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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)

Va1=Ea−Z1Ia1,Va2=−Z2Ia2,Va0=−Z0Ia0V_{a1} = E_a - Z_1I_{a1}, \qquad V_{a2} = -Z_2I_{a2}, \qquad V_{a0} = -Z_0I_{a0}

where Z0=Zg0+3ZnZ_0 = Z_{g0} + 3Z_n.

Boundary conditions (bolted fault, b and c to ground)

  a o-------------  Ia = 0
  b o-----+         Ib -->
  c o-----+         Ic -->
          |  If = Ib + Ic
        ground
Ia=0,Vb=0,Vc=0I_a = 0, \qquad V_b = 0, \qquad V_c = 0

Sequence voltages

[Va0Va1Va2]=13[1111aa21a2a][Va00]⇒Va0=Va1=Va2=Va3\begin{bmatrix} V_{a0}\\V_{a1}\\V_{a2} \end{bmatrix} = \frac13\begin{bmatrix} 1&1&1\\1&a&a^2\\1&a^2&a \end{bmatrix}\begin{bmatrix} V_a\\0\\0 \end{bmatrix} \quad\Rightarrow\quad V_{a0} = V_{a1} = V_{a2} = \frac{V_a}{3}

Equal voltages across the three networks at the fault point means they are connected in parallel at that point.

Sequence currents

Ia=Ia0+Ia1+Ia2=0I_a = I_{a0} + I_{a1} + I_{a2} = 0

So the positive sequence current Ia1I_{a1} divides into −Ia2-I_{a2} and −Ia0-I_{a0}, i.e. it returns through the negative and zero sequence networks in parallel.

Solving

From Va1=Va2=Va0V_{a1} = V_{a2} = V_{a0}:

Ia2=−Va1Z2,Ia0=−Va1Z0I_{a2} = -\frac{V_{a1}}{Z_2}, \qquad I_{a0} = -\frac{V_{a1}}{Z_0}

Substituting in Ia0+Ia1+Ia2=0I_{a0} + I_{a1} + I_{a2} = 0 with Va1=Ea−Z1Ia1V_{a1} = E_a - Z_1I_{a1}:

Ia1=(Ea−Z1Ia1)(1Z2+1Z0)  ⇒  Ia1=EaZ1+Z2Z0Z2+Z0I_{a1} = (E_a - Z_1I_{a1})\left(\frac{1}{Z_2} + \frac{1}{Z_0}\right) \;\Rightarrow\; I_{a1} = \frac{E_a}{Z_1 + \dfrac{Z_2Z_0}{Z_2 + Z_0}}

This is exactly the current in a circuit with EaE_a and Z1Z_1 in series with (Z2∥Z0Z_2 \parallel Z_0), which proves the connection.

Ia2=−Ia1Z0Z2+Z0,Ia0=−Ia1Z2Z2+Z0I_{a2} = -I_{a1}\frac{Z_0}{Z_2 + Z_0}, \qquad I_{a0} = -I_{a1}\frac{Z_2}{Z_2 + Z_0}

Fault (ground) current: If=Ib+Ic=3Ia0I_f = I_b + I_c = 3I_{a0}.

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 ZfZ_f to ground, 3Zf3Z_f is added in series with Z0Z_0.

✓ Answer checked by Claude Sonnet 5.5

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