Chapter 8: Unsymmetrical Faults

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

Generator G : 50 MVA, 20 kV, X" = X1 = X2 = 20%, X0 = 7.5%  
Motor M : 40 MVA, 20 kV, X" = X1 = X2 = 20%, X0 = 10%, Xn = 5%  
Transformer T1 : 50 MVA, 20 kV Δ /110 kVY, X = 10%  
Transformer T2 : 50 MVA, 20 kV Δ /110 kVY, X = 10%  
Transmission line: X1 = X2 = 24.2 Ω , X0 = 60.5 Ω  

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

  per unit  

Also from (8.4) we get

  per unit  

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

 

(b) Line-to-Line Fault : For a bolted LL fault, we can write from (8.16)

  per unit  

Then the fault currents are

 

Finally the sequence components of bus-2 voltages are

 

Hence faulted bus voltages are

 

(c) Double-Line-to-Ground Fault : Let us assumes that a bolted 2LG fault occurs at bus-2. Then

 

Hence from (8.24) we get the positive sequence current as

  per unit  

The zero and negative sequence currents are then computed from (8.25) and (8.26) as

  per unit

  per unit

 

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

 

 

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

 per unit  

Also

  per unit

 
  per unit  

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

 

  per unit

 
 per unit  

Similarly from Fig. 8.12, the negative sequence currents are given as

 per unit  
 per unit  

Finally notice from Fig. 8.13 that the zero sequence current flowing from the generator to the fault is 0. Then we have

  per unit

 

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 examples

Example 8.6

Example 8.7