Đang tải... (xem toàn văn)
Analysis of Electric Machinery and Drive Systems Editor(s): Paul Krause, Oleg Wasynczuk, Scott Sudhoff, Steven Pekarek
9 UNBALANCED OPERATION AND SINGLE-PHASE INDUCTION MACHINES 9.1 INTRODUCTION The method of symmetrical components, as developed by Fortescue [1], has been used to analyze unbalanced operation of symmetrical induction machines since the early 1900s This technique, which has been presented by numerous authors [2–5], is described in its traditional form in the first sections of this chapter The extension of symmetrical components to analyze unbalanced conditions, such as an open-circuited stator phase, is generally achieved by revolving field theory This approach is not used here; instead, reference-frame theory is used to establish the method of symmetrical components and to apply it to various types of unbalanced conditions [6] In particular, it is shown that unbalanced phase variables can be expressed as a series of balanced sets in the arbitrary reference frame with coefficients that may be constant or time varying This feature of the transformation to the arbitrary reference frame permits the theory of symmetrical components to be established analytically and it provides a straightforward means of applying the concept of symmetrical components to various types of unbalanced conditions In this chapter, unbalanced applied stator voltages, unbalanced stator impedances, open-circuited stator phase, and unbalanced rotor resistors of the three-phase induction Analysis of Electric Machinery and Drive Systems, Third Edition Paul Krause, Oleg Wasynczuk, Scott Sudhoff, and Steven Pekarek © 2013 Institute of Electrical and Electronics Engineers, Inc Published 2013 by John Wiley & Sons, Inc 336 337 SYMMETRICAL COMPONENT THEORY machine are considered Single-phase induction motors are analyzed, and several unbalanced and fault modes of synchronous machine operation are illustrated 9.2 SYMMETRICAL COMPONENT THEORY In 1918, C.L Fortescue [1] set forth the method of symmetrical components for the purpose of analyzing the steady-state behavior of power system apparatus during unbalanced operation Fortescue showed that the phasors representing an unbalanced set of steady-state multiphase variables could be expressed in terms of balanced sets of phasors For example, the phasors representing an unbalanced three-phase set can be expressed in terms of (1) a balanced set of phasors with an abc sequence (the positive sequence), (2) a balanced set of phasors with an acb sequence (the negative sequence), and (3) a set of three equal phasors (the zero sequence) Although the method of symmetrical components is widely used in the analysis of unbalanced static networks [2], it is perhaps most appropriate for the analysis of symmetrical induction machines during unbalanced operations Fortescue’s change of variables is a complex transformation that may be written for three-phase stationary circuits in phasor form as F+ − 0s = SFabcs (9.2-1) where the symmetrical components are (F+ − s )T = ⎡F+ s F− s F0 s ⎤ ⎣ ⎦ (9.2-2) (Fabcs )T = ⎡Fas Fbs Fcs ⎤ ⎣ ⎦ (9.2-3) The unbalanced phasors are and the transformation is expressed ⎡1 a 1⎢ S = ⎢1 a ⎢1 ⎣ a2 ⎤ a⎥ ⎥ 1⎥ ⎦ (9.2-4) The quantity a is complex, denoting a counterclockwise rotation of 2π/3 rad That is a = e j ( 2π / 3) = − + j 2 (9.2-5) a = e j ( 4π / 3) = − − j 2 (9.2-6) 338 UNBALANCED OPERATION AND SINGLE-PHASE INDUCTION MACHINES The inverse transformation is ⎡1 ( S) = ⎢ a ⎢ ⎢ ⎣a −1 1⎤ a 1⎥ ⎥ a 1⎦ ⎥ (9.2-7) The +, −, and subscripts denote the positive, negative, and zero sequences, respectively Hence, for the induction machine shown in Figure 4.2-1, F+ s, a F+ s , and aF+ s make up the positive sequence set of phasors, which we will denote as, Fas+ , Fbs+ , and Fcs+, respectively Similarly, F− s, aF− s, and a F− s are the negative sequence set, here denoted as Fas−, Fbs− , and Fcs− The zero sequence is the steady-state version of the zero quantities introduced in Chapter Although we will assume that only one frequency is present, the method of symmetrical components can be applied to each frequency present in the system [6] In order to compute the total instantaneous, steady-state power, it is first necessary to convert the phasors to the variables in sinusoidal form and then multiply phase voltage times phase current Thus Pabc = Vas I as + Vbs I bs + Vcs I cs = (Vas + + Vas − + V0 s )( I as + + I as − + I s ) + (Vbs + + Vbs − + V0 s )( I bs + + I bs − + I s ) + (Vcs + + Vcs − + V0 s )( I cs + + I cs − + I s ) (9.2-8) Uppercase letters are used to denote steady-state sinusoidal quantities We will show that the instantaneous power consists of an average value and time-varying components 9.3 SYMMETRICAL COMPONENT ANALYSIS OF INDUCTION MACHINES Although the method of symmetrical components can be used to analyze unbalanced conditions other than unbalanced stator voltages, this application of symmetrical components is perhaps the most common Once the unbalanced applied stator voltages are known, (9.2-1) can be used to determine Vas+ , Vas− , and V0 s It is clear that the currents due to the positive-sequence balanced