Chapter 3 Modeling Of a Synchronous Machine In Chapter 2 we have discussed the small-signal and transient stability of a synchronous machine, connected to an infinite bus, represented by a classical model. We have seen that there are several disadvantages of representing a synchronous generator by a classical model. In this Chapter we will look at detailed modeling of a synchronous machine. 3.1 Representation of Synchronous Machine Dynamics While modeling a synchronous machine, different ways of representation, conventions and notations are followed in the available literature. Hence, at the outset the notations and conventions used for representing a synchronous machine should be clear. In this course, IEEE standard (1110-1991) “IEEE guide to synchronous machine modeling” has been followed for representing the synchronous machine. (a) 3.1 (b) Fig. 3.1: Synchronous machine (a) sectional view (b) Stator and rotor windings with mmf along the respective axis. The conventions and notations used along with their significance will be explained in this Chapter. To model and mathematically represent a synchronous machine first all the windings that need to be included in the model should be identified. Consider sectional view of the synchronous machine shown in Fig. 3.1 (a). The synchronous machine in Fig. 3.1 is a two pole salient machine. A general model with n poles will be dealt latter in this Chapter. The conductors a and a represent the sectional view of one turn of a-phase stator winding. The dot in the conductor “a” represents current coming out of the conductor and represents in to the conductor. By applying right hand thumb rule at conductor a and a it can be observed that the mmf due to the conductors 3.2 a and a lie along axis marked A-axis. Similarly, the mmf due to b , b and c, c lie along B and C axis, respectively. As an electrical circuit the stator can be represented as three windings corresponding to three-phases, as shown in Fig. 3.1 (b). The threephase instantaneous ac voltages and currents in the stator windings are represented as va , vb , vc and ia , ib , ic . According to the generator convention, currents out of the stator windings are considered as positive where as currents into the rotor windings are considered as positive. The rotor field is excited by a dc voltage represented as v fd with a field current i fd . The mmf generated by the rotor field excitation lies normal to the pole surface, along the direct axis or d-axis. The d-axis of the rotor is at angle m with respect to the stator a a mmf axis that is A-axis. Angle m is in mechanical radians and in case of two poles machine the electrical and mechanical angle are one and the same. But in case of multiple poles the electrical angle is related to the mechanical angle through the number of poles i.e. s P m / 2 where P is the number of poles and s is the rotor angle in electrical radians. In case of two poles machine the electrical and mechanical rotor angle will be same as is the case for the synchronous machine shown in Fig. 3.1. The analysis holds true for multiple pole machine as well but with the additional condition that s P m / 2 . For rest of the Chapters we will be expressing the angle in terms of electrical radians unless specified other wise. The axis in quadrature (leading or lagging by 90 ) with respect to the d-axis is called as quadrature axis or q-axis. The q-axis can either be represented as leading d-axis or lagging d-axis. Both the conventions are followed in the literature. However, here qaxis is taken as leading d-axis according to the IEEE 1110-1991 standard. Representing damper windings needs clarification. The damper windings are copper bars placed usually in the slots of the pole face. The ends of the copper bars are shorted forming a closed path for the currents to flow. The magnetic field generated by this type of damper winding, when currents circulating through them, will be along the d-axis. However, the rotor core itself may act as closed path for induced currents during non-synchronous operations. Hence, to properly account for the action of the damper windings and damping effect of rotor core three damper windings are considered. One damper winding represented as 1d , with a voltage v1d 3.3 and current i1d , is considered whose mmf is along d-axis. Two damper windings represented as 1q, 2q , with a voltage v1q , v2 q and current i1q , i2 q are considered whose mmf is along q-axis. In the d-axis and q-axis rotor windings the current in to the winding is considered as positive. For very accurate representation of synchronous machine, even more number of damper windings may be considered along d and q axis. According to the number of windings considered along each axis a model number is give as following [1] Table 3.1: Classifications of synchronous machine model based on number of windings in each axis Number of windings in q-axis 0 Number 1 of windings 2 in d-axis 3 1 2 Model Model Model 1.0 1.1 1.2 Model Model 2.1 2.2 3 Model 3.3 The first number in the model number given in Table 3.1 represents number of windings in d axis and second number represents number of windings in q axis. There should be at least one winding, field winding, in the d axis. Hence, the first model 1.0 means that rotor is represented by one field winding, zero d-axis damper winding and zero q-axis damper windings. From the view point of complexity, in the representation of many windings along d axis and q axis, the maximum number of winding that can be represented along any axis is fixed at 3. The model which is shown in Fig. 3.1 (b) is 2.2 that is one field winding, one damper winding along daxis and two damper windings along q-axis. Model 2.2 is widely used in many industry grade transient stability simulations software. 3.4 3.1.1 Stator and rotor winding voltage equations Applying KVL at the stator windings the following equations can be written va rs ia d a dt (3.1) vb rs ib d b dt (3.2) vc rs ic d c dt (3.3) where, rs is the stator resistance and is assumed to be same in all the three phases. The flux linkages in a, b, and c phases are represented as a , b , c . The rate of change of flux linkages in phase a, b and c lead to an induced emf (electro-motive force) which is equal to the terminal phase voltage plus the drop in the stator resistance (since we are using generator convention), as can be seen from equations (3.1) to (3.3). Now applying KVL at the d and q axis rotor windings will give the following expressions v fd rfd i fd (3.4) dt d 1d dt v1d r1d i1d v1q r1q i1q d fd (3.5) d 1q v2 q r2 q i2 q (3.6) dt d 2 q (3.7) dt Where, rfd , r1d , r1q , r2 q and fd , 1d , 1q , 2 q are the rotor field, 1d, 1q and 2q winding resistances and flux linkages, respectively. 