Faults in Synchronous Generators | Power System

The synchronous generators may be subjected to different types of faults at its terminals and they are as follows- 1. Symmetrical Three-Phase (L-L-L or L-L-L-G) Fault 2. Single Line-to-Ground (L-G) Fault 3. Line-to-Line (L-L) Fault 4. Double Line-to-Ground (L-L-G) Fault.

These are listed in increasing order of severity, and the symmetrical fault follows these as the most severe. If the fault is symmetrical, it is unimportant whether there is earth connection at the fault or not.

1. Three Phase Fault:

The simplest case of a three-phase symmetrical fault is considered here. An unloaded star-connected 3-phase alternator is shorted by a 3-phase fault. The generator neutral is earthed through an impedance. The connections are shown in Fig. 5.1. The generator terminals being shorted V_a, V_b and V_c are equal to zero and the sum of the currents I_a, l_b and I_c. is zero.

As the generator develops only positive-sequence voltages, hence,

In matrix form –

Solving above equations we have, V_a0 = 0 as Z₀ is finite; I_a2 = 0 and

This means only positive-sequence network is present for the solution of three-phase fault current. The sequence network is shown in Fig. 3.10 (b).

2. Single Line-to-Ground (L-G) Fault:

An unloaded generator with its phase a grounded is shown in Fig. 5.2. The generator neutral is assumed to be grounded through an impedance Z_n. Let the fault impedance be Z_f.

Under fault condition, the currents and voltages are given as –

The symmetrical components of the fault current are given by substituting above values of currents we have-

From above equation we have –

From Eqs. (5.3) and (5.5) we have –

Substituting above values we have –

From Eqs. (5.6) and (5.70) we have –

From above Eq. (5.8) we have –

It may be recalled that Z₀ = Z_g0 + 3 Z_n

The Eqs. (5.5) and (5.8) show that the positive-, negative- and zero- sequence networks are to be connected in series for the solution of currents under fault conditions.

Sequence network representing the L-G fault on phase are given in Fig. 5.3:

In case of direct short circuit (i.e., when Z_f= 0) we get fault current from Eq. (5.10) for Z_f = 0.

Thus in this case fault current is given as –

In Fig. 5.3, the fault impedance Z_fequals zero.

In case of generator with isolated neutral, the zero-sequence impedance Z₀ becomes infinite and, therefore, from Eq. (5.11) –

The same result can be envisaged by looking at the system when the neutral is isolated- there is no return path for the current and, therefore, I_a1 = I_a2 = I_a0 = 0. This means that for such a system the fault current is zero.

If the generator was supplying balanced load during pre-fault condition, the load current was of positive-sequence only. A generator does not generate negative- or zero-sequence voltage and, therefore during pre-fault condition, there was no negative- or zero- sequence component of current to the load.

Under the fault condition, the positive-sequence current to the fault is to be added to the pre-fault current. However, only the positive-sequence current flows into the fault provided the pre-fault current into the fault is zero.

3. Line-to-Line (L-L) Fault:

An unloaded generator with line-to-line fault on its phases b and c is shown in Fig. 5.4. The generator neutral is assumed to be grounded through an impedance Z_n. The phase a is open. Let the fault impedance be Z_f.

Under fault condition, the currents and voltages are given as –

The symmetrical components of currents and voltages are given by substituting I_a = 0; I_c = -I_b and V_c = V_b – I_b Z_f in the general expressions for symmetrical components of currents and voltages at the point of fault.

from which we have –

Solving about two equations we have –

Substituting I_b from above Eq. (5.16) and (5.17) we have –

Equations (5.13) and (5.17) suggest parallel connection of positive- and negative-sequence networks through a series impedance Z_f as shown in Fig. 5.5. Since I_a0 = 0, the zero-sequence network is unconnected.

In terms of the Thevenin equivalents, from Fig. 5.5 we have

From Eq. (5.16) we have

Knowing I_a1, we can calculate V_a1and V_a2 from which voltages at the fault point can be determined.

In case of direct short circuit (i.e., when Z_f = 0) we get I_a1 and fault current I_b from Eqs. (5.18) and (5.19) respectively. Thus we have –

4. Double Line-to-Ground (L-L-G) Fault:

The circuit diagram for double line-to-ground (L-L-G) fault through an impedance Z_f on an unloaded synchronous generator is depicted in Fig. 5.6. The generator neutral is assumed to be grounded through an impedance Z_n.