CHAPTER 7

Short-Circuit Analysis—2 (Unbalanced Fault Analysis)

7.1 Introduction

If the fault involves only one or two of the three possible phases, then the fault is said to be an unbalanced or asymmetrical fault. Due to these faults, the system loses symmetry or balanced condition. Fault calculations can be easily performed by the use of a tool known as the “symmetrical component”. Using these components unbalanced vectors can be transformed into balanced vectors, through which calculations can be performed on a single-phase basis. This makes calculations easier as seen in the case of symmetrical fault analysis.

7.2 Symmetrical Components

In the year 1918, French mathematician Fortescue proposed a mathematical tool known as “symmetrical components”. According to this, a set of n-unbalanced vectors can be resolved into n sets of balanced vectors, with each set comprising of n-balanced vectors. For example, a set of 3-unbalanced vectors I_R, I_Y and I_B can be transformed into three sets of balanced vectors, with each set containing 3-balanced vectors.

As our interest is in 3-phase, let us transform the three unbalanced vectors into three sets of balanced vectors. These three sets are:

1. Positive sequence set
2. Negative sequence set
3. Zero sequence set

In terms of balanced sequence components, the unbalanced vectors can be written as follows:

 

 

Equation (7.1) can be written in the matrix form as:

 

 

In Equation (7.2) the matrix on the left hand side represents the three-phase unbalanced currents. On the right hand side, the first matrix corresponds to the positive sequence set, the second matrix corresponds to negative sequence components and the last, third matrix represents zero sequence in the three-phase system RYB. Suffix 1, 2 and 0 are assigned respectively to positive, negative and zero sequence components. The description of sequence components is as follows:

1) Positive sequence components

I_R1, I_Y1 and I_B1 corresponding to the respective unbalanced vectors I_R, I_Y and I_B are the balanced vectors having the same magnitude but displaced 120° apart. These components have the sequence of RYB of the original unbalanced vectors and are shown in Figure 7.1

2) Negative sequence components

I_R2, I_Y2 and I_B2 corresponding to the unbalanced vectors I_R, I_Y and I_B respectively are the balanced vectors having equal magnitudes, but displaced 120° apart. These components have the sequence exactly opposite to that of original unbalanced vectors. These components are shown in Figure 7.2.

In order to obtain the opposite sequence, the negative sequence components are rotated in the clockwise direction, in contrast to the conventional anticlockwise direction. This is shown in Figure 7.2(b): RBY, BYR and YRB are the negative sequences. The same negative sequence can be obtained when the vectors are rotated in the anticlockwise direction as shown in Figure 7.2(c), with the positions depicted differently.

 

 

Fig 7.1 (a) Original Unbalanced Vectors, (b) Positive Sequence Components

 

 

Fig 7.2 (a) Original Vectors, (b) Negative Sequence Components (Cw Direction), (cz) Negative Sequence Components (Acw Direction)

 

3) Zero sequence components

These components have no (zero) sequence. These components are equal in magnitude and are in phase with each other as shown in Figure 7.3.

In addition to the currents discussed earlier, voltages and powers are can also be transformed into sequence components.

 

7.2.1 Operator a

Operator ‘a’ is defined as a unit vector with magnitude equal to 1 and angle 120° and is given as:

 

 

When a vector is multiplied by the operator ‘a’, the vector shifts to 120° in the anticlockwise direction without the magnitude being changed. Similarly, when it is multiplied by a², it shifts to 240° in the anticlockwise direction.

Some important relations with operator a:

 

 

 

Fig 7.3 Zero Sequence Components

7.2.2 Sequence Components in Terms of Operator a

As sequence components have the same magnitude, the Y- and B-phase sequence components can be written in terms of R-phase.

Referring to Figure 7.1(b), and by taking I_R1 as reference:

 

 

Referring to Figure 7.2(c) and by taking I_R2 as reference:

 

 

As zero sequence components have no phase difference,

 

 

Rewriting Equation (7.1) and substituting the above relations,

 

 

Equation (7.4) may be represented in the matrix form as

 

 

Equation (7.5) may be rewritten with the operator A matrix.

 

 

where A is the operator A matrix. Similarly the voltages can be written as:

 

 

Note: For the given digital vectors, the sequence components can be obtained as:

 

 

where the inverse of the operator matrix [A] is

 

Example 7.1

Let the three-phase currents be Determine the sequence components. Assume the sequence is RYB.

Solution:

 

 

The three phase currents are balanced. Using Equation (7.8b)

 

 

Using relations on operator a,

 

 

From the above,

 

 

Note: The above numerical example illustrates that in the balanced system only positive sequence currents are present.

Example 7.2

Let the three-phase currents be The phase sequence is RYB. Find the sequence components.

Solution:

 

 

Sequence currents,

 

 

On simplification,

 

Example 7.3

Determine the neutral current in the star, neutral grounded system from the given symmetrical components

Solution:

 

7.3 Sequence Impedances

The impedance offered by the circuit component to the flow of positive sequence current is known as positive sequence impedance. Similarly, the negative and zero sequence impedances can be defined.

