xlld To+Told] P+I

exist in the direct axis there is, If there is more than one additional rotor circuit the4) = I + XfId Ild + Xf2d I2d operators G (p) and Xd (p) will be more complicated but

+ . . + Xfnd 'nd - (Xd - Xd') id may be found in the same way. The effects of externalwhere, field resistance may be found by changing the term I

Ild, I2d, etc., are the per-unit instantaneous cur- in the field voltage equation to R I. Open circuitedrents in circuits 1, 2, etc., of the direct axis, Xf1Id, Xf2d, field corresponds to R equal to infinity.

. etc., are per-unit mutual coefficients between the Similarly, there will befield and circuits 1, 2, etc., of the direct axis. Iq = [Xq - Xq (p)] iq (5)

Similar relations exist for the linkages in each of the where,additional rotor circuits except Xd - Xd' is to be replaced 2by a term xm. However, since all of these additional i, =- ia sin 0 +ib sin(6- 120) +i, sin(6+120) } (3a)circuits are closed, it follows that there is an operationalresult Xq (o) = Xq, X ( ) =q

Id = I + Ild + I2d + . + Ind So far, 10 equations have been established relating= G (p) E + H (P) id (4) the 15 quantities ea, eb, e,, ta, ib ,I a4'ay tby ,t'c lid, iqy

where E is the per-unit value of the instantaneous field Id, I, E, 0 in a general way. It follows that whenvoltage, and G (p) and H (p) are operators such that any five of the quantities are known the remaining 10

G (o) = 1 G (co) = 0 may be determined. Their determination is veryH (o) = 0 H (co) = Xd-Xd" much facilitated, however, by the introduction ofXd' = the subtransient reactance2 certain auxiliary quantities ed, e, eoy i'o 'Pd, aqV 4/0'

It will be convenient to write H (p) = Xd Xd (p) Thus letand to rewrite (4) in the form,1

to lia+ ib + ic) (3b)Id = G (p) E + [Xd-Xd (P)] id (4a) i 3{ia+tb±%}

If there are no additional rotor circuits, there is, as 2shown in Appendix I, ed= {eacos 0 + ebcos (o- 120) + e,cos (6 + 120)1

'I = I - (Xd - Xd') idE = TopT +I 2

where To is the open circuit time constant of the field e - f{ea sin 6+eb sin(0-120)+e, sin(0+120)1 (6)in radians.

There is then, 11 eO = 3j { ea + e?b + e~}

- G (p) T= -H2

XdTop +Xd )/d=3 {'{aCOS 6+PbcOs (6-120)±+Pcc9s(0+ 120)}

*Thjs definition is somewhat different from that given in 'Pq =- { 'P sin O+'Ib sin(--120) +'Pc sin(60+1l20) } (7)reference 2.

718 PARK: SYNCHRONOUS MACHINES Transactions A. I. E. E.

1 eb = ed coS(6-120) - eq sin (6-120) + eO (16)A0 { i/a + Ab + 0J' ec = ed cos(6+ 120)-eq sin (6 + 120) + eo

Referring to Fig. 2, it may be seen that when therethen from Equation (1) there is are no zero quantities, that is, when eo = i/0 = io = 0,

2 the phase quantities may be regarded as the projectioned= 3{cos 6r la+co5(6-120) p i/4+cos(6+120) P i/} of vectors e, i/i, and i on axes lagging the direct axis by

- rid Axis of Phase b

2eq=-3 {sin6p a+ sin (6- 120) pi/b\ e eq Direct Axis

+ sin (6 + 120) p J}ri- \ / edeo= p io-/ ri-but,

2 Axis of Phase aPi d = I{COS 0 Pia + COS (6- 120) p ib/

+ cos (6 + 120) p i/'}/

2-u {sin 0 i/2a+sin(- 120) p ib+sin(0+120) p i/J} p 6

ed + ri2d + /lq pGO Axis of Phase cFIG. 2

2p l =- I{sin 0 p i/a + sin (6 - 120) p Vlb angles 0, 0- 120 and 0 + 120, where taking the direct

axis as the axis of reals,+ sin(f +120) pi/} e =e+Jje0

= 41d + j ?,q-3co6O 'Pa + COS (6- 120) 'Pb + cos (6+120) V;Jp 6 =id +Ji q

eq + riq- i/d P 6 If we introduce in addition the vector quantity,hence there is I = Id +j Iq

ed = P Od-r id-/qp ° (8) the circuit equations previously obtained may be

eq=p'pq-riq+i/ dpd (9)eo = pi/-r io (10) lpIXiW

Also it may be readily verified thati/d = Id- Xd id = G (p) E- Xd (P)d (11)/iq = I-Xq iq = - Xq (p) iq (12)ii0 = - Xoio (13)

Equations (8) to (13) establish six reelatively simplerelations between the 11 quantities ed, e., eo, id, iqy i0, irectAxisi/d, i'q, i/, E, 0. In practise it is usually possible todetermine five of these quantities directlyfrom the termi- p xi d

nal conditions, after which the remaining six may becalculated with relative simplicity, After the direct, r

quadrature, and zero quantities are known the phase FIG. 3quantities may be determined from the identicalrelations

'ta ='tdcos G-iq sinG0+ io transferred into the corresponding vector forms,ib= tdCOS(0- 120)- i0zsin (6- 120) + io (14) e = pi/i=ri +[p0] ji/iiC= id COS (6 + 120)-iq sin (6 + 120) + io _o = Ix

{Pa = Pd cos 60-Pqsin 6+ {0 where, zi =XZdid+jXqiqlAb = i/d cos (6 -120) - 'P sin (60-120) + {0 (15) BFig. 3 shows these relations graphically.C= 'Pd COS (6 + 120)- i/q sin (6 + 120) + 'Ps The per-unit instantaneous power output from thee= ed cos 60-eq sin 6 + e5 armature is necessarily proportional to the sum


ea ia + eb ib + e, i. By consideration of any instant D. Constant Rotor Speedduring normal operation at unity power factor it may Suppose that the constant slip of the rotor is s.be seen that the factor of proportionality must be 2/3. Then there is,That is, ed = P ld-rid- (1-S) 'q

P = per-unit instantaneous power output eq =p V1,q-r iq + (1-s) {d= 2/3 { ea ia + eb ib + e,i, } but, 4I'd = G (p) E-Xd (p) id

Substituting from Equations (14) and (16) there q = -xq (p) iqresults the useful relation, Putting p Xd (p) + r = Zd (P)

P = edid + eqiq + eO i0 (17) PXq (p) + r = zq (p)C. Electrical Torque on Rotor there is

It is possible to determine the electrical torqueon the rotor directly from the general relation, e = p G (p) E-Zd (p) Pd+ (1-s) Xq (p) iq (20){Total power output} = eq = (1-s) [G (p) E-Xd (p) idl-Zq (p) iq (21)

{mechanical power transferred across gap} Solving gives,+ {rate of decrease of total stored magnetic energy} id = {[s Zq (p) ± (1x-)2 Xq (p)] C (p) E-Zq (p) ed- {total ohmic losses} (18) -(1-s) xq (p) eq } . D (p) (22)However, since this torque depends uniquely only i- (1-s) r G (p) E-2d (p) eq - (1-s) Xd (p) ed

on the magnitudes of the currents in every circuit of q D (p)the machine, it follows that a general formula for torque (23)may be derived by considering any special case in whicharbitrary conditions are imposed as to the way in which where, D (p) Zd (p) 2q (P) + (1 - S)2 Xd (p) Xq (p)these currents are changing as the rotor moves. E. Two Machines Connected TogetherThe simplest conditions to impose are that Id, Iq, Suppose that two machines which we will designate

id, iq, and io remain constant as the rotor moves. In respectively by the subscripts g and h, are connectedthis case there will be no change in the stored magnetic together, but not to any other machines or circuits,energy of the machine as the rotor moves, and the and assume in addition that there are no zero quantities.power output of the rotor will be just equal in magni- In this case the voltages of each machine will be equaltude and opposite in sign to the rotor losses. It followsthat under the special conditions assumed, Equation Axis Phase b(18) becomes simply, Direct Axis of{armature power output} = gq g Machine{mechanical power across gap} - {armature losses}

eegd Direct Axis of

2 r \ / Xogvh~~~~~~~~~~~~Machine2 ror, P = Tp 0- {a2+ ib2+ ic2}oh

Axis Phase a= Tp G- r { id2 + ig2 + i2 }hd

Then,T = per-unit instantaneous electrical torque

ed id + eqiq + eO io + r {td2 ± iq2 ±ji2}qpO /

but subject to the conditions imposed, Axis Phase ced = - p0G - rid FIG. 4

e, = Ad PO0-r iqeo = r io phase for phase, and it therefore follows that the voltage

vectors of each machine must coincide, as shown inIt therefore follows that, Fig. 4.