set can be determined from the voltage equations given by (6.9-11)–(6.9-13) These equations are rewritten here with the + added to the subscript to denote positive-sequence phasors ω ω ⎛ ⎞ Vas + = ⎜ rs + j e Xls ⎟ I as + + j e X M ( I as + + I ar + ) ′ ⎝ ⎠ ωb ωb (9.3-1) Var + ⎛ rr′ ω ω ′ ⎞ = ⎜ + j e Xlr ⎟ I ar + + j e X M ( I as + + I ar + ) ′ ′ ′ ⎝s ⎠ s ωb ωb (9.3-2) where ωb is the base electrical angular velocity generally selected as rated and UNBALANCED STATOR CONDITIONS OF INDUCTION MACHINES s= ωe − ωr ωe 339 (9.3-3) As mentioned earlier, the method of symmetrical components can be applied to all frequencies present in the system For example, the unbalanced applied voltages could be of any arbitrary periodic waveform whereupon the frequencies that exist in the Fourier series expression of these voltages could each be broken up into positive, negative, and zero sequences It is customary to obtain the voltage equations for the negative-sequence quantities by reasoning In particular, slip is the normalized difference between the speed of the rotating air-gap MMF and the rotor speed The negative-sequence currents establish an air-gap MMF that rotates a ωe in the clockwise direction Hence, the normalized difference between the air-gap MMF and rotor speed is (ωe + ωr)/ωe which can be written as (2 − s), where s is defined by (9.3-3) Therefore, the voltage equations for the negative-sequence quantities can be obtained by replacing s by (2 − s) in (9.3-1) and (9.3-2) In particular, ω ω ⎛ ⎞ Vas − = ⎜ rs + j e Xls ⎟ I as − + j e X M ( I as − + I ar − ) ′ ⎝ ⎠ ωb ωb (9.3-4) Var − ⎛ rr′ ω ω ′ ⎞ =⎜ + j e Xlr ⎟ I ar − + j e X M ( I as − + I ar − ) ′ ′ ′ ⎠ 2−s ⎝2−s ωb ωb (9.3-5) If the zero-sequence quantities exist in a three-phase induction machine, the steady-state variables may be determined from the phasor equivalent of (6.5-24) and (6.5-27) The electromagnetic torque may be calculated using sequence quantities; however, the derivation of the torque expression is deferred until later 9.4 UNBALANCED STATOR CONDITIONS OF INDUCTION MACHINES: REFERENCE-FRAME ANALYSIS The theory of symmetrical components set forth in the previous sections can be used to analyze most unbalanced steady-state operating conditions However, one tends to look for a more rigorous development of this theory and straightforward procedures for applying it to unbalanced conditions, such as an open stator phase or unbalanced rotor resistors Reference-frame theory can be useful in achieving these goals [6] Although simultaneous stator and rotor unbalanced conditions can be analyzed, the notation necessary to formulate such a generalized method of analysis becomes quite involved Therefore, we will consider stator and rotor unbalances separately In Reference 6, the stator variables are expressed as a series of sinusoidal functions with time-varying coefficients Such an analysis is notationally involved and somewhat difficult to follow We will not become that involved because the concept can be established by assuming a single-stator (rotor) frequency for stator (rotor) unbalances with constant coefficients We will discuss the restrictions imposed by these assumptions as we go along 340 UNBALANCED OPERATION AND SINGLE-PHASE INDUCTION MACHINES Unbalanced Machine Variables in the Arbitrary Reference Frame If we assume that the rotor is a three-wire symmetrical system and the stator applied voltages are single frequency, then unbalanced steady-state stator variables may be expressed as ⎡ Fas ⎤ ⎡ Fasα ⎢F ⎥ = ⎢F ⎢ bs ⎥ ⎢ bsα ⎢ Fcs ⎥ ⎢ Fcsα ⎣ ⎦ ⎣ Fasβ ⎤ ⎡cos ω e t ⎤ Fbsβ ⎥ ⎢ ⎥ ⎣ sin ω e t ⎥ ⎦ Fcsβ ⎥ ⎦ (9.4-1) Transforming the stator variables as expressed by (9.4-1) to the arbitrary reference frame by (3.3-1) reveals an interesting property In particular, we have ⎡ FqsA ⎤ ⎥ ⎢ ⎡ Fqs ⎤ ⎡ cos(ω e t − θ ) sin(ω e t − θ ) cos(ω e t + θ ) sin(ω e t + θ ) ⎤ ⎢ FqsB ⎥ =⎢ ⎢F ⎥ ⎥ ⎣ ds ⎦ ⎣ − sin(ω e t − θ ) cos(ω e t − θ ) sin(ω e t + θ ) − cos(ω e t + θ ) ⎦ ⎢ FqsC ⎥ ⎥ ⎢ ⎣ FqsD ⎦ (9.4-2) and F0 s = Fabcsα cos ω e t + Fabcsβ sin ω e t (9.4-3) where FqsA = ⎤ 1⎡ 1 ( Fbsβ − Fcsβ ) ⎥ ⎢ Fasα − Fbsα − Fcsα + 3⎣ 2 ⎦ = − FdsB FqsB = ⎤ 1⎡ 1 ( Fbsα − Fcsα ) ⎥ ⎢ Fasβ − Fbsβ − Fcsβ − 3⎣ 2 ⎦ = FdsA FqsC = (9.4-6) ⎤ 1⎡ 1 ( Fbsα − Fcsα ) ⎥ ⎢ Fasβ − Fbsβ − Fcsβ + 3⎣ 2 ⎦ = − FdsC Fabcsα = (9.4-5) ⎤ 1⎡ 1 ( Fbsβ − Fcsβ ) ⎥ ⎢ Fasα − Fbsα − Fcsα − 3⎣ 2 ⎦ = FdsD FqsD = (9.4-4) ( Fasα + Fbsα + Fcsα ) (9.4-7) (9.4-8) UNBALANCED STATOR CONDITIONS OF INDUCTION MACHINES Fabcsβ = ( Fasβ + Fbsβ + Fcsβ ) 341 (9.4-9) It is recalled that dθ =ω dt (9.4-10) where ω is the electrical angular velocity of the arbitrary reference frame It is interesting that the qs and ds variables form two, two-phase balanced sets in the arbitrary reference frame In order to emphasize this, (9.4-2) is written with sinusoidal functions of (ωet − θ) separated from those with the argument of (ωet + θ) It is possible to relate these balanced sets to the positive- and negative-sequence variables For this purpose, let us consider the induction machine in Figure 6.2-1 A balanced three-phase set of currents of abc sequence will produce an air-gap MMF that rotates counterclockwise at an angular