7.3.1 Sequence Impedances of Individual Components

A summary of the experimental results of sequence impedances is presented below.

a) Generator

Positive sequence impedance

Z₁ = jX″_d (if the sub-transient state of fault is of interest)

= jX′ _d (if the transient state fault is of interest)

= jX_d (if the steady-state value of fault is of interest)

Negative sequence impedance

The negative sequence impedance offered by the generator is generally written as:

 

 

Zero sequence impedance

The zero sequence impedance of the generator is usually much smaller than Z₁ and Z₂ of values. Let Z_g0 is be the zero sequence impedance of the generator. Then, the numerical values of the generator follow the relation:

 

 

b) Transformer and Transmission line

These two are said to be static devices. Since phase sequence has no effect on their operation, positive and negative sequence impedance and negative impedance are exactly be the same. For these devices, the relation between the sequence impedances is given as:

 

Example 7.4

Consider a fully transposed 3-phase transmission line, having self reactance of X_s and X_m.

Solution:

Consider the transposed transmission as shown in Figure 7.4

Let X_s = Self reactance of each line

X_m = Mutual reactance between any two lines

 

 

Fig 7.4 Power System Network of Example 7.4

 

Voltage drop across each line is due to self and mutual reactance and is given by:

 

 

In the matrix form, the above equations can be written as:

 

 

Remembering the equations

 

 

the voltage drop and current in terms of sequence components can be written as

 

 

The voltage drop of sequence components can be obtained by multiplying sequence currents with sequence impedances. Therefore, from the above equations the sequence impedance matrix is:

 

 

On simplification of the above equation, the sequence impedance can be obtained as

 

 

where is

 

 

Note: Sequence impedance and sequence network are mutually disjoint as evident from the observation that the off-diagonal terms are zeros in Equation (7.10).

7.3.1 Summary of Sequence Components

1. Three-phase unbalanced currents and voltages can be transformed into three sets of symmetrical components, namely, the positive, negative and zero sequence components.
2. Symmetrical component theory is a tool to tackle unbalance fault analysis and its use is justified more analytically than physically.
3. It may be understood that positive sequence such as currents and voltages are the original vectors while the other two sequence components are fictitious.
4. In a balanced system, only positive sequence components are present (refer Example 7.1).
5. The zero sequence currents in an unbalanced system is equal to one-third of total current in the neutral (I₀ = I_N /3).
6. The positive, negative and zero sequence networks are mutually disjoint (refer Ex: 7.4) and can be handled separately.

7.4 Sequence Networks

A single-phase per-unit equivalent reactance network is sufficient to couple three-phase fault analysis. This makes three-phase fault analysis simple and easier. The question of present interest is:

Can we complete unbalanced fault analysis on a single-phase basis?

The answer to this question is yes because of the use of symmetrical components in unbalanced fault analysis. The method of performing unbalanced fault analysis on a single-phase basis is presented below.

Consider the unbalanced current described by Equation (7.5)

 

 

The equation explains that unbalanced currents can be determined, once the three sequence currents are known. The three sequences I_R0, I_R1, I_R2 are the components of R-phase currents I_R. I_R1 can be obtained through the positive sequence network containing positive sequence component voltages and impedance. In a similar manner the negative and zero sequence currents can also be obtained through negative and zero sequence networks. These three networks are mutually disjoint. Hence, we need to convert the given single-line diagram of the power system into three mutually disjoint sequence networks for the purpose of determining the three sequence currents. Once the sequence currents are known, the fault analysis may be concluded. The following section describes the method of converting the single-line representation of a power system into three-sequence networks.

Consider the single-line diagram shown in Figure 7.5

Figure 7.1 shows a generator, transformer and transmission line. We need to convert this into three-sequence networks. We deal this problem by understanding the procedure of representing generator, transformer and transmission line individually in the three- sequence networks. Once the procedure for individual components is known, the sequence network can be easily obtained by integrating the individual elements.

7.4.1 Generator Representation in Three-Sequence Networks.

Let the generator sequence impedances be known values. Z_g1, Z_g2 and Z_g0 are the positive, negative and zero sequence impedances of the generator. In fault analysis, the generator is represented as a constant voltage source behind the reactance. As a first step the generator placed in positive, negative and zero sequence networks is represented in a similar manner as shown in Figure 7.6.

 

 

Fig 7.5 Single-Line Diagram of a Single-Power System Network

 

In Figure 7.6, E is the internal emf produced by the generator. The sequence currents, impedances and voltages are also indicated in the figure. In the second step, we verify two aspects:

 

 

Fig 7.6 Generator Representation in Three-Sequence Networks

1. Can the generator be shown as voltage source in all sequence networks?
2. The star-connected generator winding neutral may be grounded or ungrounded. Does it have any effect on generator representation?

Verification of the above aspects results in modification of the generator representation in the three-sequence networks.

For a given direction of rotation of the rotor, the generator sets up emfs in the three phases with a phase sequence of RYB. As the positive sequence of voltages is also RYB, there is no problem for the generator to produce positive sequence voltages. Hence, the generator can be represented as a voltage source in the positive sequence network. To produce negative sequence voltages, the generator rotor has to rotate in the opposite direction.