T = q Atd - id (6q (19) Referring to the figure it will be seen that the direct= vector product of s6 and z and quadrature components of voltage of the two= if X i (19a) machines are subject to the mutual relations,

a result which could have been established directly by ehd = e,d cos 6- eq sin 6physical reasoning. Formula (19) is employed by ehql = e7d sinA + egq, cos 6 (24)Dreyfus in his treatment of self-excited oscillations of e0d = ehd cos 6 + ehq sin 6synchronous machines.'4 eg,q = -ehdsin 6 + ehqcos a (25)


On the other hand, for currents there will be Xd' To p + Xdihd - - {i%d COS 6- igq sin ±} Top + 1 Xq

ihq = - {igdsin 6 + iq cos5} (26) Xd Xq To P3igd = - [ihd COS 6 + ihq sin 6} + [Xd r To + (Xd + r To) Xj p2

igq = - I-ihdsin 6 ihqCOS } (27) + [r (Xd + +(d+ To) X qTo] p

F. One Machine on an Infinite Bus [ rT+ Xd 'XT

In (E), if machine h has zero impedance, it follows + r2 + Xd Xqfrom (20) and (21) that ehd = 0, ehq = bus voltage To p + 1say = e. d (p)&

Then for machine g there is, (31)ed = e sin 6 Top-Fleq = e cos 6 (28) By the expansion theorem there is, finally,

G. Torque Angle Relations Xq EFrom Equations (11), (12), and (19), there is, 1d = r2 + Xd Xq

T d q Xd(Toa,+ 1) ((Zqan + r) ed_+ X.eq_)Xq Xd Xd Xq +a, d' (a,)

Then if the rotor leads the vector s6, by an angle 8 1there is r E

4tq --J,sin6 1q r2 + Xd X,=d COS 6 1

Iq ' Id 1 sin 6 Xd - (qXd' To a .2 + (Xd +-r To) a,, + r)eqo- (To an X d + Xd) edOT = cos6 + +F i/'2sin 28 (29) a, d' (a,,)Xq Xd 2Xd Xq

E--ant (32)A derivation of this formula for steady load con- end (32)ditions has been previously given by Doherty and where the summation is extended over the roots ofNickle.9 dH. Three-Phase Short Circuit with Constant Rotor Speed d (a) = 0 and d' (P) = dp d (p)

MaintainedSince a three-phase short circuit causes ed and eq The phase currents may, of course, be found from

to vanish suddenly, its effect with constant rotor speed -0.0018maintained may be found by impressing ed = - edo,eq = - ego in (22) and (23) where edo and eqo are the -0.0016values of ed and e, before the short circuit. The initial -0.0014currents existing before the short circuit must beadded to the currents found in this way in order to -0.0012-- Values of o~obtain the resultant current after the short circuit. -0.0010-With s = 0 and E constant there is in detail.I

Zq (p) edo + XQ (p) eqo xqE - r edo- xq eqo -0.0008D (p) r2 + Xdxq -0.0006-_

Zd (p) eqO- Xd (p) edO r E - r eqo + Xd edO -00004Aiq - 1D (p) r2 + xdxq -0.0002-

(30)I0.5 1.0 1.5 2.0 2.5 3.0

The working out of the formulas may be illustrated Values of r

by consideration of the simple case of a machine with FIG. 5no rotor circuits in addition to the field. In this case Equations (32) by the application of Equations (14).there is For the particular case

X0 (p) = xqTo = 2,000, Xd = 1.00, Xq 0.60, Xd' = 0.30_Xd' T0 p +1 Xd the roots a1,o a2, a3 of the equation d (p) = 0, were

Xd P To p +F 1 found to be as shown in Figs. 5, 6, and 7, where

ag2 = aa +F ab

D(P) = < T0 p +F 1 It wxill be noted that, as wounldl necesar^lyr be the


case, where r = 0, a, is equal to the reciprocal of theshort circuit time constant of the machine, i. e., for ° = 2 - s t, and referring to Equation (28),r = 0,

Xd 1 e4=cossta,1 -- ' T -0.001667 eq = sins t

If we now introduce a system of vectors rotatingwhile for r = O at s per-unit angular velocity there is

1l=-T=-0.000500 ed = 1.0a1=- T0 -0.000500 eTO e,=-

p ijs (33)-8 Then from (22) and (23),

-7; /td = { X (jS) + r-j (1- s) x, (js) }

-6-/ {['sI d(is) + r] [Usxq (js) + r]+ (1- )2 Xd (j S) Xq (j) }

-5- ~ 1j (1-2 s) xq (is)- r-4 lVaues of cor r2+ (1-2 s) Xd'(j S) xq'(j s) +j s r[Xd'C( S) +X,'(j S)]