velocity of ωe By definition, the positive sequence is the abc sequence, and it is often referred to as the positively rotating balanced set The negative sequence, which has the time sequence of acb, produces an air-gap MMF that rotates in the clockwise direction It is commonly referred to as the negatively rotating balanced set Let us think of the qs and ds variables as being associated with windings with their magnetic axes positioned relative to the magnetic axis of the stator windings as shown in Figure 6.2-1 Now consider the series of balanced sets formed by the qs and ds variables in (9.4-2) The two balanced sets of currents formed by the first and second column of the × matrix produce an air-gap MMF that rotates counterclockwise relative to the arbitrary reference frame whenever ω < ωe and always counterclockwise relative to the actual stator winding Hence, the balanced sets with the argument (ωet − θ) may be considered as positive sequence or positively rotating sets because they produce counterclockwise rotating air-gap MMFs relative to the stator windings It follows that the balanced sets formed by the third and fourth columns with the argument (ωet + θ) can be thought of as negative sequence or negatively rotating sets Therefore, we can express (9.4-2) as Fqs = Fqs + + Fqs − (9.4-11) Fds = Fds + + Fds − (9.4-12) The zero-sequence variables are expressed by (9.4-3) It is important to note that Fqs+ (Fds+) is the positive-sequence terms of Fqs(Fds), and that together Iqs+ with Ids+ produce the positive-sequence rotating air-gap MMF Similarly, Iqs− with Ids−, produce the negative-sequence rotating air-gap MMF It is understood that the qs and ds variables may be expressed in any reference frame by setting ω, in (9.4-2), equal to the angular velocity of the reference frame of interest For example, ω = for the stationary reference frame and ω = ωr for the rotor reference frame It is also clear that the instantaneous sequence sets may be identified in these reference frames In particular, when ω is set equal to zero, the frequency of the variables is ωe; however, the positive and negative sequence are immediately 342 UNBALANCED OPERATION AND SINGLE-PHASE INDUCTION MACHINES identifiable When ω = ωr, we see that there are two electrical angular frequencies present; ωe − ωr and ωe + ωr The latter occurs due to the air-gap MMF established by negative-sequence stator variables Because we have assumed that the rotor circuits form a three-wire symmetrical system only one positive and one negative sequence set will be present in the rotor However, the balanced negative-sequence set of rotor variables will appear in the arbitrary reference frame with the same frequency (ωet + θ) as the negative sequence set established by the stator unbalance It follows that for stator unbalances and the assumption that the rotor is a three-wire symmetrical system, (9.4-2)–(9.4-9) can be used to identify the positive- and negativesequence rotor variables We only need to (1) replace all s subscripts with r in (9.4-2)– (9.4-7), (2) add a prime to all quantities associated with the rotor variables and (3) set θ in (9.4-2) equal to ωrt Recall that the rotor variables are transformed to the arbitrary reference frame by (6.4-1), wherein β is defined by (6.4-5) Please don’t confuse the β given by (6.4-5) and the β used in the subscripts starting with (9.4-1)–(9.4-9) The instantaneous electromagnetic torque may be expressed in arbitrary referenceframe variables as ⎛ 3⎞ ⎛ P ⎞ X Te = ⎜ ⎟ ⎜ ⎟ M ( I qs I dr − I ds I qr ) ′ ′ ⎝ ⎠ ⎝ ⎠ ωb (9.4-13) If we use (9.4-2) to express Iqs and Ids and the equivalent expression for I qr and I dr , the ′ ′ torque, for a stator unbalance, may be expressed as ⎛ 3⎞ ⎛ P ⎞ ⎛ X ⎞ Te = ⎜ ⎟ ⎜ ⎟ ⎜ M ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ωb ⎠ ′ ′ ′ ( I qsA I qrB − I qsB I qrA − I qsC I q′rD + I qsD I qrC ) + ( − I qsA I qrD + I qsC I qrB − I qsB I qrC + I qsD I qrA ) cos 2ω e t o ′ ′ ′ ′ + ( I qsA I qrC − I qsB I qrD − I qsC I qrA + I qsD I qrB ) sin 2ω e t ′ ′ ′ ′ (9.4-14) Torque may be expressed in per unit by expressing XM in per unit and eliminating the factor (3/2)(P/2)(1/ωb) Phasor Relationships The phasors representing the instantaneous sequence quantities Fqs and Fds for a stator unbalance given in (9.4-2)–(9.4-7) may be written as Fqs + = FqsA − jFqsB (9.4-15) Fds + = FqsB + jFqsA = j Fqs + (9.4-16) UNBALANCED STATOR CONDITIONS OF INDUCTION MACHINES Fqs − = FqsC − jFqsD 343 (9.4-17) Fds − = − FqsD − jFqsC = − j Fqs − (9.4-18) If (9.4-15)–(9.4-18) are used to express (9.4-11) and (9.4-12) in phasor form, we have ⎡ Fqs ⎤ ⎡1 ⎤ ⎡ Fqs + ⎤ ⎢ ⎥ = ⎢ j − j ⎥ ⎢F ⎥ ⎦ ⎣ qs − ⎦ ⎣ Fds ⎦ ⎣ (9.4-19) Solving (9.4-19) for Fqs+ and Fqs− yields ⎡ Fqs + ⎤ ⎡1 − j ⎤ ⎡ Fqs ⎤ ⎢ ⎥= ⎢ ⎥ ⎢ ⎥ ⎣ Fqs − ⎦ ⎣1 j ⎦ ⎣ Fds ⎦ (9.4-20) Equation (9.4-20) is the symmetrical component transformation for a two-phase system and (9.4-19) is the inverse The zero sequence may be expressed in phasor form from (9.4-3) as F0 s = Fabcsα − jFabcsβ (9.4-21) If we add the zero sequence to (9.4-20), we can write ⎡ Fqs + ⎤ ⎡1 − j ⎤ ⎡ Fqs ⎤ ⎢ ⎥ 1⎢ ⎥ ⎢ ⎥ ⎢ Fqs − ⎥ = ⎢1 j ⎥ ⎢ Fds ⎥ ⎢ F0 s ⎥ ⎥ ⎣ ⎦ ⎢ ⎣0 ⎦ ⎢ F0 