SYMMETRICAL FAULTS Revised: 10/8/13 1:49 PM

Transcription

1 SYMMETRICAL FAULTS Revised: 10/8/13 1:49 PM 10/8/13 Symmetrical Faults 1

2 What is a fault? A faults is any failure which interferes with the normal flow of current. Most faults on transmission lines of 115 kv and higher are caused by lightning, which results in the flashover of insulators causing a low impedance path to ground. Line- to- line faults not involving ground are less common. Experience shows that 70 to 80% of transmission line failures are single line- to- ground faults. Permanent faults are caused by lines being on the ground, insulator strings breaking, ice loads, and equipment failure. 10/8/13 Symmetrical Faults 2

3 What is a fault? Roughly 5% of all faults involve all three phases and these are called symmetrical three- phase faults or just symmetrical faults. Line- to- line faults that do not involve ground and double line- to- ground faults are called unsymmetrical faults since they cause an imbalance between the phases. The currents which flow in a power system immediately a\er the occurrence of a fault differ from those flowing a few cycles later just before the circuit breakers are called upon to open the line on both sides of the fault. 10/8/13 Symmetrical Faults 3

4 What is a fault? Two factors which determine the proper selec^on of circuit breakers are the current flowing immediately a\er the fault occurs and the current which the breaker must interrupt. In fault analysis values of these currents are calculated for different types of faults at different loca^ons in the system. The informa^on obtained form these calcula^ons are used to determine the relay se_ng that control the circuit breakers. 10/8/13 Symmetrical Faults 4

5 Transients in Series RL Circuits The selec^on of a circuit breaker depends on: normal opera^ng current the current it has to interrupt the maximum current it may have to momentarily carry How does an over- current occur? To approach the problem of calcula^ng the ini^al current when a system is short- circuited, consider what happens when an ac voltage is applied to an RL circuit. 10/8/13 Symmetrical Faults 5

6 Mo^va^on Transients in Series RL Circuits t = 0 L R V max sin( ωt + α ) i( t) Differen^al equa^on: Solu^on: i( t) = V max Z V max sin( ωt + α ) = Ri + L di dt sin( ωt + α θ ) e Rt L sin( α θ ) Z = R 2 + ω 2 L 2, θ = tan 1 ( ω L R) 10/8/13 Symmetrical Faults 6

7 Mo^va^on Transients in Series RL Circuits ( ) = V max i t Z α θ = 0 ( ) e Rt L sin( α θ ) sin ωt + α θ ac- component constant amplitude dc- component exponen^ally decaying α θ = π 2 What about when a real synchronous generator is shorted? 10/8/13 Symmetrical Faults 7

8 Review of Synchronous Generators. 10/8/13 Symmetrical Faults 8

9 The Synchronous Generator The synchronous generator is driven by a turbine to convert mechanical energy into electrical energy. The windings of the synchronous machine consdtute a group of inducdvely coupled electric circuits, some of which rotate reladve to others so that the mutual inductances are variable. The models developed for the the various windings are applicable to both steady- state and transient analysis. Only linear magnedc circuits are considered. 10/8/13 Symmetrical Faults 9

10 The Synchronous Generator The two principal parts of a synchronous machine are ferromagnedc structures. The stadonary part (a hollow cylinder), called the armature (see next slide), has longitudinal slots in which there are coils of the armature windings. These carry the current to an electrical load. The rotor is the part of the machine which is mounted on the sham and rotates ins ide the hollow stator. The winding on the rotor, called the field winding, is supplied with dc current. The flux across the air gap between the armature and rotor generates voltages in the coils of the armature windings. 10/8/13 Symmetrical Faults 10

11 The Synchronous Generator Two Types Armature Armature Gap S Rotor N Gap Rotor X X The cylindrical rotor is called a non-salient pole machine. The non-cylindrical rotor is called a salient pole machine. 10/8/13 Symmetrical Faults 11

12 The Synchronous Generator For the two- pole machine, one cycle of voltage is generated for each revoludon of the two- pole rotor. In the four- pole machine, two cycles are generated in each armature coil per revoludon. Since the number of cycles of revoludon equals the number of pairs of poles, the frequency of the generated voltage is: f = P 2 N 60 = P 2 f m Hertz where f is the electrical frequency in Hertz, P is the number of poles, N is the rotor speed in RPM, and f m = N/60 is the mechanical frequency in rpm. We see that a two- pole, 60 Hz machine operates at 3600 rpm, while a four- pole machine operates at 1800 rpm. 10/8/13 Symmetrical Faults 12

13 The Synchronous Generator Cylindrical Rotor, Non- Salient Pole Machine Gap Armature S Rotor N d- axis Direct Axis X X 10/8/13 Symmetrical Faults 13

14 The Synchronous Generator Non- cylindrical Rotor, Salient Pole Machine Armature WE LL COME BACK TO THIS SHORTLY. Gap S N Rotor N d- axis Direct Axis S q- axis Quadrature Axis 10/8/13 Symmetrical Faults 14

15 The Synchronous Generator - Cylindrical Rotor Model d- axis R, L aa a- axis i a a + v a Sta^onary Armature d- axis R, L ff i f rota^on + v f R, L f bb R, L cc b- axis + v b v c + i b i b c c c- axis 10/8/13 Symmetrical Faults 15

16 The Synchronous Generator - Cylindrical Rotor Model R, L aa a- axis i a a + d- axis v a Sta^onary Armature L af R, L ff i f rota^on + v f R, L bb f R, L cc L cf b- axis L bf + v b v c + i b i b c c c- axis 10/8/13 Symmetrical Faults 16

17 The Synchronous Generator - Cylindrical Rotor Model R, L aa a- axis i a a + d- axis v a Sta^onary Armature L af L ab = L ba = M L = L = M s ac ca s R, L ff L aa = L bb = L cc = L s i f rota^on + v f R, L bb f R, L cc L cf b- axis L bf + v b v L i bc = L cb = M c + s b i b c c c- axis 10/8/13 Symmetrical Faults 17

18 The Synchronous Generator - Cylindrical Rotor Model θ d R, L aa a- axis i a a + d- axis v a Sta^onary Armature L af L ab = L ba = M L = L = M s ac ca s R, L ff L aa = L bb = L cc = L s i f rota^on + v f R, L bb f R, L cc L cf b- axis L bf + v b v L i bc = L cb = M c + s b i b c c c- axis 10/8/13 Symmetrical Faults 18

19 The Synchronous Generator - Cylindrical Rotor Model L af = M f cosθ d L bf = M f cos( θ d 120 ) L cf = M f cos( θ d 240 ) 10/8/13 Symmetrical Faults 19

