chapter2.UnsymmetricalFaultCalculations

2

Unsymmetrical Fault Calculations

Chapter 1 discussed the nature of sequence networks and how three distinct sequencenetworks can be constructed as seen from the fault point. Each of these networks canbe reduced to a single Thevenin positive, negative, or zero sequence impedance. Onlythe positive sequence network is active and has a voltage source which is the prefaultvoltage. For unsymmetrical fault current calculations, the three separate networkscan be connected in a certain manner, depending on the type of fault.

Unsymmetrical fault types involving one or two phases and ground are:

. A single line-to-ground fault

. A double line-to-ground fault

. A line-to-line fault

These are called shunt faults. A three-phase fault may also involve ground. Theunsymmetrical series type faults are:

. One conductor opens

. Two conductors open

The broken conductors may be grounded on one side or on both sides of thebreak. An open conductor fault can occur due to operation of a fuse in one ofthe phases.

Unsymmetrical faults are more common. The most common type is a line-to-ground fault. Approximately 70% of the faults in power systems are single line-to-ground faults.

While applying symmetrical component method to fault analysis, we willignore the load currents. This makes the positive sequence voltages of all the gen-erators in the system identical and equal to the prefault voltage.

In the analysis to follow, Z1, Z2, and Z0 are the positive, negative, and zerosequence impedances as seen from the fault point; Va, Vb, and Vc are the phase to

Copyright 2002 by Marcel Dekker, Inc. All Rights Reserved.

ground voltages at the fault point, prior to fault, i.e., if the fault does not exist; andV1, V2, and V0 are corresponding sequence component voltages. Similarly, Ia, Ib, andIc are the line currents and I1, I2, and I0 their sequence components. A fault impe-dance of Zf is assumed in every case. For a bolted fault Zf ¼ 0.

2.1 LINE-TO-GROUND FAULT

Figure 2-1(a) shows that phase a of a three-phase system goes to ground through animpedance Zf . The flow of ground fault current depends on the method of systemgrounding. A solidly grounded system with zero ground resistance is assumed. Therewill be some impedance to flow of fault current in the form of impedance of thereturn ground conductor or the grounding grid resistance. A ground resistance canbe added in series with the fault impedance Zf . The ground fault current must have areturn path through the grounded neutrals of generators or transformers. If there isno return path for the ground current, Z0 ¼ 1 and the ground fault current is zero.This is an obvious conclusion.

Phase a is faulted in Fig. 2-1(a). As the load current is neglected, currents inphases b and c are zero, and the voltage at the fault point, Va ¼ IaZf . The sequencecomponents of the currents are given by

I0

I1

I2

��

�� ¼1

3

1 1 1

1 a a2

1 a2 a

��

��Ia

0

0

��

�� ¼1

3

Ia

Ia

Ia

��

�� ð2:1Þ

Also,

I0 ¼ I1 ¼ I2 ¼1

3Ia ð2:2Þ

3I0Zf ¼ V0 þ V1 þ V2 ¼ �I0Z0 þ Va � I1Z1ð Þ � I2Z2 ð2:3Þ

which gives

I0 ¼Va

Z0 þ Z1 þ Z2 þ 3Zf

ð2:4Þ

The fault current Ia is

Ia ¼ 3I0 ¼3Va

Z1 þ Z2 þ Z0ð Þ þ 3Zf

ð2:5Þ

This shows that the equivalent fault circuit using sequence impedances can be con-structed as shown in Fig. 2-1(b). In terms of sequence impedances’ network blocksthe connections are shown in Fig. 2-1(c).

This result could also have been arrived at from Fig. 2-1(b):

Va � I1Z1ð Þ þ �I2Z2ð Þ þ �I0Z0ð Þ � 3ZfI0 ¼ 0

which gives the same equations (2.4) and (2.5). The voltage of phase b to groundunder fault conditions is

40 Chapter 2


Vb ¼ a2V1 þ aV2 þ V0

¼ Va

3a2Zf þ Z2 a2 � a� �þ Z0 a2 � 1

� �Z1 þ Z2 þ Z0ð Þ þ 3Zf

ð2:6Þ

Similarly, the voltage of phase c can be calculated.An expression for the ground fault current for use in grounding grid designs

and system grounding is as follows:

Ia ¼3Va

R0 þ R1 þ R2 þ 3Rf þ 3RGÞ þ j X0 þ X1 þ X2ð Þð ð2:7Þ

Unsymmetrical Fault Calculations 41

Figure 2-1 (a) Line-to-ground fault in a three-phase system; (b) line-to-ground fault equiva-lent circuit; (c) sequence network interconnections.


where Rf is the fault resistance and RG is the resistance of the grounding grid;R0, R1, and R2 are the sequence resistances and X0, X1, and X2 are sequencereactances.