-1-2 / { 2 Xd(+ S) Xq) S)

o k [r j S§X d(j+12ri( 1S) + Xq (j )340 0.5 1.0 1.5 2.0 2.5 3.0 J- 2 [ (

Values of r

FIG. 6 [jS Xd (j S) + r] (- j) - (1- s) Xd (j s)

The root aa is found to be almost exactly equal to q r2A+(1-2 s) Xd (i S)Xq (i s)A+j s r[xd (js) +AXq (js)]the value which it would have were To = c, i. e.,

r (Xd' + Xq) X d ( S) + { Xd ( ) Xq s)aa = 2xd' x approximately { - 2s

r1+ 1 2s [r +j-s (Xd (jS) + Xq (js)) (35)

x,, 1.0

'70V of The expressions for average power and torque then>0.6 - Values of o(bC).4 - \become,

E Pa0 = 1/2 [ed id + eq.. i0.2- 1 \ 2 3 Tav = 1/2 [iq. l'd- id {jq]

064 t \where the dot indicates the scalar product, or0.8 - \Pav = 1/2 [1 . id- j *iq]

_ -.2- ~ \ = 1/2 [Real of id -Imaginary of iq] (36)i 1.4- - \:alues of r There is in general,1.6-1.8- - \ ed + rtd = P4'd(1d- S) /q

2.2 \ eq + r q = (1-s) l/'d + P 4I'2.4

FIG. 7 ed + r id - (1 - s)eqe+riq p

Thus, in the special case considered this approximate &d e A-1 s)formula gives (1-s) P

(0.30 A- 0.60) r_= 2 XO.@30 X 0.60 -*0 p (ed A- rid) A- (1- s) (q A- r i,,) (7

which checks the result found by the exact solution i -(of the cubic. p(e,, A- riq)-(1- s) (ed A- r id)I. Starting Torque )1/ = 2 1S2(38)On infinite bus and with slip s, there will be, choosing P' -(


j s (ed + r ?td) + (1- s) (eq ± r iq) -j .i It follows that the vector amounts of forward41d = 1 2 and backward m. m. f. or current are1- 2s1

j s (eq + riq)- (1- s) (ed + rid) forward current = 2f= 2 (ic +i q)Y'2 1-2 s

with ed = 1.0, eq = j- backward current = ib =2 (id j iq) (42)

js +jsrid + (1- s) (-j) + (1- s) r iqWd 1-2s If we define by analogy,

-(1- 2s) j + r [ S id + (1-s) iq] forward voltage = 2 (ed + j eq)1- 2s

-i + 1- 2 [jsid + (1-s) i]q (39) backward voltage = 2 (ed-jel) (43)

j s (-j + r iq) - (1- s) - r (1-s) id There is,= 1- 2s 1 { -2r )

-(1 - 2s) + r[jsiq- (1-s) id]

Xd (js) X s)+ 12(r + js [XdQ s)r

=r+1 2 1s i- (12 S) id] (40) r+Xq (iS)]) }

Thus,

pi~~ ~~ --i s2 (is +( )i)| ib =2 i[xq (j s) -Xd (j s)] ;. Xd (i s) z' (i s

Tav=1/2 r- id (1) id -js si(l (1 S)i) r

r (1 - s) (iq2 + id2)-=Pav+2(1 2 s)

1 ef = 1.0 (44)L+25stq id

deb =0 (45)

pa r (iq2 r2) + s [ d] PaV = ef . if = real of if (46)

Pa,rq2+id ± ~ 7Tav Pav + rijf2 + 1-2si2 (47)

=Pav +r 2 + 21-2 (iq +iiid)2 (41)2 2(1-2 s) J. Zero Armature Resistance, One Machine ConnectedMr. Ralph Hammar, who has been engaged in the to an Infinite Bus

application of the general method of calculation out- Assume that a machine of negligible armaturelined above, to the predetermination of the starting resistance is operating from an infinite bus of per-unittorque of practical synchronous motors, has suggested voltage e, at synchronous speed, with a steady excita-an interesting modification of formulas (36) and (41), tion voltage Eo, and displacement angle 8o. At thebased upon the fact that, since the total m. m. f. con- instant t = 0, let a and E change.sists of direct and quadrature components pulsating at There is,slip frequency, it may be resolved into two components, 1 Gone moving forward at a per-unit speed 1-s+s=1.0, E- AdO 1G+ () A

and the other moving backward at a per-unit speed Xdi Xd (P) Xd (P)1- s-s = 1-2 s. Thus from this standpoint halfof both the direct and quadrature components will move t __ qO 1 ,forward, and half backward. Since the quadrature 1' Xq -Xq (p)axis is ahead of the direct it follows that as far as con- Ad = e cos 8cerns the for-ward component the quadrature current iq ~ iis equivalent to a d-c. j iq, while as regards backward A - lcomponent it is equivalent to a direct component Fromwhich there is, by obvious re-arrangement,