20 Armature Flux Linkages λ a = L aa i a + L ab i b + L ac i c + L af i f = L s i a M s λ b = L ba i a + L bb i b + L bc i c + L bf i f = L s i b M s λ c = L ca i a + L cb i b + L cc i c + L cf i f = L s i c M s Armature Flux Linkages ( i b + i ) c + L af i f ( i a + i ) c + L bf i f ( i a + i ) b + L cf i f Balanced Three- Phase System λ f = L af i a + L bf i b + L cf i c + L ff i f i a + i b + i c = 0 10/8/13 Symmetrical Faults 20

21 Algebra λ a = ( L s + M s )i a + L af i f λ b = ( L s + M s )i b + L bf i f λ c = ( L s + M s )i c + L cf i f Since the field current is dc and the field rotates with constant angular velocity, dθ d dt = ω, θ d = ωt +θ o, i f = I f λ a = ( L s + M s )i a + M f I f cos( ωt +θ ) o λ b = ( L s + M s )i b + M f I f cos( ωt 120 +θ ) o λ c = ( L s + M s )i c + M f I f cos( ωt 240 +θ ) o 10/8/13 Symmetrical Faults 21

22 Circuit Equa^ons dλ v a = R a i a a dt generator = R a i a ( L s + M ) di a s dt + ( ) ω M I sin ωt +θ f f o ( ) e a = 2 E i sin ωt +θ o Internal emf: e a = 2 E i sin( ωt +θ ) o (or synchronous emf) where: E i = ω M f I f 2 For convenience (it s arbitrary anyway) set θ o = δ + 90 We have the circuit model 10/8/13 Symmetrical Faults 22

23 The Synchronous Generator - Cylindrical Rotor Model v a = R a i a ( L s + M ) di a s dt + ω M f I f sin( ωt +θ ) o ~ + _ R L s + M s e b e c i a e a = 2 E i cos( ωt + δ ) a + v a _ L s + M s L s + M s R R 10/8/13 Symmetrical Faults 23 i c v c + c i b b v b +

24 The Synchronous Generator - Cylindrical Rotor Model Also: i a = 2 I a cos( ωt + δ θ ) a i b = 2 I a cos( ωt + δ θ a 120 ) i c = 2 I a cos( ωt + δ θ a 240 ) where θ a is the phase angle of the lag of the current i a with respect to e a. 10/8/13 Symmetrical Faults 24

25 The Synchronous Generator - Cylindrical Rotor Model L s + M s R i a = 2 I a cos( ωt + δ θ ) a + ± e a = 2 E i cos( ωt + δ ) v a = 2 V a cosωt This is a steady- state model that misses details of the field current needed for transient analysis. 10/8/13 Symmetrical Faults 25

26 The Synchronous Generator - Cylindrical Rotor Model To facilitate transient analysis (as with a fault) it is necessary to recast this model into a different form. Recall: λ f = L af i a + L bf i b + L cf i c + L ff i f and: Subs^tu^ng: λ f = M f L af = M f cosθ d L bf = M f cos( θ d 120 ) L cf = M f cos( θ d 240 ) i a cosθ + i cos θ 120 d b d i a cosθ d = 2 I a cosθ d cos ωt + δ θ a ( ) + i c cos( θ d 240 ) ( ) + L i ff f = 2 I a cos( ωt + δ + 90 )cos( ωt + δ θ ) a 10/8/13 Symmetrical Faults 26

27 To facilitate a symmetric fault it helps to recast this model a bit. i a cosθ d = 2 I a cosθ d cos( ωt + δ θ ) a = 2 I a cos( ωt + δ + 90 )cos( ωt + δ θ ) a = I a 2 cos θ + 90 a ( ) + cos( 2ωt + 2δ θ a + 90 ) ( ) = I a 2 sinθ sin 2ωt + 2δ θ a a 2cosα cosβ = cos( α β ) + cos( α + β ) 10/8/13 Symmetrical Faults 27

28 Similarly for the other two terms: Subs^tu^ng: i a cosθ d + i b cos( θ d 120 ) + i c cos( θ d 240 ) = I a 2 = 3 I a 2 sinθ a i b cos( θ d 120 ) = I a i c cos( θ d 240 ) = I a ( ) 2 sinθ sin 2ωt + 2δ θ 120 a a ( ) 2 sinθ sin 2ωt + 2δ θ 240 a a 3sinθ a +sin( 2ωt + 2δ θ a ) + sin( 2ωt + 2δ θ a 120 ) + sin( 2ωt + 2δ θ a 240 ) = 0 This is a balanced second harmonic terms that sums to zero at every point in Dme. 10/8/13 Symmetrical Faults 28

29 Subs^tu^ng: λ f = M f i a cosθ d + i b cos θ d 120 ( ) + i c cos( θ d 240 ) = L ff i f 3 M f I a 2 sinθ a = L ff i f M f i d + L i ff f where i d = 3 I a sinθ a or i d = 2 3 i cosθ + i cos θ 120 a d b d ( ) + i c cos( θ d 240 ) 10/8/13 Symmetrical Faults 29

30 Observa^ons: λ f = L ff i f M f i d, i d = 3 I a sinθ a The flux linkages with the field winding that are due to a combina^on of the the three line currents do not vary with ^me. They can thus be regarded as coming from a steady dc current i d in a fic^^ous dc circuit coincident with the d- axis and thus sta^onary with respect to the field circuit. The two circuits rotate together in synchronism and have mutual inductance (3/2) 1/2 M f between them. Circuit model 10/8/13 Symmetrical Faults 30

31 Alternate Circuit Model Suitable for Transient Analysis a- axis θ d d- axis i d 3 2 M f Armature equivalent winding rota^ng with rotor i f R, L ff b- axis + v f f Field winding rota^ng with rotor c- axis 10/8/13 Symmetrical Faults 31

32 Example A 60- Hz three- phase synchronous generator with negligible armature resistance has the following inductance parameters: L aa = L s = mH, L ab = M s = mH L ff = mH, M f = mH The machine is rated at 635 MVA, 0,9 power- factor lagging, 3600 rpm, 24 kv. When opera^ng under rated condi^ons, the line- to- neutral terminal voltage and the line current of phase a are v a = 19596cosωt V, i a = 21603cos( ωt ) Determine the magnitude of the synchronous internal voltage, the field current I f, and the flux linkages with the field winding. Repeat these calcula^ons when a load of 635 MVA is served at rated voltage and unity power factor. What is the field current for rated armature voltage on a open circuit? 10/8/13 Symmetrical Faults 32

33 Example v amax = 2 24,000 i amax 3 = 19596V = VA V = 2 635, = 21603A θ = cos = , lagging Synchronous Internal Voltage: e a = 2 E i cos( ωt + δ ) =0 = R ia a + v a + ( L s + M ) di a s dt = v a di + ( )10 3 a dt = 19596cosωt ( d )10 3 dt cos( ωt ) = 19596cosωt ω 21603( )10 3 sin ωt ( ) 10/8/13 Symmetrical Faults 33