A single rotor cannot rotate simultaneously in both the directions. Thus, a generator cannot generate negative sequence voltages. Hence it should not be shown as a voltage source in the negative sequence network. Finally, when the rotor starts rotating, it sets up a sequence. The zero-sequence voltages have no sequence. Therefore, the generator should generate zero-sequence voltages under still condition of the rotor. As this is not possible, it can be concluded that the generator should not be shown as a voltage source in a zero-sequence network.

Discussion on the effect of neutral grounding in generator representation

The following discussion is also valid for negative sequence network.

Let the generator neutral current I_n be resolved into components as: I_n = I_n1 + I_n2 + I_n0 where I_n1, I_n2 and I_n0 are the sum of positive, negative and zero sequence current components of R, Y and B phases as:

 

 

 

Fig 7.7 Generator Neutral Condition in the Positive Sequence Network

 

In Figure 7.3, the algebraic sum of currents leaving junction N is I_R1 + I_Y1 + I_B1, which is always zero. Since positive sequence currents are balanced and have a 120° phase shift,

 

 

Let the switch S in Figure 7.3 be open. That is, the generator neutral is ungrounded. For this condition, the current entering into junction N (i.e. I_n1) is zero and Kirchhoff's current law (KCL) is satisfied. Hence, there is no problem for the flow of positive sequence currents when the neutral is in open condition. Now, when the neutral is solidly grounded a current of I_n1 can flow through it. But as I_n1 = I_R1 + I_Y1 + I_B1 = 0, the sum of incoming currents is zero and KCL is satisfied. However, if the neutral is reactance grounded with Z_n, the value of I_n1 cannot be limited to zero current. It can therefore be concluded that neutral grounding has no effect on the representation of a generator in positive and negative sequence networks, and that the condition of the neutral need not be verified.

Zero sequence Network

Let the generator neutral be ungrounded. Referring to Figure 7.8(a), the incoming current into junction N is zero, whereas the algebraic sum of outgoing currents is I_R0 + I_Y0 + I_B0 = 3 × I_R0. In this ground neutral condition, KCL is not satisfied and therefore such currents do not exist. This effect can be represented as an open circuit between neutral (ZPB) and junction N as shown in Figure 7.8(b).

 

 

Fig 7.8(a) Generator Neutral Condition in the Zero-Sequence Network

 

 

Fig 7.8(b) Neutral Ungrounded Condition of the Generator

 

Next, let the generator neutral be grounded through an impedance Z_n. In this condition, the circuit is closed, and incoming current to junction N can flow through it. The value of incoming current is I_no = 3I_R0, and as the sum of outgoing current is also equal to 3I_R0, KCL is satisfied. Therefore such currents do exist.

Due to flow of 3I_R0 current into neutral, a voltage drop of 3I_R0 Z_n takes place. This raises the potential of neutral from zero to 3I_R0 Z_n. Now the zero potential point (ZPB) shifts down from neutral to the ground. Referring to Figure 7.8(b), the total voltage drop from ZPB to Bus-1 is:

 

 

where E = 0 in this case.

Let us consider a single-phase current component I_R0 flowing into terminals. Equation (7.6) can be rewritten to consider this effect as:

 

 

Hence Z_n is multiplied by 3 to get voltage drop 3I_R0 Z_n. The zero-sequence network of the generator is shown in Figure 7.8(d).

 

 

Fig 7.8(c) Neutral Grounded Condition of the Generator

 

 

Fig 7.8(d) Generator Representation in the Zero-Sequence Network

In Figure 7.8(d), the circuit component 3Z_n can be replaced with:

• Open circuit (Z_n = ∞), when neutral is ungrounded.
• Short circuit (Z_n = 0), when neutral is solidly grounded.
• With numerical value 3Z_n, if neutral is grounded with an impedance Z_n.

Summary of generator representation

• The generator should be shown as a voltage source only in a positive sequence network.
• The condition of the neutral has no effect in the representation of a generator, both in positive and negative sequence networks.
• The condition of the neutral has a significant effect on generator representation in zero-sequence networks.

Generator representation in the three sequence networks is summarized in Figure 7.9

 

 

 

Fig 7.9. Generator Representation in the Three Sequence Networks

 

 

where, Z₀ in the Equation (7.10) is

 

7.4.2 Transformer Representation in the Three Sequence Networks

Power transformer in symmetrical fault analysis is represented as a series impedance element, provided per unit values are considered. Referring to Figure 7.1, the power transformer is connected between buses 1 and 2. Let the transformer be connected as a series impedance element in the three sequence networks, as in symmetrical fault analysis. We need to verify two aspects before finalizing this representation.

• The primary and secondary windings of a power transformer may be star or delta connected. We need to verify whether the type of connection has an effect on representation of the transformer or not.
• If the transformer winding is star connected the neutral point may be ungrounded, grounded or impedance grounded and we need to verify if this condition of the neutral has an effect on representation of the transformer.

Verification of the above, results in modifications to the representation of the transformer in the there sequence networks. First, we verify the effect of the type of windings on transformer representations. The following explanation is valid for positive and negative sequence networks. Consider Figure 7.10, in which a star (ungrounded neutral)-delta transformer is connected between buses 1 and 2.

