Example 8.4
Consider the network shown in Fig. 8.10. The system parameters are given below
Generator G : 50 MVA, 20 kV, X" = X_{1} = X_{2} = 20%, X_{0} = 7.5% 

Motor M : 40 MVA, 20 kV, X" = X_{1} = X_{2} = 20%, X_{0} = 10%, X_{n} = 5% 

Transformer T_{1} : 50 MVA, 20 kV Δ /110 kVY, X = 10% 

Transformer T_{2} : 50 MVA, 20 kV Δ /110 kVY, X = 10% 

Transmission line: X_{1} = X_{2} = 24.2 Ω , X_{0} = 60.5 Ω 

We shall find the fault current for when a (a) 1LG, (b) LL and (c) 2LG fault occurs at bus2.
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. 8.118.13.
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 bus2, we have the following Thevenin impedances
Alternatively we find from Figs. 8.11 and 8.12 that
(a) SingleLinetoGround Fault : Let a bolted 1LG fault occurs at bus2 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) LinetoLine Fault : For a bolted LL fault, we can write from (8.16)
per unit 

Then the fault currents are
Finally the sequence components of bus2 voltages are
Hence faulted bus voltages are
(c) DoubleLinetoGround Fault : Let us assumes that a bolted 2LG fault occurs at bus2. 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 bus2 voltages are
Therefore voltages at the faulted bus are