3.5 3.1.2 Stator and rotor windings flux linkage equations The flux linkages of different windings can be expressed in terms of current through the windings and inductance of the windings as: i fd a Laa Lab Lac ia Lafd La1d La1q La 2 q i1d L L L i L L L L b ba bb bc b bfd b d b q b q 1 1 2 i c Lca Lcb Lcc ic Lcfd Lc1d Lc1q Lc 2 q 1q i2 q iabc abc Lss Lsr (3.8) abc Lss iabc Lsr irotor (3.9) irotor In equation (3.8) the diagonal elements of the matrix Lss represent the self inductance of a, b, c windings and off-diagonal elements represent the mutual inductance among a, b, c phases. The matrix Lsr represents the mutual inductance between the stator and rotor windings. A similar expression for flux linkage of the rotor windings can be written as 0 i fd L fdfd L fd 1d 0 fd L fda L fdb L fdc i a 0 i1d 1d L1da L1db L1dc i L1dfd L1d 1d 0 1q L1qa L1qb L1qc b 0 0 L1q1q L1q 2 q i1q ic i2 q 2 q L2 qa L2 qb L2 qc 0 0 L L 2 q1q 2q 2q iabc rotor Lrs Lrr (3.10) irotor ……… rotor Lrs iabc Lrr irotor (3.11) In the matrix Lrr , the mutual inductance between the d-axis windings ( fd ,1d ) and the q-axis windings (1q, 2q ) is zero as the flux due to these windings are in quadrature. The elements of inductance matrices Lss , Lsr , Lrr , and Lrs are dependent on the angle s . The dependency of stator and rotor inductances on the angle s can be understood from the way in which the air gap between the stator and rotor varies with respect to time. 3.6 It can be observed from Fig. 3.1 at s 0 the rotor d-axis is aligned along the magnetic flux axis of a a winding. The air gap that has to be traversed by the magnetic flux produced by the a a is minimum, that is twice the air gap length between the pole face and stator along the d -axis, in this position and hence the permeance is maximum. As the angle s increases, the air gap that has to be traversed by the magnetic field of the a a starts increasing. At s 90 the air gap is maximum, that is twice the air gap length from stator to the rotor along q -axis, and permeance is minimum. The variation of the permeance with respect to angle s is shown in Fig. 3.2. Fig. 3.2: Variation of permeance with respect to angle s It can be observed from Fig. 3.2 that there is a positive average permeance which is constant and there is double frequency oscillating component. This is because for a two pole machine when the rotor rotates 180 there will be two permeance peaks at s 0 and s 180 which indicates that the permeance is varying at double the rotor speed for a two pole machine. In view of this observation the permeance of the path that a a magnetic flux has to pass can be written as 3.7 s Pa Pavg Pp cos(2 s ) (3.12) mmf f a N a ia of a a winding, where N a is number of turns of phase-a winding, can be split into two components that is d -axis and q -axis component. f ad f a cos( s ) (3.13) f aq f a sin( s ) (3.14) Now these two mmf f ad , f aq complete their path through two air gaps and hence two permeance Pd , Pq (permeance along d and q axis). The flux along d -axis is f ad Pd and along q -axis is f aq Pq . These fluxes along the mmf axis of a a is given as aa f ad Pd cos( s ) f aq Pq sin( s ) f a Pd cos 2 ( s ) f a Pq sin 2 ( s ) P (1 cos(2 s )) Pq (1 cos(2 s )) N a ia d 2 2 P Pq ( Pd Pq ) cos(2 s ) N a ia d 2 2 if the flux linkage of phase-a due to current ia is represented as aa then laa aa ia Pd Pq N aaa N a2 ia 2 P Pq 2 d Na 2 cos(2 s ) (3.15) if we assume that there is a leakage inductance lls because of flux of phase-a linking itself then, (3.15) can be written as laa laao laap cos(2 s ) . Pd Pq laao lls N a2 2 P Pq 2 d , laap N a 2 Where, . Similarly we can derive an expression for the mutual inductance between aphase and b-phase. The mmf of b-phase can be split into two components along d- 3.8 axis and q-axis. The mmf along these axes can be multiplied with the permeance along d-axis and q-axis then effective flux linkage along a a mmf axis can be found. ba fbd Pd cos( s ) fbq Pq sin( s ) f b Pd cos( s ) cos( s 120) fb Pq sin( s ) sin( s 120) Pq P N b ib d (cos(2 s 120) cos(120)) (cos(120) cos(2 s 120)) 2 2 P Pq ( Pd Pq ) N b ib d cos(2 s 120) 4 2 (3.16) if the flux linkage of phase-a due to current ib represented as ba then Pd Pq ( Pd Pq ) cos(2 s 120) 4 2 ba N aba N a N bib (3.17) if the number of turns of a and b phase are same, that is N a N b . Also, assuming leakage inductance lls is same for all the phases, (3.17) can be approximately written as, 1 lab laao laap cos(2 s 120) 2 (3.18) Similarly it can be proved that mutual inductance between phase-a and phase-c will be 1 lac laao laap cos(2 s 120) 2 and that 1 lbc laao laap cos(2 s 180) 2 Hence, 3.9 between phase-b and phase-c is 1 laao laap cos(2 s 120) lls laao laap cos(2 s ) 2 1 Lss laao laap cos(2 s 120) lls laao laap cos(2 s 120) 2 1 laao laap cos(2 s 120) 1 laao laap cos(2 s 180) 2 2 1 laao laap cos(2 s 120) 2 1 laao laap cos(2 s 180) 2 lls laao laap cos(2 s 120) …………. (3.19) The mutual inductance between the rotor windings and the stator windings is straight forward as the air gap that has to be traversed by the d-axis winding and qaxis windings mmf to link the stator windings is fixed. The rotor windings flux along the mmf axis of a a will vary only according to the angle s . Hence, lafd cos( s ) Lsr lbfd cos( s 120) l cos( 120) s cfd la1d cos( s ) la1q sin( s ) la 2 q sin( s ) lb1d cos( s 120) lb1q sin( s 120) lb 2 q sin( s 120) lc1d cos( s 120) lc1q sin( s 120) lc 2 q sin( s 120) ………………… (3.20) Now let, lafd lbfd lcfd lsfd la1d lb1d lc1d ls1d la1q lb1q lc1q ls1q la 2 q lb 2 q lc 2 q ls 2 q then (3.20) can be written in a simplified form as lsfd cos( s ) Lsr lsfd cos( s 120) lsfd cos( s 120) ls1d cos( s ) ls1q sin( s ) ls 2 q sin( s ) ls1d cos( s 120) ls1q sin( s 120) ls 2 q sin( s 120) ls1d cos( s 120) ls1q sin( s 120) ls 2 q sin( s 120) ... 3.10 …….. (3.21) The complexity of analyzing the system of equations, (3.1)-(3.12) along with (3.19)-(3.20), arises from the fact that inductances are a function of the angle s and they vary with respect to time. R. H. Park [2]-[3] has proposed a method of changing the time varying ac quantities into time independent quantities through a transformation. This transformation is called as Park’s transformation. Next we will discuss about the Park’s transformation. 3.2 Synchronous Machine Dynamics in Synchronous Reference Frame The synchronous machine dynamics depend on the rotor angle with respect to the a a mmf axis, which varies with time. Due to this time varying nature of the parameters of the synchronous machine it becomes very difficult to analyze the system. R. H. Park has proposed a method to transform time varying ac quantities into time invariant quantities [2], [3]. Park’s transformation is explained below. Parks transformation The transformation matrix proposed by R. H. Park [3], also called as dq 0 transformation, is as given below: Tdqo cos( s ) cos( s 120) cos( s 120) k1 sin( s ) sin( s 120) sin( s 120) k2 k2 k2 The choice of k1 , k2 is arbitrary. In standard practice however a value of (3.22) k1 2 / 3 and k2 1 / 2 are used. In this course the same value mentioned above are used. There is an alternative choice which is also used. The alternative choice is k1 2 / 3 and k2 1 / 2 . We will discuss the effect of these choices on the modeling later. Let us apply Park’s transformation to the three-phase stator currents of a synchronous generator given below: 3.11 ia I m cos(s t ) (3.23) ib I m cos(s t 120) (3.24) ic I m cos(s t 120) (3.25) Now let us apply the Park’s transformation on the stator currents given in (3.23) to (3.25) id ia I m cos(s t s ) iq Tdqo ib I m sin(s t s ) i ic 0 0 (3.26) But s is the rotor angle with respect to the stationary a a mmf axis, in electrical radians. Instead of taking s with respect to a stationary reference if we take a synchronously rotating reference, same speed as the rotor then, we can express