Let the positive sequence current of the three phases come through the generation terminals before Bus-1. The transformer can be shown as a series reactance element, if this current appears beyond Bus-2. Positive sequence currents can flow into the star winding, provided KCL is satisfied at junction N. The sum of incoming currents of the balanced positive sequence currents is I_R1 + I_Y1 + I_B1 = 0. Similarly, the current leaving N is also zero (since N is not grounded). This verifies that KCL is valid and therefore it can be concluded such currents do exist. The flow of sequence currents into the primary winding causes the magnetic core of the power transformer to get magnetized. The magnetized core establishes secondary currents in the delta-connected winding. The line current in the delta winding is the vector difference of the two phase currents, and it is not equal to zero due to the 120° phase difference. The line current flows beyond Bus-2 and this proves that the transformer acts as a series impedance element to pass positive sequence currents from Bus-1 (generator side) to Bus-2 (transmission line side). The currents in other types of transformer windings can be verified in the same way.

 

 

Fig 7.10 Star-Delta Power Transformer Connected Between Buses 1 and 2

 

Summarizing the above, it can be stated that a transformer can be simply represented as series impedance on a reactance element without verifying the type of transformer winding or the condition of the neutral. Moreover, as Z₁ = Z₂ for a transformer, both positive and negative sequence networks appear quite similar.

The zero-sequence equivalent circuit of a transformer is not like the others. It is influenced by the type of winding and the condition of the neutral. We shall consider two cases and leave the third to the reader as an exercise.

Case-1: Let the transformer primary be star connected (neutral grounded) and the secondary be delta connected.

 

 

Fig 7.11(a) A Star-Delta Transformer for a Zero-Sequence Network

Consider Figure 7.11(a). The transformer is connected between buses 1 and 2. Let the zero-sequence current flow out from generator terminals. These currents can flow into primary winding, provided KCL is satisfied at junction N. The sum of incoming currents to junction N is I_R0 + I_Y0 + I_B0 = 3I_R0 and the out flowing current is I_N0 = 3I_R0. This verifies that KCL is valid and that such currents do exist. These primary currents induce secondary currents in the delta connected windings (phases). But the line current on the delta side is zero as the difference between any two phase currents, say for example, I_R0 – I_Y0 = 0. Therefore currents can appear up to the secondary side and return. This is shown in Figure 7.11(b).

 

 

Fig 7.11(b) A Star-Delta Transformer Representation in the Zero Sequence Network

 

Case-2: Now consider a star-star transformer. The neutral on both sides are grounded solidly. This is shown in Figure 7.12(a).

 

 

Fig 7.12(a) A Star-Star Transformer for a Zero Sequence Network

 

In this case, KCL is satisfied at the junctions on both sides and current can pass from Bus-1 to Bus-2 through the power transformer. In other words, the transformer can act like a series impedance element as shown in Figure 7.12(b).

A current of 3I_R0 at the neutral produces a drop of 3I_R0 Z_n. This effect can be shown in the network by adding 3Z_n to Z₀ of the transformer.

Switch diagram for the representation of transformer in a zero-sequence network

The switch diagram shown in Figure 7.13 can be used to represent a transformer in the zero-sequence network. The circuit consists of two series switches (1 and 2) and two shunt switches (1¹ and 2¹). 1–1¹ corresponds to the primary and 2–2¹ corresponds to the secondary side of the transformer. Z₀ is the zero sequence impedance of the transformer.

 

 

Fig 7.12(b) Representation of a Star-Star Transformer for a Zero Sequence Network

 

 

Fig 7.13 Switch Diagram for Representation of Transformer in the Zero-Sequence Network

Rules for using the switch diagram:

1. Series switches are closed when the winding is star connected and neutral grounded.
2. Shunt switches are closed when the winding is delta connected.

Example: Let the primary side of the transformer be delta connected and the secondary side, star connected with the neutral grounded. Then, switches 1¹ and 2 may be closed. Other types of connections are given in Example 7.5.

Example 7.5

Give the zero-sequence network equivalent for the transformer connected between buses 1 and 2. Use switch diagram.

 

 

7.4.3 Transmission Line Representation

Transmission lines can be represented as series impedance elements in all the three sequence networks. Transmission lines between buses 2 and 3 may be depicted as shown in Figure 7.14. Similar representations may be made for all the three sequence networks. In Figure 7.14, Z₁, Z₂ and Z₀ are the per unit sequence impedance of the respective networks.

 

 

Fig 7.14 Transmission Line Representation for the Three Sequence Networks

 

Knowledge of representation of individual components helps to convert any power system network into three-sequence networks for the calculation of fault currents in unbalanced fault analysis.

