In the previous chapter, attention was confined

4 4 Principles of Power System CHAPTE CHAPTE 8 Unsymmetrical Fault Calculations 8. Usymmetrical Faults on -Phase System 8. Symmetrical Components Method 8. Operator a 8.4 Symmetrical Components in Terms of Phase Currents 8.5 Some Facts about Sequence Currents 8.6 Sequence Impedances 8.7 Sequence Impedances of Power System Elements 8.8 Analysis of Unsymmetrical Faults 8.9 Single Line-to-Ground Fault 8.0 Line-to-Line Fault 8. Double Line-to-Ground Fault 8. Sequence Networks 8. eference Bus for Sequence Networks Introduction In the previous chapter, attention was confined to the analysis of symmetrical faults e.g. all three lines short-circuited (L L L) or all three lines short-circuited with an earth connection at the fault (L L L G). When such a fault occurs, it gives rise to symmetrical fault currents i.e. fault currents in the three lines are equal in magnitude and displaced 0º electrical from one another. Although symmetrical faults are the most severe and impose heavy duty on the circuit breakers, yet the analysis of such faults can be made with a fair degree of ease. It is because the balanced nature of fault permits to consider only one phase in calculations ; the conditions in the other two phases being similar. The great majority of faults on the power system are of unsymmetrical nature; the most common type being a short-circuit from one line to ground. When such a fault occurs, it gives rise to unsymmetrical currents i.e. the magnitude of fault currents in the three lines are different having unequal phase displacement. The calculation procedure known as method of symmetrical components is used to determine the currents and voltages on the occurrence of an unsymmetrical fault. In this chapter, we shall focus our attention on the analysis of unsymmetrical faults. 4

Unsymmetrical Fault Calculations 4 8. Unsymmetrical Faults on -Phase System Those faults on the power system which give rise to unsymmetrical fault currents (i.e. unequal fault currents in the lines with unequal phase displacement) are known as unsymmetrical faults. On the occurrence of an unsymmetrical fault, the currents in the three lines become unequal and so is the phase displacement among them. It may be noted that the term unsymmetry applies only to the fault itself and the resulting line currents. However, the system impedances and the source voltages are always symmetrical* through its main elements viz. generators, transmission lines, synchoronous reactors etc. There are three ways in which unsymmetrical faults may occur in a power system (see Fig. 8.). (i) Single line-to-ground fault (L G) (ii) Line-to-line fault (L L) (iii) Doube line-to-ground fault (L L G) The solution of unsymmetrical fault problems can be obtained by either (a) Kirchhoff s laws or (b) Symmetrical components method. The latter method is preferred because of the following reasons : (i) It is a simple method and gives more generality to be given to fault performance studies. (ii) It provides a useful tool for the protection engineers, particularly in connection with tracing out of fault currents. Electronic earth fault indicator 8. Symmetrical Components Method In 98, Dr. C.L. Fortescue, an American scientist, showed that any unbalanced system of -phase currents (or voltages) may be regarded as being composed** of three separate sets of balanced vectors viz. * In other words, no piece of equipment ever has a red phase impedance which differs from a yellow phase impedance. ** This has come to be known as symmetrical component theory. This is a general theory and is applicable to any three vector system whose resultant is zero.

44 Principles of Power System (i) a balanced *system of -phase currents having positive (or normal) phase sequence. These are called positive phase sequence components. (ii) a balanced system of -phase currents having the opposite or negative phase sequence. These are called negative phase sequence components. (iii) a system of three currents equal in magnitude and having zero phase displacement. These are called zero phase sequence components. The positive, negative and zero phase sequence components are called the symmetrical components of the original unbalanced system. The term symmetrical is appropriate because the unbalanced -phase system has been resolved into three sets of balanced (or symmetrical) components. The subscripts, and 0 are generally used to indicate positive, negative and zero phase sequence components respectively. For instance, I 0 indicates the zero phase sequence component of the current in the red phase. Similarly, I implies the positive phase sequence component of current in the yellow phase. Illustration. Let us now apply the symmetrical components theory to an unbalanced -phase system. Suppose an unsymmetrical fault occurs on a -phase system having phase sequence B. According to symmetrical components theory, the resulting unbalanced currentesi, I and I B (see Fig. 8.) can be resolved into : (i) a balanced system of -phase currents, I, I and I B having positive phase sequence (i.e. B) as shown in Fig. 8. (i). These are the positive phase sequence components. (ii) a balanced system of -phase currents I, I and I B having negative phase sequence (i.e. B) as shown in Fig. 8. (ii). These are the negative phase sequence components. (iii) a system of three currents I 0, I 0 and I B0 equal in magnitude with zero phase displacement from each other as shown in Fig. 8. (iii). These are the zero phase sequence components. * A balanced system of -phase currents implies that three currents are equal in magnitude having 0º displacement from each other. Positive phase sequence means that phase sequence is the same as that of the original -phase system.

Unsymmetrical Fault Calculations 45 The current in any phase is equal to the vector sum of positive, negative and zero phase sequence currents in that *phase as shown in Fig. 8.4. The following points may be noted : I I I I 0 I I I I 0 I B I I I B B B0 (i) The positive phase sequence currents ( I, I and I B ), negative phase sequence currents (I, I and I B ) and zero phase sequence currents (I 0, I 0 and I B0 ) separately form balanced system of currents. Hence, they are called symmetrical components of the unbalanced system. (ii) The symmetrical component theory applies equally to -phase currents and voltages both phase and line values. (iii) The symmetrical components do not have separate existence. They are only mathematical components of unbalanced currents (or voltages) which actually flow in the system. (iv) In a balanced -phase system, negative and zero phase sequence currents are zero. This is demonstrated in example 8.7. 8. Operator a As the symmetrical component theory involves the concept of 0º displacement in the positive sequence set and negative sequence set, therefore, it is desirable to evolve some operator which should cause 0º rotation. For this purpose, operator a (symbols h or λ are sometimes used instead of a ) is used. It is defined as under : The **operator a is one, which when multiplied to a vector rotates the vector through 0º in the anticlockwise direction. Consider a vector I represented by OA as shown in Fig. 8.5. If this vector is multiplied by operator a, the vector is rotated through 0º in the anticlockwise direction and assumes the position OB. a I I 0º I (cos 0º j sin 0º) * Star connected system being considered in Fig. 8.4. ** Just as the operator j rotates a vector through 90º in the anticlockwise direction.