as s s t (3.27) Substituting (3.27) in (3.26) lead to id I m cos( ) iq I m sin( ) i 0 0 (3.28) Which mean that if the rotor, rotating at synchronous speed, has a fixed angle difference of with respect to a synchronously rotating reference then ac quantities ia , ib , ic can be transformed to dc quantities id , iq , io . It can also be observed that for a balanced system I m id2 iq2 (3.29) 3.12 1 I 0 (ia ib ic ) 0 3 (3.30) The reason for choosing k1 2 / 3 and k2 1 / 2 can be understood from (3.28) to (3.30). Due this choice the peak value of the time independent currents id , iq is equal to the peak value of the stator current that is there is a direct correlation between the stator currents and the Park’s transformed time independent currents. The physical meaning of this transformation can be understood from Blondel two-reaction theory [2]. Blondel two-reaction theory says that the traveling wave of mmf, created in the air gap of the synchronous machine due to the combined effect of three-phase stator mmfs, can be split into two sinusoidal components in such a way that the peak of one component is always aligned along d axis and the peak of other component aligned along q axis. The currents id , iq in (3.28) produce the same mmf along d axis and q axis as suggested by Blondel two-reaction theory. The currents id , iq can also be understood as the currents through two fictitious windings, rotating at synchronous speed, producing same mmf as that of the fixed stator windings along d axis and q axis, respectively. In fact from (3.29) it can also be observed that id , iq can act as real and imaginary parts of a current phasor. We can now convert all three-phase ac quantities into dc quantities through dq 0 transformation. Just like (3.28) we can represent three-phase stator voltages and flux linkage in terms of dq 0 components as vd vd va va 1 vq Tdqo vb or vb Tdq 0 vq v v vc vc 0 0 (3.31) d d a a 1 q Tdqo b or b Tdq 0 q c c 0 0 (3.32) where, 3.13 1 Tdqo cos( s ) cos( s 120) cos( s 120) sin( s ) 1 sin( s 120) 1 sin( s 120) 1 (3.33) Now in order to check the effect of dq 0 transformation of voltages and current on the instantaneous power we can find the relation between the complex power in terms of abc components and dq 0 as va S vb vc T T vd ia i T 1 v b dqo q v ic o T id 1 Tdqo iq i o vd id vd 1 T 1 vq Tdqo (Tdqo ) iq vq v i v o o o T 3 2 0 0 0 0 id 3 0 iq 2 0 3 io (3.34) or 3 S va ia vb ib vc ic (vd id vq iq 2vo io ) 2 Hence, the transformation (3.35) is not power invariant that is va ia vbib vc ic (vd id vq iq vo io ) . The readers can verify that by choosing k1 2 / 3 and k2 1/ 2 we can get a power invariant transformation i.e. va ia vbib vc ic (vd id vq iq vo io ) Now applying dq0 transformation to equations (3.1)-(3.3), we can get 3.14 vd id d rs 0 0 d 1 1 vq Tdqo 0 rs 0 Tdqo iq Tdqo dt Tdqo q v i 0 0 rs o o o d d rs 0 0 id d 1 1 d 0 rs 0 iq Tdqo q Tdqo TdqoTdqo q dt dt 0 0 rs io o o d rs 0 0 id 0 0 d d 0 rs 0 iq 0 0 q q dt 0 0 rs io 0 0 0 o o (3.36) Hence, we can write the stator and rotor equations in terms of dq 0 as vd rs id q vq rs iq d vo rs io d d dt (3.37) d q (3.38) dt d o dt v fd r fd i fd v1d r1d i1d v1q r1q i1q (3.39) d fd dt (3.40) d 1d dt (3.41) d 1q v2 q r2 q i2 q (3.42) dt d 2 q (3.43) dt In equation (3.37) and (3.38), the terms q , d are called as speed induced voltages in the stator and these induced voltages are due to the variation of the flux with respect to space. Similarly, the terms d d d q , are called as transformer dt dt induced voltages and these are induced due to the variation of the flux with respect to 3.15 time. In steady state the transformer induced voltages will be zero and only speed induced voltages will be present. It has to be observed that dq 0 transformation is not required for rotor side parameters as they are already defined either along d or q - axis. Applying dq 0 transformation to flux linkage equation (3.9) and (3.11), we get i fd d id i1d 1 q Tdqo LssTdqo iq Tdqo Lsr i i 1q 0 0 i2 q (3.44) Substituting equation (3.19) and (3.21) along with equation (3.1) in (3.44) will lead to, after simplification, 3 2 (laao laap ) d q 0 0 0 lsfd 0 0 Let ld 0 0 3 (laao laap ) 2 0 ls1d 0 0 ls1q 0 0 0 lls id i q i0 (3.45) i fd 0 i1d ls 2 q i1q 0 i2 q 3 3 laao laap and lq laao laap 2 2 . Define, lmd = ld - lls lmq = lq - lls , then the following equations can be written. 3.16 and d (lls lmd )id lsfd i fd ls1d i1d (3.46) q (lls lmq )iq ls1q i1q ls 2 q i2 q (3.47) o lls io (3.48) 3 2 (3.49) 3 2 (3.50) 3 2 (3.51) 3 2 (3.52) fd lsfd id l fdfd i fd l fd 1d i1d 1d lsfd id l1dfd i fd l1d 1d i1d 1q ls1q iq l1q1q i1q l1q 2 q i2 q 2 q ls 2 q iq l2 q1q i1q l2 q 2 q i2 q It can be observed from (3.46) and (3.52) that the mutual inductances, though now are not a function of time, are not same. In order to make all the mutual inductances between the windings equal, for sake of reducing complexity in analysis, a proper per unit system can be chosen. Hence, let us look at per unit representation of the synchronous machine model developed so far. 3.3 Per Unit Representation Let stator line to neutral rms voltage be taken as the base voltage and the generator three-phase apparent power be taken as the base power. With these two bases defined the other base quantities can be defined as given in equation (3.53). These are the base quantities in stator reference frame. It can be observed from equation (3.28) that the voltage and current in terms of dq 0 are same as the peak values of abc components and the rated MVA expressed in terms of dq0 components has an additional factor 3 / 2 involved. Hence, the stator base quantities in terms of dq 0 components can be expressed as given in (3.54). 3.17 Vbase V rms (l n) kV, Sbase Rated MVA 3Vbase I base Sbase kA, I base 3Vbase V Z base base , I base base s 2 f elec.rad, f 50 or 60 Hz 2 mbase base mech.rad P Z base Lbase H base V 1 base base Wb-turn, tbase s base base Vdqo _ base 2 Vbase kV 3 Sbase Vdqo _ base I dqo _ base MVA 2 S 2 Sbase 2 Sbase I dqo _ base 2 base 3 Vdqo _ base 3 2 Vbase 3Vbase Z dqo _ base Vdqo _ base Ldqo _ base Z dqo _ base dqo _ base Vdqo _ base I dqo _ base base base H Wb-turn 2 I base kA (3.53) (3.54) So, far we have defined stator base quantities in terms of abc and dq 0 components. Now we can define rotor base quantities. The rotor base quantities should be chosen in such a way that the mutual inductances between the stator and rotor windings and between rotor windings themselves should be same. To do this we can take the same power base, Sbase , as the power base for the rotor windings also but define current bases in a such a way that the mutual inductances among different windings become same. The field winding base quantities are shown below: 3.18 Sbase V fd _ base I fd _ base lmd I fd _ base I dqo _ base lsfd Sbase 3 lsfd V fd _ base Vdqo _ base I fd _ base 2 lmd V fd _ base fd _ base base V fd _ base Z fd _ base I fd _ base V fd _ base L fd _ base base I fd _ base (3.55) Similarly we can define the base quantities of damper windings along d-axis as following: l 3 ls1d Vdqo _ base , I1d _ base md I dqo _ base I1d _ base 2 lmd ls1d V1d _ base V1d _ base , L1d _ base I1d _ base base I1d _ base Sbase V1d _ base I1d _ base , V1d _ base V1d _ base 1d _ base , Z1d _ base base Sbase (3.56) The base quantities of windings along