7.4.4 Summary of Sequence Networks

1. Only positive sequence network contains voltage sources.
2. For equal numerical values of positive and negative sequence impedances, the sequence networks appear quite similar except for the voltage sources present in the positive sequence network.
3. Power transformer connection types and the neutral of the generator and the transformer need not be verified while drawing positive and negative sequence networks.
4. For a generator, the condition of the neutral may be depicted as given:
   • Neutral ungrounded: show open circuit
   • Neutral solidly grounded: show short circuit
   • Neutral impedance grounded with neutral-to-ground impedance Z_n: show generator as impedance element with value Z_0. = Z_g0. + 0.3Z_n, where Z_g0 is the zero-sequence impedance of the generator.
5. The transformer is represented as a series impedance element both in positive and negative sequence networks. (Remember Z₁= Z₂ for a transformer)
6. Care should be taken while representing a transformer in the zero-sequence network. (Refer Figure 7.13)
7. Zero-sequence currents cannot flow if the return path is not available.
8. The transmission line is represented as a series impedance element in all sequence networks. (Remember Z₁= Z₂ for a transmission line)

Example 7.6

Figure 7.15 represents a simple power system. Draw its positive, negative and zero-sequence networks.

 

 

Fig 7.15 Simple Power System for Example 7. 6

Solution:

Positive sequence network

 

 

Negative sequence network

 

 

Zero-sequence network

 

7.5 Unbalanced or Unsymmetrical Fault Analysis

The following are the types of unbalanced faults:

1. Single line-to-ground fault (SLG fault)
2. Double line fault (LL fault)
3. Double line-to-ground fault (LLG fault)

It has been mentioned in Section 7.4 that fault analysis is also carried out on a single-phase basis, similar to symmetrical fault analysis, with the difference lying in the fact that the former involves working with three-sequence networks instead of with a single network. Though this type of analysis may call for more effort, it is advantageous to work on a single-phase basis. The method of drawing the three-sequence diagram for a given power system network consisting of generators, transformers and transmission lines has also been discussed in earlier sections.

The following steps are involved in unsymmetrical fault analysis. These steps are common to all the three types of balanced faults.

Step 1: Convert the given power system network into three equivalent p. u. sequence networks.

Step 2: Look for the fault location and reduce the three-sequence networks into three Thevenin's equivalents across the fault terminal F and ZPB. Get the Thevenin's positive, negative and zero sequence impedances. Also, obtain pre-fault open circuit voltage E from the Thevenin's equivalent circuit.

Step 3: Obtain the values of sequence currents.

Step 4: Obtain the unbalanced currents using the equation:

 

 

Step 5: Obtain the sequence voltages from

 

 

where, I_R1, I_R2, I_R0 are previously calculated and Z₁, Z₂, Z₀ are Thevenin's equivalent impedances obtained earlier.

Step 6: Obtain phase voltages from:

 

 

This concludes the fault analysis.

7.5.1 Single Line-to-Ground Fault (SLG Fault)

Assume initially that the system is isolated from the load end. The pre-fault balanced condition of the system is:

Now, consider an SLG fault occurring at point F as shown in Figure 7.16.

 

 

Fig 7.16 SLG Fault

 

I_f is the fault current. The fault has an impedance of Z_f. If the fault is a dead short-circuit, then Z_f = 0 or V_R.= 0 with a fault impedance of Z_f. During the fault, the condition of the system is:

 

 

To complete the fault analysis, we need the relation between sequence currents and their magnitudes.

Relation between sequence currents:

 

 

From the above,

 

 

The relation between sequence currents for a SLG fault is:

 

 

i.e., all the sequence currents are equal. Now, we shall find the magnitudes of the sequence currents:

Fault voltage V_f = V_R = I_f Z_f = I_RZ_f

In terms of sequence components,

Using Thevenin's equivalent circuits obtained across F and ZPB, the sequence voltages can be written as:

 

 

As all the sequence currents are equal,

 

 

From the above equation, the magnitude of sequence current I_R1 is

 

 

Since all the sequence currents are equal,

 

 

and the fault current is: I_f = I_R = 3I_R1

Therefore,

 

 

The relations represented by Equations (7.12), (7.13) and (7.14) can be easily obtained by connecting Thevenin's equivalent sequence networks in series as shown in Figure 7.17. Further, the sequence currents may also be computed by using this series circuit.

The next step is finding the sequence voltages as:

 

 

 

Fig 7.17 Sequence Network Connected in Series for a SLG Fault

 

and three-phase voltages as:

 

 

This concludes SLG fault analysis.

7.5.2 Double Line Fault (LL Fault)

Assume initially that the power system is isolated from the rest of the system, such that I_R = I_Y = I_B = 0 and that the three-phase voltages are balanced. Now, assume that a double line fault with a fault impedance of Z_f has occurred across Y and B phases as shown in Figure 7.18.

 

 

Fig 7.18 LL Fault

During the fault condition of the system:

I_R = 0 (due to healthy condition);

Fault current I_f = I_Y – I_B (for the direction shown)

Relation between sequence currents:

The sequence currents at the fault point are:

 

 

From the above equation,

 

 

and

Therefore the relation between sequence currents for an LL fault is:

 

 

Magnitude of sequence currents:

Replacing symmetrical components for V_Y, V_B or I_f (=I_Y) in the above equation,

 

 

Using Equation (7.16)

 

 

Substituting equations for sequence voltages,

 

 

As I_R2 – I_R1, the above equation modifies to:

 

 

From the above, the magnitude of I_R1 is:

 

 

where E, Z₁, Z₂ are Thevenin's equivalent sequence network parameters.