E- ecosa Xd Xd (P) Xd - Xd"flau did Xd + e (Cos Cos° ) + e c Y adn Ead E sin 8 (u) 5 (u) d u

Xd (p) - G(p) Xd AE 1- bn e t4' En A E (u) d u (48a)Xd Xd (P) Xd °0

e sina5 X(? Xq (P) e sin a= e + e x )-e (sina- sin8o) (48) i

Xq XrjXq (P) q Xq

Then,Xq - Xq" a ICiE e sin a Xd-Xq -F+ e -, a., tLqn EJ naqu cos5 (u) a' (u) d u

T ~~+2 X1e2 sin2a Xq X, 0Xd 2 Xd X,iEesin6 e2(xd-Xq)

Xq - Xq (p) T+ sin28+ e2cosa Xq Xq (P) (sin a- sin0o) (49) Xd 2XdXq

+ e2 sin a Xd - Xd (p) ) - a) +e2 Xd Xd sin 32adn E-adnI eadnu sin 8(u) &'(u) du

e sin , d(p(P).E+2zc'Cos6ag0_gtea8Cos()o') Xd XXd Xd (p) (co

Xdinx(P) -XdG (p) AE-e qrEXdXd (P) Xqe2 Xqt saUZ aq azqntf ECqucs5()5()d

But quantities adn aqn, ad, aqqn bn,,, [,, may be foundsuch that- e sina b ,,nbE14'r- A E' (u) d u (49a)

Xq Xq (p) Xq Xqt cxd tXq(p) Xq- X e Formula (49a) may be used to determine starting

torque and current with zero armature resistance, by(50) introducing a (t) = s t, a' (t) = s. Thus the average

Xd - Xd (p) Xd -Xdfl at component of torque is found to be,*1~~~~~ a,2n E-qn,

Xd (Y} Xd I Xd- Xd a adn S

Tav =2 Xd Xd' adn adn2 + S2Xd (p)-XdG(P) - ad- 6F-nt

Xd (p) 1= b,En 1 Xq Xq"_ a~(qn SXif Xq ()+ 2 Xq Xqy a0qn aq 2 + s2 (52)

Xd"=Xd (cx) Sincead,,= 1.0 asoE adn = 1.0 as2- s2 is never greater than 1 2^ and

bn = 1.0 E adn = aqn = 1.0It therefore follows from the operational rule that, it follows that Tav is never greater than

J (p) F (t) = F (o) 4 (t) +-Ff) (t- u) F' (u) d u (51) 1 Xd Xd Xq, Xq"0 4 Xd Xd -Ft Xq Xq" 53

where, Equation (53) thus provides a very simple criterion4 (t) = f (P) . 1 of the maximum possible starting torque of a syn-

that if chronous motor of given dimensions, when armaturea = a (t) resistance is neglected.

p a = a' (t) The same formula may also be used to obtainA E = A\ E (t) a- simple expression for the damping and synchronizing

p A\ E = A\ E' (t) components of pulsating torque due to a given smallEquations (48) and (49) may be rewritten in the form, angular pulsation of the rotor.

E - e cos8 ~~~Thus if the angular pulsation iSE-e cos a /\6~~~~A = [Ab] sin (s t)- Xd and if the punlscation of torquel is! expresose in the form


d Xq -Xq"- anq p'A T = TsA6±+ Td d t A + , e2 cos sin

d t ~~~~~~~~XqXq'p+althere results, Therefore in the case under consideration there is

Xd - Xd" adn 82 for machine a,T= Tso ± e2sin20 X 2 (cd )2 + s2 T = eIa (Xda - Xqa)

IXda ±e Xda Xqa J6

+elcs6XXq Xqll aqn2S (54) +

- Xqa2 Ea P

XqXq" ~~~(a~~n)2 + 82

+Xqa eqa 2

nqa p + anqa .6a (55)

sTd=e2in26oXd-Xd' ad, scadnS Td= e2sin2Xd Xd" (ad n+ 82 where: e - per-unit bus voltage

Ia = per-unit excitation of machine a, etc.+el cos2 o Xq - Xq" aqns8 cqn This equation can be represented by Fig. 8, in which

+ e2 COS2 6 Xq Xq Pt (aq,)2 + s2 the charge through the circuit represents (da) and the

where, Rla R2a Rna

eIdO COS60 e2 (Xd Xq)-

c +± cos 2 6o Co i C-a-CnX,j ~~Xd XqCl C2Cn

bo = average angular displacement, i. e., total FIG. 8angle = 6 = 5o + A 6.

