Chapter 9 Balanced Faults, Part II. 9.4 Systematic Fault Analysis Using Bus Impedance Matrix

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Polly Atkinson
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1 Chapter 9 Balanced Faults, Part II 9.4 Systematic Fault Analysis Using Bus Impedance Matrix In the previous analysis we employed the Thevenin model and ound the Thevenin voltage and impedance by means o circuit manipulation. That method wors or small power circuits only: It is impractical or a real large power system. A systematic method is presented in this section which applies to power systems o any size. The method can also be programmed on a computer. Consider a typical o an n- power system networ as shown below. Here we show es i and. The generator shown is a voltage source behind a reactance which may be Xd, Xd,or Xd. Transmission lines are represented by their equivalent π model and all impedances are pu on a common MVA base. A balanced three-phase ault is to be applied at through an impedance Z. The preault voltages are obtained rom a i S S i Z power low solution and represented by the vector V1 V = V V n (1.1)

2 Generally the short circuit currents are so large compared to the steady state loads, the loads may be neglected. However, the loads can be represented as impedances using the above voltages or example, the load S i may be approximated by the impedance 2 Vi Zi = (1.2) * Si The changes in networ voltage due to the ault are ound rom the system with all volt- V 0 applied as shown in the igure be- ages shorted to ground and the preault voltage ( ) low: i Z i Z V th =V (0) I (F) Z Note the capacitors or the pi equivalent o line i. Also note the loads are replaced by the approximation o impedances. The voltage source at i is grounded and its impedance remains as shown above. The voltage changes caused by the ault in this circuit are represented by the column vector V1 V = V (1.3) Vn From Thevenin's theorem the voltages during the ault are ound rom V F = V 0 + V (1.4) ( ) ( ) We also now or the nodal equations that I = Y V (1.5) where I is the current entering the and Y is the admittance matrix. The diagonal elements o each is the sum o the admittances connected to that, i.e. ii m ij (1.6) j = 0 Y = y j i 2

3 The o-diagonal element is equal to the negative o the admittance between the nodes, i.e. Yij = Yji = yij (1.7) where y ij is the actual admittance between nodes i and j. For the Thevenin circuit above, the nodal equations are 0 Y11 Y1 Y1n V1 I = Y1 Y Yn V (1.8) 0 Yn1 Yn Ynn V n Note that the minus sign is due to the act the ault current is shown leaving node. The above matrix equation can be written as I F = Y V (1.9) ( ) which can be solved or the voltage change thus V = Z I F (1.10) 1 Y where Z = is nown as the impedance matrix (not to be mixed-up with the impedance matrix in mesh equations o circuit theory!) Using (1.10) in (1.4) we have: V F = V 0 + Z I F (1.11) ( ) Which can be expanded into matrix orm: V1 V1 Z11 Z1 Z1n 0 V = V +Z1 Z Zn I V n V n Zn1 Zn Znn 0 (1.12) Using the th equation we have: V F = V 0 Z I F (1.13) Also it is clear that From (1.13) and (1.14) we have: ( ) ( ) V F = Z I F (1.14) I V = Z + Z (1.15) Thus we can solve or any voltage using (1.12) or row i: V F = V 0 Z I F (1.16) And using (1.15) in (1.16) we have: i i i ( ) V V i F = V i Z i Z Z + (1.17) 3

4 Knowing all the voltages during a ault we can solve or all the ault currents, in particular, or line current rom i to j we have: Vi Vj Iij = (1.18) z It is noted that this method wors or any under ault (i.e. the aulted could be any one o the es in the system). [Note that z ij is the impedance between i and j. It is not the i-j-th element o Z.] The method suggested above to ind the impedance matrix by inversion o the admittance matrix is not easible or very large power systems. An alternate method or the direct ormation (or building) o the matrix Z will be discussed in the next section. Example 9.2 ij A three-phase ault with a ault impedance o Z = j0.16 pu occurs at 3 in the networ or Example 9.1. Using the impedance matrix method, compute the ault current, the voltages, and the line currents during the ault. -j5 -j2.5 -j First the networ in Example 9.1 is redrawn to the right using impedances. By inspection we can ind the Y thus: -j j2.5 j8.75 j1.25 j2.5 Y = j1.25 j6.25 j2.5 j2.5 j2.5 j5 Note that this is the Y beore the ault occurs. By inversion using Matlab we have: Y = j*[ ]; Z = inv(y) I 3 (F) V th =V 3 (0) Z = j0.16 Z = i i i i i i i i i From equation (1.15) we have: Z = j*.16; V0=[1; 1; 1]; I3F = V0(1)/(Z(3,3)+Z) I3F = 4

5 i and rom (1.16) we have: VF = V0-I3F*Z(:,3) VF = Finally rom (1.17) we have: z12 = j*.8; z13 = j*.4; z23 = j*.4; % I12 = (VF(1) - VF(2))/z12 I13 = (VF(1) - VF(3))/z13 I23 = (VF(2) - VF(3))/z23 I12 = I13 = I23 = i i i These results are identical to those ound earlier in Example 1. Note that the need or the repeated simpliication to ind the Thevenin impedance is removed, but a matrix inversion is introduced. This inversion will also be eliminated by the direct building method o the impedance matrix to be discussed in the next section. 5

4.10 Unbalanced fault analysis using Z BUS matrix:

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