s ⎥ ⎣ ⎦ (9.4-22) Recall from our work in Chapter that for balanced sets, the phasors are the same in all asynchronously rotating reference frames The frequency of the balance set in a specific reference frame need to be considered only when expressing the instantaneous balanced set Therefore, except for F0 s, all phasors in (9.4-22) are valid in any asynchronous reference frame, and it is not necessary to use a superscript, although we will for clarity Let us write (9.4-22) as ⎡ Fqss+ ⎤ ⎡1 − j ⎤ ⎡ Fas ⎤ ⎢ s ⎥ 1⎢ ⎥ s⎢ ⎥ ⎢ Fqs − ⎥ = ⎢1 j ⎥ K s ⎢ Fbs ⎥ ⎢ F0 s ⎥ ⎢0 ⎥ ⎢ Fcs ⎥ ⎦ ⎣ ⎦ ⎣ ⎣ ⎦ (9.4-23) What have we done? Well, first we added a superscript to the variables given in (9.4-22) to give us the sense of being in the stationary reference frame Next, we transs s formed Fqs , Fds , and F0 s to Fas , Fbs, and Fcs by K s With θ = 0, K s becomes s s 344 UNBALANCED OPERATION AND SINGLE-PHASE INDUCTION MACHINES ⎡ ⎢1 ⎢ 2⎢ s Ks = ⎢ ⎢ ⎢1 ⎢2 ⎣ − 2 − 1⎤ − ⎥ ⎥ 3⎥ ⎥ ⎥ ⎥ ⎥ ⎦ (9.4-24) Because K s is not a function of time, it is permissible to use it to transform phasors It s can be shown that ⎡1 − j ⎤ 1⎢ S = j 0⎥ Ks ⎥ s 2⎢ ⎢0 ⎥ ⎣ ⎦ (9.4-25) s Fqs ± s = SFabcs (9.4-26) Thus (9.4-23) may be written as where S is the symmetrical component transformation given by (9.2-4) Therefore, (9.4-26) is (9.2-1), and we have s Fqs + = F+ s (9.4-27) = F− s (9.4-28) s qs − F Now that we are working in the stationary reference frame, let us rewrite (9.4-20) as s s ⎡ Fqs + ⎤ ⎡1 − j ⎤ ⎡ Fqs ⎤ ⎢ s ⎥ = ⎢1 j ⎥ ⎢ s ⎥ ⎣ ⎦ ⎣ Fds ⎦ ⎣ Fqs − ⎦ (9.4-29) For stator unbalances with symmetrical three-wire rotor circuits, (9.4-29) also applies to rotor phasors in the stationary reference frame That is ′ ⎡ Fqrs+ ⎤ ⎡1 − j ⎤ ⎡ Fqrs ⎤ ′ ⎢ s ⎥ = ⎢1 j ⎥ ⎢ s ⎥ ′ ′ ⎣ ⎦ ⎣ Fdr ⎦ ⎣ Fqr − ⎦ (9.4-30) The steady-state voltage equations in the stationary reference frame may be obtained from (6.5-34) by setting ω = and p = jωe Thus ωe ⎡ ⎢rs + j ω b X ss ⎢ s ⎡ Vqs ⎤ ⎢ ⎢ s⎥ ⎢ ⎢ Vds ⎥ = ⎢ ⎢Vqrs ⎥ ⎢ ω e ′ XM ⎢ s⎥ ⎢ j ′ ⎣Vdr ⎦ ⎢ ω b ⎢ ωr ⎢ ω XM ⎣ b rs + j ωe X ss ωb ω − r XM ωb ωe j XM ωb j ωe XM ωb ω rr + j e Xrr ′ ωb ωr Xrr ′ ωb ⎤ ⎥ ⎥ s ωe ⎥ ⎡ I qs ⎤ j XM ⎥ ⎢ s ⎥ ωb I ⎥ ⎢ dss ⎥ ωr ′ ⎥ ⎢ I qr ⎥ − Xrr ⎥ ⎢ ⎥ ′ ωb ′s ⎥ ⎣ I dr ⎦ ω ⎥ rr′ + j e Xrr ⎥ ′ ωb ⎦ (9.4-31) 345 UNBALANCED STATOR CONDITIONS OF INDUCTION MACHINES If we now incorporate (9.4-29) and (9.4-30) into (9.4-31), the sequence voltage equations become ⎡ rs ⎡V ⎤ ⎢ ⎢ ⎢ s ⎥ ′ ⎢ Vqr + ⎥ ⎢ ⎢ s ⎥ ⎢ ⎢ s ⎥=⎢ ⎢ Vqs − ⎥ ⎢ ⎢ Vqrs− ⎥ ⎢ ′ ⎢ ⎥ ⎢ ⎣2 − s⎦ ⎢ ⎢ ⎣ s qs + ωe XM ωb ωe XM ωb ω rr′ + j e Xrr ′ ωb s 0 0 +j j ωe X ss ωb j 0 ωe X ss ωb ω j e XM ωb rs + ⎤ ⎥ ⎥ s ⎥ ⎡ I qs + ⎤ ⎥ ⎢ I qrs+ ⎥ ′ ⎥⎢ s ⎥ ⎢ I qs − ⎥ ω ⎥ j e XM ⎥ ⎢I ′s ⎥ ωb ⎥ ⎣ qr − ⎦ ωe rr′ ⎥ +j Xrr ⎥ ′ 2−s ωb ⎦ (9.4-32) where s= ωe − ωr ωe (9.4-33) The steady-state electromagnet torque, for a stator unbalance, may be expressed ⎛ P⎞ X s s *s *s Te = ⎜ ⎟ M Re ⎡ j ( I qs + I qrs+ − I qs − I qrs− )⎤ + Re ⎡ j ( − I qs + I qrs− + I qs − I qrs+ )⎤ cos 2ω e t ′ ′ ⎦ ′ ′ ⎦ ⎣ ⎣ ⎝ ⎠ ωb s s + Re ⎡I qs + I qrs− − I qs − I qrs+ ⎤sin 2ω e t ′ ⎦ ′ ⎣ (9.4-34) where the asterisk denotes the conjugate Steady-state instantaneous stator variables, Fas, Fbs, and Fqs, may be obtained from the phasors Fas , Fbs, and Fcs determined from (9.4-23) With the assumption of a symmetrical rotor, the frequency of the stator variables is ωe; however, as has been mentioned the rotor variables Far , Fbr , and Fcr each contain two frequencies (ωe − ωr) and ′ ′ ′ (ωe + ωr) The instantaneous rotor variables can be obtained by using the rotor equivalent of (9.4-15)–(9.4-18) to identify FqrA through FqrD, which can be substituted into the ′ ′ rotor equivalent of (9.4-2) To obtain Fqrr and Fdrr, θ in (9.4-2) must be set to ωrt, where′ ′ upon K r may be used to obtain Far , Fbr , and Fcr Since we have assumed symmetrical ′ ′ ′ r three-wire stator and rotor circuits, neither F0s nor F0′r exist; however, we have included the zero-sequence notation in order to show the equivalence to the symmetrical component transformation If a zero sequence were present in the rotor circuits, it would consist of two frequencies, unlike the stator zero sequence, which would contain only ωe 362 UNBALANCED OPERATION AND SINGLE-PHASE INDUCTION MACHINES ( Fasα + Fbsβ ) = FdsD (9.8-23) ( Fasβ + Fbsα ) = − FdsC (9.8-24) FqsC = FqsD = The sinusoidal variations with the argument of (ωet − θ) are the positive-sequence variables; the negative-sequence variables have the argument (ωet + θ) We can write (9.8-20) as Fqs = Fqs + + Fqs − (9.8-25) Fds = Fds + + Fds − (9.8-26) The torque equations given by (9.4-13) and (9.4-14) can be used to calculate the torque for a two-phase machine if the (3/2) factor is eliminated, and XM is replaced by Xms, and ωb is generally the rated angular velocity Phasor Relationships If we express the variables in the stationary reference frame in phasor form wherein θ s s is set equal to zero; then Fas = Fqs and Fbs = − Fds , and we can write from (9.4-20) s ⎡ Fqs+ ⎤ ⎡1 j ⎤ ⎡ Fas ⎤ ⎢ s ⎥ = ⎢1 − j ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ Fbs ⎦ ⎣ Fqs− ⎦ (9.8-27) where the symmetrical component transformation for a two-phase system is S= ⎡1 j ⎤ ⎢1 − j ⎥ ⎣ ⎦ ⎡ 1⎤ (S)−1 = ⎢ ⎥ ⎣− j j ⎦ (9.8-28) (9.8-29) Following a similar procedure, we can write Fqrs = Fqrs+ + Fqrs− ′ ′ ′ (9.8-30) F′ = F′ + F′ (9.8-31) s Fdr + = jFqrs+ ′ (9.8-32) = − jF ′ (9.8-33) s dr s dr + s dr − and s dr − F s qr − 363 SINGLE-PHASE INDUCTION MACHINES The steady-state torque may be calculated using (9.4-34) if the three multiplier is replaced by two and XM with Xms Unbalanced Stator Impedances During the capacitor-start and capacitor-start, capacitor-run modes of operation, a capacitor is placed in series with one of the stator windings For this purpose, let ega = ias z( p) + vas (9.8-34) egb = vbs (9.8-35) where νas and νbs are the voltages across the stator phases and ega and egb are source voltages, and z(p) is the operational impedance in series with the as-winding In phasor form, we have Vas = Ega − I as Z (9.8-36) Vbs = Egb (9.8-37) where Z is the impedance Following a procedure similar to that for the three-phase induction machine, we can write ωe ⎡1 ⎢ Z + rs + j ω b X ss ⎢ ⎡ E1 ⎤ ⎢ ω j e X ms ⎢ ⎥ ⎢ ωb ⎢0 ⎥=⎢ ⎢ E2 ⎥ ⎢ Z ⎢ ⎥ ⎢ ⎣0⎦ ⎢ ⎢ ⎢ ⎣ ωe X ms ωb rr′ ω + j e Xrr ′ s ωb j 0 Z ω Z + rs + j e X ss ωb ωe j X ms ωb ⎤ ⎥ ⎥ s ⎥ ⎡ I qs + ⎤ ⎥ ⎢I ′s ⎥ ⎥ ⎢ qr + ⎥ s ωe ⎥ ⎢ I qs − ⎥ j X ms ⎢ s ⎥ ⎥ ωb ′ ⎥ ⎣ I qr − ⎦ rr′ ωe ⎥ +j Xrr ⎥ ′ 2−s ωb ⎦ (9.8-38) where E1 = ( Ega + jEgb ) (9.8-39) E2 = ( Ega − jEgb ) (9.8-40) 364 UNBALANCED OPERATION AND SINGLE-PHASE INDUCTION MACHINES Open-Circuited Stator Phase In the capacitor-start mode of operation, the induction machine operates with only one winding energized once the rotor has reached 60–80% of rated speed If we assume it s is the as-winding, which is open-circuited then because vas is vqs with θ = s s vqs = pλqs (9.8-41) s Because ias is zero, iqs is zero, and from (9.8-15), we can write s λqs = Lms iqrs ′ (9.8-42) Thus, during open-circuit operation, we have vas = X ms piqrs ′ ωb (9.8-43) vbs = egb (9.8-44) s Vqs + = X ms s j I qr + jEgb ′ ωb (9.8-45) s Vqs − = X ms s j I qr − jEgb ′ ωb (9.8-46) I qs − = − I qs + (9.8-47) Substituting into (9.8-27) yields and because I as = 0, we have Following a procedure similar to that for the three-phase machine, we can write ωe ⎡ ⎡ ⎤ ⎢rs + j ω X ss b ⎢ − jEgb ⎥ ⎢ ⎢ ⎥ ⎢ ωe ⎢ ⎥ = ⎢ j ω X ms b ⎢ ⎥ ⎢ ⎢ ⎥ ⎢ ω ⎣ ⎦ ⎢ − j e X ms ωb ⎣ ωe X ms ωb ω rr′ + j e Xrr ′ ωb s j ωe ⎤ X ms ⎥ ωb s ⎥ ⎡ I qs + ⎤ ⎥⎢ s ⎥ ′ ⎥ ⎢ I qr + ⎥ ⎥ ⎢ I qrs− ⎥ ′ ⎦ ω rr′ ⎥⎣ + j e Xrr ⎥ ′ ωb 2−s ⎦ −j (9.8-48) 365 SINGLE-PHASE INDUCTION MACHINES Torque, N•m Te,pul Te,av –188.5 –94.2 94.2 188.5 –1 wrm , rad/s –2 –3 Figure 9.8-2 Steady-state torque versus speed characteristics for a single-phase induction motor Operating Characteristics of Single-Phase Induction Motors The steady-state torque versus speed characteristics are shown in Figure 9.8-2 for a symmetrical two-phase induction motor with rated voltage applied to one phase and the other phase open circuited The symmetrical two-phase induction machine is a four-pole, (1/4)-hp, 110-V, 60-Hz motor with the following parameters: rs = 2.02 Ω, Xls = 2.79 Ω, Xms = 66.8 Ω, rr′ = 4.12 Ω, and Xlr = 2.12 Ω The total inertia is ′ J = 1.46 × 10−2 kg · m2 The average steady-state electromagnetic torque Te,av = Te+ − Te− and the magnitude of the double-frequency component of the torque |Te,pul| are plotted in Figure 9.8-2 There are at least two features worth mentioning First, the plot of the average torque Te,av for ωrm < is the negative mirror image of that for ωrm > Second, the plot of the pulsating torque |Te,pul| is symmetrical about the zero speed axis Finally, we see that the torque is zero at ωrm = If we take a two-phase symmetrical induction motor and apply the same singlephase voltage to both phases, the net torque at stall will be zero because the winding currents will be instantaneously equal and the air-gap MMF will pulsate along an axis midway between the as- and bs-axes Consequently, two equal and oppositely rotating air-gap MMFs result If, however, we cause the current in one of the phases to be different instantaneously from that in the other phase, a starting torque can be developed because this would make one of the rotating air-gap MMFs larger than the other One way of doing this is to place a capacitor in series with one of the windings of a twophase symmetrical induction motor This will cause the current in the phase with the 366 UNBALANCED OPERATION AND SINGLE-PHASE INDUCTION MACHINES Figure 9.8-3 Steady-state torque versus speed characteristics with a capacitor in series with one winding of the two-phase induction machine series capacitor to lead the current in the other winding when the same voltage is applied to both We have already derived the equations necessary to calculate the component currents with an impedance in series with the as winding In particular, (9.8-38) can be used to calculate the component currents with a capacitor in series with the as winding If we set Z = −j1/ωeC, and let Ega = Egb , the single-phase source voltage, counterclockwise rotation will occur because I as will lead I bs Recall that for the assumed positive direction of the magnetic axes and for a balanced two-phase set, we have had I as leading I bs by 90° for counterclockwise