Therefore, the relation may be given by

 

 

The fault current I_f = I_Y can be obtained as:

 

 

but

Substituting a² – a value, the equation for I_f is:

 

 

and the magnitude of fault current is:

 

 

Knowing the values of sequence currents and sequence voltages, three-phase voltages may be obtained by a similar procedure as explained for SLG fault in Section 7.5.1. Sequence networks for the fault can be drawn in series opposition as shown in Figure 7.19

 

 

Fig 7.19 Sequence Networks for LL Fault

Figure 7.15 is drawn in accordance with Equation (7.18) and can be used to find and establish sequence currents.

7.5.3 Double Line-to-Ground (LLG) Fault

Let the pre-fault condition of the system be: I_R = I_Y = I_B = 0 and the three-phase voltages be balanced. Now, a double line-to-ground fault occurs between Y and B phases to the ground. The fault impedance is Z_f. This is shown in Figure 7.20(a).

 

 

Fig 7.20(a) LLG Fault on the System

When fault occurs, the condition of the system may be written as:

I_R = 0 (due to healthy condition)

I_f = fault current = I_Y + I_B

 

 

Equation (7.18) in sequence components can be written as:

 

 

From the above,

 

 

From Equation (7.18)

 

 

Replacing V_Y, I_Y, I_B in terms of sequence components,

 

 

As

 

 

The above equation can be written as

 

 

During fault condition,

 

 

Substituting the above condition in Equation (7.20),

 

 

Equation (7.19, 7.21, and 7.22) can be obtained by connecting the sequence networks in parallel. This is shown in Figure 7.20(b):,

 

 

Fig 7.20(b) Connection of Sequence Networks for LLG Fault

In Figure 7.20(a), Equations (7.19, 7.21 and 7.22) can be proved.

Relation between sequence currents:

 

 

If Z_f = 1 (dead short-circuit), all the sequence voltages are equal. This means all the sequence networks are connected in parallel for an LLG fault. Using Figure 7.20(a), the magnitude of sequence currents can be found as:

I_R1 = Total current in the circuit

 

 

where E, Z₁, Z₂ and Z₃ are Thevenin's equivalences of positive, negative and zero sequence networks. Using current division formula

 

 

Equations (7.24–7.26) can be used to determine sequence currents. Using these sequence currents, the fault current and sequence voltages can be obtained as:

 

 

Using the sequence voltages, all the three phase voltages can be obtained as:

 

 

This concludes LLG fault analysis.

7.5.4 Three-Phase Symmetrical Fault in Terms of Sequence Components

Under balanced condition

This means that, symmetrical fault analysis requires only the positive sequence network as shown in Figure 7.21.

 

 

Fig 7.21 Three-Phase Fault

Figure 7.21 is the Thevenin's equivalent circuit obtained using positive sequence network across faulted point F and ZPB.

Observations:

• The three phase fault current

   

  

   

• The positive sequence voltage at the point of fault location is V_R1 = 0 (voltage across short circuit)

Summary of Faults

1. All types of faults consist of positive sequence networks.
2. All types of unsymmetrical faults consist of negative sequence network
3. If the fault involves the ground (return path available) then
   • zero sequence network is present
   • zero sequence currents can flow.
4. All the sequence networks are connected in series for an SLG fault and in parallel for an LLG fault.
5. All sequence currents are equal for an SLG fault and all sequence voltages are equal for an LLG fault.
6. The current flowing through the fault in the case of a three-phase fault is zero because of symmetry. However, the fault across the three phases raises the current in all the three phases.
7. All sequence voltages, including positive sequence voltages are zero for a three-phase fault.

7.6 Comparison of SLG and 3-Phase Faults

When the fault involves only one phase, fault current is expected to be the least. Where as, involvement of all the three phases gives rise to the highest fault current. A comparison of these two faults, when they occur on the terminals of various equipments is presented in this section. Let the sequence impedance Z₁, Z₂ and Z₀ be pure reactances, and consider a dead short-circuit with Z_f = 0. The three-phase and SLG fault currents are given by:

 

 

Case-1: Ungrounded neutral alternator

When the fault is at generator terminals with the generator neutral grounded, Z₀ = ∞.

For this case

Case-2 : Solidly grounded neutral alternator

In this case, the zero-sequence reactance of the generator is X₀ = X_g0 + 3X_n = X_go (as X_n = 0 for solidly grounded neutral case)

The following relations are valid for the generator:

(for the sub-transient state)

Substituting the above relations,

 

 

From the above,

 

 

It is seen that for the solidly grounded neutral alternator, an SLG fault is more than a three-phase fault.

Case-3: Generator neutral reactance grounded

Generator neutral is grounded through reactance X_n. X_n has no effect on the magnitude of the three-phase fault, but it is designed as Substituting X_n value of the above and following the relations between sequences of the generator, the SLG fault current is:

 

 

This means, for the designed value of X_n, the SLG fault current is equal to the three-phase fault current. For different values of X_n, the relations are shown below:

Xn value	Relation

Case-4 Faults on Transformer / Transmission line

For the transformer and transmission lines of static devices, X₀ >> X₁. For this relation of reactances, a three-phase fault is always the reverse of an SLG fault.