2.2 LINE-TO-LINE FAULT

Figure 2-2(a) shows a line-to-line fault. A short-circuit occurs between phases b andc, through a fault impedance Zf . The fault current circulates between phases b and c,flowing back to source through phase b and returning through phase c; Ia ¼ 0,Ib ¼ �Ic. The sequence components of the currents are

I0

I1

I2

��

�� ¼1

3

1 1 1

1 a a2

1 a2 a

��

��0

�Ic

Ic

��

�� ¼1

3

0

�aþ a2

�a2 þ a

��

�� ð2:8Þ

From Eq. (2.8), I0 ¼ 0 and I1 ¼ �I2.

42 Chapter 2

Figure 2-2 (a) Line-to-line fault in a three-phase system; (b) line-to-line fault equivalentcircuit; (c) sequence network interconnections.


Vb � Vc ¼ 0 1 �1��

Va

Vb

Vc

��

�� ¼ 0 1 �1��

1 1 1

1 a2 a

1 a a2

��

��V0

V1

V2

��

��

¼ 0 a2 � a a� a2��

V0

V1

V2

��

��

ð2:9Þ

Therefore,

Vb � Vc ¼ a2 � a� �

V1 � V2ð Þ¼ a2I1 þ aI2� �

Zf

¼ a2 � a� �

I1Zf

ð2:10Þ

This gives

V1 � V2ð Þ ¼ I1Zf ð2:11ÞThe equivalent circuit is shown in Fig. 2-2(b) and (c).

Also

Ib ¼ a2 � a� �

I1 ¼ �jffiffiffi3

pI1 ð2:12Þ

and,

I1 ¼Va

Z1 þ Z2 þ Zf

ð2:13Þ

The fault current is

Ib ¼ �Ic ¼�j

ffiffiffi3

pVa

Z1 þ Z2 þ Zf

ð2:14Þ

2.3 DOUBLE LINE-TO-GROUND FAULT

A double line-to-ground fault is shown in Fig. 2-3(a). Phases b and c go to groundthrough a fault impedance Zf . The current in the ungrounded phase is zero, i.e.,Ia ¼ 0. Therefore, I1 þ I2 þ I0 ¼ 0.

Vb ¼ Vc ¼ Ib þ Icð ÞZf ð2:15ÞThus,

V0

V1

V2

��

�� ¼1

3

1 1 1

1 a a2

1 a2 a

��

��Va

Vb

Vb

��

�� ¼1

3

Va þ 2Vb

Va þ ðaþ a2ÞVb

Va þ ðaþ a2ÞVb

��

�� ð2:16Þ



Figure 2-3 (a) Double line-to-ground fault in a three-phase system; (b) double line-to-

ground fault equivalent circuit; (c) sequence network interconnections.


which gives V1 ¼ V2 and

V0 ¼1

3Va þ 2Vbð Þ

¼ 1

3V0 þ V1 þ V2ð Þ þ 2 Ib þ Icð ÞZf½ �

¼ 1

3V0 þ 2V1ð Þ þ 2 3I0ð ÞZf½ �

¼ V1 þ 3ZfI0

ð2:17Þ

This gives the equivalent circuit of Fig. 2-3(b) and (c).The fault current is

I1 ¼Va

Z1 þ ½Z2kðZ0 þ 3Zf Þ�

¼ Va

Z1 þZ2ðZ0 þ 3Zf ÞZ2 þ Z0 þ 3Zf

ð2:18Þ

2.4 THREE-PHASE FAULT

The three phases are short-circuited through equal fault impedances Zf , Fig. 2-4(a).The vectorial sum of fault currents is zero, as a symmetrical fault is considered andthere is no path to ground.


Figure 2-4 (a) Three-phase symmetrical fault; (b) equivalent circuit; (c) sequence network.


I0 ¼ 0 Ia þ Ib þ Ic ¼ 0 ð2:19Þ

As the fault is symmetrical:

Va

Vb

Vc

��

�� ¼Zf 0 0

0 Zf 0

0 0 Zf

��

��Ia

Ib

Ic

��

�� ð2:20Þ

The sequence voltages are given by

V0

V1

V2

��

�� ¼ ½Ts��1

Zf 0 0

0 Zf 0

0 0 Zf

��

��½Ts�I0

I1

I2

��

�� ¼Zf 0 0

0 Zf 0

0 0 Zf

��

��I0

I1

I2

��

�� ð2:21Þ

This gives the equivalent circuit of Fig. 2-4(b) and (c).

Ia ¼ I1 ¼Va

Z1 þ Zf

Ib ¼ a2I1

Ic ¼ aI1

ð2:22Þ

2.5 PHASE SHIFT IN THREE-PHASE TRANSFORMERS

2.5.1 Transformer Connections

Transformer windings can be connected in wye, delta, zigzag, or open delta. Thetransformers may be three-phase units, or three-phase banks can be formed fromsingle-phase units. Autotransformer connections should also be considered. Thevariety of winding connections is, therefore, large [1]. It is not the intention todescribe these connections completely. The characteristics of a connection can beestimated from the vector diagrams of the primary and secondary emfs. There is aphase shift in the secondary voltages with respect to the primary voltages, depend-ing on the connection. This is of importance when paralleling transformers. Avector diagram of the transformer connections can be constructed based on thefollowing:

1. The voltages of primary and secondary windings on the same leg of thetransformer are in opposition, while the induced emfs are in the samedirection. (Refer to Appendix C for further explanation.)

2. The induced emfs in three phases are equal, balanced, and displacedmutually by a one-third period in time. These have a definite phasesequence.

Delta–wye connections are discussed, as these are most commonly used. Figure2-5 shows polarity markings and connections of delta–wye transformers. For allliquid immersed transformers the polarity is subtractive according to ANSI

46 Chapter 2


(American National Standard Institute) standard [2]. (Refer to Appendix C for anexplanation.) Two-winding transformers have their windings designated as highvoltage (H) and low voltage (X). Transformers with more than two windings havetheir windings designated as H, X, Y, and Z. External terminals are distinguishedfrom each other by marking with a capital letter, followed by a subscript number,i.e., H1, H2, and H3.

2.5.2 Phase Shifts in Winding Connections

The angular displacement of a polyphase transformer is the time angle expressed indegrees between the line-to-neutral voltage of the reference identified terminal andthe line-to-neutral voltage of the corresponding identified low-voltage terminal. InFig. 2-5(a), wye-connected side voltage vectors lead the delta-connected side voltagevectors by 30�, for counterclockwise rotation of phasors. In Fig. 2-5(b) the delta-connected side leads the wye-connected side by 30�. For transformers manufacturedaccording to the ANSI/IEEE (Institute of Electrical and Electronics Engineers, Inc.,USA), standard [3], the low-voltage side, whether in wye or delta connection, has a


Figure 2-5 Winding connections and phase displacement for voltage vectors for delta–wyeconnected transformers.


phase shift of 30� lagging with respect to the high-voltage side phase-to-neutralvoltage vectors. Figure 2-6 shows ANSI/IEEE [3] transformer connections and aphasor diagram of the delta side and wye side voltages. These relations and phasedisplacements are applicable to positive sequence voltages.

The International Electrotechnical Commission (IEC) allocates vector groups,giving the type of phase connection and the angle of advance turned though inpassing from the vector representing the high-voltage side emf to that representingthe low-voltage side emf at the corresponding terminals. The angle is indicated muchlike the hands of a clock, the high-voltage vector being at 12 o’clock (zero) and thecorresponding low-voltage vector being represented by the hour hand. The totalrotation corresponding to hour hand of the clock is 360�. Thus, Dy11 and Yd11symbols specify 30� lead (11 being the hour hand of the clock) and Dy1 and Yd1signify 30� lag. Table 2-1 shows some IEC vector groups of transformers and theirwinding connections.

2.5.3 Phase Shift for Negative Sequence Components

The phase shifts described above are applicable to positive sequence voltages orcurrents. If a voltage of negative phase sequence is applied to a delta–wye connectedtransformer, the phase angle displacement will be equal to the positive sequencephasors, but in the opposite direction. Therefore, when the positive sequence cur-rents and voltages on one side lead the positive sequence current and voltages on theother side by 30�, the corresponding negative sequence currents and voltages will lagby 30�. In general, if the positive sequence voltages and currents on one side lag thepositive sequence voltages and currents on the other side by 30�, the negativesequence voltages and currents will lead by 30�.

Example 2.1

Consider a balanced three-phase delta load connected across an unbalanced three-phase supply system, as shown in Fig. 2-7. The currents in lines a and b are given.

48 Chapter 2

Figure 2-6 Phase designations of terminal markings in three-phase transformers according

to ANSI/IEEE standard.



Table 2-1 Transformer Vector Groups, Winding Connections, and Vector Diagrams

(continued)


50 Chapter 2

Table 2-1 (continued)


The currents in the delta-connected load and also the symmetrical components ofline and delta currents are required to be calculated. From these calculations, thephase shifts of positive and negative sequence components in delta windings and linecurrents can be established.