34 Example Synchronous Internal Voltage: ω = 120π ( ) = sin xcos y cos xsin y sin x y 2 E i cos( ωt + δ ) = 19596cosωt ω 21603( )10 3 sin ωt ( ) = 19596cosωt 33875sin ωt = 34323cosωt 30407sinωt ( ) = 45855cos ωt E i = 45855, δ = ( ) Field Current: E i = ω M f I f 2 I f = 2 E i ω M f = 3838A 10/8/13 Symmetrical Faults 34

35 Example Flux linkages with field windings: λ f = L ff i f 3 M f θ a is the angle of lag measured wrt e a. Since i a lags o behind v a, which lags o behind e a, then θ a = o o = o I a 2 sinθ a Thus: I a sinθ a = sin = A 2 λ f = L ff i f 3 M f I a 2 sinθ a = = = Weber-turns 10/8/13 Symmetrical Faults 35

36 Example Repea^ng for unity power factor: di e a = 2 E i cos( ωt + δ ) = v a + ( )10 3 a dt = 19596cosωt 33785sinωt = 39057cos( ωt ) I f = 2 E i ω M f = = 3269A The current i a is in- phase with v a, and lags e a by o I a sinθ a = 15276sin = 13214A λ f = = = Weber-turns /8/13 Symmetrical Faults 36

37 Example We see that when the power factor of the load goes from 0.9 lagging to 1.0 under rated mega- voltamperes loading and voltage condidons, the field current is reduced from 3838 to 3269 A. Also, the net air- gap flux linking the field winding of the generator is reduced along with the demagnedzing influence of armature reacdon. The field current required to maintain rated terminal voltage in the machine under open- circuit condidons (i a = 0) is I f = 2 E i ω M f = π = 1640A 10/8/13 Symmetrical Faults 37

38 The Synchronous Generator - Salient Rotor Model The round- rotor theory already just considered gives good results for the steady- state performance of the synchronous machine. However, for transient analysis we need to consider a two- axis model. We now introduce the two- axis model by means of the equadons of the salient- pole machine in which the air gap varies between poles. The largest generadng units are steam- turbine driven with round- rotor construcdon; fossil- fired units have two poles and nuclear units have four pole s for reasons of economical design and efficiency. Hydroelectric generators usually have more pole- pairs and are of salient- pole construcdon. These units run at lower speeds so as to avoid mechanical damage due to centrifugal forces. 10/8/13 Symmetrical Faults 38

39 The Synchronous Generator - Salient Rotor Model The three- phase salient- pole machine, like its round- rotor counterpart, has three symmetrically distributed armature windings a, b, and c, and a field winding f on the rotor which produces a sinusoidal flux distribudon around the air gap. In both types of machines the field sees the same air gap and magnedzing paths in the stator regardless of the rotor posidon. Thus, the field winding has constant self- inductance L ff. Moreover, both machine have the same sinusoidal varying mutual inductances as before. The difference is that the self- inductances L aa, L bb, and L cc and the mutual inductances L ab, L bc, and L ca between them are no longer constant but also vary as a funcdon of the rotor angular displacement. 10/8/13 Symmetrical Faults 39

40 The Synchronous Generator - Salient Rotor Model Flux Linkages: λ a = L aa i a + L ab i b + L ac i c + L af i f λ b = L ba i a + L bb i b + L bc i c + L bf i f λ c = L ca i a + L cb i b + L cc i c + L cf i f These look the same as seen before (Slide 19), but unlike the case of the round rotor the L s are no longer well- approximated by constants. X X 10/8/13 Symmetrical Faults 40

41 The Synchronous Generator - Salient Rotor Model For the armature, Self- inductances: L s > L m > 0 Mutual- inductances: L aa = L s + L m cos2θ d ( ) ( ) L bb = L s + L m cos2 θ d 2π 3 L cc = L s + L m cos2 θ d + 2π 3 M s > L m > 0 ( ) ( ) ( ) L ab = L ba = M s L m cos2 θ d + π 6 L bc = L cb = M s L m cos2 θ d π 2 L ca = L ac = M s L m cos2 θ d + 5π 6 Mostly based on geometrical considera^ons 10/8/13 Symmetrical Faults 41

42 The Synchronous Generator - Salient Rotor Model For the rotor, Self- inductances: Field winding: D- Damper winding: Q- Damper winding: L ff L D L Q Mutual- inductances: Field/D- winding: Field/Q- winding: D- winding/q- winding: Armature/Field: M r 0 0 L af = L fa = M f cos2θ d L bf = L fb = M f cos2( θ d 2π 3) ( ) L cf = L fc = M f cos2 θ d 4π 3 Mostly based on geometrical considera^ons 10/8/13 Symmetrical Faults 42

43 The Synchronous Generator - Salient Rotor Model Armature- damper winding mutual- inductances, Armature/D- winding: What are these? Armature/Q- winding: L ad = L Da = M D cos2θ d ( ) ( ) L bd = L Db = M D cos2 θ d 2π 3 L cd = L Dc = M D cos2 θ d 4π 3 L aq = L Qa = M Q cos2θ d L bq = L Qb = M Q cos2( θ d 2π 3) L cq = L Qc = M Q cos2( θ d 4π 3) Mostly based on geometrical considera^ons 10/8/13 Symmetrical Faults 43

44 The Synchronous Generator - Salient Rotor Model Damper Windings or Amortisseur Windings Shorting Bar Field Windings 10/8/13 Symmetrical Faults 44

45 Two- Axis Machine Model Clearly the equadons for the flux linkages of the salient- pole machine are more complicated than their round- rotor counterparts. Fortunately, the equadons of the salient- pole machine can be expressed in a simple form by transforming the a, b, and c variables of the stator into corresponding sets of new variables, called the direct- axis, quadrature- axis, and zero- sequence quanddes which are disdnguished by the subscripts d, q, and 0: respecdvely. The three stator currents i a, i b, and i c can be transformed into three equivalent currents, called the direct- axis current i d, the quadrature- axis current i q and the zero- sequence current i o. The transformadon is made by a matrix P called Park s Transforma^on. 10/8/13 Symmetrical Faults 45

46 Park s Transforma^on Recall for the round rotor machine we found (Slide 30) i d, what we will now call the direct current as: i d = 2 3 i cosθ + i cos θ 120 a d b d ( ) + i c cos( θ d 240 ) q- axis Quadrature Axis S d- axis N S N N Direct Axis X X rota^on S d- axis Note how the q- axis lags the d- axis by 90- degrees. 10/8/13 Symmetrical Faults 46

47 Park s Transforma^on Note how the q- axis lags the d- axis by 90- degrees. i d = 2 3 i cosθ + i cos θ 120 a d b d i q = 2 3 i sinθ + i sin θ 120 a d b d ( ) + i c cos( θ d 240 ) ( ) + i c sin( θ d 240 ) P = 2 3 cosθ d cos( θ d 120 ) cos( θ d 240 ) sinθ d sin( θ d 120 ) sin( θ d 240 )??? Define P as a unitary matrix: P 1 = P T The unitary property assures that the power is unaltered by P. 10/8/13 Symmetrical Faults 47