It can be shown that for the case of no additional voltage across the circuit represents the electricalrotor circuits, Equations (54) are exactly equivalent to torque of the machine (Ta).Equations (24) and (25) in Doherty and Nickle's paper, The capacitances and resistances must be chosenSynchronous Machines III. The new formulas herein so thatdeveloped are, however, very much simpler in form, Xda Xqaespecially since in the case which Doherty and Nickle Coa e Ia Xqa + e2 (Xda Xqa)have treated, there is only one term in the summation;that is, n = 1, and a is merely the reciprocal of the short Xqa Xqa"circuit time constant of the machine, expressed in Cna e2anQa (Xqa-Xqa")radians.K. The Equivalent Circuit of Synchronous Machines 1

Operating in Parallel at No Load, Neglecting the na Cna anqaEffect of Armature Resistance

The equation for the mechanical torque isLet, ba = angle of rotor a and bus Tsa Ta ± MaPSa (57)

Oo = angle of rotor a in space where:In general, the shaft torque of a machine depends Ma = inertia factor of machine a in radians

on its acceleration and speed in space, and the magni-tude and rate of change of the bus voltage as a vector. 2_X_stored_mech._energy_at_normal_speeIf all of the machines are operating at no load and if base powerthere is no armature resistance, a small displacementof any one machine will change the magnitude of the 0.462 W R2 (rev, per mm.bus voltage only by a second order quantity; conse- -2 r f 1000 /quently for small displacements the magnitude of the base kw.bus voltage may be regarded as fixed, and only theangle of the bus and rotor need be considered. Further- 5a per-unit speed of machine amore, the electrical torque may be found in terms t time in seconds dof (6) by employing an infinite bus formula. But = dt JEquation (49a) implies the alternative general opera- But, 5a = p Ga

tional form, ~~~~~~~~Thus there is

T =e Idosin 6 +e2(Xd-Xq) sin26 Tsa = Ta+±Map2 Ga (57a)Xd + 2 Xd Xq which corresponds to the equivalent circuit of Fig. 9,

,, ~~~~~~~inwhich change - GaXdd'P The machine opeoratingr on axn infinite bus can be


represented by the equivalent circuit of Fig. 10, since in the inductive branch of the circuit. Thus a gover-the condition nor which acts through a single time constant may be

Oa = fa = 0 represented by the circuit of Fig. 14, whereis fulfilled.

Several machines in parallel on the same bus may be CRa Rna Cob

La Tsa Lb TsbLa

FIG. 11

Ta

Tsa

FIG. 9C

represented by the diagram of Fig. 11, since the con- FIG. 12ditionsOa -ba = Ob = . . . (= bus angle in space)

Ta + Tb + TC, etc. = bus power output = 0A transmission line may be represented by a,

condenser.Thus two machines connected by a line of reactance L

(x) would be represented by the circuit of Fig. 12, wherex

C= e2 (58) RShaft torques are, of course, represented by voltages.

Tsa

CoaFIG. 13

Sa POa LaCia

Ta

Cna I

FIG. 10

Mechanical damping, such as that due to a fan on a FIG. 14motor shaft or that due to the prime mover, is repre-sented by resistance in series with the inductance (L) 1Las in Fig. 13. (R) must be chosen equal to the rate R=regulationof decrease in available driving torque with increasein speed. time constant of governor in elec. radians

Governors and other prime mover characteristics Cg = R(9may also be represented by connecting their circuits Rg(9


An induction motor is represented by the simple circuit A iq =

of Fig. 15 and is precisely the circuit of a synchronous Zd (p) (-A eq + 4'dO p 6)-Xd (p) (- ed - O0 P A 6)machine with only one time constant and C = on D ()account of I = 0.