46 Principles of Power System I ( 0 5 j 0 866) a 0 5 j 0 866... (i) If the vector assuming position OB is multiplied by operator a, the vector is further rotated through 0º in the anticlockwise direction and assumes the position OC. a I I 40º I (cos 40º j sin 40º) I ( 0 5 j 0 866) a 0 5 j 0 866... (ii) Thus the operator a will turn the vector through 40º in the anticlockwise direction. This is the same as turning the vector through 0º in clockwise direction. a I I 0º Similarly, a I I 60º I (cos 60º j sin 60º) a... (iii) Properties of Operator a (i) Adding exps. (i) and (ii), we get, a a ( 0 5 j 0 866) ( 0 5 j 0 866) a a 0 (ii) Subtracting exp. (ii) from exp. (i), we get, a a ( 0 5 j 0 866) ( 0 5 j 0 866) j 7 a a j 8.4 Symmetrical Components in Ter erms of Phase Currents The unbalanced phase currents in a -phase system can be expressed in terms of symmetrical components as under : I I I I 0 I I I I 0 I B I I I B B B0

Unsymmetrical Fault Calculations 47 Fig. 8.6 shows the vector representation of symmetrical components. It is usually profitable in calculations to express the symmetrical components in terms of unbalanced phase currents. Let us express the symmetrical components of -phase in terms of phase currents I, I and I B. For this purpose, express all symmetrical components of and B phases in terms of the symmetrical components of -phase by means of operator a as shown in Fig. 8.6. Note that the positive sequence set shown in Fig. 8.6 (i) can be expressed in terms of I by means of operator a. Thus positive sequence current I B in phase B leads I by 0º and, therefore, IB ai. Similarly, positive sequence current in phase is 40º ahead of I so that I a I. In an exactly similar manner, the negative sequence set can be expressed in terms of I by means of operator a as shown in Fig. 8.6(ii). It is clear from Fig. 8.6 that : I I I I0...(i) I I I I 0 a I a I I0...(ii) I B I I I B B B0 0 a I a I I...(iii) (i) Zero sequence current. By adding exps. (i), (ii) and (iii), we get, I I IB I ( a a) I ( a a ) I0 I ( 0) I ( 0) I I (Œ a a 0) I 0 I I I 0 0 e B As the red phase is always taken as the reference phase, therefore, subscript is usually omitted. I 0 I I I e B (ii) Positive sequence current. Multiply exp.(ii) by a and exp. (iii) by a and then adding these exps. to exp. (i), we get, I ai a IB I ( a a ) I ( a a ) I 0 ( a a ) I I (0 I (0) I I Omitting the subscript, we have, j j 4 *) 0 I a I a IB e e I I a I a IB (iii) Negative sequence current. Multiply exp. (ii) by a and exp. (iii) by a and then adding these exps. to (i), we get, B I a I a I I or I * a 4 a a a a a a 4 a a 0 j j 4 0 I ( a a ) I ( a a ) I ( a a) I ( 0) I ( ) I ( 0) I 0 I a I aib e e B I a I ai j j

48 Principles of Power System The following points may be noted carefully : (i) The currents I, I and I 0 are the symmetrical components of -phase. Because of the symmetry of each set, the symmetrical components of yellow and blue phases can be easily known. (ii) Although the treatment has been made considering currents, the method applies equally to voltages. Thus the symmetrical voltage components of -phase in terms of phase voltages shall be : e e e E 0 E E E E B B E a E a E E E a E a EB 8.5 Some Facts about Sequence Currents It is now desirable to get the readers acquainted with the following facts about positive, negative and zero phase sequence currents : (i) A balanced -phase system consists of positive sequence components only; the negative and zero sequence components being zero. (ii) The presence of negative or zero sequence currents in a -phase system introduces unsymmetry and is indicative of an abnormal condition of the circuit in which these components are found. (iii) The vector sum of the positive and negative sequence currents of an unbalanced -phase system is zero. The resultant solely consists of three zero sequence currents i.e. Vector sum of all sequence currents in -phase unbalanced system I0 I0 IB0 (iv) In a -phase, 4 wire unbalanced system, the magnitude of zero sequence components is onethird of the current in the neutral wire i.e. Zero sequence current [Current in neutral wire] In the absence of path through the neutral of a -phase system, the neutral current is zero and the line currents contain no zero -sequence components. A delta-connected load provides no path to the neutral and the line currents flowing to delta-connected load can contain no zero-sequence components. (v) In a -phase unbalanced system, the magnitude of negative sequence components cannot exceed that of the positive sequence components. If the negative sequence components were the greater, the phase sequence of the resultant system would be reversed. (vi) The current of a single phase load drawn from a -phase system comprises equal positive, negative and zero sequence components. Example 8.. Prove that : a (i) a (ii) a a a a a j j j Solution. (i) a a a ( a)( a) a a( a) a a a ( a a 0) a

Unsymmetrical Fault Calculations 49 (ii) a a a ( a)( a ) a a a ( a)( a ) a a Example 8.. In a -phase, 4-wire system, the currents in, and B lines under abnormal conditions of loading are as under : I 00 0º A ; I 50 00º A ; I B 0 80º A Calculate the positive, negative and zero sequence currents in the -line and return current in the neutral wire. Solution. Let I0, I and I be the zero, positive and negative sequence currents respectively of the line current in red line. I 0 I I I B [00 0º 50 00º 0 80º] *[(86 60 j 50) (5 j 4 ) ( 0 j0)] [ 8 6 j 6 7] (7 j ) 7 9 4 68º A I B I a I a I [ 00 0º 0º 50 00º 0º 0 80º] [00 0º 50 60º 0 60º] [(86 6 j 50) (5 j 4 ) (5 j 5 98)] [6 6 j 9 8] (4 j 9 76) 57 98 4 º A I [ I a I a IB] [00 0º 0º 50 00º 0º 0 80º] [00 0º 50 80º 0 00º] [(86 6 j 50) ( 50 j 0) (5 j 5 98)] [5 6 j 4 0] (7 j 8) 8 96 4 9º A Current in the neutral wire I I I (8 6 j 6 7) 8 87 4 7º A B Example 8.. The currents in a -phase unbalanced system are : I ( j 6) A ; I ( j ) A ; I B ( 5 j 0) A The phase sequence in B. Calculate the zero, positive and negative sequence components of the currents. Solution. ed phase Zero phase sequence component, I 0 I I I B * With the help of scientific calculator, polar form can be directly changed to rectangular form and viceversa.

40 Principles of Power System [( j 6) ( j ) ( 5 j 0)] Positive phase sequence component is [9 j 4] ( j ) A I B I a I a I [ ( j 6) ( 0 5 j 0 866) ( j ) *( 0 5 j 0 866) ( 5 j 0)] Negative phase sequence component is [ 55 j 0 9] (0 85 j 0 ) A I I a I a IB [( j 6) ( 0 5 j 0 866) ( j ) ( 0 5 j 0 866) ( 5 j 0)] [ 5 55 j 6 4] ( 85 j 5 47) A ellow phase Zero phase sequence component is I0 I 0 ( j ) A Positive phase sequence component is I a I ( 0 5 j 0 866) (0 85 j 0 ) ( 5 j 4 4) A Negative phase sequence component is I ai ( 0 5 j 0 866) ( 85 j 5 47) (5 7 j ) A Blue phase Zero phase sequence component is IB 0 I0 I ( j ) A 0 Positive phase sequence component is IB a I ( 0 5 j 0 866) (0 85 j 0 ) ( 4 j 4 ) A Negative phase sequence component is I a I ( 0 5 j 0 866) ( 85 j 5 47) ( 8 j 4 4) A B Example 8.4. The sequence voltages in the red phase are as under : E 0 00 V ; E (00 j 00) V ; E 00 V Find the phase voltages E, E and EB. Solution. In the polar form, we have, E 0 00 0º V; E 6 6 56º V; E 00 80º V E E E E 0 * a 0 5 j 0 866 and a 0 5 j 0 866 00 (00 j 00) ( 00) 00 j 00 6 6 56º volts

Unsymmetrical Fault Calculations 4 0 E E a E a E 00 0º 40º 6 6 56º 0º 00 80º 00 0º 6 44º 00 00º (00 j 0) ( 86 58 j ) (50 j 86 6) 6 58 j 09 8 99 89º volts 0 E B E ae a E 00 0º 0º 6 6 56º 40º 00 80º 00 0º 6 9 44º 00 40º (00 j 0) ( 4 j ) (50 j 86 6) 6 6 j 09 8 8 57 66 º volts Example 8.5. The zero and positive sequence components of red phase are as under : E 0 (0 5 j 0 866) V ; If the phase voltage E phase voltages E and E B. Solution. or E 0º V 0º V, find the negative sequence component of red phase and the E E E E 0 (0 5 j 0 866) E Negative sequence component in -phase is E 0 5 j 0 866 60º volts In polar form, E 0 0 5 j 0 866 60º 0 Now E E a E a E [ 60º] [ 40º 0º] [ 0º 60º] 60º 40º 80º (0 5 j 0 866) ( j 7) ( j 0) 5 j 598 0º volts 0 E B E ae a E [ 60º] [ 0º 0º] [ 40º 60º] 60º 0º 00º (0 5 j 0 866) ( j 7) (0 5 j 0 866) 0 volt Example 8.6. The current from neutral to ground connection is A. Calculate the zero phase sequence components in phases. Solution. We know that zero sequence components in all phases have the same value and that each component is equal to one-third the current in the neutral wire. Zero sequence current in each phase 4 A

4 Principles of Power System Example 8.7. A balanced star connected load takes 90 A from a balanced -phase, 4-wire supply. If the fuses in the and B phases are removed, find the symmetrical components of the line currents (i) before the fuses are removed (ii) after the fuses are removed Solution. Fig. 8.7. shows the star-connected system with fuses in phases B and. (i) Before removal of fuses. Before fuses are removed from and B lines, the system is balanced and current in each line is 90 A. I 90 0ºA ; I 90 40º A ; I B 90 0ºA Since the system is balanced, it will have only positive sequence currents i.e., negative sequence and zero sequence components will be zero in the three lines. This can be readily established. I 0 I I 0 B0 I I I B [90 0º 90 40º 90 0º] [90 0º 90 0º 90 0º] 0 A Hence zero sequence components in three lines are zero. I B I a I a I [ 90 0º 0º 90 40º 0º 90 0º] [90 0º 90 0º 90 40º] [ 90 0º 90 0º 90 0º] 0 A Also I a I 0º 0 0 A and I B a I 40º 0 0 A Hence negative sequence components in the three lines are also zero. It can be easily shown that three positive sequence components will have the following values : I I 90 0º A ; I I 90 40º A; IB IB 90 0º A

Unsymmetrical Fault Calculations 4 (ii) After removal of fuses. When the fuses are removed in and B phases, the system becomes unbalanced with line currents as under : I 90 0ºA ; I IB 0 A The sequence currents in the three lines can be found out as under : I 0 I I 0 B0 I I I B [90 0º 0 0] 0 0º A i.e. zero sequence current in each line is 0 0º A. I B I a I a I [90 0º 0 0] 0 0ºA I a I 40º 0 0º 0 40ºA I B I a I 0º 0 0º 0 0ºA B I a I a I [90 0º 0 0] 0 0ºA I a I 0º 0 0º 0 0º A I B a I 40º 0 0º 0 40º A The reader may wonder how sequence currents can flow in the yellow and blue lines when fuses are removed in them. The answer is that these components do not have separate existence. They are only the mathematical components of the current which does exist. Thus the current in the yellow line is zero and this can be readily established from its sequence components : I I0 I I 0 0º 0 40º 0 0º 0 0º 0 0º 0 0º 0 A Similary, it can be proved that sum of sequence currents in the blue line is zero and that is what the circuit reveals. Example 8.8. A -φ, 4-wire-system supplies loads which are unequally distributed in the three phases. An analysis of the current flowing in, and B lines shows that in line, positive phase sequence component is 00 0º A and the negative phase sequence component is 00 60ºA. The total observed current flowing back to the supply in the neutral conductor is 00 00º A. Calculate the currents in the three lines. Solution. Zero phase sequence current in -line is I 0 Current in neutral wire 00 00º 00 00ºA Positive phase sequence current in -line is I 00 0ºA

44 Principles of Power System Negative phase sequence current in -line is I 00 60ºA Current in the -line, I I I I 00 00º 00 0º 00 60º 0 Current in the -line, I I a I a I Current in B line, IB (50 j 86 6) (00 j 0) (50 j 86 6) 00 0º A 0 00 00º 40º 00 0º 0º 00 60º 00 00º 00 40º 00 80º (50 j 86 6) ( 00 j 7 ) ( 00 j 0) 50 j 59 8 00 0º A 0 I ai a I 00 00º 0º 00 0º 40º 00 60º 00 00º 00 0º 00 00º (50 j 86 6) ( 00 j 7 ) (50 j 86 6) 0 A Example 8.9. One conductor of a -phase line is open. The current flowing to the -connected load through the line is 0 A. With the current in line [See Fig. 8.8] as reference and assuming that line B is open, find the symmetrical components of the line currents. Solution. The line currents are : I 0 0º A ; I 0 80º A ; I B 0 A -line I 0 I I I B [ 0 0º 0 80º 0] 0 A I B I a I a I [ 0 0º 0º 0 80º 0] 5 j 89 5 78 0ºA I B I a I a I [ 0 0º 40º 0 80º 0] -line I 0 5 j 89 5 78 0ºA I 0 0 A I a I 40º 5 78 0º 5 78 50º A