rotation of the air-gap MMF These steady-state torque versus speed characteristics are shown in Figure 9.8-3 for C = 530.5 μF In capacitor-start, single-phase induction motors, the winding with the series capacitor is disconnected from the source after the rotor has reached 60–80% of synchronous speed This is generally accomplished by a centrifugal switching mechanism located inside the housing of the motor Once the winding with the series capacitor is disconnected, the device then operates as a single-phase induction motor In Figure 9.8-4, the plot of average torque versus speed with a series capacitor in one phase (Fig 9.8-3) is superimposed upon the plot of average torque versus speed with a single-phase winding (Fig 9.8-2) The transition from capacitor-start to single-phase operation at 75% of synchronous speed is illustrated The free acceleration characteristics of the example capacitor-start single-phase induction motor are shown in Figure 9.8-5 The variables νas, ias, νbs, ibs, νc, Te, and ωrm are plotted The voltage vc is the instantaneous voltage across the capacitor that is connected in series with the bs winding, which is disconnected from the source at a normal current zero once the rotor reaches 75% of synchronous speed The voltage across the capacitor is shown to remain constant at its value when the bs winding is disconnected from the source In practice, this voltage would slowly decay owing to leakage currents within the capacitor, which are not considered in this analysis The SINGLE-PHASE INDUCTION MACHINES 367 Figure 9.8-4 Average steady-state torque versus speed characteristics of a capacitor-start single-phase induction motor Figure 9.8-5 Free-acceleration characteristics of a capacitor-start single-phase induction motor 368 UNBALANCED OPERATION AND SINGLE-PHASE INDUCTION MACHINES Figure 9.8-6 Torque versus speed characteristics for Figure 9.8-5 torque versus speed characteristics given in Figure 9.8-6 are for the free acceleration shown in Figure 9.8-5 9.9 ASYNCHRONOUS AND UNBALANCED OPERATION OF SYNCHRONOUS MACHINES The analysis of unbalanced operation of synchronous machines is quite involved, requiring manipulation of a large number of equations to calculate even steady-state performance This complexity is actually anticipated because the machine is electrically unsymmetrical in the rotor, and when a stator unbalance occurs, such as a lineto-ground fault, a series of harmonic voltages is induced due to the asymmetry of the rotor circuits and the network external to the machine Although the transient performance of a synchronous machine during unbalanced stator conditions is of interest, the analysis of steady-state performance is of limited value, since in practice, an unbalanced condition due to a system fault is generally removed before the machine reaches steady-state operation On the other hand, the analysis of steady-state performance of a synchronous machine during asynchronous operation (pole slipping) is of importance Over the years, there have been several approximate methods set forth to analyze unbalanced operation of synchronous machines [3, 9] For the most part, this work has been directed toward simplified representations of a synchronous machine to approximate its dynamic performance during unbalanced faults in the same way that the voltage behind transient reactance representation (Chapter 5) is used to approximate the dynamic behavior during symmetrical disturbances We will not become involved in a lengthy analyses because, as we have mentioned, the solutions are of limited value Instead, we will use computer solutions to demonstrate various modes of unbalanced operation ASYNCHRONOUS AND UNBALANCED OPERATION OF SYNCHRONOUS MACHINES ias, pu –5 ibs, pu –5 ics, pu 369 –5 0.1 second i¢ ifdr, pu –1 –2 Te, pu –1 –2 1.0 wr wb 0.5 Figure 9.9-1 Free acceleration of synchronous motor with H = 1.0 second and no load Motor Starting The free acceleration characteristics of a synchronous motor with the field winding short circuited are shown in Figure 9.9-1 The per unit parameters of this 60-Hz, eightpole, 4-kV, 6000-hp machine are rs Xls rk′q Xlkq ′ = 0.0121 X md = 1.03 rkd = 0.0302 ′ = 0.140 X mq = 0.75 Xlkd = 0.092 ′ = 0.039 H =1s rfd = 0.00145 ′ = 0.115 Xlfd = 0.267 ′ In Figure 9.9-1, ias, ibs, ics, i ′ r , Te, and ωr/ωb are plotted with the peak value of rated fd stator current taken as 1.0 pu The voltages applied to the stator windings are balanced, and ias is assumed positive into the machine and Te is positive for motor action During the acceleration period shown in Figure 9.9-1, the field winding is short-circuited through a discharge resistor, making the total field circuit resistance 0.3 pu The friction and windage losses are neglected, and since there is no mechanical load on the shaft of the machine, it accelerates to synchronous speed without applying a voltage to the field winding as the machine approaches synchronous speed The steady-state torque speed characteristics of this machine are shown in Figure 9.9-2 Therein the average torque Te,av and the zero-to-peak amplitude of the pulsating torque |Te,pul| are plotted These plots are