Electric Supply Systems

EEE 333

Power System Analysis

Unsymmetrical Fault Analysis

Lecture Contents

 Unsymmetrical Faults

 Different Types of Faults (S-L-G, L-L and L-G) Analysis

Learning Outcomes

The students will be able to…

 Understand different types of unsymmetrical faults

 Analyze single line-to-ground fault, line-to-line fault and

double line-to-ground fault

Single Line-to-Ground Fault Figure shows a 3-phase generator with neutral grounded through a impedance Zn.

Suppose a line-to-ground fault occurs on phase

a

through impedance Zf.

Assuming the generator is initially on no-load, the boundary conditions at the fault point are:

Substituting for I_b=I_c=0, the symmetrical components of currents are:

Single Line-to-Ground Fault

Then, we find that

Phase a voltage in terms of symmetrical components is:

Then, we have

�

 

_� ⁰ =0 − � ⁰ � _� ⁰

Substituting and , we get

 

The fault current is,

Single Line-to-Ground Fault

Substituting for the symmetrical components of currents in the given equations, the symmetrical components of voltage and phase voltages at the point

of fault are obtained.

�

 

_� ⁰ =0 − � ⁰ � _� ⁰

Sequence network connection (in series) for line-to-ground fault as below.

For line-to-ground faults, the Thevenin impedance to the point of fault is obtained for each sequence networks are placed in series. In general, the positive- and negative-sequence impedances are equal. If the generator neutral is solidly grounded, and for bolted fault .

 

Line-to-Line Fault

Figure shows a 3-phase generator with a fault through an impedance Z

_f

between phases b and c.

Assuming the generator is initially on no-load, the boundary conditions at the fault point are:

Substituting for and , symmetrical components of currents are:

 

Line-to-Line Fault

Again we get,

Substituting for and , considering , we get

 

Then, we find that, From here, we get

� _� ⁰ =0 − � ⁰ � _� ⁰

 

Substituting for , we get

 

Since , solving for results in

 

Line-to-Line Fault

The fault current is

Substituting the symmetrical components of currents, the symmetrical components of voltage and phase voltages at the point of fault are obtained

The phase currents are:

�

_�

=− �

_�

= ( _�

²

_{− �} ) _�

_�¹

 

� _� ⁰ =0 − � ⁰ � _� ⁰

 

Line-to-Line Fault

By connecting the positive- and negative-sequence networks in opposite as shown in the equivalent circuit. In practical, the positive- and negative- sequence impedances are found to be equal. For a bolted fault,

 

Double Line-to-Ground Fault Figure shows a 3-phase generator with a fault on phases b and c through an impedance Z

_f

to ground.

Assuming the generator is initially on no-load, the boundary conditions at the fault point are:

The phase voltages and are:

 

Double Line-to-Ground Fault

Substituting the symmetrical components of currents, then we have

Substituting and then we have

 

Substituting symmetrical components of voltage and solving for then we get

 

� _� ⁰ =0 − � ⁰ � _� ⁰

 

Similarly, for we get

 

………… (i)

1+�+�²=0

 

�

  _�^�

=�

_�^�

Double Line-to-Ground Fault And finally, from we have

 

Observing equations (i), (ii) & (iii) it can be concluded that the positive- sequence impedance in series with the parallel combination of the negative- and zero-sequence networks as shown in the equivalent circuit in the Figure below. The phase currents can be calculated and finally the fault current is obtained from

………… (iii)

Example 10.5

The one-line diagram of a simple power system is shown below. The neutral of each generator is grounded through a current limiting reactor of 0.25/3 per unig on a 100- MVA base. The generators are running on no-load at their rated voltage and rated frequency with their emfs in phase. Determine fault current for the following faults:

(a) A balanced 3-phase fault at bus 3 through a fault impedance per unit (b) A single line-to-ground fault at bus 3 through a fault impedance per unit (c) A line-to-line fault at bus 3 through a fault impedance per unit

(d) A double line-to-ground fault at bus 3 through a fault impedance per unit

 

Example 10.5

The positive-sequence impedance network is shown in Figure below.

Combining parallel branches, the positive-sequence Thevenin impedance is

Example 10.5

Since the negative-sequence impedance of each element is the same as the positive-sequence impedance, we have

�₃₃² =�₃₃¹ = �0.22

 

Example 10.5

The equivalent circuit for zero-sequence network is constructed according to transformer winding connection and given in Figure below.

Combining the parallel branches, the zero-sequence Thevenin impedance is

Delta to Wye conversion

Example 10.5

(a) Balanced 3-phase fault at bus 3:

The fault current is

Assuming no-load generated emfs are equal to 1.0 per unit, the fault current is:

(b) Single line-to-ground fault at bus 3:

The sequence components of the fault currents are:

Example 10.5

(c) Line-to-line fault at bus 3:

The zero-sequence component of current is zero,

 

(d) Double line-to-ground fault at bus 3:

The positive-sequence components of the fault currents is:

The positive-and negative-sequence components of the fault current are

The fault current is

Example 10.5

The zero-sequence component of current is:

The fault current is

And the phase currents are: