4.10 Unbalanced fault analysis using Z BUS matrix:

Transcription

1 4.10 Unbalanced fault analysis using Z BUS matrix: In the previous section, it is observed that, for fault calculations the Thevenin s equivalent networs, at the fault point, are needed for the three sequence networs. Since the three sequence networs are independent, the Z BUS matrices of these sequence networs can be found seperately. The diagonal elements of the three sequence Z BUS matrices infact, are the Thevenin s equivalent impedances of the sequence networs as seen from the various buses. Let, the three sequence bus impedance matrices for zero sequence, positive sequence and negative sequence networs be represented as [ Z (0) BUS ], [ Z BUS ] and [ Z ] BUS respectively. If the fault is at the th bus, then Z (0), Z and Z of the sequence bus impedance matrices are the zero, positive and negative Thevenin s equivalent impedances, respectively, as seen from the faulted bus. Hence, the first step in the fault analysis using Z BUS matrix is the determination of the three sequence networs and subsequently, finding the bus impedance matrix for each sequence networ. To illustrate this step, consider the single line diagram of the power system shown in Fig Figure 4.60: Single line diagram of Power System The positive sequence equivalent networ for the system is shown in Fig In this figure all the elements of the system have been represented by their positive sequence equivalents. Similarly by representing all elements by their negative sequence impedances, the negative sequence equivalent networ can be obtained. The negative sequence networ is shown in Fig For the zero sequence equivalent networ, the generator neutral connections and transformer connections have to be considered. The zero sequence equivalent networ is shown in Fig In the next step, the [ Z ] matrix for the three sequence networs is found using [ Z ] building algorithm. Once [ Z (0) ], [ Z ] and [ Z ] matrices are nown, the following procedure is followed for the fault analysis of the given networ. 170

2 Figure 4.61: The positive sequence equivalent networ Figure 4.62: The negative sequence equivalent networ (a) LG fault: Let the fault be on phase a of bus with a fault impedance Z f as shown in Fig From equations (4.98) and (4.101), it can be seen that the three equivalent sequence networs are in series for calculating the sequence components of the fault currents. Hence, generalizing equation (4.101) for fault at th bus, the expression for sequence component of fault current can be written as: = Ī = Ī = V (0) Z (0) () () + 3 Z (4.121) () f (0) Z, Z and Z are the th diagonal elements of [ Z (0) ], [ Z ] and [ Z ] matrices respectively. V (0) is the prefault voltage of th bus, usually taen as 1 0 pu. 171

3 Figure 4.63: The zero sequence equivalent networ Figure 4.64: LG fault on phase a of th bus The fault current is given by: [Ī (abc) (F)] = [Ā] [Ī (012) (F)] (4.122) (b) LL fault: Let the fault be between phases phase b and phase c of bus through an impedance Z f as shown in Fig From equation (4.109) and Fig it is observed that the positive sequence and negative sequence equivalent networs are connected in phase opposition. Thus, the expression of equation (4.109) for the sequence components of fault current at bus can be generalized as: = 0 172

4 Figure 4.65: LL fault between phase b and phase c of th bus and Ī = Z (0) V (0) = Ī (4.123) The phase components of fault current is the calculated from equation (4.122) I = I (b) (c) = I (4.124) (c) LLG fault: Fig shows an LLG fault involving phases phase b and phase c of bus through an impedance Z f. Referring to equation (4.119) and Fig.4.66, the generalized expression Figure 4.66: LLG fault involving phase b and phase c of th bus for sequence components of fault current at bus can be written as Ī = Z V (0) + Z (Z(0) Z + 3 Z f ) + 3 Z f (0) 173

5 Ī = V (0) Z Ī Z (4.125) Z (0) = V (0) Z Ī + 3 Z f The phase currents can be obtained from equation (4.122), the fault current is then calculated as I = I (b) (c) + I (4.126) Calculation of voltages and Line currents during fault: To calculate the voltages of buses during fault equation (4.94) can be generalized as: V (0) (0) i = Z i V i = V i (0) V i = Z Ī i The pre fault voltage V 1 i (0)is usually set as pu. Z i Ī (4.127) The bus phase voltage during fault is calculated from the following relation. V (abc) i = [Ā] [ V (012) i ] (4.128) where [Ā] is the symmetrical component transformation matrix. To calculate the symmetrical components of line currents in the line from bus i to bus j the following relation is used: ij = V (0) (0) i V j z (0) ij Ī ij = Ī ij = V i V j z ij V i V j z ij (4.129) where z (0) ij, z ij and z ij are the zero, positive and negative sequence impedance respectively of the line between bus i and bus j. The phase currents for the line can be calculated from the symmetrical components using the relation: 174

6 [Ī abc ij ] = [Ā][Ī 012 ij ] (4.130) The process of fault analysis of a power system networ is illustrated in the next lecture with the help of an example. 175