FAULT CURRENT COMPUTATION USING SEQUENCE NETWORKS
In this section we shall demonstrate the use of sequence networks in the calculation of fault currents using sequence network through some examples.
Example 8.4 Consider the network shown in Fig. 8.10. The system parameters are given below
We shall find the fault current for when a (a) 1LG, (b) LL and (c) 2LG fault occurs at bus-2.
Fig. 8.10 Radial power system of Example 8.4. Let us choose a base in the circuit of the generator. Then the per unit impedances of the generator are:
The per unit impedances of the two transformers are
The MVA base of the motor is 40, while the base MVA of the total circuit is 50. Therefore the per unit impedances of the motor are
For the transmission line
Therefore
Let us neglect the phase shift associated with the Y/ Δ transformers. Then the positive, negative and zero sequence networks are as shown in Figs. 8.11-8.13.
Fig. 8.11 Positive sequence network of the power system of Fig. 8.10.
Fig. 8.12 Negative sequence network of the power system of Fig. 8.10.
Fig. 8.13 Zero sequence network of the power system of Fig. 8.10. From Figs. 8.11 and 8.12 we get the following Ybus matrix for both positive and negative sequences
Inverting the above matrix we get the following Zbus matrix
Again from Fig. 8.13 we get the following Ybus matrix for the zero sequence
Inverting the above matrix we get
Hence for a fault in bus-2, we have the following Thevenin impedances
Alternatively we find from Figs. 8.11 and 8.12 that
(a) Single-Line-to-Ground Fault : Let a bolted 1LG fault occurs at bus-2 when the system is unloaded with bus voltages being 1.0 per unit. Then from (8.7) we get
Also from (8.4) we get
Also I fb = I fc = 0. From (8.5) we get the sequence components of the voltages as
Therefore the voltages at the faulted bus are
Then the fault currents are
Finally the sequence components of bus-2 voltages are
Hence faulted bus voltages are
Hence from (8.24) we get the positive sequence current as
The zero and negative sequence currents are then computed from (8.25) and (8.26) as
Therefore the fault currents flowing in the line are
Furthermore the sequence components of bus-2 voltages are
Therefore voltages at the faulted bus are
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Example 8.5 Let us now assume that a 2LG fault has occurred in bus-4 instead of the one in bus-2. Therefore
Also we have
Hence
Also
Therefore the fault currents flowing in the line are
We shall now compute the currents contributed by the generator and the motor to the fault. Let us denote the current flowing to the fault from the generator side by Ig , while that flowing from the motor by Im . Then from Fig. 8.11 using the current divider principle, the positive sequence currents contributed by the two buses are
Similarly from Fig. 8.12, the negative sequence currents are given as
Finally notice from Fig. 8.13 that the zero sequence current flowing from the generator to the fault is 0. Then we have
Therefore the fault currents flowing from the generator side are
and those flowing from the motor are
It can be easily verified that adding Ig and Im we get If given above. |
In the above two examples we have neglected the phase shifts of the Y/ Δ transformers. However according to the American standard, the positive sequence components of the high tension side lead those of the low tension side by 30° , while the negative sequence behavior is reverse of the positive sequence behavior. Usually the high tension side of a Y/ Δ transformer is Y-connected. Therefore as we have seen in Fig. 7.16, the positive sequence component of Y side leads the positive sequence component of the Δ side by 30° while the negative sequence component of Y side lags that of the Δ side by 30° . We shall now use this principle to compute the fault current for an unsymmetrical fault.
Let us do some more examplesExample 8.6
Let us consider the same system as given in Example 8.5. Since the phase shift does not alter the zero sequence, the circuit of Fig. 8.13 remains unchanged. The positive and the negative sequence circuits must however include the respective phase shifts. These circuits are redrawn as shown in Figs. 8.14 and 8.15.
Note from Figs. 8.14 and 8.15 that we have dropped the √3 α vis-à-vis that of Fig. 7.16. This is because the per unit impedances remain unchanged when referred to the either high tension or low tension side of an ideal transformer. Therefore the per unit impedances will also not be altered.
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Fig. 8.14 Positive sequence network of the power system of Fig. 8.10 including transformer phase shift.
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Fig. 8.15 Negative sequence network of the power system of Fig. 8.10 including transformer phase shift.
Since the zero sequence remains unaltered, these currents will not change from those computed in Example 8.6. Thus
 and  | per unit |
Now the positive sequence fault current from the generator Iga1 , being on the Y-side of the Y/ Δ transformer will lead I ma1 by 30° . Therefore
 per unit |
 per unit |
Finally the negative sequence current I ga2 will lag I ma2 by 30° . Hence we have
 per unit |
 per unit |
Therefore
 |
Also the fault currents flowing from the motor remain unaltered. Also note that the currents flowing into the fault remain unchanged. This implies that the phase shift of the Y/ Δ transformers does not affect the fault currents.
Example 8.7
Let us consider the same power system as given in Example 1.2, the sequence diagrams of which are given in Figs. 7.18 to 7.20. With respect to Fig. 7.17, let us define the system parameters as:
Generator G1 : 200 MVA, 20 kV, X" = 20%, X0 = 10%
Generator G2 : 300 MVA, 18 kV, X" = 20%, X0 = 10%
Generator G3 : 300 MVA, 20 kV, X " = 25%, X0 = 15%
Transformer T1 : 300 MVA, 220Y/22 kV, X = 10%
Transformer T2 : Three single-phase units each rated 100 MVA, 130Y/25 kV, X = 10%
Transformer T3 : 300 MVA, 220/22 kV, X = 10%
Line B-C : X1 = X2 = 75 Ω , X0 = 100 Ω
Line C-D : X1 = X2 = 75 Ω , X0 = 100 Ω
Line C-F : X1 = X2 = 50 Ω , X0 = 75 Ω
Let us choose the circuit of Generator 3 as the base, the base MVA for the circuit is 300. The base voltages are then same as those shown in Fig. 1.23. Per unit reactances are then computed as shown below.
| Generator G1 : |  , X0 = 0.15 |
| Generator G2 : |  , X0 = 0.0656 |
| Generator G3 : |  , X0 = 0.15 |
| Transformer T1 : |  |
| Transformer T2 : |  |
| Transformer T3 : |  |
| Line B-C : |  ,  |
| Line C-D : | , |
| Line C-F : |  , |
Neglecting the phase shifts of Y/ Δ connected transformers and assuming that the system is unloaded, we shall find the fault current for a 1LG fault at bus-1 (point C of Fig. 7.17).
From Figs. 7.18 and 7.19, we can obtain the positive and negative sequence Thevenin impedance at point C as (verify)
| X1 = X2 = j 0.2723 per unit |
Similarly from Fig. 7.20, the Thevenin equivalent of the zero sequence impedance is
| X0 = j 0.4369 per unit |
Therefore from (8.7) we get
 per unit |
Then the fault current is Ifa = 3 Ifa0 = 3.0565 per unit.