48 Park s Transforma^on P = 2 3 cosθ d cos( θ d 120 ) cos( θ d 240 ) sinθ d sin( θ d 120 ) sin( θ d 240 ) Park s transformadon gives what is known as the Two- Axis Model of a salient pole generator. 10/8/13 Symmetrical Faults 48

49 Park s Transforma^on i d i q i 0 = P i a i b i c v d v q v 0 = P v a v b v c λ d λ q λ 0 = P λ a λ b λ c The P- transformadon defines a set of currents, voltages, and flux linkages for three ficddous coils, one of which is the stadonary 0- coil. The other two are the d- coil and the q- coil, which rotate in synchronism with the rotor. The d- and q- coils have constant flux linkages with the field and any other windings which may exist on the rotor. The resuldng d, q, and 0 flux- linkage equadons are 10/8/13 Symmetrical Faults 49

50 Park s Transforma^on For example, the resuldng d, q, and 0 flux- linkage equadons are from the stator flux- linkages as follows: λ a = L aa i a + L ab i b + L ac i c + L af i f λ b = L ba i a + L bb i b + L bc i c + L bf i f λ c = L ca i a + L cb i b + L cc i c + L cf i f λ d λ a λ a λ d λ q = P λ b λ b = P 1 λ q λ 0 λ c λ c λ 0 P 1 λ d λ q λ 0 λ d λ q λ 0 = = P L aa L ab L ac L ab L bb L bc L ac L bc L cc L aa L ab L ac L ab L bb L bc L ac L bc L cc P 1 P 1 λ d λ q λ 0 λ a λ b λ c λ d λ q λ 0 = + + P L aa L ab L ac L ab L bb L bc L ac L bc L cc L af L bf L cf L af L bf L cf i f i f i a i b i c + L af L bf L cf i f 10/8/13 Symmetrical Faults 50

51 Park s Transforma^on SubsDtuDng for the L s from the previous table: = L aa L ab L ac L ab L bb L bc L ac L bc L cc ( ) M s L m cos2( θ d + 5π 6) ( ) L s + L m cos2( θ d 2π 3) M s L m cos2( θ d π 2) ( ) M s L m cos2( θ d π 2) L s + L m cos2( θ d + 2π 3) L s + L m cos2θ d M s L m cos2 θ d + π 6 M s L m cos2 θ d + π 6 M s L m cos2 θ d + 5π 6 L s M s M s = M s L s M s M s M s L s ( ) = L s + M s ( ) cos2( θ d + 5π 6) ( ) cos2( θ d 2π 3) cos2( θ d π 2) ( ) cos2( θ d + 2π 3) cos2θ d cos2 θ d + π 6 L m cos2 θ d + π 6 cos2( θ d + 5π 6) cos2 θ d π 2 M ( ) cos2( θ d + 5π 6) ( ) cos2( θ d 2π 3) cos2( θ d π 2) ( ) cos2( θ d + 2π 3) cos2θ d cos2 θ d + π 6 L m cos2 θ d + π 6 cos2( θ d + 5π 6) cos2 θ d π 2 10/8/13 Symmetrical Faults 51

52 Park s Transforma^on Now for the tedious part subsdtute all this L aa L ab L ac L ab L bb L bc L ac L bc L cc = L s + M s ( ) M ( ) cos2( θ d + 5π 6) ( ) cos2( θ d 2π 3) cos2( θ d π 2) ( ) cos2( θ d + 2π 3) cos2θ d cos2 θ d + π 6 L m cos2 θ d + π 6 cos2( θ d + 5π 6) cos2 θ d π 2 λ d λ q λ 0 = P L aa L ab L ac L ab L bb L bc L ac L bc L cc P 1 λ d λ q λ 0 + P M f cos2θ d ( ) ( ) M f cos2 θ d 2π 3 M f cos2 θ d 4π 3 i f P = 2 3 cosθ d cos θ d 120 sinθ d sin θ d 120 ( ) cos( θ d 240 ) ( ) sin( θ d 240 ) , P 1 = P T to get 10/8/13 Symmetrical Faults 52

53 Park s Transforma^on to get λ d λ q λ 0 = L d L q L 0 i d i q i M f 0 0 i f where: Direct- axis inductance: L d = L s + M s L m Quadrature- axis inductance: L q = L s + M s 3 2 L m Zero- sequence inductance: L 0 = L s 2M s all constants! 10/8/13 Symmetrical Faults 53

54 Park s Transforma^on The equadon for the flux linkages of the field circuit remain unchanged (Slide 28): λ f = L ff I f M f i d 10/8/13 Symmetrical Faults 54

55 Park s Transforma^on P 10/8/13 Symmetrical Faults 55

56 Two- Axis Machine Model θ d a- axis Ignore damper winding for the moment. q- axis d- axis i d 3 2 M f R, L q i q R, L d Rota^on i f R, L ff + v f f b- axis All coils rotate together. c- axis 10/8/13 Symmetrical Faults 56

57 Two- Axis Machine Model The constant inductance coefficients make for quite simple to use. Physically, these simpler flux- linkage equadons show that L d is the self- inductance of an equivalent d- axis armature winding which rotates at the same speed as the field and which carries current i d to produce the same mmf on the d- axis as do the actual stator currents i a, i b, and i c. Similarly for L q and i q and the q- axis. The ficddous d- axis winding and the field winding represendng the physical field can be considered to act like two coupled coils which are stadonary with respect to each other as they rotate together sharing the mutual inductance between them. 10/8/13 Symmetrical Faults 57

58 Two- Axis Machine Model Furthermore, the field and the d- axis coil do not couple magnedcally with ficddous q winding since it lags the d- axis in space by 90. The zero- sequence inductance L 0 is associated with a stadonary ficddous armature coil with no coupling to any other coils. Under balanced condidons this coil carries no current, and therefore we omit it from our discussion. 10/8/13 Symmetrical Faults 58

59 Example To get a feeling for Park s Transforma^on Under steady- state operadng condidons the armature of the salient- pole synchronous generator carries symmetrical sinusoidal three- phase currents: i a = i b = i c = 2 I a sin( θ d θ ) a 2 I a sin( θ d 120 θ ) a 2 I a sin( θ d 240 θ ) a Find the corresponding d- q- 0 currents of the armature. 10/8/13 Symmetrical Faults 59

60 Example To get a feeling for Park s Transforma^on i d i q i 0 = 2 3 cosθ d cos( θ d 120 ) cos( θ d 240 ) sinθ d sin( θ d 120 ) sin( θ d 240 ) i a i b i c MulDply through: i d = 2 3 cosθ i + cos ( θ 120 d a d )i b + cos( θ d 240 )i c i q = 2 3 sinθ i + sin ( θ 120 d a d )i b + sin( θ d 240 )i c i 0 = ( ) 2 i a + i b + i c = 0 balanced three- phase 10/8/13 Symmetrical Faults 60