7.7 Consideration of Pre-Fault Load Currents

The exact value of current during the faulty condition of a system can be obtained by superimposing pre-fault load currents on to the fault currents. The following presents the procedural steps:

1. Calculate the pre-fault load current.
2. Calculate the pre-fault voltage at the fault point (this voltage value is to be considered in the positive sequence network).
3. For the given single-line diagram, draw the three-sequence networks.
4. Obtain the Thevenin's equivalent sequence networks and interconnect the sequence networks depending upon the type of fault.
5. Calculate sequence component currents of fault at different locations in the network.
6. Sequence components of fault current at different locations are added to the same type of sequence components of pre-fault load currents at the same locations. If the load is balanced, only the positive sequence currents need to be superimposed.

Example 7.7

An alternator rated 11 kV, 50 MVA with neutral solidly grounded has X₁ = X₂ = j0.1 p. u. and X₀ = j0.05 p.u. It is supplying power to an 11 KVA, 25 MVA motor which has X₁ = X₂ = j0.1 p.u and X₀ = j0.05 p.u. through a short transmission line. Line reactances are X₁ = X₂ = j 0.01 p.u and X₀ = j0.4 p.u. All the equivalent reactances are pre-calculated on a 50 MVA, 11 KV base. The motor is drawing a power of 25 MW at 0.866 lag. The motor terminal voltage at loaded condition is 0.95 p. u. An SLG fault occurs on the phase-R of the generator. Calculate the fault current in phase-R of the generator under the given fault conditions.

Solution:

 

 

 

 

Taking the motor voltage as reference, the load current lags it by 30°.

 

 

Considering the positive sequence network of the system

 

 

Fig 7.22(a) Positive Sequence Network

Terminal voltage of generator = V

 

 

Fig 7.22(b) shows the interconnection of sequence networks for an SLG fault

 

 

Fig 7.22(b) Interconnection of Sequence Networks for an SLG Fault

The sequence components of the fault are

 

 

The component of positive sequence fault current flowing through the generator is

 

 

As the load current is a fully positive sequence, the total positive sequence component current during the fault is:

 

 

Negative sequence component current through generator (0.018 – j3.345) p.u. Zero sequence component current through the generator

 

 

The total current through Generator is

 

Example 7.8

Figure 7.23 shows a simple power system network. Draw the positive, negative and zero sequence networks.

 

 

Fig 7.23 A Simple Power System Network

The numerical data of various components in the power system is given as under: (all values are in p. u.)

Generator G₁: 50 MVA, 11 KV, X₁ = X₂ = 0.2; X₀ = 0.01;

Generator G2: 100 MVA, 11 KV, X₁ = X₂ = 0.25; X₀ = 0.02;

Transformer T₁: 50 MVA, 11/132 KV, X₁ = X₂ = 0.2; X₀ = 0.4

Transformer T₂: 100 MVA, 11/132 KV, X₁ = X₂ = 0.1; X₀ = 0.1

132 KV Line: 100 MVA, 132 KV, X₁ = X₂ = 0.1; X₀ = 0.25

Solution:

Common base MVA = 100 MVA;

Base voltage on LT side = 11 KV

Base voltage on HT side = 132 KV

Equipment per unit sequence impedances and common base values are completed below.

 

 

132 KV Line: X₁ = X₂ = 0.1; X₀ = 0.25 p.u

G₂ (neutral-ground reactance)

 

 

The pre-fault induced emf of G₁ and G₂ are assumed as 1 p. u.

a) Positive sequence network:

 

 

b) Negative sequence network:

 

 

c) Zero-sequence network:

 

Example 7.9

For the network in Example 7.8, simulate fault on Bus-3. Obtain Thevenin's equivalent positive, negative and zero sequence networks.

Solution:

Thevenin's equivalent circuit of positive, negative and zero sequence networks across Bus-3 and ZPB are obtained as follows:

a) Thevenin's equivalent positive and negative sequence networks:

 

 

b) Thevenin's equivalent zero-sequence network

 

 

 

The part of the network on the left hand side of Bus-3 is not considered due to discontinuity at Bus-2.

Example 7.10

A single line-to-ground fault occurs at Bus-3 of the network of Example 7.8; using Thevenin's equivalent sequence networks obtained in Example 7.9, determine the faulty phase and healthy phase:

a)  Currents

b)  Phase voltages

c)  Line-to-line voltages

Solution:

(a) Thevenin's equivalent sequence reactance are obtained in Example 7.7. X₁ = X₂ = j0.252 p. u, X₀ = j0.1 p. u.

Sequence currents are equal for an LG fault

 

 

Assuming a fault in an R-phase

 

 

 

The magnitude of SLG fault current

 

 

Current in all phases

 

 

(b) Sequence voltages at the fault point are:

 

 

The three-phase voltage (phase) values are:

 

 

 

Thus, line-to-neutral voltages at the fault point are:

 

 

(c) Line-to-line voltages

 

Example 7.11

A synchronous generator rated 3-phase 11 KV, 100 MVA has X₁ = X₂ = j0.1 p.u and X₀ = j0.04 p.u. Determine the fault current and line-to-line voltages during the fault condition: (a) if a single line-to-ground (SLG) fault occurs on the generator terminals, (b) if the generator neutral is solidly grounded, (c) if the generator is operating at no-load and rated voltage prior to the fault.

