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 Y_bus matrix for both positive and negative sequences

Inverting the above matrix we get the following Z_bus matrix

Again from Fig. 8.13 we get the following Y_bus 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

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

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 I_g , while that flowing from the motor by I_m . 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

Therefore the fault currents flowing from the generator side are

and those flowing from the motor are

It can be easily verified that adding I_g and I_m we get I_f given above.