61 Example To get a feeling for Park s Transforma^on Simplifying 2sin xcos y = sin( x + y) + sin( x y) i a cosθ d = 2 I a sin( θ d θ a )cosθ d = I a 2 sin 2θ θ d a i b cos( θ d 120 ) = 2 I a cos( θ d 120 )sin( θ d 120 θ ) d ( ) sinθ a = I a 2 sin 2θ 240 θ d a i c cos( θ d 240 ) = 2 I a cos( θ d 240 )sin( θ d 240 θ ) d ( ) sinθ a = I a 2 sin 2θ 480 θ d a ( ) sinθ a 10/8/13 Symmetrical Faults 61

62 Example To get a feeling for Park s Transforma^on SubsDtuDng i d = 2 3 I a 2 sin ( 2θ θ d a ) + sin( 2θ d 240 θ a ) + sin( 2θ d 480 θ a ) 3sinθ a = 0 = 3 I a sinθ a It should come as no surprise that i q = 3 I a cosθ a Note how the expression for i d is EXACTLY the same as for the round- rotor machine (see Slide 30). 10/8/13 Symmetrical Faults 62

63 VOLTAGE EQUATIONS Remarkably simple in d, q, 0 variables. As before, the line to neutral voltages are: dλ v a = Ri a a dt generator v b = Ri b dλ b dt v c = Ri c dλ c dt These would be EXTREMELY difficult to deal with if lem in terms of a, b, c, but simplify TREMENDOUSLY using Park s transformadon, but going from a, b, c to d, q, 0 is a bit tedious. 10/8/13 Symmetrical Faults 63

64 Voltage Equa^ons General procedure lots of algebra: MulDply both sides by P v a v b v c v d v q v 0 = R = R i a i b i c i d i q i 0 d dt P d dt λ a λ b λ c P 1 P λ d λ q λ 0 v a v b v c = RP i a i b i c P d dt λ a λ b λ c Note: P = P( t) since θ d = ωt +θ o 10/8/13 Symmetrical Faults 64

65 Voltage Equa^ons General procedure lots of algebra: v d v q v 0 = R = R = R i d i q i 0 i d i q i 0 i d i q i 0 P d dt PP 1 d dt d dt P 1 λ d λ q λ 0 λ d λ q λ 0 λ d λ q λ 0 P P d dt PT d dt P 1 10/8/13 Symmetrical Faults 65 λ d λ q λ 0 λ d λ q λ 0 This is the work.

66 Voltage Equa^ons It all simplifies to: v d v q v 0 = R i d i q i 0 dλ d dt dλ q dt dλ 0 dt + ωλ q ωλ d 0 where ω = dθ d dt 10/8/13 Symmetrical Faults 66

67 Voltage Equa^ons Summary: d- axis v d = Ri d dλ d dt ωλ q λ f = L ff i f M f i d λ d = L d i d M f i f q- axis v q = Ri q dλ q dt λ q = L q i q + ωλ d and since the field winding is not subject to the P- transformadon v f f = R f i f + dλ f dt 10/8/13 Symmetrical Faults 67

68 Circuit Model i f v d = Ri d L d di d dt ωλ q v f f R f L ff Field winding 3 2 M f L d R i d + v d d- axis armature equivalent winding ωλ q + v f f = R f i f + dλ f dt L q R i q + v q q- axis armature equivalent winding + ωλ d v q = Ri q L q di q dt ωλ d 10/8/13 Symmetrical Faults 68

69 Circuit Model We see that the field coil is mutually coupled to the d- coil on the d- axis. The q- coil is magnedcally uncoupled from the other two windings since the d- axis and the q- axis are spadally in quadrature with one another. However, there is interacdon between the two axes by means of the voltage sources which are rotadonal emfs or speed voltages internal to the machine due to the rotadon of the rotor. Note that the speed voltage in the d- axis depends on λ q, and similarly, the speed voltage in the q- axis depends on λ d. These sources represent ongoing electromechanical energy conversion. 10/8/13 Symmetrical Faults 69

70 Example No numbers, but again to appreciate the model A direct current I f is supplied to the field winding of an unloaded salient- pole synchronous generator rotadng with constant angular velocity ω. Determine the open- circuit armature voltages and their d- q- 0 components. Since open circuited: SubsDtuDng these into: i d i q i 0 = P i a i b i c = v d = Ri d dλ d dt ωλ q, λ f = L ff i f M f i d, λ d = L d i d M f i f v q = Ri q dλ q dt + ωλ d, λ q = L q i q, v f f = R f i f + dλ f dt gives 10/8/13 Symmetrical Faults 70

71 Example No numbers, but again to appreciate the model gives λ d = 3 2 M f i f v d = dλ d dt ωλ q = 0 Thus: λ q = 0 λ 0 = 0 v d v q v 0 = P v a v b v c v q = dλ q dt + ωλ d = ω 3 2 M f i f v 0 = Ri 0 dλ 0 dt v a v b v c = 0 v d = P T v q v 0 10/8/13 Symmetrical Faults 71

72 Example No numbers, but again to appreciate the model v a v b v c v d = P T v q v 0 = 3 2 = 2 3 sinθ d ( ) ( ) sin θ d 120 sin θ d 240 cosθ d sinθ d 1 cos( θ d 120 ) sin( θ d 120 ) cos( θ d 240 ) sin( θ d 240 ) ω M f i f ω M f i f 0 Again this is idendcal to the round- rotor machine. 10/8/13 Symmetrical Faults 72

73 Summary Park's transformadon replaces the physical stadonary windings of the armature by: 1. A direct- axis circuit which rotates with the field circuit and is mutually coupled to it. 2. A quadrature- axis circuit which is displaced 90 from the d- axis, and thus has no mutual inductance with the field or other d- axis circuits although it rotates in synchronism with them, and 3. A stadonary stand- alone 0- coil with no coupling to any other circuit, and thus not shown. This model is most useful in analyzing the performance of the synchronous machine under short- circuit condidons, which we now (finally!) consider. 10/8/13 Symmetrical Faults 73

74 TRANSIENT AND SUBTRANSIENT EFFECTS When a fault occurs in a power network, the current flowing is determined by the internal emfs of the machines in the network, by their impedances, and by the impedances in the network between the machines and the fault. The current flowing in a synchronous machine immediately amer the occurrence of a fault differs from that flowing a few cycles later and from the sustained, or steady- state, value of the fault current. This is because of the effect of the fault current in the armature on the flux generadng the voltage in the machine. The current changes reladvely slowly from its inidal value to its steady- state value owing to the changes in reactance of the synchronous machine. 10/8/13 Symmetrical Faults 74