Solution:

It is usual practice for the rating of the equipment to be taken as base values.

 

 

For an SLG fault

 

 

Sequence voltages at fault point are:

 

 

Phase voltages at fault point are:

 

 

 

Line-to-line voltages

 

 

since, the generator line-to-neutral voltage is taken as 1 p.u.

 

Example 7.12

Consider the power system shown in Figure 7.24.

 

 

Fig 7.24 Power System Network of Example 7.12

 

An SLG fault occurs at Bus-2 at the far end of the line. Consider the numerical data:

 

 

Determine the fault current magnitude:

1. If the generator neutral is solidly grounded and fault impedance is zero
2. If the generator neutral is solidly grounded and fault impedance is 0.1 p. u on 50 MVA, 33 KV.
3. If the generator neutral is reactance grounded with X_n = 0.1 p. u on 50 MVA, 33 KV and fault impedance is 0.1 p. u on 50 MVA, 33 KV.

Solution:

Let Base MVA = 50 MVA

Base voltage = 33 KV

 

 

Line sequence reactance in p.u

 

 

1. Total sequence reactance upto the fault location:

    

   

2.  

   

3.  

   

Example 7.13

Consider the power system of Example 7.12.

1. If a three-phase fault occurs at the end of the line, compute the fault current.
2. Compare the three-phase fault current with an SLG fault for the conditions (a) to (c)

Solution:

1. 
2. 1. 
   2. 
   3. 

Example 7.14

Consider the synchronous generator of Example 7.11. Determine the fault current and line-to-line voltages if a double line fault occurs on the generator terminals.

Solution:

 

 

For LL fault,

 

 

Now, the line current is:

 

 

For LL fault,

 

 

Line-to-neutral phase voltages are:

 

 

While calculating the sequence [1 + a + a² = 0] the current and line-to-neutral voltage (phase voltage) are taken as 1 p.u

 

 

Line-to-line voltages

 

Example 7.15

Consider the synchronous generator of Example 7.11. Determine the fault current and line-to-line voltages if a double line-to-ground fault occurs at the generator terminals

Solution:

 

 

Positive, negative and zero sequence networks are connected in parallel as shown in Figure 7.25.

 

 

Fig 7.25 Thevenin's Equivalent Sequence Networks Connected in Parallel for an LLG Fault

From the figure,

 

 

Using the current division formula,

 

 

The line currents in R, Y and B are:

 

 

Base current = 5248.8 A

 

 

For an LLG fault, all the sequence voltages are equal.

 

 

Line-to-neutral voltages of the generator are:

 

 

Since 1 p.u voltage KV

 

 

Line voltages

 

Example 7.16

Consider the power system of Example 7.8 and Thevenin's equivalent sequence networks obtained in Example 7.9. Determine the magnitude of fault current if a double line (LL) fault occurs on Bus-3.

Solution:

Thevenin's equivalent sequence networks for LL fault are connected as shown in Figure 7.26

 

 

Fig 7.26 Sequence Networks for Double Line Fault

Considering data E = 1 X₁ = X₂ = 0.252 p.u

 

 

Sequence voltages

 

 

Phase voltages at the location of fault are:

 

 

Line voltages at the location of fault are:

 

 

Since sequence currents are computed by considering line-to-neutral voltage as 1 p. u, the base voltage taken in the problem is:

 

Example 7.17

Consider the power system of Example 7.8 and Thevenin's equivalent sequence networks obtained in Example 7.9. Determine the magnitude of fault current if a double line- to-ground fault (LLG) occurs on Bus-3. Assume fault impedance as zero.

Solution:

All the three sequence networks are connected in parallel for an LLG fault as shown in Figure 7.27.

 

 

Fig 7.27 Sequence Networks Connected in Parallel for an LLG Fault

 

 

Line current in the three phases are:

 

 

 

Base current = 437.4 Amp (ref Example 7.10)

 

 

Fault current:

 

 

Sequence voltages

 

 

Phase voltages at fault location

 

 

Line voltages at fault location

 

 

While calculating sequence voltages, phase voltage E is taken as 1 p.u.

Therefore, 1 p.u voltage at fault location

 

Example 7.18

Consider the power system of Example 7.7 and Thevenin's equivalent sequence networks obtained in Example 7.9. Determine the magnitude of fault current if a symmetrical three-phase fault current occurs on Bus-3.

Solution:

For symmetrical faults, only positive sequence network is considered.

 

 

Fault currents in lines are:

 

 

Line voltages at the fault location

Since V_R1 = V_R2 = V_R0 = 0 all line voltages are zero at the fault location.

Example 7.19

Three faults simulated on a power system give the following results:

 

 

The base values at the fault location are: 100 MVA, 132 KV. Find the p. u values of all sequence impedances of Thevenin's equivalent networks.

Solution:

 

 

From the above,