Two-conductor open circuit fault

In a two-conductor open fault, two phases (say b and c) are broken between points $p$ and $q$ while phase a stays healthy. Writing the fault conditions in symmetrical components shows that the three sequence networks, seen between $p$ and $q$, must be joined in series.

Fault conditions

  p                        q
 a ---------------------------  (healthy, Ia flows)
 b -----x            x--------  (open, Ib = 0)
 c -----x            x--------  (open, Ic = 0)

Let $V_{a}, V_{b}, V_{c}$ be the series voltage drops across the break (from $p$ to $q$) in each phase.

• Phases b and c are open: $I_b = 0,\ I_c = 0$
• Phase a is healthy (no break): $V_a = 0$ (drop across the break in phase a is zero)

Sequence currents

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

so

$$I_{a0} = I_{a1} = I_{a2} = \frac{I_a}{3}$$

The same current flows through all three sequence networks.

Sequence voltages

Since $V_a = V_{a0} + V_{a1} + V_{a2} = 0$:

$$V_{a0} + V_{a1} + V_{a2} = 0$$

Sequence network equations

Seen from the break terminals $p$-$q$, each sequence network has a Thevenin form. Only the positive sequence network contains a source (the pre-fault open-circuit voltage $V_{pq}^{0}$ or the equivalent driving voltage $E$):

$$\begin{aligned} V_{a1} &= E - Z_1 I_{a1} \\ V_{a2} &= -Z_2 I_{a2} \\ V_{a0} &= -Z_0 I_{a0} \end{aligned}$$

Adding and using $V_{a0}+V_{a1}+V_{a2}=0$ and $I_{a0}=I_{a1}=I_{a2}$:

$$I_{a1} = \frac{E}{Z_1 + Z_2 + Z_0}$$

Interpretation

Equal currents in all three networks and voltages that add to zero are exactly the conditions of three elements in series in a closed loop:

  p1 +--[ +ve seq N/W, Z1, E ]--+ q1
     |                          |
  p2 +--[ -ve seq N/W, Z2 ]-----+ q2   (series loop:
     |                          |      q1->p2, q2->p0,
  p0 +--[ zero seq N/W, Z0 ]----+ q0   q0->p1)

Hence the positive, negative and zero sequence networks are connected in series between the fault points to simulate a two-conductor open fault, and the healthy-phase current is $I_a = 3I_{a1} = \dfrac{3E}{Z_1+Z_2+Z_0}$. This is the series-fault counterpart of the single line-to-ground shunt fault.