75 TRANSIENT AND SUBTRANSIENT EFFECTS Our interest now is in the inductance effecdve in the armature of the synchronous machine when a three- phase short circuit suddenly occurs at its terminals. Before the fault occurs, suppose that the armature voltages are v a, v b, and v c, and that these give rise to the voltages v d, v q, and v 0 according to Park s TransformaDon. The short circuit of phases a, b, and c imposes the condidons v a = v b = v c = 0 which lead to the condidons v d = v q = 0. Thus, to simulate short- circuit condidons, the terminals of the d- axis and q- axis circuits of the circuit model must also be shorted. 10/8/13 Symmetrical Faults 75

76 TRANSIENT AND SUBTRANSIENT EFFECTS i f i d v f f R f L ff 3 Field winding 2 M f L d R d- axis + v d S + v d + v d The switches S should be interpreted in a symbolic sense; namely, when the switches are both open, the sources - V d and - V q are in the circuit, and when the switches are closed, the two sources are removed from the circuit. ωλ q + L q R q- axis + ωλ d + v q i q S + v q + v q 10/8/13 Symmetrical Faults 76

77 TRANSIENT AND SUBTRANSIENT EFFECTS Since the model is linear we now use superposidon. Assume that the rotor speed ω remains at its pre- fault steady- state value. With both switches closed we have the steady- state operadon of the machine since the added sources v d and v q do nothing. Suddenly opening the switches adds the series voltage sources v d and v q producing the required short circuits. Thus, the sources v d and v q determine the instantaneous changes from the steady state due to the sudden short- circuit fault. 10/8/13 Symmetrical Faults 77

78 TRANSIENT AND SUBTRANSIENT EFFECTS We can calculate the fault- induced changes of all variables by sewng the external sources v ff, v d and v q equal to zero and suddenly applying the voltages v d and v q to the unexcited rotadng machine, as shown in the next slide. The internal speed voltages ωλ q and ωλ d are inidally zero because flux linkages with all coils are zero in the (next) figure before applying the voltages v d and v q. 10/8/13 Symmetrical Faults 78

79 TRANSIENT AND SUBTRANSIENT EFFECTS i f i d v f f = 0 R f Field winding L ff 3 2 M f L d R d- axis t = 0 + v d ωλ q + = 0 i q R t = 0 + L q q- axis v q + ωλ d = 0 10/8/13 Symmetrical Faults 79

80 TRANSIENT AND SUBTRANSIENT EFFECTS From the d- axis summary: λ f = L ff i f M f i d λ d = L d i d M f i f we compute the change Δλ f = L ff Δi f M f Δi d Δλ d = L d Δi d M f Δi f But since the field winding is a closed, physical winding its flux linkages (field current) cannot change instantaneously, so set Δλ f = 0 Δi f = 1 L ff 3 2 M f Δi d 10/8/13 Symmetrical Faults 80

81 TRANSIENT AND SUBTRANSIENT EFFECTS SubsDtuDng: Δi f = 1 L ff 3 2 M f Δi d into: Δλ d = L d Δi d M f Δi f gives: Δλ d = L d 3 2 M f 2 L ff Δi d The flux linkage per unit current defines the d- axis transient inductance: L d = Δλ d = L Δi d 3 d 2 M f 2 L ff 10/8/13 Symmetrical Faults 81

82 TRANSIENT AND SUBTRANSIENT EFFECTS The d- axis transient inductance: Since 3 2 M f 2 L ff > 0 L d = L d 3 2 M f 2 the direct- axis transient reactance X d = ωl d is always less than the direct- axis synchronous reactance X d = ωl d. Thus, following abrupt changes at its terminals, the synchronous machine reflects in its armature a transient reactance which is less than its steady- state reactance. L ff 10/8/13 Symmetrical Faults 82

83 TRANSIENT AND SUBTRANSIENT EFFECTS Further Considera^ons In defining X d, we assume that the field is the only physical rotor winding. As previously mendoned, most salient- pole machines of pracdcal importance have damper windings consisdng of shorted copper bars through the pole faces of the rotor; and even in a round- rotor machine, under short- circuit condidons eddy currents are induced in the solid rotor as if in damper windings. The effects of the eddy- current damping circuits are represented by direct- axis and quadrature- axis closed coils, which are treated in very much the same way as the field winding except that they have no applied voltage. 10/8/13 Symmetrical Faults 83

84 TRANSIENT AND SUBTRANSIENT EFFECTS Further Considera^ons To account for the addidon of damper windings, we need only add to our model the closed D- circuit and Q- circuit shown on the next slide, which have self- inductances L D and L Q and mutual inductances with the other windings as shown. In the steady state the flux linkages are constant between all circuits on the same rotor axis. The D- and Q- circuits are then passive (having neither induced nor applied voltages) and do not enter into steady- state analysis. 10/8/13 Symmetrical Faults 84

85 TRANSIENT AND SUBTRANSIENT EFFECTS Further Considera^ons + v f f i f R f Lff Field winding M R 3 2 M i f R d L d d- axis + v d + v D = 0 i D R D D- damper winding L D ωλ q M D + v Q = 0 i Q R Q Q- damper winding 3 2 M Q L Q L q R q- axis i q + v q + ωλ d 10/8/13 Symmetrical Faults 85

86 TRANSIENT AND SUBTRANSIENT EFFECTS Further Considera^ons Under short- circuit condidons, however, we can determine from the inidal d- axis flux- linkage changes resuldng from sudden shordng of the synchronous machine with damper- winding effects. The procedure is the same as already discussed. The field and D- damper circuits represendng closed physical windings are mutually coupled to each other and to the d- coil represendng the armature along the direct axis. There cannot be sudden change in the flux linkages of the closed windings, and so we can write the flux- linkage changes along the d- axis by modifying our earlier results as follows: 10/8/13 Symmetrical Faults 86

87 TRANSIENT AND SUBTRANSIENT EFFECTS Further Considera^ons Δλ f = L ff Δi f M f Δi d + M r Δi D = 0 Δλ d = L d Δi d M f Δi f M D Δi D Δλ D = 3 2 M D Δi d + M R Δi f + L D Δi D = 0 Note how these are similar to our previous equadons but have extra terms because of the addidonal self- and mutual inductances associated with the D- damper circuit. 10/8/13 Symmetrical Faults 87

88 TRANSIENT AND SUBTRANSIENT EFFECTS Further Considera^ons Solving: Δi f = 3 2 M f L D 3 2 M r M D L ff L D M r 2 Δi d and subsdtudng: Δi D = 3 2 M D L ff 3 2 M r M f L ff L D M r 2 Δi d Δλ d Δi d = L d = L d 3 2 M f 2 L D + M D 2 L ff 2M f M r M D L ff L D M r 2 This is the direct- axis sub- transient inductance. A similar inductance can be defined for the q- axis. 10/8/13 Symmetrical Faults 88