Results similar to these have been previously shown where,by Arnold, Nickle,10 and others, but simpler and more D (p) = Zd (p) Zq (P) + Xd (p) Xq (p)approximate circuits were used, the branches of the but from Equations (28),several circuits were not directly evaluated in terms of edo + A ed = e sin (6o + A 6)machine constants, and the derivation was incomplete eqO +.,A eq = e cos (o + A 5)in that the limitation to no load and zero resistance wasnot appreciated. A ed = e cos 6o A 6L. Torque Angle Relations of a Synchronous Machine A eq =- e sin 6o A 6 (62)

Connected to an Infinite Bus, for Small Angular A id =Deviations from an Average Operating Angle -(e cos 6 + j'qp)Zq(p)±(e sino6 + IdOp)Xq(p)

There is, in general, Do .A6T = To + A T = (4do+ A'd) (iqO + A iq)

- (ido + A id) ('q00 + t'q) iqFor small angular deviations, (e sin 6o +-4dO p) Zd (p) +. (e cos 6o +1-0o P) Xd (P)

A T = iqO A 'ld + PdO A iq- idO A 41q - 'qO A id D (p) *A6

={dO + idO Xq (p) A ii- {{qO+ iqo Xd(p)} A id (60) (63)d (e sin 6 + ltdO P) Zd (P) }

[4ldO+idO Xq (p)] ¢±(e cos 6o + VIqo P) Xd (P)

AT= (64)

{ (e cos 6o + 4'qO p) Zq (p)L o + [4qO+iqO Xd (P)]

- (e sin 6o + 4'dO p) Xq (p)JcC- I gR D(p)

say,A T = f (p) . -A

From (57a) the equation for shaft torque becomesA T, = (M p2+ f (p)).6

Thus,

A68 = Mp2 + f (p) .A T, (65)

AppendixFormula for Linkages and Voltage in Field Circuit with

FIG. 15 no Additional Rotor CircuitsIn this case the per-unit field linkages will depend

ed0+A\ ed=p A 'Pd-r(ido+A id>- ('Pq9+A 'q)(l+P A 6) linearly on the armature and field currents. That is,eqo +,A eq =p A 'Pq-r(iqo+,A iq) +(q)dO±+ Ad d) (l+P A 6) in general,

A ed= P A d -r A id- 'qO p A 6 A-'q a=aI- bid,A eq = p A V1P- r A iq + 'PdO P A 6 + A 'Pd Then if normal linkages are defined as those existing

from which there is at no load there must be a = 1.0.Zd (p) A cid- Xq (p) A iq =- A ed - 'qo p A 6 The quantity b may be found by suddenly impressingZq (p) A iq + Xd (p) A id =- A eq + 'PdO p A 6 terminal linkages 'Pd with no initial currents in the

machines and E = 0.A ~~~~~~ci = ~~~~~~~~By definition there is, initially

Zq (p)(- A\ed-q'poo /6) + Xq (p)(- A eq + 'PdOP A6)

D (p) id = 'Pd

(61) Xd


but also there must be from the definition of Xd2 12. Dreyfus, L., "Ausgleichvorgange in der SymmetrischenMehrphasenmaschine," Elektrotech. u. Maschinenbau, 30 S 25,

I- d 121,139,1912.Xd

13. Dreyfus, L., "Freie Magnetische Energie zwischenVerketteten Mehrphasensystemen," Elektrotech u. Maschinenbau,

hence there must be an initial induced field current of 29 S. 891, 1911.amount 14. Dreyfus, L., "Einfuhrung in die Theorie der Selbster-

Xdregten Sehwingungen Synchroner Maschinen," Elektrotech. u.

I l=d Maschinenbau, 29 S. 323, 345, 1911.Xd' ________

But, initially the field linkages are zero, thus Diseussion-FXd b H. C. Specht: I should think Mr. Park's theory could be

= IXd + X' 0 applied just as well to the so-called synchronous inductionL Xd motor, that is an induction motor in which the rotor teeth

between the poles are cut out for a distance of about one-thirdhence b = Xd - Xd' of the pole pitch. Such a motor runs at synchronous speed.

Similarly, there will be However, the pull-out torque is much less than that of an induc-E = per-unit field voltage

tion motor with the full number of teeth.

C. MacMillan: There was one statement in the first page= c p I + d I of Mr. Park's paper to the effect that "Idealization is resorted

Normal field voltage will be here defined as those to, to the extent that saturation and hysteresis in every magneticexisting at no load and normal voltage. This requires circuit and eddy currents in the armature iron are neglected. . .existing at no load and normal voltage. This requires And with regard to Fig. 5, Mr. Park remarked that it represented

that d = 1. The quantity c may then be recognized a rigorous solution. Perhaps Mr. Park could give us a little moreas the time constant of the field in radians when the insight into the effect of taking into account saturation, and givearmature is open circuited, since with the field shorted other cases in which certain elements have been neglected withunder these conditions there is more or less effect upon the final results.