calculated using a procedure similar to that in the case of unbalanced rotor resistors of an induction machine It is interesting to compare the plot of average 370 UNBALANCED OPERATION AND SINGLE-PHASE INDUCTION MACHINES 1.6 1.4 1.2 Te,av Torque, per unit 1.0 0.8 0.6 Te,pul 0.4 0.2 0.2 0.4 0.6 wr wb 0.8 1.0 Figure 9.9-2 Steady-state torque-speed characteristics of the same synchronous motor depicted in Figure 9.9-1 torque with that in Figure 9.7-3 where the “one-half speed dip” is more pronounced The pulling into step after applying the field voltage is illustrated in Figure 9.9-3 In this case, the machine is loaded with a small shaft load (0.2 pu) to simulate friction and windage losses and/or a small external mechanical load The total shaft inertia, including the mechanical load, is seconds The machine fails to pull into synchronism and instead operates as an induction motor at a speed slightly less than synchronous Only the very last portion of the acceleration period is shown At a normal field current zero, the discharge resistor is disconnected or bypassed and simultaneously, 1.0 pu field voltage ( E xfd ) is applied to the field winding The machine pulls into step and operates ′r as a synchronous motor ASYNCHRONOUS AND UNBALANCED OPERATION OF SYNCHRONOUS MACHINES 371 ics, pu –2 ′ ifdr, pu –1 Te, pu 1.0 pu field voltage applied 1.0 second wr wb 1.00 0.95 Figure 9.9-3 Pulling into step after applying 1.0 pu field voltage with H = 4.0 seconds and TL = 0.2 pu Pole Slipping Asynchronous operation of a hydro turbine generator is demonstrated in Figure 9.9-4 The parameters for the hydro turbine generator are given in Section 5.10 The machine is initially operating steadily at rated conditions as in Figure 5.10-8 In particular, the input torque is constant at 0.85 pu and E xfd fixed at 1.6 pu The stator and field cur′r rents are plotted along with Te, ωr/ωb, and δ A three-phase fault occurs at the terminals, which reduces the voltages instantaneously to zero at the instant νas passes through zero going positive In Figure 5.10-8, the fault was removed in 0.466 seconds; the machine returned to synchronous speed with the rotor speed and angle reaching maximum of approximately 2.2% (above synchronous speed) and 164°, respectively In Figure 9.9-4, the three-phase fault remains on the terminals of the machine for approximately 0.8 second When the fault is removed, the rotor speed is approximately 4% above synchronous speed and the rotor angle is approximately 400° In other words, the rotor has advanced more than one revolution ahead of the system Here, it is helpful to recall the definition of δ given by (5.7-1) When the rotor advances 360° from its original operating point, it is said to have “slipped a pole.” In Figure 9.9-4, the machine slips a pole before returning to synchronous speed If the fault had remained for only slightly longer, the machine would not have been able to return to synchronous speed Due to the possibility of damage to the mechanical system resulting from large average torques (in excess of pu in Fig 9.9-4), and due to the large stator currents that occur during pole slipping, power systems are generally designed to protect against its occurrence In particular, overspeed protection would prevent reclosing the machine back onto the system if the rotor speed exceeds a predetermined value of generally between 3% and 5% of synchronous speed of the rotor 372 UNBALANCED OPERATION AND SINGLE-PHASE INDUCTION MACHINES 10 –5 –10 10 ibs, pu –5 –10 10 ics, pu –5 –10 ias, pu ifdr, pu i¢ Te, pu 0.1 second –5 3-phase fault Fault cleared 1.04 wr 1.02 wb 1.00 0.98 0.96 200 100 d, degrees –100 –200 200+ Figure 9.9-4 Pole slipping of hydro turbine generator Line-to-Neutral Fault The behavior of the steam turbine generator during a line-to-neutral fault is shown in Figure 9.9-5 The parameters are given in Section 5.10 The machine is initially operating at rated conditions with the input torque equal to 2.48 pu for the steam turbine generator The line-to-neutral fault was applied to phase a at the instant νas passed through zero going positive It was assumed that the neutral of the wye-connected source voltage is connected to the neutral of the wye-connected stator windings Thus, during the fault vas = (9.9-1) vbs = egb (9.9-2) vcs = egc (9.9-3) where egb and egc are the source voltages After 0.25 second, the fault was removed and the source voltage (ega) instantaneously reapplied The currents ias, ibs, ics, and i ′ r are fd plotted along with Te, ωr/ωb and δ Following the fault, the initial steady-state operating condition is reestablished With the exception of the high value of ias, this fault is not severe; in fact, the machine would remain in synchronism during a sustained line-toneutral fault of this type ASYNCHRONOUS AND UNBALANCED OPERATION OF SYNCHRONOUS MACHINES ias, pu 10 –5 –10 ibs, pu –1 ics, pu –1 –2 373 ′r ifd, pu Te, pu –2 –4 1.002 wr 1.001 1.000 wb 0.999 0.998 d, degrees 0.1 second Line-toFault cleared neutral fault 40 20 Figure 9.9-5 Dynamic performance of a steam turbine generator during a line-to-neutral fault at the terminals Line-to-Line Fault The