Unsymmetrical Fault Calculations 45 B-line I a I 0º 5 78 0º 5 78 50ºA I B0 I B I 0 0 A a I 0º 5 78 0º 5 78 90º A I B a I 40º 5 78 0º 5 78 90ºA Note that components I B and I B have finite values although the line B is open and can carry no net current. As expected, the sum of I B and I B is zero. However, the sum of components in line is 0 0ºA and the sum of components in line is 0 80ºA. Example 8.0. Three resistors of 5Ω, 0Ω and 0Ω are connected in delta across the three phases of a balanced 00 volts supply. What are the sequence components in the resistors and in supply lines? Solution. Let the voltages across 5Ω, 0Ω and 0Ω be E corresponding currents in the resistors be I vector diagram shown in Fig. 8.8 (ii)., E and E B respectively and the, I and I B. These voltages can be represented by the E 00 0º V ; E 00 60º V ; E B 00 60ºV Current in 5Ω, I E 5 00 0 º 0 0º A 5 Current in 0Ω, 00 60º I E 0 0 60ºA 0 Current in 0Ω, 00 60º IB EB 0 5 60ºA 0 Sequence currents in resistors Zero sequence component of I is I 0 I I I B [ 0 0º 0 60º 5 60º] [( 0 j 0) (5 j 8 66) ( 5 j 4 )]

46 Principles of Power System [ 5 j 4 ] 4 7 j 44 4 4 60 9ºA Positive sequence component of I is I B I a I a I [ 0 0º 0º 0 60º 40º 5 60º] [ 0 0º 0 80º 5 80º] [( 0 j 0) ( 0 j 0) ( 5 j 0)] [ 5 j 0] 66 j 0 66 80ºA Negative sequence component of I is I B I a I a I [ 0 0º 40º 0 60º 0º 5 60º] [ 0 0º 0 00º 5 60º] [( 0 j 0) (5 j 8 66) ( 5 j 4 )] [ 5 j 4 ] 4 7 j 44 4 4 60 9º A The sequence components of I and I B can be found as under : I I 4 4 60 9º A 0 0 I a I 40º 66 80º 66 60º A I ai 0º 4 4 60 9º 4 4 40 9º A I I I 4 4 60 9º A B0 0 ai 0º 66 80º 66 00º A B B I a I 40º 4 4 60 9º 4 4 79 º A Sequence currents in supply lines Line current in -line, I I I r 5 60º 0 60º ( 5 j 4 ) (5 j 8 66) 5 j 99 00 9º A Line current in -line, I y I IB 0 0º 5 60º ( 0 j 0) ( 5 j 4 ) 5 j 4 9 69º A Line current in B-line, I b I I 0 60º ( 0 0º) (5 j 8 66) ( 0 j 0) 5 j 8 66 6 45 9 ºA B

Unsymmetrical Fault Calculations 47 Zero sequence component of I r is I r0 * I I I Positive sequence component of I r is I r e r y b j 0 0A** eir a Iy a Ibj eib Ij aei IBj a ei Ij a I a I a IB e j aei a I a IBj ea ajei a I a IBj Now a a j and I ai a I I d ie j B I r j I j I j ( 66 j 0) Negative sequence component of I r is j 0 0 90º A I r Ir a Iy a Ib eib Ij a ei IBj aei Ij e j e j e Bj B B a I a I ai a I a I ai ( a a) I a I a I B Now a a j and I a I a I I I r j I j I j ( 4 7 j 44) e j 5 j 7 7 6 09 ºA [ a and a 4 a] Note. Incidentally, we have the formulas for relation among sequence components in the phases and lines. I r j I ; I r j I Example 8.. A delta connected load is supplied from a -phase supply. The fuse in the B line is removed and current in the other two lines is 0 A. Find the symmetrical components of line currents. Solution. Let, and B be the supply lines. When fuse in the line B is removed, the various line currents are : * Since vector sum of I I I 0, I 0 r y b r0 ** This shows that in delta formation, the zero sequence currents are present in phases but they disappear in line currents. As line current is the difference of two phase currents, therefore, the zero sequence components cancel out.

48 Principles of Power System -line I r 0 0º A ; I y 0 80º A ; I b 0 A I r0 I r I y I b [0 0º 0 80º 0] [(0 j0) ( 0 j0) 0] [0] 0A I r r y b I a I a I [0 0º 0º 0 80º 0] [0 0º 0 00º] [(0 j0) (0 j 7 )] I r [0 j 7 ] 0 j 5 77 54 0ºA r y b I a I a I [0 0º 40º 0 80º 0] [0 0º 0 60º] [(0 j0) (0 j 7 )] -line B-line I y0 [0 j 7 ] 0 j 5 77 54 0ºA I r0 0A I y a I r 40º 54 0º 54 0º A I y I bo a I r 0º 54 0º 54 50º A I ro 0A I b a I r 0º 54 0º 54 90ºA I b a I r 40º 54 0º 54 70º A Example 8.. Three impedances of 5 j0, 6 j 5 and j5 ohms are connected in star to red, yellow and blue lines of a 00 V, -phase, -wire supply. The phase sequence is B. Calculate the line current I. Solution. This is a case of unbalanced -phase star connected load supplied from a balanced - phase supply. Since the phase sequence is B, I V 00 0º V ; V a V 00 40º V Let V, V and V B be the voltages across impedances in, and B phases respectively and, I and I B the resulting line currents. V V V 00 j 0 and V VB VB 00( 0 5 j 0 866) Since I I IB 0A B

Unsymmetrical Fault Calculations 49 I I I 0A 0 0 B0 e 0 j e j...(i) e 0 j e j e j...(ii) e B0 B Bj e B Bj e j...(iii) V (5 j0) I (5 j0) I I I (5 j0) I I V (6 j5) I (6 j5) I I I (6 j5) I I Subtracting exp. (ii) from exp. (i), we get, V (6 j5) a I a I V B ( j5) I B ( j5) I I I ( j5) I I ( j5) ai a I V I I 5 j 0 a I a I 6 j 5 e jb g e jb g or 00 ( 67 j )I ( j 7) I...(iv) Subtracting exp. (iii) from exp. (ii), we get, or V V a I a I 6 j 5 a I a I j 5 B e jb g e jb g 00( 0 5 j 0 866) (5 8 j 8)I (8 84 j 8)I or 650 j 858 (5 8 j 8) I (8 84 j 8)I...(v) Solving exps. (iv) and (v), we get, I 4 j 65 and I 95 j 4