W. J. Lyon: In a paper of this description, certain premises(To p + 1) I 0O should be chosen and, with these always in mind, the mathe-c p T ± I = 0 matical development should be rigorous. The paper may then

= be criticized either because of insufficient premises or becauseof incorrect mathematical development. I believe that the

c = To= time constant of field with armature former is the kinder method; it is the one I shall employ.open circuited. The premises that Mr. Park chooses are that the field and

armature windings are symmetrical, that saturation and hystere-Bibliography sis are neglected and that the armature windings are in effect

1. Doherty, R. E. and Nickle, C. A., Synchronous Machines sinusoidally distributed. I take this last to mean that the air-V, A. I. E. E. Quarterly TRANS., Vol. 48, No. 2, April, 1929. gap flux due to the armature currents is sinusoidally distributed,

2. Park, R. H. and Robertson, B. L., The Reactance of for if the armature windings themselves were sinusoidally distrib-Synchronous Machines, A. I. E. E. Quarterly TRANS., Vol. 47, uted, there would be produced space harmonics in the air-gapNo. 2, April, 1928, p. 514. flux distribution due to the saliency of the poles, which, as we all

3. Park, R. H., "Definition of an Ideal Synchronous know, would complicate the problem tremendously. In orderMachine and Formula for the Armature Flux Linkages," General that the mathematical method used by Mr. Park shall be rigor-Elec. Rev., JuIne, 1928, Vol. 31, pp. 332-334. ous, I believe it is necessary to make one further assumption.

4. Alger, P. L., The Calculation of the Reactance of I think I can best explain this by asking you to consider theSynchronous Machines, A. I. E. E. Quarterly TRANS., Vol. 47, result of supplying the field winding with a sinusoidal currentNo. 2, April, 1928, p. 493. while the armature rotates at some speed which may be called

5. Doherty, R. E. and Nickle, C. A., Synchronous synchronous. Under these conditions, there will first be pro-Machines IV, A. I. E. E. Quarterly TRANS., Vol. 47, No. 2, duced in the armature windings two sets of balanced currentsApril, 1928, p. 457, Discussion p. 487. each of which will produce 3 component flux distributions in the

6. Wieseman, R. W., Graphical Determination of Magnetic gap. The first of these is what would be produced if the air-Fields; Practical Application to Salient-Pole Synchronous Machine gap were uniform, and is proportional to 1/2 (Xd + Xq - Xa),Design, A. I. E. E. TRANS., Vol. XLVI, 1927, p. 141. where Xa equals the armature leakage reactance. The second

7. Doherty, R. E. and Nickle, C. A., Synchronous of these components is proportional to 1/2 (Xd - xq). TheMachines III, Torque-Angle Characteristics Under Transient third component is of the same size as the second. Using theConditions, A. I. E. E. TRANS., Vol. XLVI, 1927, p. 1. values that Mr. Park gives under Section H of his paper, the

8. Bekku, S., "Sudden Short Circuit of Alternator," relative magnitudes of these components would be (0.8 - xa)Researches of the Electrotechnical Laboratory No. 203, June, and 0.2. The first and second components react on the field,1927. and produce in it a current of the impressed field frequency.

9. Doherty, R. E. and Nickle, C. A., Synchronous These are the components that Mr. Park has recognized, but theMachines I and II, An Extension of Blondel's Two-Reaction third component produces an entirely different frequency in theTheory, A. I. E. E. TRANS., VOl. XLV, PP 912-47. field, which will then be reflected into the armature and the

10. Nickle, C. A., Oscillographic Solution of Electro- process will be repeated. That is, in this respect, it is similarmechanical Systems, A. I. E. E. TRANS., VOl. XLIV, 1925, PP. to the condition that exists in a single-phase alternator. As far844-856. as I am aware, the Heaviside operational method cannot be used

11. Dreyfus, L., "Ausgleichvorgange Beim Plotzlichen to obtain a rigorous solution for the single-phase alternator.Kurzschluss von Synchron Generatoren," Archiv f. Electrotech., In spite of this, I think the objection that I have raised is of no5S103, 1916. more importance than the effect of neglecting saturation or

Two Reaction Theory of Synchronous Machines Generalized Method of Analysis I

Documents

Transcript of Two Reaction Theory of Synchronous Machines Generalized Method of Analysis I