performance of the steam turbine generator during and following a line-to-line fault is shown in Figure 9.9-6 Again, the machine was initially operating at rated conditions Phases a and b were short-circuited at the terminals at the instant νas passed through zero going positive The fault was removed and the voltages reapplied after 0.25 second In each case, the neutral of the machine and source were connected, therefore, during the fault, we have vas = vbs (9.9-4) vcs = egc (9.9-5) Line-to-Line-to-Neutral Fault The behavior of the steam generator during and following a line-to-line-to-neutral (double line-to neutral) fault is demonstrated in Figure 9.9-7 With the machine operating at rated conditions, the terminals of phases a and b are short circuited to the neutral at the instant νas passes through zero going positive It is again assumed that the neutrals of the source voltages and the stator windings are connected Thus, during the fault we have 374 UNBALANCED OPERATION AND SINGLE-PHASE INDUCTION MACHINES 10 –5 –10 ias, pu ibs, pu 10 –5 –10 ics, pu –1 –2 ′r ifd, put 0.1 second Fault cleared Line-to-line fault –2 –4 1.008 1.004 1.000 0.996 0.992 Te, pu wr wb d, degrees 60 40 20 Figure 9.9-6 Dynamic performance of a steam turbine generator during a line-to-line fault at the terminals ias, pu 10 –5 –10 ibs, pu 10 –5 –10 ics, pu –1 –2 pu Te, pu –2 –4 1.008 1.004 wr 1.000 wb 0.996 0.992 0.1 second ′r ifd, Fault cleared Line-to-line-to-neutral fault 60 d, 40 degrees 20 Figure 9.9-7 Dynamic performance of a steam turbine generator during a line-to-line-toneutral fault at the terminals 375 PROBLEMS vas = vbs = (9.9-6) vcs = egc (9.9-7) REFERENCES [1] C.L Fortescue, “Method of Symmetrical Co-Ordinates Applied to the Solution of Polyphase Networks,” AIEE Trans., Vol 37, 1918, pp 1027–1115 [2] E Clarke, Circuit Analysis of A-C Power Systems, Vols I and II, John Wiley and Sons, New York, 1943 and 1950 [3] W.V Lyon, Transient Analysis of Alternating-Current Machinery, Technology Press of MIT and John Wiley and Sons, New York, 1954 [4] D.C White and H.H Woodson, Electromechanical Energy Conversion, John Wiley and Sons, New York, 1959 [5] Y.H Ku, Electric Energy Conversion, Ronald Press, New York, 1959 [6] P.C Krause, “The Method of Symmetrical Components Derived by Reference Frame Theory,” IEEE Trans Power App Syst., Vol 104, June 1985, pp 1492–1499 [7] P.C Krause, O Wasynczuk, and S Sudhoff, Analysis of Electric Machinery and Drive Systems, 2nd ed., IEEE Press, Piscataway, NJ, and John Wiley & Sons, New York, 2002 [8] H.L Garbarino and E.T.B Gross, “The Goerges Phenomenon—Induction Motors with Unbalanced Rotor Impedances,” AIEE Trans., Vol 69, 1950, pp 1569–1575 [9] C Concordia, Synchronous Machines, John Wiley and Sons, New York, 1951 PROBLEMS Show that, for a given source frequency, |Te,pul| is zero for ωr = when the stator voltages are unbalanced and the induction machine is singly fed Consider the stator circuit shown in Figure 9.5-1 Assume ega = cos 377t egb = 2π ⎞ ⎛ cos ⎜ 377t − ⎟ ⎝ 3⎠ π⎞ ⎛ egc = cos ⎜ 377t + ⎟ ⎝ 2⎠ The induction machine is the 10-hp machine given in Section 9.7 Calculate the torque for ωr/ωb of 0, 0.1, 0.2, , 1.0 Plot Te+, Te−, and |Te,pul| A dc voltage of 1.0 pu is applied across two stator terminals of the 10-hp machine given in Section 9.7 Plot the steady-state torque-speed characteristics for −1.0 ≤ ωr / ωb ≤ 1.0 376 UNBALANCED OPERATION AND SINGLE-PHASE INDUCTION MACHINES Figure 9P-1 Equivalent circuit for single-phase winding Determine the set of voltage equations similar to (9.5-18) that can be used to analyze the steady-state performance of an induction machine with an impedance z(p) in series with (a) the bs winding and (b) the cs winding Determine the set of voltage equations similar to (9.5-18) that can be used to analyze the steady-state performance of an induction machine with an impedance z(p) in series with the as winding and supplied from an unbalanced set of source voltages Express νsa in Figure 9.5-3 in terms of the source voltages and νas, then write a phasor expression for Vsa Derive the voltage equations in a form similar to (9.8-48) for (a) ibs = and (b) ics = Derive the voltage equations for a three-phase induction machine with iar = ′ What modifications must be made to (9.4-32) in order to analyze unbalanced stator voltages of a two-phase symmetrical induction motor 10 The equivalent circuit given in Figure 9P-1 is for an induction motor with only one stator winding Show that this equivalent circuit is the same as (9.8-48) ... (9. 5-21)– (9. 5-23) and solving the result for νng yields 350 UNBALANCED OPERATION AND SINGLE- PHASE INDUCTION MACHINES vng = 1 (egb + egc ) + vas 2 (9. 5-24) Substituting (9. 5-24) into (9. 5-22) and (9. 5-23)... (9. 5-8)– (9. 5-10), and we have vng = − ias z( p) (9. 5-11) 348 UNBALANCED OPERATION AND SINGLE- PHASE INDUCTION MACHINES Equation (9. 5-11) is valid as long as ega + egb + egc is zero Substituting (9. 5-11)... only ωe 346 UNBALANCED OPERATION AND SINGLE- PHASE INDUCTION MACHINES 9. 5 TYPICAL UNBALANCED STATOR CONDITIONS OF INDUCTION MACHINES Although it is not practical to consider all stator unbalanced