q-axis are given as: Sbase V1q _ base I1q _ base V2 q _ base I 2 q _ base I1q _ base lmq l s1 q I dqo _ base , I 2 q _ base lmq ls 2 q I dqo _ base V1q _ base 3 ls1q 3 ls 2 q Vdqo _ base , V2 q _ base Vdqo _ base 2 lmq 2 lmq 1q _ base V1q _ base Z 2 q _ base V2 q _ base base , 2 q _ base I 2 q _ base , L1q _ base V2 q _ base base , Z1q _ base V1q _ base base I1q _ base V1q _ base I1q _ base , L2 q _ base Let, 3.19 , V2 q _ base base I 2 q _ base (3.57) Vdqo _ base Vdqo _ base Vdqo _ base iq id i0 , Iq , I0 Id I dqo _ base I dqo _ base I dqo _ base v fd v1q v1d V fd , V1d , V1q , V fd _ base V1d _ base V1q _ base v2 q i fd i1d V2 q , I fd , I1d , V2 q _ base I fd _ base I1d _ base i2 q i1q d , I 2q , d , I1q I 2 q _ base dqo _ base I1q _ base q fd 1d q , fd , 1d , dqo _ base fd _ base 1d _ base rs 1q 1q , 2 q 2 q , Rs , Z dqo _ base 1q _ base 2 q _ base rfd r1q r2 q r1d R fd , R1d , R1q , R2 q Z fd _ base Z1d _ base Z1q _ base Z 2 q _ base Vd vd , Vq vq , V0 v0 , (3.58) 3.3.1 Stator and rotor winding voltage equations in per units The left hand side parameters in equation (3.58) are nothing but synchronous generator parameters in per units. Now equations (3.37)-(3.43) can be expressed in per units, with base quantities defined by (3.53)-(3.57) and per unit synchronous machine parameters represented as given in (3.58), as Vd Rs I d 1 d d q base base dt (3.59) Vq Rs I q 1 d q d base base dt (3.60) Vo Rs I o 1 d 0 base dt (3.61) V fd R fd I fd 1 d fd base dt (3.62) 3.20 V1d R1d I1d V1q R1q I1q base d 1d dt 1 d 1q 1 (3.63) (3.64) base dt V2 q R2 q I 2 q 1 d 2 q (3.65) base dt The instantaneous stator power in dq 0 , given in (3.35), can also be expressed in per units. Before expressing instantaneous stator power in per unit there is an important aspect which can be understood from instantaneous stator power when expressed in terms of stator voltage equations, given in (3.37) to (3.39). Hence, by substituting equations (3.37) to (3.39) in (3.35), and rearranging the terms we get S d d 3 rs id q dt 2 d q d o i r i iq 2 rs io d s q d dt dt io 3 d d q d 3 3 d id iq 2 o io d iq q id rs id 2 rs iq 2 2rs io 2 (3.66) dt dt 2 2 2 dt For a balanced system, v0 , i0 are zero hence equation (3.66) can be written as 3 d d 3 3 S d id q iq d iq q id rs id 2 rs iq 2 dt 2 2 2 dt (3.67) It can be observed from equation (3.67) that there are three terms in the equation. The first term, d q 3 d d id iq , corresponds to the rate of change of magnetic energy dt 2 dt in the stator coils. The second term, gap. The third term, 3 d iq q id , is the power transferred over air 2 3 rs id 2 rs iq 2 , is stator copper losses. The power transferred 2 over the air gap appears as torque, at the shaft speed, either opposing the motion of 3.21 the rotor in case of generator or aiding the motion of the rotor in case of motor. Hence, we can find the electrical torque, at shaft speed from the power transferred over the air gap as 2 3 tem te d iq q id P 2 3P or te d iq qid N.m 22 (3.68) Here, te is the electrical torque at the shaft with shaft speed m in mech.rad/s with P number of poles. The electrical torque can now be expressed in per units. Before that we need to define the electrical torque base at shaft speed and is given as Sbase 1 3 Vdq 0 _ base I dq 0 _ base 2 2 base base 2 P P 3 P Vdq 0 _ base 3P I dq 0 _ base dq 0 _ base I dq 0 _ base 2 2 base 22 Tbase (3.69) Dividing the electrical torque given in (3.68) by the base torque defined in (3.69) leads to 3P d iq qid te Te 22 d I q q I d Tbase 3 P dq 0 _ base I dq 0 _ base 22 (3.70) 3.3.2 Stator and rotor flux linkage equations in per units In section 3.3.1 the per unit representation of stator and rotor winding voltage equations, in synchronous reference frame, were explained. Here, the per unit representation of the flux linkage equations will be explained. Let us start with the windings along d-axis d (lls lmd )id lsfd i fd ls1d i1d (3.71) 3.22 3 2 (3.72) 3 2 (3.73) fd lsfd id l fdfd i fd l fd 1d i1d 1d lsfd id l1dfd i fd l1d 1d i1d From equation (3.58) we know that Id id I dqo _ base , I fd i fd I fd _ base , I1d i1d (3.74) I1d _ base We can express equation (3.74) in a different form as id I d I dqo _ base , i fd I fd I fd _ base , i1d I1d _ base I1d (3.75) Since, we need to represent equation (3.71) to (3.73) in per units we need to divide them by their respective base quantities and substituting equation (3.75) we get d dqo _ base fd fd _ base 1d 1d _ base (lls lmd )( I d I dqo _ base ) dqo _ base lsfd I fd I fd _ base dqo _ base ls1d I1d _ base I1d dqo _ base (3.76) 3 lsfd ( I d I dqo _ base ) l fdfd I fd I fd _ base l fd 1d I1d _ base I1d fd _ base fd _ base fd _ base 2 (3.77) 3 lsfd ( I d I dqo _ base ) l1dfd I fd I fd _ base l1d 1d I1d _ base I1d 1d _ base 1d _ base 1d _ base 2 (3.78) It can be observed from equation (3.54)-(3.56) that dqo _ base Vdqo _ base / base , fd _ base V fd _ base / base , 1d _ base V1d _ base / base . Substituting these and also using equation (3.58) in equation (3.76)-(3.78), we get d base (lls lmd )( I d I dqo _ base ) baselsfd I fd I fd _ base basels1d I1d _ base I1d Vdqo _ base Vdqo _ base Vdqo _ base 3.23 (3.79) fd 3 baselsfd ( I d I dqo _ base ) basel fdfd I fd I fd _ base basel fd 1d I1d _ base I1d 2 V fd _ base V fd _ base V fd _ base (3.80) 1d 3 baselsfd ( I d I dqo _ base ) basel1dfd I fd I fd _ base basel1d 1d I1d _ base I1d 2 V1d _ base V1d _ base V1d _ base (3.81) In order to simplify equations (3.79) to (3.81) for ease of analysis, we have defined base quantities of rotor windings, given in (3.55) to (3.57), as I fd _ base S lmd 3 lsfd Vdqo _ base I dqo _ base , V fd _ base base I fd _ base 2 lmd lsfd (3.82) I1d _ base lmd S 3l I dqo _ base , V1d _ base base s1d Vdqo _ base ls1d I1d _ base 2 lmd (3.83) It is important to understand the choice of base quantities for rotor windings [4]. The rotor winding base quantities are chosen such that most of the mutual inductance terms in equation (3.79) to (3.81) become same, lmd . For example if we substitute the field current base given in (3.82) in (3.79), the mutual inductance term lsfd gets cancelled and replaced by lmd . We can also express the flux linkage equation of q axis windings in per units as base (lls lmq )( I q I dqo _ base ) basels1q I1q I1q _ base basels 2 q I 2 q _ base I 2 q Vdqo _ base Vdqo _ base Vdqo _ base (3.84) 1q 3 basels1q ( I q I dqo _ base ) basel1q1q I1q I1q _ base basel1q 2 q I 2 q I 2 q _ base 2 V1q _ base V1q _ base V1q _ base (3.85) 2q 3 basels 2 q ( I q I dqo _ base ) basel2 q1q I1q I1q _ base basel2 q 2 q I 2 q I 2 q _ base 2 V2 q _ base V2 q _ base V2 q _ base (3.86) q 3.24 Let us define some more quantities to further simplify equations (3.79) to (3.81) and (3.84) to (3.86) X ls xls Z dq 0 _ base X md X mq X fd X 1d X 1q X 2q xmd xmq Z dq 0 _ base Z fd _ base x1q1q Z1q _ base x2 q 2 q Z 2 q _ base Z dq 0 _ base baselmq Z dq 0 _ base basel fdfd Z fd _ base baselmd I dq 0 _ base Vdq 0 _ base baselmq I dq 0 _ base Vdq 0 _ base basel fdfd I fd _ base V fd _ base basel1q1q Z1q _ base basel2 q 2 q Z 2 q _ base basel fd 1d I1d _ base V fd _ base basel1q1q I1q _ base V1q _ base basel2 q 2 q I 2 q _ base V2 q _ base basel fd 1d lsfd V fd _ base ls1d (3.88) (3.89) (3.90) (3.91) (3.92) (3.93) I fd _ base (3.94) basel fd 1d lsfd Z fd _ base ls1d X 1q 2 q baselmd (3.87) Vdq 0 _ base l I x1d 1d l base 1d 1d base 1d 1d 1d _ base Z1d _ base Z1d _ base V1d _ base X fd 1d basells I dq 0 _ base Z dq 0 _ base Z dq 0 _ base x fdfd basells basel1q 2 q I 2 q _ base V1q _ base basel1q 2 q ls1q V1q _ base ls 2 q I1q _ base (3.95) basel1q 2 q lsfd Z1q _ base ls1d Let, X d X ls X md , X q X ls X mq , X lfd X fd X md , X l1d X 1d X md , X l1q X 1q X mq , X l 2 q X 2 q X mq , 3.25 (3.96) The reactance X d , X q are called as direct axis and quadrature axis synchronous reactance. Substituting equations (3.87) to (3.96) in equations (3.79) to (3.81), (3.84) to (3.86) and taking K d L fd 1d Lmd , Kq L1q 2 q Lmq , we get d X d ( I d ) X md I fd X md I1d (3.97) fd X md ( I d ) X fd I fd K d X md I1d (3.98) 1d X md ( I d ) K d X md I fd X 1d I1d (3.99) q X q ( I q ) X mq I1q X mq I 2 q (3.100) 1q X mq ( I q ) X 1q I1q K q X mq I 2 q (3.101) 2 q X mq ( I q ) K q X mq I1q X 2 q I 2 q (3.102) We can observe from equations (3.97) to (3.102) the flux linkage equations in per unit system are less complex than the flux linkage equations in actual values. The second advantage is that all the mutual inductances/reactances along d-axis are equal if K d is assumed to be unity. Similarly, all mutual inductances/reactances along qaxis are equal if K q is assumed to be unity. In fact in actual synchronous generators the values of K d and K q are very near to unity. The per unit representation of the voltage equation and flux linkage equations of a synchronous generator are summarized below: Stator voltage equations Vd Rs I d 1 d d q base base dt (3.103) Vq Rs I q 1 d q d base base dt (3.104) 3.26 Vo Rs I o 1 d 0 base dt (3.105) Rotor voltage equations 1 V fd R fd I fd V1d R1d I1d V1q R1q I1q d fd base dt (3.106) base d 1d dt (3.107) 1 d 1q 1 (3.108) base dt V2 q R2 q I 2 q 1 d 2 q (3.109) base dt Stator flux linkage equations d X d ( I d ) X md I fd X md I1d (3.110) fd X md ( I d ) X fd I fd X md I1d (3.111) 1d X md ( I d ) X md I fd X 1d I1d (3.112) Rotor flux linkage equations q X q ( I q ) X mq I1q X mq I 2 q (3.113) 1q X mq ( I q ) X 1q I1q X mq I 2 q (3.114) 2 q X mq ( I q ) X mq I1q X 2 q I 2 q (3.115) 3.4 Synchronous Machine Parameters It is customary to represent the voltage equation and flux linkage equations in terms of sub-transient and transient reactances, open circuit sub-transient and transient time constants. Parameters of the industry grade synchronous generators are given in terms of above parameters. Let us define the parameters 3.27 3.4.1 Sub-transient and transient reactance X d" X ls 1 1 1 1 X md X lfd X l1d (3.116) X d" is called as the direct axis sub-transient reactance. The expression given in (3.116) can be derived. Suppose a voltage is applied at the stator terminals with all other rotor circuits short circuited such that only current I d flows then immediately after the voltage is applied ( t 0 ) the flux linkages fd and 1d will be zero. Hence we can write equations (3.110) to (3.112) as d X d ( I d ) X md I fd X md I1d (3.117) fd 0 X md ( I d ) X fd I fd X md I1d (3.118) 1d 0 X md ( I d ) X md I fd X 1d I1d (3.119) Solving for I fd , I1d in term of I d , also considering equation (3.96), we get I fd I1d X md X lfd X md X l1d Id X md X l1d X lfd X l1d X md X lfd X md X lfd X md X l1d X lfd X l1d (3.120) (3.121) Id Substituting equation (3.120) and (3.121) in (3.117), we get 2 X md ( X lfd X l1d ) d Xd X md X lfd X md X l1d X lfd X l1d I d Since, X d X ls X md , equation (3.122) can be further simplified as 3.28 (3.122) d X ls X md X lfd X l1d I d X md X lfd X md X l1d X lfd X l1d 1 I d X d" I d X ls 1 1 1 X md X lfd X l1d (3.123) Similarly the quadrature axis sub-transient reactance can be expressed as X q" X ls 1 1 1 1 X mq X l1q X l 2 q (3.124) The direct axis transient reactance is defined as X d' X ls 1 1 1 X md X lfd (3.125) For deriving equation (3.125) the same logic applied for finding sub-transient reactance can be used with an additional assumption that the damper winding transients settle down faster as compared to field winding transients hence I1d can be assumed to be zero. With this assumption, by solving equations (3.117) and (3.118), we get 2 X md X lfd X md I d X ls X md X lfd X md X lfd 1 I d X d' I d X ls 1 1 X X md lfd d Xd 3.29 I d (3.126) Similarly, quadrature axis transient reactance can be defined as X q' X ls 1 1 1 X mq X l1q (3.127) 3.4.2 Open circuit sub-transient and transient time constants For finding the sub-transient open circuit time constant a voltage is applied to the field winding with the stator terminals open circuited so I d is zero. At time t 0 the flux linkage 1d is zero. Substituting these assumption in (3.107) and (3.112) we get I fd X 1d I1d X md (3.128) dI fd dI 1 X 1d 1d R1d I1d V1d 0 X md base dt dt (3.129) or 1 dI fd 1 X 1d dI1d R1d I base dt base X md dt X md 1d (3.130) Substituting equations (3.130) and (3.129) in (3.106) the following expression can be obtained V fd R fd 2 X1d 1 X md X fd X 1d I1d X md base X md dI 1d dt (3.131) Substituting equation (3.111), (3.130) in (3.106) a first order differential equation in terms of I1d with a forcing function V fd , can be obtained as 3.30 X fd dI fd X md dI1d R fd I fd V fd base dt base dt X fd X 1d dI1d X fd R1d X dI R fd X 1d I1d md 1d I V fd base dt X md 1d X md base X md dt 2 1 X md X fd X 1d base X md (3.132) dI X fd R1d R fd X 1d I1d V fd 1d dt X md 2 1 X md X fd X1d base X fd R1d X1d R fd dI X md V fd 1d I1d dt X R R X fd 1 d fd 1 d In a practical generator R1d R fd hence X1d R fd can be neglected as compared to X fd R1d in equation (3.132). With this assumption and further simplification we get X md X lfd X ls base R1d X md X lfd 1 dI X md V fd 1d I1d dt X fd R1d (3.133) Hence, the open circuit direct axis sub-transient time constant, in seconds, is given as X md X lfd 1 Tdo" X l1d base R1d X md X lfd 1 base R1d 1 X l1d 1 1 X md X lfd (3.134) Similarly, the open circuit quadrature axis sub-transient time constant, in seconds, can be express as 1 1 " X l 2q Tqo 1 1 base R1q X mq X l1q (3.135) 3.31 for finding the open circuit direct axis transient time constant, along with the assumption taken for computing Tdo" , the current in the damper winding I1d is assumed to be zero then from equation (3.106) and (3.111) fd X fd I fd X fd dI fd base R fd dt (3.136) I fd 1 V fd R fd (3.137) Hence, the open circuit direct axis transient time constant, in seconds, is defined as Tdo' X fd (3.138) base R fd Similarly the open circuit quadrature axis transient time constant, in seconds is defined as Tqo' X 1q (3.139) base R1q Also, defining new variables as Eq' X X md X fd , E fd md V fd , Ed' mq 1q X fd R fd X 1q (3.140) We can express equations (3.103) to (3.115) using the sub-transient and transient reactances and open circuit time constant along with equation (3.140). The rotor currents along d -axis can be eliminated from equation (3.111) and (3.112) as 1 1 I fd X fd X md fd X fd X md X md Id I1d X md X 1d 1d X md X 1d X md 3.32 (3.141) 2 X lfd X l1d X lfd X md X l1d X md , then equation (3.141) can be let, X fd X 1d X md expressed as 1 2 X 1d fd X md 1d X 1d X md X md Id 1 2 I1d X md fd X fd 1d X fd X md X md Id I fd (3.142) The coefficients of fd , 1d , I d can be expressed in terms of sub-transient, transient and steady state reactances as given below 2 X md X md 1 X md 1 ( X d X d' )( X d' X