440 Principles of Power System I I I (4 j 65) (95 j 4) 9 j 76 4 8 4º A Example 8.. A star connected load consists of three equal resistors of Ω resistance. The load is assumed to be connected to an unsymmetrical -phase supply, the line voltages are 00 V, 46 V and 400 V. Find the magnitude of current in any phase by the method of symmetrical components. Solution. This is a case of a balanced star-connected load supplied from an unbalanced -phase supply. Fig. 8. (i) shows the balanced star-connected load receiving unbalanced supply. Fig. 8. (ii) shows the vector diagram. Since the vector sum of three voltages is zero, these can be represented by the three sides of a triangle as shown in Fig. 8.. eferring to Fig. 8., it is clear that : () ( 75 cos θ) ( 75 sin θ) or 4 ( 75) (cos θ sin θ) 75 cos θ or 4 5 cos θ cos θ (4 4)/ 5 0 θ 90º and cos α 75cos θ 0 0 5 α 60º As the phase sequence is B, therefore, various line voltages are : 00 80º ( 00 j0) V V V B 46 80º 90º 46 90º (0 j 46) V V B 400 60º (00 j 46) V The current in any phase (or line) is equal to phase voltage divided by resistance in that phase. 00 80º Line current, I 5 47 80º A 46 90º Line current, I 99 77 90º A

Unsymmetrical Fault Calculations 44 400 60º Line current, I B 0 94 60ºA Sequence components in red phase are : I 0 I I I B I [5 47 80º 99 77 90º 0 94 60º] [( 5 47 j 0) (0 j 99 77) (5 47 j 99 99)] [0] 0A B I a I a I [5 47 80º 0º 99 99 90º 40º 0 94 60º] [5 47 80º 99 99 0º 0 94 80º] [( 5 47 j0) ( 7 j 99 99) ( 0 94 j0)] I [ 59 4 j 99 99] 7 j 76 69ºA B I a I a I [5 47 80º 40º 99 99 90º 0º 0 94 60º] [5 47 80º 99 99 0º 0 94 60º] [( 5 47 j0) (7 j 99 99) (5 47 j 99 99)] [7 j 00] 57 66 j 66 58 0º A TUTOIAL POBLEMS. In a -phase, 4-wire system, currents in, and B lines under abnormal conditions of loading are: I 50 45º A ; I 50 50º A ; I B 00 00º A Calculate the zero, positive and negative phase sequence currents in the -line and return current in the neutral connection. [I 0 5 7º A ; I 48 0 87 6º A; I 6 40 45º A; I N 56 6 7º A]. In a -phase system, the phase voltages are as under : E 0º V ; E B 0º V ; E 0 V Find the zero, positive and negative phase sequence components in the -phase. [E 0 0 0ºV; E 0 40ºV; E 0 66 0º V]. The currents in a -phase unbalanced system are : I (80 j 0) A ; I ( 0 j 60) A ; I B (70 j 60) A The phase sequence is B. Calculate the zero, positive and negative sequence components of the red line current and determine the current in the neutral wire. [I 0 0A; I 76 58 ºA ; I 8 7 6ºA; I N 0A] 4. A -phase, 4-wire system supplies loads which are unequally distributed in the three phases. An analysis of the circuit shows that positive and negative phase sequence components of the current in the red line are as under :

44 Principles of Power System I (7 89 j 0 7) A ; I ( j 7) A The total observed current flowing back to supply in the neutral conductor is zero. Calculate the current in the three lines. [I (0 j )A; I ( j 4) A; I B ( 8 j 6) A] 8.6 Sequence Impedances Each element of power system will offer impedance to different phase sequence components of current which may not be the same. For example, the impedance which any piece of equipment offers to positive sequence current will not necessarily be the same as offered to negative sequence current or zero sequence current. Therefore, in unsymmetrical fault calculations, each piece of equipment will have three values of impedance one corresponding to each sequence current viz. (i) Positive sequence impedance (Z ) (ii) Negative sequence impedance (Z ) (iii) Zero sequence impedance (Z 0 ) The impedance offered by an equipment or circuit to positive sequence current is called positive sequence impedance and is represented by Z. Similarly, impedances offered by any circuit or equipment to negative and zero sequence currents are respectively called negative sequence impedance (Z ) and zero sequence impedance (Z 0 ). The following points may be noted : (a) In a -phase balanced system, each piece of equipment or circuit offers only one impedance the one offered to positive or normal sequence current. This is expected because of the absence of negative and zero sequence currents in the -phase balanced system. (b) In a -phase unbalanced system, each piece of equipment or circuit will have three values of impedance viz. positive sequence impedance, negative sequence impedance and zero sequence impedance. (c) The positive and negative sequence impedances of linear, symmetrical and static circuits (e.g. transmission lines, cables, transformers and static loads) are equal and are the same as those used in the analysis of balanced conditions. This is due to the fact that impedance of such circuits is independent of the phase order, provided the applied voltages are balanced. It may be noted that positive and negative sequence impedances of rotating machines (e.g. synchronous and induction motors) are normally different. (d) The zero sequence impedance depends upon the path taken by the zero sequence current. As this path is generally different from the path taken by the positive and negative sequence currents, therefore, zero sequence impedance is usually different from positive or negative sequence impedance. 8.7 Sequence Impedances of Power System Elements The concept of impedances of various elements of power system (e.g. generators, transformers, transmission lines etc.) to positive, negative and zero sequence currents is of considerable importance in determining the fault currents in a -phase unbalanced system. A complete consideration of this topic does not fall within the scope of this book, but a short preliminary explanation may be of interest here. The following three main pieces of equipment will be considered : (i) Synchronous generators (ii) Transformers (iii) Transmission lines (i) Synchronous generators. The positive, negative and zero sequence impedances of rotating machines are generally different. The positive sequence impedance of a synchronous generator is equal to the synchronous impedance of the machine. The negative sequence impedance is much less

