SLG fault: sequence networks in series

In a single line to ground (L-G) fault, the three sequence currents of the faulted phase are equal and the sequence voltages add to $3Z_fI_{a1}$, so the positive, negative and zero sequence networks are connected in series, and $I_f = 3E_a/(Z_1+Z_2+Z_0+3Z_f)$.

Starting point: an unloaded star-connected generator with neutral grounded through $Z_n$, generating balanced EMFs $E_a$, $a^2E_a$, $aE_a$; fault on phase a through fault impedance $Z_f$.

Sequence networks of an unloaded generator

The generator EMFs are balanced (positive sequence only), so with $Z_0 = Z_{g0} + 3Z_n$:

$$V_{a0} = -Z_0I_{a0}, \qquad V_{a1} = E_a - Z_1I_{a1}, \qquad V_{a2} = -Z_2I_{a2}$$

 Positive        Negative        Zero
  F1 o            F2 o            F0 o
     |Ia1 ^          |Ia2 ^          |Ia0 ^
    Z1              Z2              Z0
     |               |               |
   (Ea)              |               |
     |               |               |
  N1 o            N2 o            N0 o

Boundary conditions (fault on phase a through $Z_f$)

  a o----------+
  b o---  (Ib=0)
  c o---  (Ic=0)
               |  Ia = If
              Zf
               |
            ground

$$I_b = 0, \qquad I_c = 0, \qquad V_a = Z_fI_a$$

Sequence currents

$$\begin{bmatrix} I_{a0}\\I_{a1}\\I_{a2} \end{bmatrix} = \frac13\begin{bmatrix} 1&1&1\\1&a&a^2\\1&a^2&a \end{bmatrix}\begin{bmatrix} I_a\\0\\0 \end{bmatrix} \quad\Rightarrow\quad I_{a0} = I_{a1} = I_{a2} = \frac{I_a}{3}$$

So the same current flows in all three sequence networks, which means they are in series.

Voltage condition

$$V_a = V_{a0} + V_{a1} + V_{a2} = Z_fI_a = 3Z_fI_{a1}$$

Substituting the network equations:

$$-Z_0I_{a1} + E_a - Z_1I_{a1} - Z_2I_{a1} = 3Z_fI_{a1}$$ $$I_{a1} = I_{a2} = I_{a0} = \frac{E_a}{Z_1 + Z_2 + Z_0 + 3Z_f}$$

Fault current

$$I_f = I_a = 3I_{a1} = \frac{3E_a}{Z_1 + Z_2 + Z_0 + 3Z_f}$$

For a bolted fault ($Z_f = 0$): $I_f = \dfrac{3E_a}{Z_1+Z_2+Z_0}$.

Interconnection of sequence networks

   F1 --(Ea)--[Z1]-- N1 == F2 --[Z2]-- N2 == F0 --[Z0]-- N0
   |                                                       |
   +------------------------[ 3Zf ]------------------------+
              common current  Ia1 = Ia2 = Ia0

The three networks are connected in series with $3Z_f$: $N_1$ is joined to $F_2$ (the "==" links), $N_2$ is joined to $F_0$, and $N_0$ returns to $F_1$ through $3Z_f$. The common current $I_{a1}$ flows through all of them. Both conditions ($I_{a0} = I_{a1} = I_{a2}$ and $V_{a0}+V_{a1}+V_{a2} = 3Z_fI_{a1}$) are satisfied only by this series connection.

Remarks

• The neutral current is $I_n = 3I_{a0} = I_f$; it flows only if the neutral is grounded. For an ungrounded neutral, $Z_0 = \infty$ and $I_f = 0$.
• If $Z_0$ is small (solidly grounded), the L-G fault current can exceed the three-phase fault current $E_a/Z_1$.