Line-to-line fault current calculation

In a line-to-line (L-L) fault the zero sequence current is zero, $I_{a1} = -I_{a2}$, and the positive and negative sequence networks are connected in parallel (opposition) through $Z_f$ at the fault point. Fault current is $I_f = -j\sqrt3\,E_a/(Z_1+Z_2+Z_f)$.

Model: the network is represented at the fault bus k by its Thevenin sequence equivalents: prefault voltage $V_f = V_k(0)$ and impedances $Z_1 = Z_{kk}^{(1)}$, $Z_2 = Z_{kk}^{(2)}$, $Z_0 = Z_{kk}^{(0)}$ (from the sequence Z-bus matrices). For a generator terminal fault, $V_f = E_a$ and $Z$ are the machine's own values.

Boundary conditions (fault between phases b and c through $Z_f$)

  a o-----------  Ia = 0
  b o-----+        Ib = If -->
          Zf
  c o-----+        Ic = -Ib

$$I_a = 0, \qquad I_b + I_c = 0, \qquad V_b - V_c = Z_fI_b$$

Sequence currents

$$\begin{aligned} I_{a0} &= \tfrac13(I_a + I_b + I_c) = \tfrac13(0 + I_b - I_b) = 0 \\ I_{a1} &= \tfrac13(I_a + aI_b + a^2I_c) = \tfrac13(a - a^2)I_b \\ I_{a2} &= \tfrac13(I_a + a^2I_b + aI_c) = \tfrac13(a^2 - a)I_b \end{aligned}$$

Hence

$$I_{a0} = 0, \qquad I_{a2} = -I_{a1}$$

Voltage condition

$$\begin{aligned} V_b - V_c &= (a^2 - a)V_{a1} + (a - a^2)V_{a2} = (a^2 - a)(V_{a1} - V_{a2}) \\ Z_fI_b &= Z_f\,(a^2I_{a1} + aI_{a2}) = Z_f(a^2 - a)I_{a1} \end{aligned}$$

so

$$V_{a1} - V_{a2} = Z_fI_{a1}$$

With $V_{a1} = V_f - Z_1I_{a1}$ and $V_{a2} = -Z_2I_{a2} = Z_2I_{a1}$:

$$I_{a1} = -I_{a2} = \frac{V_f}{Z_1 + Z_2 + Z_f}$$

Fault current

$$\begin{aligned} I_b &= I_{a0} + a^2I_{a1} + aI_{a2} = (a^2 - a)I_{a1} = -j\sqrt3\,I_{a1} \\ I_f &= I_b = -I_c = \frac{-j\sqrt3\,V_f}{Z_1 + Z_2 + Z_f} \end{aligned}$$

Voltages at the fault bus

$$\begin{aligned} V_{a1} &= V_f - Z_1I_{a1}, \quad V_{a2} = Z_2I_{a1}, \quad V_{a0} = 0 \\ V_a &= V_{a1} + V_{a2}, \quad V_b = a^2V_{a1} + aV_{a2}, \quad V_c = aV_{a1} + a^2V_{a2} \end{aligned}$$

For a bolted fault ($Z_f = 0$), $V_{a1} = V_{a2}$, so $V_a = 2V_{a1}$ and $V_b = V_c = -V_{a1}$.

Voltages at other buses (i) of a network

Using the sequence bus impedance matrices, with the fault at bus k:

$$V_i^{(1)} = V_i(0) - Z_{ik}^{(1)}I_{a1}, \qquad V_i^{(2)} = -Z_{ik}^{(2)}I_{a2}, \qquad V_i^{(0)} = 0$$

Phase voltages follow from $V_{abc} = AV_{012}$ (with the $\pm30^\circ$ phase shift of Δ-Y transformers applied to the positive and negative sequence values when crossing them). Line currents then follow from $I_{ij} = (V_i - V_j)/z_{ij}$ in each sequence network.

Interconnection of sequence networks

        Ia1 -->           <-- Ia2
  +--(Ea)--Z1---o F1--Zf--o F2---Z2--+
  |                                  |
  N1 ------------------------------- N2

  Zero sequence network: not connected (Ia0 = 0)

The positive and negative sequence networks are connected in parallel (in opposition) through $Z_f$; the zero sequence network is left open.

Steps in summary

1. Form $Z_1$, $Z_2$ at the fault point (Thevenin), and take $V_f$ = prefault voltage.
2. $I_{a1} = -I_{a2} = V_f/(Z_1+Z_2+Z_f)$, $I_{a0} = 0$.
3. $I_f = -j\sqrt3 I_{a1}$.
4. Find sequence voltages, then phase voltages at the fault and other buses.