Unsymmetrical Fault Calculations 44 than the positive sequence impedance. The zero sequence impedance is a variable item and if its value is not given, it may be assumed to be equal to the positive sequence impedance. In short : Negative sequence impedance < Positive sequence impedance Zero sequence impedance Variable item may be taken equal to ve sequence impedance if its value is not given It may be worthwhile to mention here that any impedance Z e in the earth connection of a starconnected system has the effect to introduce an impedance of Z e per phase. It is because the three equal zero-sequence currents, being in phase, do not sum to zero at the star point, but they flow back along the neutral earth connection. (ii) Transformers. Since transformers have the same impedance with reversed phase rotation, their positive and negative sequence impedances are equal; this value being equal to the impedance of the transformer. However, the zero sequence impedance depends upon earth connection. If there is a through circuit for earth current, zero sequence impedance will be equal to positive sequence impedance otherwise it will be infinite. In short, Positive sequence impedance Negative sequence impedance Impedance of Transformer Zero sequence impedance Positive sequence impedance, if there is circuit for earth current Infinite, if there is no through circuit for earth current. (iii) Transmission lines. The positive sequence and negative sequence impedance of a line are the same; this value being equal to the normal impedance of the line. This is expected because the phase rotation of the currents does not make any difference in the constants of the line. However, the zero sequence impedance is usually much greater than the positive or negative sequence impedance. In short : Positive sequence impedance Zero sequence impedance Negative sequence impedance Impedance of the line Variable item may be taken as three times the ve sequence impedance if its value is not given 8.8 Analysis of Unsymmetrical Faults In the analysis of unsymmetrical faults, the following assumptions will be made : (i) The generated e.m.f. system is of positive sequence only. (ii) No current flows in the network other than due to fault i.e. load currents are neglected. (iii) The impedance of the fault is zero. (iv) Phase shall be taken as the reference phase. In each case of unsymmetrical fault, e.m.f.s per phase are denoted by E, E and E B and the terminal p.d. per phase by V, V and V B. 8.9 Single Line-to-Ground ound Fault Consider a -phase system with an earthed neutral. Let a single line-to-ground fault occur on the red phase as shown in Fig. 8.. It is clear from this figure that : *V 0 and I I 0 B * Note that V is the terminal potential of phase i.e. p.d. between N and. Under line-to-ground fault, it will obviously be zero.

444 Principles of Power System The sequence currents in the red phase in terms of line currents shall be : I 0 ei I IBj I I ei a I a IBj I I ei a I a IBj I I 0 I I I Fault current. First of all expression for fault current I will be derived. Let Z, Z and Z 0 be the positive, negative and zero sequence impedances of the generator respectively. Consider the closed loop NEN. As the sequence currents produce voltage drops due only to their respective sequence impedances, therefore, we have, E I Z I Z I 0 Z 0 V As V 0 and I I I0 E I Z Z Z or I 0 e 0 0 E Z Z Z Fault current, I I 0 0 j E Z Z Z 0...(i) Examination of exp. (i) shows that the equivalent circuit from which fault current may be calculated is as given in Fig. 8.4. It is clear that fault current is obtained by connecting the phase sequence impedances in series across an imaginary generator of voltage E. This is a wonderful part of the method of symmetrical components and makes the analysis easy and interesting. In fact, this method permits to bring any unsymmetrical fault into a simple circuit of

Unsymmetrical Fault Calculations 445 interconnection of sequence impedances appropriate to the fault condition prevailing. The assumption made in arriving at exp. (i) is that the fault impedance is zero. However, if the fault impedance is Z e, then expression for fault current becomes : E I Z Z Z0 Ze It may be added here that if the neutral is not grounded, then zero sequence impedance will be infinite and the fault current is zero. This is expected because now no path exists for the flow of fault current. Phase voltages at fault. Now let us calculate the phase voltages at fault (i.e. voltage between each line and fault). Since the generated e.m.f. system is of positive sequence only, the sequence components of e.m.f. in -phase are : E 0 0 ; E 0 and E E The sequence voltages at the fault for -phase are : V E Z V E I Z E Z Z Z Z Z0 Z Z Z E 0 V 0 Z I V 0 0 I0 Z0 Z Z Z Z 0 Z0 Z Z Z 0 0 It can be readily seen that V V V0 0. This is expected because -phase is shorted to ground. The phase voltages at fault are : V V0 V V 0 V V a V av 0 0 V B V av a V Summary of esults. For line (-phase)-to-ground fault : (i) I E Fault current Z Z Z (ii) V 0 0 0 V V a V av V B V av a V 0 E E ; I 0 ; I 0 8.0 Line to Line Fault Consider a line-to-line fault between the blue (B) and yellow () lines as shown in Fig. 8.5. The conditions created by this fault lead to : V VB ; I 0 and I IB 0 Again taking -phase as the reference, we have, B

446 Principles of Power System e Bj 0 I 0 I I I Now V V B Expressing in terms of sequence components of red line, we have, V a V av V av a V 0 0 or V ( a a) V ( a a) V V...(i) Also I I 0 B e 0 j e 0 j 0 e j 0 0 or I a I a I I a I a I or ( a a) I I I or I I 0 [ I 0 0]...(ii) Fault current. Examination of exp. (i) and exp (ii) reveals that sequence impedances should be connected as shown in Fig. 8.6. It is clear from the figure that : E I I Z Z Fault current, I I a I a I 0 0 a F HG E ( a a) Z Z I F HG E Z ZKJ E a Z Z j E IB Z Z Phase voltages. Since the generated e.m.f. system is of positive phase sequence only, the sequence components of e.m.f. in -phase are : E 0 0 ; E 0 and E E I KJ

Unsymmetrical Fault Calculations 447 The sequence voltages at the fault for -phase are : V V E I Z E Z Z Z E V 0 I Z V 0 0 I0 Z0 0 The phase voltages at fault are : V V V0 V V Z F HG Z Z Z E Z E I KJ Z 0 Z Z E Z Z Z E Z Z Z E 0 V V a V av 0 a F HG F HG ( a a) I F HG Z Z Z Z E KJ a Z Z Z V Z Z Z Z E E V B V av a V F HG 0 I KJ Z Z E Z 0 a Z Z E KJ a Z Z Z E ( a a) F HG I Z Z Z E Z V B E Z Z Summary of esults. For line-to-line fault (Blue and ellow lines) : j E (i) I I IB 0 ; Z Z (ii) V Z V Z Z E B and V 8. Double Line-to-Ground ound Fault Z Z Z E I KJ F HG I KJ I KJ ( a a ) Consider the double line-to-ground fault involving B lines and earth as shown in Fig. 8.7. The