89 TRANSIENT AND SUBTRANSIENT EFFECTS Further Considera^ons The direct- axis sub- transient reactance is X d = ωl d X d is considerably smaller than X d, hence X d < X d < X d We have shown that the synchronous machine has different reactances when it is subjected to short- circuit faults at its terminals. Immediately upon occurrence of the short circuit, the armature of the machine behaves with an effecdve reactance X d, which combines with an effecdve resistance determined by the damping circuits to define a direct- axis, short- circuit sub- transient Dme constant T d, typically in the range of 0.03 seconds. 10/8/13 Symmetrical Faults 89

90 TRANSIENT AND SUBTRANSIENT EFFECTS Further Considera^ons The direct- axis sub- transient reactance is X d = ωl d The period over which X d is effecdve is called the sub- transient period, and this is typically 3 to 4 cycles of system frequency in duradon. When the damper- winding currents decay to negligible levels, the D- and Q- circuits are no longer needed and the model reverts to original one obtained. The machine currents decay more slowly with a direct- axis, short- circuit transient Dme- constant T d determined by X d and a machine resistance which depends on R f of the field. 10/8/13 Symmetrical Faults 90

91 TRANSIENT AND SUBTRANSIENT EFFECTS Further Considera^ons The period of effecdveness of X d is called the transient period and T d is of the order of 1 second. Finally, for sustained steady- state condidons the d- and q- axis reactances X d = ωl d and X q = ωl q determine the performance of the saient- pole machine, just as the synchronous reactance X d applies to the round- rotor synchronous machine in the steady state. The various reactances are supplied by the machine manufacturers. 10/8/13 Symmetrical Faults 91

92 Example Same Example as from Slide 32. Calculate the per- unit value of X d. Use the radngs of 635 MVA and 24 kv as base quanddes. As before: L d = L s + M f = = mH The transient inductance is calculated from: L d = L d 3 2 M f 2 The transient reactance is: ( ) 2 = L ff = mH X d = ω L d = 120π mH = 0.254Ω 10/8/13 Symmetrical Faults 92

93 Example Same Example as from Slide 33. The impedance base on the machine radngs is Z base = V 2 base = 242 VI base 635 = 0.907Ω Thus: X d = = 0.28 per unit Note: X d = ω L d = 120π mH = Ω 1.72 per unit Hence X d X d 10/8/13 Symmetrical Faults 93

94 Short- Circuit Currents As seen in the beginning, when an ac voltage is applied suddenly across a series R- L circuit the current generally has two components: a dc component, which decays according to the Dme constant L/R of the circuit, and a steady- state sinusoidally varying component of constant amplitude. A similar but more complex phenomenon occurs when a short circuit appears suddenly across the terminals of a synchronous machine. We have the model do the short- circuit analysis. 10/8/13 Symmetrical Faults 94

95 Short- Circuit Currents The resuldng phase currents in the machine will have dc components, which cause them to be offset or asymmetrical when ployed as a funcdon of Dme as we have seen. We generally neglect the dc- components of the currents. If we were to examine the current in one of the phases we would find that the ac- component varied as: i( t) = 2 E i cosωt 1 X d X d X d e t T d X d X d e t T d We see the influence of the transient and sub- transient reactances. 10/8/13 Symmetrical Faults 95

96 TRANSIENT AND SUBTRANSIENT EFFECTS i( t) t e T d e t T d c b a 2 E i cosωt X d t = 0 Short occurs at t = 0. Time 2 E i cosωt X d 10/8/13 Symmetrical Faults 96

97 TRANSIENT AND SUBTRANSIENT EFFECTS With the dc levels removed (which wouldn t make through the transformer anyway), the armature phase current has three componetns, two of which decay at different rates over the sub- transient and transient periods. NeglecDng the small armature resistance, the height at point a is the maximum value of the sustained short- circuit current, with rms value given by: I = E i X d = E i X d 10/8/13 Symmetrical Faults 97

98 TRANSIENT AND SUBTRANSIENT EFFECTS If the envelope of the current wave is extended back to zero Dme and the first few cycles where the falloff is very rapid were neglected, the intercept is at height b. The rms value of this current is known as the transient current: I = E i X d Similarly, the rms value of the current whose height is at c is known as the sub- transient current: I = E i X d 10/8/13 Symmetrical Faults 98

99 TRANSIENT AND SUBTRANSIENT EFFECTS The sub- transient current is omen called the inidal symmetrical rms current, a more descripdve name because it conveys the idea of neglecdng the dc component and taking the rms value of the ac component of the current immediately amer the occurrence of the fault. The simple equadons: I = E i X d indicate a method of determining the fault current in a generator when its reactances are known. I = E i X d 10/8/13 Symmetrical Faults 99

100 TRANSIENT AND SUBTRANSIENT EFFECTS If the generator is unloaded when the fault occurs, the machine is represented by the no- load voltage to neutral in series with the proper reactance: j X d j X d jx d ± E i ± E i ± E i Used to calculate currents for sub- transient condidons. Used to calculate currents for transient condidons. Used to calculate currents for steady- state condidons. 10/8/13 Symmetrical Faults 100

101 Example Two generators are connected in parallel to the low- voltage side of a three- phase Δ - Y transformer as shown. G 1 G 2 Δ- Y Generator 1 is rated 50,000 kva, 13.8 kv. Generator 2 is rated 25,000 kva, 13.8 kv. Each generator has a sub- transient reactance of 25% on its own own base. The transformer is rated 75,000 kva, 13.8Δ /69Y kv, with a reactance of 10%. Before the fault occurs, the voltage on the high- voltage side of the transformer is 66 kv. The transformer is unloaded and there is no circuladng current between the generators. Find the sub- transient current in each generator when a three- phase short circuit occurs on the high- voltage side of the transformer. 10/8/13 Symmetrical Faults 101

102 Example Select 69 kv, 75,000 kva as the base in the high- voltage circuit. Then, the base voltage on the low- voltage side is 13.8 kv. E i1 = E i2 = = p.u. G 1 G 2 50,000 kva 13.8 kv 25,000 kva 13.8 kv VA Base = 75,000 kva V Base = 13.8 kv VA Base = 75,000 kva V Base = 69 kv Δ- Y 75,000 kva 13.8Δ /69Y kv For Generator 1: X d1 = ,000 50,000 = p.u. Note: How did I get this? Z p.u. = Z rated V rated V base 2 VABase VA rated. 10/8/13 Symmetrical Faults 102

103 Example Select 69 kv, 75,000 kva as the base in the high- voltage circuit. Then, the base voltage on the low- voltage side is 13.8 kv. E i1 = E i2 = = p.u. G 1 G 2 50,000 kva 13.8 kv 25,000 kva 13.8 kv VA Base = 75,000 kva V Base = 13.8 kv VA Base = 75,000 kva V Base = 69 kv Δ- Y 75,000 kva 13.8Δ /69Y kv For Generator 2: Transformer: X d 2 = ,000 25,000 X t = 0.1 p.u. = p.u. 10/8/13 Symmetrical Faults 103

