L-G fault through Zf: currents and voltages

Difference between positive, negative and zero sequence components

L-G fault through $Z_f$ on phase a (solidly grounded, no load)

Generator equations: $V_{a1} = E_a - Z_1I_{a1}$, $V_{a2} = -Z_2I_{a2}$, $V_{a0} = -Z_0I_{a0}$ (solid grounding: $Z_0 = Z_{g0}$).

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 ($Z_f = 0$): $I_f = \dfrac{3E_a}{Z_1+Z_2+Z_0}$. The currents in the healthy phases are $I_b = I_c = 0$.

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.

Line-to-ground voltages at the fault

Let $I_1 = I_{a1} = I_{a2} = I_{a0} = \dfrac{E_a}{Z_1+Z_2+Z_0+3Z_f}$.

Phase a:

$$V_a = Z_fI_a = 3Z_fI_1 = \frac{3Z_fE_a}{Z_1 + Z_2 + Z_0 + 3Z_f}$$

Phase b:

$$\begin{aligned} V_b &= V_{a0} + a^2V_{a1} + aV_{a2} \\ &= -Z_0I_1 + a^2(E_a - Z_1I_1) - aZ_2I_1 \end{aligned}$$

Replacing $a^2E_a = a^2I_1(Z_1 + Z_2 + Z_0 + 3Z_f)$:

$$V_b = I_1\left[(a^2 - a)Z_2 + (a^2 - 1)Z_0 + 3a^2Z_f\right]$$

Phase c: similarly, with $aE_a = aI_1(Z_1+Z_2+Z_0+3Z_f)$:

$$V_c = I_1\left[(a - a^2)Z_2 + (a - 1)Z_0 + 3aZ_f\right]$$

For a bolted fault ($Z_f = 0$): $V_a = 0$ and

$$V_b = \frac{E_a\left[(a^2 - a)Z_2 + (a^2 - 1)Z_0\right]}{Z_1 + Z_2 + Z_0}, \quad V_c = \frac{E_a\left[(a - a^2)Z_2 + (a - 1)Z_0\right]}{Z_1 + Z_2 + Z_0}$$