448 Principles of Power System conditions created by this fault lead to : I 0 ; V V 0 B Since V V B 0, it is implied that : V V V V...(i) 0 Also I I I I 0 (given)...(ii) 0 Fault current. Examination of exp. (i) and exp. (ii) reveals that sequence impedances should be *connected as shown in Fig. 8.8. It is clear that : I I Z I E Z Z0 Z Z Z 0 Z 0 Z 0 Z I 0 I Z Z0 Fault current, I F I IB Z I ** I 0 Z Z Z Z Z 0 0 E ZZ 0 Z Z Z 0 Z E Z Z Z Z Z Z 0 0 * Since V V V V, sequence impedances must be in parallel. ** I 0 ei I IBj (0 Fault Current) Fault current I 0 0

449 Unsymmetrical Fault Calculations 449 Phase Voltages. The sequence voltages for phase are : V E I Z ; V 0 I Z ; V 0 I Z 0 0 0 0 Now V V V V V V V V V 0 V a V av V0 ( a a ) V (Q V V V ) 0 0 V 0 (Q a a 0) V B av a V V ( a a ) V 0 0 Example 8.4. A -phase, 0 MVA, kv generator with a solidly earthed neutral point supplies a feeder. The relevant impedances of the generator and feeder in ohms are as under : Generator feeder Positive sequence impedance j j 0 Negative sequence impedance j 0 9 j 0 Zero sequence impedance j 0 4 j 0 If a fault from one phase to earth occurs on the far end of the feeder, calculate (i) the magnitude of fault current (ii) line to neutral voltage at the generator terminal Solution. The circuit diagram is shown in Fig. 8.9. The fault is assumed to occur on the red phase. Taking red phase as the reference, Phase e.m.f. of -phase, E 0 650 V (i) The total impedance to any sequence current is the sum of generator and feeder impedances to that sequence current. Total Z j j 0 j Ω Total Z j 0 9 j 0 j 9 Ω Total Z 0 j 0 4 j 0 j 4 Ω For a line-to-ground fault, we have, I I I 0 E 650 Z Z Z j j 9 j 4 0

450 Principles of Power System 650 j 7 5 j 846 A Fault current, I I 0 ( j 846) j 58 A (ii) Line-to-neutral voltage of -phase, V E I Z I Z I0 Z0 where Z0, Zand Z are the sequence impedances of generator. e E I Z Z Z 0 0 650 ( j 846) (j j 0 9 j 0 4) 650 j 846 ( j 5) 650 5 45 V Example 8.5. A -phase, kv, 0 MVA alternator has sequence reactances of X 0 0 05 p.u., X 0 5 p.u. and X 0 5 p.u. If the generator is on no load, find the ratio of fault currents for L-G fault to that when all the -phases are dead short-circuited. Solution. Taking red phase as the reference, let its phase e.m.f. be E p.u. Line-to-ground fault. Suppose the fault occurs on the red phase. Then, j I I I 0 E X X X 0 I 0 j 0 5 j 0 5 j 0 05 j 05 j 85 Fault current, I I 0 ( j 85) j 8 55 A Three phase fault. When a dead short circuit occurs on all the three phases, it gives rise to symmetrical fault currents. Therefore, the fault current (say I sh ) is limited by the positive sequence reactance (i.e. X ) only. Fault current, I sh E X j 0 5 j 6 66 atio of two fault currents I j 855 I j 666 84 sh i.e. single line-to-ground fault current is 84 times that due to dead short circuit on the - phases. Example 8.6. A -phase, kv, 5 MVA generator with X 0 0 05 p.u., X 0 p.u. and X 0 p.u. is grounded through a reactance of 0 Ω. Calculate the fault current for a single line to ground fault. Solution. Fig. 8.0 shows the circuit diagram. The fault is assumed to occur on the red phase. Taking red phase as the reference, let its phase e.m.f. be E p.u. First of all, convert the reactance X n into p.u. value from the following relation : kva rating *p.u. value of X n X n in ohms ( kv) 000 kva rating * % X n X n in ohms. If this value is divided by 00, we get p.u. value. (kv) 0

45 Unsymmetrical Fault Calculations 45 0 5, 000 0 06 p.u. ( ) 000 For a line-to-ground fault, we have, I I I 0 E X X ( X X ) 0 j 0 j 0 j( 005 006 ) j 57 p.u. j 0 66 Fault current, I I 0 ( j 57) j 4 76 p.u. Fault current in amperes ated current p.u. value 6 5 0 4 76 688 A 0 Example 8.7. A -phase, -wire system has a normal voltage of 0 4 kv between the lines. It is supplied by a generator having positive, negative and zero sequence reactances of 0 6, 0 5 and 0 Ω per phase respectively. Calculate the fault current which flows when a line-to-line fault occurs at the generator terminals. Solution. Suppose the short circuit fault occurs between yellow and blue phases. Taking red phase as the reference, its phase e.m.f. is : Phase e.m.f. of -phase, E 0 4 0 6000 V Now X j 0 6 Ω ; X j 0 5 Ω ; X 0 j 0 Ω For line-to-line fault, we have, Fault current, I F E X X (in magnitude) 6000 ( 06 05 ) 9447 5 A Example 8.8. The per unit values of positive, negative and zero sequence reactances of a network at fault are 0 08, 0 07 and 0 05. Determine the fault current if the fault is double line-toground. n

45 45 Principles of Power System Solution. Suppose the fault involves yellow and blue phases and the ground. Taking red phase as the reference, let its phase e.m.f. be E p.u. Now, X j 0 08 p.u. ; X j 0 07 p.u. ; X 0 j 0 05 p.u. For a double line-to-ground fault, we have, X E Fault current, I F I IB X X X X0 X X0 j 0 07 j 008 j 007 j 008 j 005 j 007 j 005 j 0 ( 56 40 5) 0 4 4 j 0 0 j 6 p.u. Example 8.9. A 0 MVA, kv, -phase, 50 Hz generator has its neutral earthed through a 5% reactor. It is in parallel with another identical generator having isolated neutral. Each generator has a positive sequence reactance of 0%, negative sequence reactance of 0% and zero sequence reactance of 5%. If a line to ground short circuit occurs in the common bus-bar, determine the fault current. Solution. Fig. 8. shows the two generators in parallel. The generator has its neutral earthed through a reactance ( 5%) whereas generator has ungrounded neutral. The earth fault is assumed to occur on the red phase. Taking red phase as the reference, its phase e.m.f. E 0 65 V. For a line to ground fault, the *equivalent circuit will be as shown in Fig. 8. (i) which further reduces to the circuit shown in Fig. 8. (ii). * Note the equivalent circuit diagram. The positive sequence reactances (0%) of two generators are in parallel and so are their negative sequence reactances (0%). The zero sequence reactance of generator is zero because its neutral is ungrounded. However, the zero sequence reactance of generator 5% 5% 0%.

