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DLG fault sequence network connection
Short answer
Boundary conditions yield V_a1 = V_a2 = V_a0 and I_a0+I_a1+I_a2=0. Substituting network equations gives I_a1 = E_a / [Z_1 + (Z_2 Z_0)/(Z_2+Z_0)]. This corresponds to the positive sequence network in series with the parallel combination of negative and zero networks.
For a double line to ground fault (bolted, zero fault impedance) in an unloaded generator (at alternator terminal), obtain the sequence network and show that the positive sequence network is connected in series with the parallel connection of negative sequence and zero sequence networks.
In a double line to ground (L-L-G) fault on phases b and c, all three sequence voltages at the fault are equal and the sequence currents add to zero, so the positive sequence network is in series with the parallel combination of the negative and zero sequence networks.
Generator sequence equations (unloaded)
where .
Boundary conditions (bolted fault, b and c to ground)
a o------------- Ia = 0
b o-----+ Ib -->
c o-----+ Ic -->
| If = Ib + Ic
ground
Sequence voltages
Equal voltages across the three networks at the fault point means they are connected in parallel at that point.
Sequence currents
So the positive sequence current divides into and , i.e. it returns through the negative and zero sequence networks in parallel.
Solving
From :
Substituting in with :
This is exactly the current in a circuit with and in series with (), which proves the connection.
Fault (ground) current: .
Interconnection
Ia1 --> F
+--(Ea)--Z1----------o------+---------+
| | |
| Z2 Z0
| | Ia2 | Ia0
N1 -------------------------+---------+
N2 N0
Positive sequence network in series with the parallel combination of negative and zero sequence networks. With fault impedance to ground, is added in series with .
✓ Answer checked by Claude Sonnet 5.5
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