d" ) ' fd 1 1 Eq fd X md ( X d' X ls ) 2 X fd X fd X md 2 X md ( X d X d' )( X d' X d" ) ' X md fd , Eq ' 2 X fd ( X d X ls ) ' 2 1 1 X md 1 ( X d X d )( X d' X d" ) 1d X md 1d 1d ( X d' X ls ) 2 X md X md X X 1 X 1d fd fd 1d X md 1 1 X lfd X l1d X md 1 ( X d X d' )( X d" X ls ) 2 X 1d X md X md I d X I d X ( X ' X ) I d lfd md d ls ' X lfd X l1d X md ( X d X ls ) " , X X X X lfd md d ls ( X d X d' ) ( X d' X d" ) ' X X 1 X md fd fd md fd E ( X ' X )2 q X fd d ls X fd ( X d' X d" ) X , Eq' md fd ' 2 ( X d X ls ) X fd ' " (Xd Xd ) 1 1d X fd 1d ( X d' X ls ) 2 X X 1 2 X fd X md X md I d lfd md ( X d' X d" ) I Id d ( X d' X ls ) Replacing the coefficient in (3.142) with new coefficients, equation (3.142) can be expressed as 3.33 I fd 1 X md ' ( X d X d' )( X d' X d" ) ' ( X d' X ls )( X d" X ls ) I d 1d Eq Eq ' 2 ' " ( X d X ls ) (Xd Xd ) 1 Eq' ( X d X d' ) I d I1d X md (3.143) I1d ( X d' X d" ) Eq' ( X d' X ls ) I d 1d ( X d' X ls ) 2 (3.144) Substituting equation (3.143) and (3.44) in equation (3.110) the stator d -axis flux can be expressed as d X d ( I d ) X md I fd X md I1d Eq' X d' I d X d' X ls I1d X d" I d (3.145) ( X d" X ls ) ' ( X d' X d" ) Eq ' 1d ( X d' X ls ) ( X d X ls ) In a similar way the q -axis flux linkage equation along with current I1q , I 2 q can be expressed as q X q" I q I1q I 2q ( X q" X ls ) ( X q' X ls ) Ed' ( X q' X q" ) ( X q' X ls ) 1 Ed' ( X q X q' ) I q I 2 q X mq ( X q' X q" ) ( X X ls ) ' q 2 E ' d 2q (3.146) (3.147) ( X q' X ls ) I q 2 q (3.148) So far the flux linkage equations were expressed in terms of sub-transient, transient and steady state reactances. In order to express the voltage equations in terms of these new parameters some simplification can be considered. In the equations (3.103) and (3.104), the stator voltage equations, the rate of change of stator fluxing linkage along d , q -axis can be neglected. Since, the stator is connected to the rest of the network electrically, if its dynamics are considered then the entire network 3.34 dynamics i.e. transformers, transmission lines, load etc have to be considered which will increase the computational burden immensely. Also the stator as well as network transients are much faster as compared to the rotor dynamics and hence as compared to rotor dynamics the stator and network transients can be considered as instantaneous changes. Second assumption is, in the speed induced voltage terms, d , q the speed is assumed to be synchronous speed that is base . This assumption does not mean the speed is constant but the effect of speed variation on the induced voltage is insignificant and one more advantage is that this assumption counter balances the first assumption thereby nullifying effect of both assumptions. With this assumptions and substituting (3.143) to (3.148) in equations (3.106) to (3.109), the following expression can be obtained. Note: Tdo" X md X lfd X l1d base R1d X md X lfd 1 1 base R1d X lfd X l1d X lfd X md X l1d X md X fd 1 1 1 1 base R1d X fd base R1d X fd base R1d 1 1 Tqo" ' base R2 q ( X q X q" ) ( X q' X ls ) 2 1 ' ( X d X d" ) ' 2 ( X d X ls ) 1 d d Vd Rs I d base dt base q (3.149) 1 d q Vq Rs I q base dt base d (3.150) 1 d 0 Vo Rs I o base dt (3.151) Tdo' ( X ' X d" ) Eq' ( X d X d' ) I d ' d Eq' ( X d' X ls ) I d 1d E fd 2 dt ( X d X ls ) dEq' (3.152) 3.35 Tdo" d 1d Eq' ( X d' X ls ) I d 1d dt Tqo' ( X q' X q" ) dEd' ' ' Ed' ( X q' X ls ) I q 2 q Ed ( X q X q ) I q ' 2 dt ( X X ) q ls " Tqo (3.153) d 2 q dt Ed' ( X q' X ls ) I q 2 q d X d" I d q X q" I q (3.155) ( X d" X ls ) ' ( X d' X d" ) 1d Eq ' ( X d' X ls ) ( X d X ls ) ( X q" X ls ) ( X q' X ls ) Ed' ( X q' X q" ) ( X q' X ls ) (3.154) 2q (3.156) (3.157) Apart from equations (3.149) to (3.157), equation of motion or swing equation also needs to be considered for synchronous machine. The equation of motion is given as, given in Chapter 2, equation (2.4), 2 2 d s J t m te 2 P dt (3.158) Where, tm , te are the mechanical input torque from the prime mover and the electrical output torque. J kg.m 2 is the inertia constant of the rotating machine. Now defining machine inertia constant as H 1 2 J base 2 P 2 Sbase MJ/MVA . Replace J with the defined machine inertia constant in equation (3.158), then HSbase d 2 s t m te 2 2 dt 2 base P (3.159) Rearranging we can express (3.159) as 3.36 tm te t t 2 H d 2 s m e Tm Te 2 Sbase Sbase base dt Tbase Tbase 2 2 base base P P (3.160) The rotor angle s is with respect to fixed stator reference frame but instead it can be expressed with respect to synchronously rotating reference frame, as given in equation (3.27) s baset , also including the damping torque as given in equation (2.32) Chapter 2 in (3.160) lead to d d s base base dt dt (3.161) 2 H d 2 2 H d Tm Te D base 2 base dt base dt (3.162) Equations (3.149) to (3.157), along with equations of motion (3.161) to (3.162), describe the behavior of a synchronous machine. 3.5 Effect of Saturation on Synchronous Machine Modeling The modeling of synchronous machine discussed so far assumed that there is no saturation either in the stator or rotor core. Before considering the effect of saturation some assumptions are made [5], they are 1. The leakage inductances of stator and rotor have negligible effect due to saturation of the stator and rotor core as the leakage flux path is mostly through the air. 2. The leakage fluxes of stator and rotor have negligible effect on saturation of the stator and rotor core and the saturation mostly depends on the airgap flux linkage. 3. Hence, most of the effect of saturation is on the mutual inductances X md , X mq . 3.37 4. The effect of saturation on mutual inductances can be computed through the open circuit characteristics of the synchronous generator and the same effect can be considered to be applicable even when the generator is loaded. 5. There will be no coupling inductances between the d-axis and q-axis windings due to the nonlinearity introduced by the saturation. In order to discuss the effect of saturation the open circuit characteristic should be understood. From equation (3.110) and (3.113) it can be observed that at no load open circuit condition the following expression hold Vt d X md I fd , q 0 (3.163) It can be observed that in per unit the stator flux linkage and the stator terminal voltage in no load condition are same. A stator flux linkage can be defined in case of open circuit as t d2 q2 d Vt (3.164) It can be understood from equations (3.163) and (3.164) that the terminal voltage or the stator flux linkage is directly proportional to the field current in case saturation is not considered as can be seen from the air-gap line shown in Fig. 3.3. With saturation of stator and rotor core considered the relation between the terminal voltage and the field current during open circuit is no longer linear. In case of no saturation a field current I fag produces 1.0 pu terminal voltage but with saturation to produce the same terminal voltage the filed current required is I fnl which is higher than I fag as can be seen from Fig. 3.3. In the same figure the short circuit characteristics are also given. The field current to produce 1.0 pu short circuit current is I fsc . Because of the armature reaction in case of short circuit the field current I fsc will correspond to an internal voltage which will lie on the linear portion of the open circuit characteristics. 