104 Example Pictured below is the reactance diagram before the fault. A three- phase fault is simulated by closing the switch. The internal voltages of the two machines may be considered to be in parallel since they are idendcal in magnitude and phase and no circuladng current flows between them. G 1 E i1 j X d1 = j0.375 jx t = j0.1 P (fault) G 2 E i2 j X d 2 = j0.75 S 10/8/13 Symmetrical Faults 104

105 Example The equivalent sub- transient reactance is: X d = X d1 j X d 2 = X d1 X d 2 = X d1 + X d = 0.25 p.u. With E i1 = E i2 = E i the current in the short- circuit is: = X d1 j X d 2 = I = E i X d1 X d 2 = X d1 + X d 2 ( ) = j( ) j X d + X t = 0.25 p.u. = j2.735 p.u. 10/8/13 Symmetrical Faults 105

106 Example The voltage on the Δ- side of the transformer is: V t = I jx t = j2.735 j0.1= p.u. In generators 1 and 2: I 1 = E V i 1 t = j X d1 I 2 = E V i 2 t = j X d j j0.75 = j1.823 p.u. = j0.912 p.u. 10/8/13 Symmetrical Faults 106

107 Summary The steady- state performance of the synchronous machine relies on the concept of synchronous reactance X d, which is the basis of the steady- state equivalent circuit of the machine. Transient analysis of the synchronous generator requires a two- axis machine model. We have seen that the corresponding equadons involving physical a- b- c phase variables can be simplified by Park s transformadon, which introduces d,q,0 currents, voltages, and flux linkages. Simplified equivalent circuits which follow from the d- q- 0 equadons of the machine allow definidons of the subtransient reactance X d and transient reactance X d. The transient reactance is also important for system stability analysis. 10/8/13 Symmetrical Faults 107

108 Power Systems Three- Phase Short Circuits AssumpDons made to calculate the sub- transient fault current for a three- phase short circuit in a power system: 1. Transformers are represented by their leakage reactances. Winding resistances, shunt admiyances, and Δ Y phase shims are neglected. 2. Transmission lines are represented by their equivalent series reactances. Series resistances and shunt admiyances are neglected. 3. Synchronous machines are represented by constant- voltage sources behind subtransient reactances. Armature resistance, saliency, and saturadon are neglected. 10/8/13 Symmetrical Faults 108

109 Power Systems Three- Phase Short Circuits These assumpdons are made for simplicity, and may not always apply. For example, in distribudon systems, the resistances of primary and secondary distribudon lines may in some cases significantly reduce fault current magnitudes. Saliency: The word saliency is used as a short expression for the fact that the rotor of a synchronous machine has different electric and magnedc properdes on two axes 90 o apart; the direct axis, or axis of symmetry of a field pole, and the quadrature axis, or axis of symmetry midway between two field poles. This difference between the two axes is present not only in salient- pole machines but also, to a lesser extent, in round- rotor machines, because of the presence of the field winding on the direct axis only. 10/8/13 Symmetrical Faults 109

110 Power Systems Three- Phase Short Circuits Consider a generator that is loaded when a fault occurs. Pre- Fault Model Synchronous reactance No- Load voltage jx dg + + E g V t Z ext Pre- fault current Terminal voltage Neutral Fault loca^on P I L + Z L V L = V f This is a steady- state model. It does not capture transient behavior. If a three- phase short circuit fault occurs from P to neutral, the equivalent circuit shown above does not sadsfy the condidons for calculadng sub- transient current. 10/8/13 Symmetrical Faults 110

111 Power Systems Three- Phase Short Circuits AMer the fault occurs, here is the correct circuit: Z ext P I L + j X dg + + E g V t Neutral S V f Z L Fault Model ( X ) dg I L E g = V t + j X dg I L = V f + Z ext + j 10/8/13 Symmetrical Faults 111

112 Power Systems Three- Phase Short Circuits With the switch open: ( X ) dg I L E g = V t + j X dg I L = V f + Z ext + j This equadons defines E g, the sub- transient internal voltage, is used to calculate the subtransient current I. Similarly, to calculate the transient current I, it must be supplied through the transient reactance X dg and the transient internal voltage E g : ( X ) dg I L E g = V t + j X dg I L = V f + Z ext + j 10/8/13 Symmetrical Faults 112

113 Power Systems Three- Phase Short Circuits Thus, the value of the load current I L determines the values of the voltages E g and E g, which are both equal to the no- load voltage E g only when I L is zero so that E g is then equal to V t. It is important to note that the pardcular value of E g in series with X g represents the generator immediately before and immediately amer the fault occurs only if the prefault current in the generator has the corresponding value of I L. On the other hand, E g in series with the synchronous reactance X dg is the equivalent circuit of the machine under steady- state condidons for any value of the load current. The magnitude of E g is determined by the field current of the machine, and so for a different value of I L in the pre- fault circuit le g I would remain the same but a new value of E g would be required. 10/8/13 Symmetrical Faults 113

114 Power Systems Three- Phase Short Circuits Synchronous motors have reactances of the same type as generators. When a motor is short- circuited, it no longer receives electric energy from the power line, but its field remains energized and the inerda of its rotor and connected load keeps it rotadng for a short period of Dme. 10/8/13 Symmetrical Faults 114

115 Power Systems Three- Phase Short Circuits The internal voltage of a synchronous motor causes it to contribute current to the system, for it is then acdng like a generator. By comparison with the corresponding formulas for a generator the subtransient internal voltage E m and transient internal voltage E m for a synchronous motor are given by E m = V t j X dm I L E m = V t j X dm I L where V t is now the terminal voltage of the motor. 10/8/13 Symmetrical Faults 115

116 Power Systems Three- Phase Short Circuits Fault currents in systems containing generators and motors under load may be solved in one of two ways: 1. by calculadng the subtransient (or transient) internal voltages of the machines or 2. by using Thévenin's theorem. An example will illustrate. 10/8/13 Symmetrical Faults 116

117 Power Systems Three- Phase Short Circuits Suppose that a synchronous generator is connected to a synchronous motor by a line of external impedance Z ext. The motor is drawing load current I L from the generator when a symmetrical three- phase fault occurs at the motor terminals. The equivalent circuits and current flows of the system immediately before and immediately amer the fault occurs are 10/8/13 Symmetrical Faults 117

118 Power Systems Three- Phase Short Circuits Before the Fault: Z ext P I L + + j X dg j X dm + E g V t V f + E m Neutral 10/8/13 Symmetrical Faults 118

119 Power Systems Three- Phase Short Circuits A\er the Fault: Z ext I g P I m + j X dg + I '' f j X dm + E g V t V f + E m Neutral 10/8/13 Symmetrical Faults 119