·        Phase-to-earth fault

The conditions for a solid fault from line a to earth
are represented by the equations Ib=0, Ic =0 and V_a =0,

As in the previous equations, it can easily be deduced
 that I_a1= I_a2= I_ao = E_a/ (Z₁ +Z₂+ Z_o). Therefore,
the sequence networks will be connected in series,
as indicated in Figure 2.10a. The current and voltage
conditions are the same when considering an open-circuit
fault in phases b and c, and thus the treatment
and connection of the sequence networks will be similar.

·        Phase-to-Phase fault

The conditions for a solid fault between
lines h and c are represented by the equations

 
I_a = 0, I_b = –I_c and V_b = V_c.
Equally, it can be shown that
I_ao= 0 and I_a1 = E_a/(Z₁+Z₂) = I_a2.  
For this case, with no zero-sequence current,
the zero-sequence network is not involved and the overall sequence network is composed of the positive- and negative-sequence networks in parallel as indicated in Figure 2.10b.

·        Phase-to-Phase-to-earth fault
 

The conditions for a fault between lines b and c and earth are represented by the equations 1_a = 0 and V_b=Vc =0. From these equations it can be proved that:

The three sequence networks are connected in parallel
as shown in Figure 2.10c.

2.3 Equivalent impedances for a power system.

When it is necessary to study the effect of any change on
the power system, the system must first of all be represented
by its corresponding sequence impedances.
The equivalent positive- and negative-sequence impedances
can be calculated directly from:

Z= V²/P

Where:

  Z  = Equivalent positive and negative-sequence impedances
 V  =nominal phase-to-phase voltage

  P = three-phase short circuit power

The equivalent zero-sequence of a system can be derived from the expressions of sequence components referred to for a single-phase fault, i.e.

Ia1=Ia2=Ia3 = VLN/ (Z1 + Z₂ + Z₀)

Where:
          VLN = the line-to-neutral voltage.

For lines and cables the positive and negative ímpedances are equal.

Thus, on the basis that the generator ímpedances are not significant in most distribution-network fault studies, it may be assumed that overall

Ζ₂ = Z₁ which simplifies the calculations.

Thus, the above formula reduces to Ia = 3I₀ = 3 VLN / (2Z₁ + Zo),

Where V_LN = line-to-neutral voltage and Zo= (3VLN / Ia) - 2Z1      

3 Supplying the current and voltage signals to protection systems

In the presence of a fault the current transformers (CTs) circulate current proportional to the fault current to the protection equipment without distinguishing between the vectorial magnitudes of the Sequence components.
 

Figure 10 Connection of sequence networks for a3ymmetrical faults

a Phase-to-earth fault

b Phase-to-phase fault

c Double phase-to-earth fault

Therefore, in the majority of cases, the relays operate on the basis of the corresponding values of fault current and / or voltages, regardless of the values of the sequence components. It is very important to emphasise that, given this, the advantage of using symmetrical components is that they facilitate the calculation of fault levels even though the relays in the majority of cases do not distinguish between the various values of the symmetrical components.