45 Unsymmetrical Fault Calculations 45 The percentage reactances in Fig. 8. (ii) can be converted into ohmic values as under : X % reactance (Voltage in kv) 0 Base kva ( ) 0 0 0 605 Ω 0 0 X ( ) 0 5 0 05 Ω 0 0 ( ) 0 X 0 0 85 Ω 0 0 E 65 Fault current, I X 0 605 0 05 85 X X j j j 0 905 j 6998 A j 75 Example 8.0. A 50 MVA, kv three-phase alternator was subjected to different types of faults. The fault currents are as under : -phase fault 000 A ; Line-to-Line fault 600 A ; Line-to-ground fault 400 A The generator neutral is solidly grounded. Find the values of the three sequence reactances of the alternator. Ignore resistances. Solution. Let X, X and X 0 be the positive, negative and zero sequence reactances respectively of the alternaor. For -phase fault, Fault current E ph X or 000 000 X (magnitude)

454 454 Principles of Power System X For line-to-line fault, we have, Fault current or 600 000 000 E X X ph 000 X X 75 Ω or X X 000 600 4 Ω X 4 X 4 75 056 Ω For line-to-ground fault, we have, Eph Fault current X X X0 or 400 000 X X X 0 or X X X 0 000 4 56 Ω 400 X 0 4 56 X X 4 56 75 056 0 05 Ω TUTOIAL POBLEMS (magnitude) (magnitude). A -phase, 75 MVA, 0 8 p.f. (lagging), 8 kv star-connected alternator having its star point solidly earthed supplies a feeder. The relevant per-unit (p.u.) impedances, based on the rated phase voltage and phase current of the alternator are as follows : Generator Feeder Positive sequence impedance (p.u.) j 7 j 0 Negative sequence impedance (p.u.) j 0 8 j 0 Zero sequence impedance (p.u.) j 0 j 0 Determine the fault current for a one line-to-earth fault occuring at the far end of the feeder. The generated e.m.f. per phase is of positive sequence only and is equal to the rated voltage per phase. [4400 A]. A -phase, 75 MVA, 8 kv star-connected alternator with a solidly earthed neutral point has the following p.u. impedances based on rated phase voltage and rated phase current : Positive phase sequence impedance j p.u. Negative phase sequence impedance j 0 6 p.u. Zero phase sequence impedance j 0 08 p.u. Determine the steady-state fault current for the following : (i) -phase symmetrical short-circuit (ii) one line-to-earth fault (iii) two line-to-earth fault. The generated e.m.f. per phase is equal to the rated voltage. [(i)840 A (ii) 490 A (iii) 580 A]. The per unit values of positive, negative and zero sequence reactances of a network at fault are 0 08, 0 07 and 0 05 respectively. Determine the fault current if fault is line-to-line-to-ground. [j 6 p.u.] 8. Sequence Networks The analysis of an unsymmetrical fault by symmetrical components method can be conveniently done by drawing sequence networks. A sequence network of a particular sequence current in a given power system is the path for the flow of that sequence current in the system. It is composed of impedances offered to that sequence current in the system. Since there are three sequence currents

455 Unsymmetrical Fault Calculations 455 (viz. positive sequence current, negative sequence current and zero sequence current), there will be three sequence networks for a given power system, namely ;. Positive sequence network. Negative sequence network. Zero sequence network. Positive sequence network. The positive sequence network for a given power system shows all the paths for the flow of positive sequence currents in the system. It is represented by oneline diagram and is composed of impedances offered to the positive sequence currents. While drawing the positive sequence network of a given power system, the following points may be kept in view: (i) Each generator in the system is represented by the generated voltage in series with appropriate reactance and resistance. (ii) Current limiting impedances between the generator s neutral and ground pass no positive sequence current and hence are not included in the positive sequence network. (iii) All resistances and magnetising currents for each transformer are neglected as a matter of simplicity. (iv) For transmission lines, the shunt capacitances and resistances are generally neglected. (v) Motor loads are included in the network as generated e.m.f. in series with appropriate reactance.. Negative sequence network. The negative sequence network for a given power system shows all the paths for the flow of negative sequence currents in the system. It is also represented by one-line diagram and is composed of impedances offered to the negative sequence currents. The negative sequence network can be readily obtained from positive sequence network with the following modifications : (i) Omit the e.m.fs. of -phase generators and motors in the positive sequence network. It is because these devices have only positive sequence-generated voltages. (ii) Change, if necessary, the impedances that represent rotating machinery in the positive sequence network. It is because negative sequence impedance of rotating machinery is generally different from that of positive sequence impedance. (iii) Current limiting impedances between generator s neutral and ground pass no negative sequence current and hence are not included in the negative sequence network. (iv) For static devices such as transmission lines and transformers, the negative sequence impedances have the same value as the corresponding positive sequence impedances.. Zero sequence network. The zero sequence network for a given power system shows all the paths for the flow of zero sequence currents. The zero sequence network of a system depends upon the nature of connections of the -phase windings of the components in the system. The following points may be noted about zero sequence network : (i) The zero sequence currents will flow only if there is a return path i.e. path from neutral to ground or to another neutral point in the circuit. (ii) In the case of a system with no return path for zero sequence currents, these currents cannot exist. 8. eference ence Bus for Sequence Networks While drawing the sequence networks, it is necessary to specify the reference potential w.r.t. which all sequence voltage drops are to be taken. For this purpose, the reader may keep in mind the following points : (i) For positive or negative sequence networks, the neutral of the generator is taken as the

456 456 Principles of Power System reference bus. This is logical because positive or negative sequence components represent balanced sets and hence all the neutral points must be at the same potential for either positive or negative sequence currents. (ii) For zero sequence network, the reference bus is the ground at the generator. Example 8.. An unloaded generator is grounded through a reactance Z n as shown in Fig. 8.. If a single line-to-ground fault occurs, draw (i) the positive sequence network (ii) negative sequence network and (iii) zero sequence network. Solution. Fig. 8. shows the unloaded generator with single line-to-ground fault. We shall now draw the sequence networks for this system. (i) Positive sequence network. The generated voltages are of positive sequence only becasue the generator is designed to supply -phase balanced voltages. Therefore, the positive sequence network is composed of phase e.m.fs. in series with positive sequence impedance of the generator. Fig. 8.4 (i) shows the positive sequence current paths whereas Fig. 8.4 (ii) shows the singlephase positive sequence network.