3.38 Vt ( pu ) Air-gap line 1.0 I sc ( pu ) 1.0 I fag I fnl I fsc If Fig. 3.3: Open circuit characteristics of a synchronous generator If we assume that the stator resistance is negligible and taking the unsaturated synchronous reactance as X s _ unsat then we can write the following expression KI fsc 1.0 X s _ unsat (3.165) Where, K is the slope of the air-gap line. Equation (3.165) is written on the basis that during a short circuit the terminal voltage is zero and the internal voltage is fully applied across the synchronous reactance and hence multiplying the synchronous reactance with the short circuit current gives the internal voltage. Similarly, for producing 1.0 pu terminal voltage during no load an expression can be written as KI fag 1.0 (3.166) Hence, X s _ unsat I fsc (3.167) I fsag The synchronous reactance unsaturated can now be defined as the ratio of field current required to produce rated short circuit current to the field current required to 3.39 produced 1.0 pu terminal voltage without saturation or from air-gap line. On similar reasoning we can define the saturated synchronous reactance as X s _ sat I fsc (3.168) I fsnl Short circuit ratio (SCR) is defined as the field current required to produced the rated terminal voltage at rated speed at no load considering the effect of saturation to the field current required to produce the rated short circuit current SCR I fsnl I fsc 1 (3.169) X s _ sat Now in order to estimate the effect of saturation on the synchronous reactance Fig. 3.4 can be used. Air-gap line t _ ag flux linkage J t lin I fag I fnl If Fig. 3.4: Open circuit characteristics to compute the effect of saturation 3.40 From equation (3.163) and (3.164) and the assumption made that the saturation does not effect the leakage inductance the saturated and unsaturated d-axis inductance can be written as t X md _ unsat X md _ sat I fag tag (3.170) I fnl t (3.171) I fnl Dividing equation (3.171) by (3.170) leads to X md _ sat t t X md _ unsat X tag t J md _ unsat (3.172) From Fig. 3.4 it can be observed that if t is below or equal to lin then there is no nonlinearity and hence J is zero so from equation (3.172) the saturated and unsaturated mutual inductances are same. If t is greater than lin then J can be approximated by the exponential mathematical expression given in equation (3.173) J Asat e B sat t lin (3.173) In case of cylindrical rotor it can be assumed that the effect of saturation is same along both the d and q axis hence X mq _ sat t X t J mq _ unsat (3.174) In case of salient pole rotor since the air-gap along the q-axis is significantly more we can assume that the effect of saturation is negligible. Equation (3.173) and (3.174) are defined with respect no load condition but it can be extended to loaded condition as well. This can be done as following 3.41 d X ls X md ( I d ) X md I fd X md I1d X ls I d md q X ls X mq ( I q ) X mq I1q X mq I 2 q X ls I q mq (3.175) Substituting equation (3.103) and (3.104), in steady state, in equation (3.175) gives md d X ls I d Vq Rs I q X ls I d (3.176) mq q X ls I q Vd Rs I d X ls I q or md j mq Vq Rs I q X ls I d j Vd Rs I d X ls I q j Vd jVq Rs jX ls I d jI q (3.177) Now if we define an internal voltage behind the leakage reactance as El Vt Rs jX ls I t (3.178) Then we can write observe from equation (3.177) and (3.178) that 2 2 t md mq El (3.179) The flux linkage computed in equation (3.179) can be used in equation (3.172) and (3.174) to compute the saturated mutual inductances along d and q axis. The saturated reactances X md _ sat , X mq _ sat have to be used to compute the sub-transient and transient reactances, and time constants. The saturated values of sub-transient and transient reactances and time constants should be used in equations (3.149) to (3.157). 3.6 Estimation of Synchronous Machine Parameters through Operational Impedance The synchronous machine parameters, sub-transient and transient inductances and time constants, should be found out through some experiments. An operational 3.42 impedance can be defined which can be computed through test and then this operational impedance can be related to the synchronous machine parameters. This can be done through standstill frequency response [6], [7]. Before going to the details of the test let us first define operational impedance. The stator equations given in equations (3.103) to (3.105) can be written in the Laplace domain (with d replaced by s j0 ) gives dt Vd Rs I d r s q base base d (3.180) Vq Rs I q r s d base base q (3.181) Vo Rs I o s base 0 (3.182) If the flux linkage can be defined as d X dop ( s) I d Gop ( s)V fd (3.183) q X doq ( s ) I q (3.184) 0 X oop ( s) I o (3.185) where, X dop ( s ) , X doq ( s ) are the operational reactances along d and q axis. Now suppose a test is done by applying a variable frequency stator voltage with the rotor at stand still such that only the d – axis voltage and current will be present, due to the position of the rotor with respect to the stator, then substituting equation (3.183) in equation (3.180) with r 0 will lead to s s Vd Rs X dop ( s ) I d G (s)V fd base base op 3.43 (3.186) If in the rotor the field is shorted then V fd is zero and hence with Vd , I d known from the test at different frequencies the operational d – axis reactance X dop ( s ) can be found. Similarly, we can repeat the test by applying a test voltage to the stator with rotor at standstill and positioned in such a way with respect to the rotor that only q axis voltage and current are present then substituting equation (3.184) in (3.181), with r 0 lead to equation (3.187) from which the operational reactance X doq ( s ) can be found for different frequencies. s Vq Rs X qop ( s ) I q base (3.187) Theoretically we can express X dop ( s ) , X doq ( s ) in terms of the sub-transient and transient inductances and time constants. This can be done as following, with the damper windings voltages V1d 0 , by expressing voltage equations of the d-axis windings in Laplace domain as s base s base fd R fd I fd V fd (3.188) 1d R1d I1d (3.189) Substituting rotor flux linkage equations in equations (3.188) and (3.189) and elimination of rotor current lead to 3.44 I fd 2 s s s X md R1d X 1d X md base base base I 2 d s s s R 1d X 1d R fd X fd X md base base base s R X 1d base 1d V 2 fd s s R s X X fd X md 1d R fd 1d base base base I1d X md s base s R1d base X 1d I d (3.190) I fd (3.191) Substituting equations (3.190) and (3.191) in the d-axis stator flux linkage equations lead to equation (3.183) where s s s s 2 X md X 1d 2 X md R fd X fd R1d base base base X dop ( s ) X d base 2 s s s X 1d R fd X fd X md R1d base base base (3.192) s s X md R1d X 1d X md base base Gop ( s ) 2 R s X R s X s X 1d fd 1d base fd base md base (3.193) Equation (3.192) and (3.193) can also be expressed in factored form in terms of time constants as X dop ( s ) X d Gop ( s ) Go 1 sT 1 sT 1 sT 1 sT ' d " d ' do " do (3.194) 1 sTkd 1 sT 1 sT ' do " do 3.45 Where, Tdo' , Tdo" are the open circuit transient and sub-transient time constant as given in equations (3.185) and (3.186). Td' , Td" are called the short circuit transient and subtransient time constants and given as X md X lfd X ls X l1d X md X ls X lfd X ls X md X lfd base R1d X md X ls 1 Td' X lfd X md X ls base R fd X X Go md , Tkd l1d R fd R1d Td" 1 (3.195) In a similar way the q-axis operational reactance can also be defined as X qop ( s ) X q 1 sT 1 sT 1 sT 1 sT ' q " q ' qo " qo (3.196) The open circuit time constant Tqo' , Tqo" are already defined. The short circuit time constant are given as Tq" X mq X l1q X ls X l 2q base R2 q X mq X ls X l1q X ls X mq X l1q 1 X mq X ls 1 Tq' X l1q base R1q X mq X ls (3.197) It can be observed equations (3.94) and (3.96) that for s 0 X dop ( s ) X d , X qop ( s ) X q (3.198) As s , like immediately after a disturbance the effect on stator flux linkage, 3.46 X dop ( s ) X d X qop ( s ) X q T T X , T T T T X T T ' d " d ' do " do ' q " q ' qo " qo " d " q (3.199) In case the damper windings are absent then as s X dop ( s ) X d X qop ( s ) X q T X , T T X T ' d ' do ' q ' qo ' d (3.200) ' q In a synchronous generator Tdo' Td' Tdo" Td" and Tqo' Tq' Tqo" Tq" . Hence, if the gain variation of the transfer function X dop ( s ) and X qop ( s ) with respect to frequency variation is plotted then the corner frequencies will correspond to the time constants Tdo' , Td' , Tdo" , Td" and Tqo' , Tq' , Tqo" , Tq" with Tdo' , Tqo' corresponding to lowest corner frequency and Td" , Tq" corresponding to highest corner frequency. With this all the synchronous generator parameters can be found out. But how to apply a test voltage such that only d-axis or q-axis, voltage and current will be present to find the operational impedance along either d-axis or q-axis has to be explained. Let a test voltage, vtest , be applied across b-phase and c-phase with a-phase open and the standstill rotor is positioned such that its d-axis lead the mmf axis of a-phase of the stator by 90 so that s in parks transformation will be fixed at 90 since the rotor is standstill, then vtest vb vc , va 0, vb vc (3.201) itest ib ic , ia 0, ib ic 3.47 3 3 3 3 3 vb vc vtest , id ib ic 3 itest 2 2 2 2 2 1 1 1 1 vq vb vc 0, iq ib ic 0 2 2 2 2 vd (3.202) let the line-line test voltage and test current be defined as vtest 3 2 Vt cos ot o and itest 3 2 I t cos ot o then 3 3 2 Vt cos ot o , 2 id 3 3 2 I t cos o t o vd (3.203) In per units equation (3.203) can be written as 3 Vt _ pu cos ot o , 2 id 3 I t _ pu cos ot o vd (3.204) Hence, the d-axis operational impedance as a function of the variable frequency o is given as Z dop o Vt _ pu I t _ pu 2 Vd Id (3.205) 1 Vd 1 Vt _ pu Rs j o X dop Z dop o base Id 2 I t _ pu 2 X dop 1 Vt _ pu j base Rs o 2 I t _ pu (3.206) Similarly for finding the q-axis operational reactance test voltage has to be applied like before with the standstill rotor positioned such that the angle between d-axis of the rotor and a-phase is zero then 3.48 1 1 1 1 vd vb vc 0, id ib ic 0 2 2 2 2 3 3 3 3 3 vq vb vc vtest , iq ib ic 3 itest 2 2 2 2 2 (3.207) Hence, the operational reactance of q-axis can be written as 1 1 Vt _ pu Rs j o X qop Z qop o base Iq 2 I t _ pu 2 Vq X qop 1 Vt _ pu j base Rs o 2 I t _ pu 3.49 (3.208) Example Problems E1. A three-phase 300 MVA, 20 kV, 0.9 pf, 50 Hz, 2 pole synchronous generator has the following parameters. laa = 4.675 + 0.0534 cos (2q ) mH æ pö lab = -2.3375 - 0.0534 cos çç2q + ÷÷÷ mH çè 3ø lls = 0.5792 mH lafd = 67.2 cos (2q ) mH l fd = 1084.08 mH rs = 0.0014 W r fd = 0.0635 W Define the base quantities and express all the generator parameters in per units in dq 0 reference frame. Sol: The d-axis and q-axis inductances are given as 3 ld = ´(4.675 + 0.0534) = 7.0926 mH 2 (E1.1) 3 lq = ´(4.675 - 0.0534) = 6.9324 mH 2 (E1.2) lmd = ld - lls = 6.5134 mH (E1.3) lmq = lq - lls = 6.3532 mH (E1.4) 3.50 The base quantities are defined as üï ïï ïï ïï ïï ïï S ïï Ibase = base = 8.665 kA, ïï 3Vbase ïï ïï wbase = 2´ p ´50 = 314.159 rad / s ïï ïï Vdq 0 _ base = 2Vbase = 16.32 kV , ïï ïï I dq 0 _ base = 2 Ibase = 12.25 kA, ïï ïï Vdq 0 _ base Z dq 0 _ base = = 1.3317 W, ïï ïï I dq 0 _ base ïý ïï Z dq 0 _ base Ldq 0 _ base = = 4.2389 mH , ïï ïï wbase ïï lmd ï I fd _ base = I dq 0 _ base = 1.1873 kA,ïï ïï lafd ïï ïï Sbase = 252.66 kV V fd _ base = ïï I fd _ base ïï ïï V fd _ base ïï = 212.86 W, Z fd _ base = ïï I fd _ base ïï ïï Z fd _ base ï = 677.55 mH , ïï L fd _ base = ïïþ wbase 20 = 11.54 kV , 3 Sbase = 300 MVA, Vbase = (E1.5) The per unit generator parameters can now be computed as Ld = ld Ldq 0 _ base Lmd = Lls = = 1.673 pu, lmd Ldq 0 _ base lls Ldq 0 _ base Lq = r fd Z fd _ base Ldq 0 _ base = 1.5365 pu, Lmq = = 0.1366 pu, L fd = Llfd = L fd - Lmd = 0.0635 pu , Rs = R fd = lq = 1.6354 pu, lmq Ldq 0 _ base l fd L fd _ base rs Z dq 0 _ base = 0.0002983 pu , 3.51 = 1.4987 pu, = 1.6 pu , = 0.00105 pu, (E1.6) The transient d-axis inductance and open circuit time constant can be calculated as L'd = Lls + Td' 0 = Lmd Llfd Lmd + Llfd = 0.1975 pu , (E1.7) L'd = 2.1074 s wbase R fd It has to be noted here that in per units the inductance and inductance reactance are same. E2. A synchronous generator has the following parameters in per units X d = 1.508, X q = 1.489, X md = 1.371 X mq = 1.352, X ls = 0.1366, X fd = 1.6 X lfd = 0.229, Td' 0 = 8 s, X q' = 0.65 Tq' 0 = 1.0 s, X d" = 0.23, Td"0 = 0.03 s With these parameters defined find the following generator parameters: R fd , R1d , R1q , X d' , X l1q , X1q , X l1d , X1d Sol: The direct axis transient reactance is given in (3.125), substituting the given parameters in (3.125) lead to X d' X ls 1 1 1 X md X lfd 0.1366 1 1 1 1.371 0.229 0.3328 The direct axis transient open circuit time constant is given as Tdo' (E2.1) X fd base R fd which the per unit field winding resistance can be found as given in (E2. 2) 3.52 , from R fd X fd T ' base do 0.3328 0.0001324 314.159 8 (E2.2) From equation (3.127) the following expression can be written for finding the leakage inductance of the quadrature axis winding 1q X mq X mq X l 1q X q' X ls 0.8277 (E2.3) 1 Similarly the rest of the parameters can be computed as following X1q = X l1q + X mq = 2.179 R1q = X q' wbaseTq' 0 X l1d = (E2.4) = 0.006935 (E2.5) X md + X lfd = 0.6775 X md X lfd -1 X d" - X ls (E2.6) X1d = X l1d + X md = 2.0485 R1d = X d" wbaseTd"0 (E2.7) = 0.0927 (E2.8) 3.53 References 1. IEEE Task Force, “Current usage and suggested practices in power system stability simulations for synchronous machines”, IEEE Trans. On Energy Conversion, Vol. EC-1, No. 1, 1986, pp. 77-93. 2. A. Blondel, “The two-reaction method for study of oscillatory phenomena in coupled alternators,” Revue Generale de L Electricite, Vol. 13, pp.235-251, February 1923, March 1923. 3. R. H. Park, “Two-reaction theory of synchronous machines-generalized method of analysis-part I,” AIEE Trans., Vol. 48, pp.716-727, 1929; part II, Vol. 52,pp. 352-355, 1933. 4. A.W. Rankin, “Per-Unit Impedances of Synchronous Machines,” AIEE Trans., Vol. 64, Part I, pp. 569-573, August 1945; Part II, pp. 839-845, December 1945. 5. G. Shackshaft and P. B. Henser, “Model of generator saturation for use in power system studies”, Proc. IEE, Vol. 126, No. 8, pp. 759-763, 1979. 6. M. E. Coultes and W. Watson, “Synchronous machine models by standstill frequency response tests,” IEEE Trans. Power Appar. Syst., PAS-100, 4, Apr. 1981, 1480-1489. 7. IEEE Std 115A, “IEEE trail used standard procedure for obtaining synchronous machine parameters by standstill frequency response testing,” Supplement to ANSI/IEEE std. 115-1983, IEEE, New York, 1984. 3.54