The line current in c is given by

Ic ¼ � Ia þ Ibð Þ¼ �30þ j6:0A

The currents in delta windings are

IAB ¼ 1

3Ia � Ibð Þ ¼ �3:33þ j4:67 ¼ 5:735 < 144:51�A

IBC ¼ 1

3Ib � Icð Þ ¼ 16:67� j5:33� 17:50 < �17:7�A

ICA ¼ 1

3Ic � Iað Þ ¼ �13:33þ j0:67 ¼ 13:34 < 177:12�A

Calculate the sequence component of the currents IAB. This calculation gives

IAB1 ¼ 9:43 < 89:57�A

IAB2 ¼ 7:181 < 241:76�A

IAB0 ¼ 0A

Calculate the sequence component of current Ia. This calculation gives

Ia1 ¼ 16:33 < 59:57�A

Ia2 ¼ 12:437 < 271:76�A

Ia0 ¼ 0A

This shows that the positive sequence current in the delta winding is 1=ffiffiffi3

ptimes the

line positive sequence current, and the phase displacement is þ30�, i.e.,

IAB1 ¼ 9:43 < 89:57� ¼ Ia1ffiffiffi3

p < 30� ¼ 16:33ffiffiffi3

p < ð59:57� þ 30�ÞA


Figure 2-7 Balanced delta-connected load on an unbalanced three-phase power supply.


The negative sequence current in the delta winding is 1=ffiffiffi3

ptimes the line negative

sequence current, and the phase displacement is �30�, i.e.,

IAB2 ¼ 7:181 < 241:76� ¼ Ia2ffiffiffi3

p < �30� ¼ 12:437ffiffiffi3

p < ð271:76� � 30�ÞA

This example illustrates that the negative sequence currents and voltages undergo aphase shift which is the reverse of the positive sequence currents and voltages.

The relative magnitudes of fault currents in two winding transformers forsecondary faults are shown in Fig. 2-8, on a per unit basis. The reader can verifythe fault current flows shown in this figure.

52 Chapter 2


2.6 UNSYMMETRICAL FAULT CALCULATIONS

Example 2.2

The calculations using symmetrical components can best be illustrated with anexample. Consider a subtransmission system as shown in Fig. 2-9. A 13.8-kVgenerator G1 voltage is stepped up to 138 kV. At the consumer end the voltage


Figure 2-8 Three-phase transformer connections and fault current distribution for second-ary faults.


is stepped down to 13.8 kV, and generator G2 operates in synchronism with thesupply system. Bus B has a 10,000-hp motor load. A line-to-ground fault occurs atbus B. It is required to calculate the fault current distribution throughout thesystem and also the fault voltages. The resistance of the system components isignored in the calculations.

Impedance Data

The impedance data for the system components are shown in Table 2-2. GeneratorsG1 and G2 are shown solidly grounded, which will not be the case in a practicalinstallation. A high-impedance grounding system is used by utilities for groundinggenerators in step-up transformer configurations. Generators in industrial facilities,directly connected to the load buses are low-resistance grounded, and the groundfault currents are limited to 200–400 A. The simplifying assumptions in the exampleare not applicable to a practical installation, but clearly illustrate the procedure ofcalculations.

The first step is to examine the given impedance data. Generator-saturatedsubtransient reactance is used in the short-circuit calculations and this is termedpositive sequence reactance; 138-kV transmission line reactance is calculated fromthe given data for conductor size and equivalent conductor spacing. The zerosequence impedance of the transmission line cannot be completely calculated fromthe given data and is estimated on the basis of certain assumptions, i.e., a soilresistivity of 100�m.

Compiling the impedance data for the system under study from the givenparameters, from manufacturers’ data, or by calculation and estimation can betime consuming. Most computer-based analysis programs have extensive datalibraries and companion programs for calculation of system impedance data andline constants, which has partially removed the onus of generating the data fromstep-by-step analytical calculations. Appendix B provides models of line constantsfor coupled transmission lines, bundle conductors, and line capacitances. References3 and 4 provide analytical and tabulated data.

Next, the impedance data are converted to a common MVA base. Afamiliarity with the per unit system is assumed. The voltage transformation ratio

54 Chapter 2

Figure 2-9 A single line diagram of power system for Example 2.2.


of transformer T2 is 138–13.2 kV, while a bus voltage of 13.8 kV is specified, whichshould be considered in transforming impedance data on a common MVA base.Table 2-1 shows raw impedance data and their conversion into sequenceimpedances.

For a single line-to-ground fault at bus B, the sequence impedance net-work connections are shown in Fig. 2-10, with the impedance data for compo-nents clearly marked. This figure is based on the fault equivalent circuit shownin Fig. 2-1(b), with fault impedance Zf ¼ 0. The calculation is carried out perunit, and the units are not stated in every step of the calculation.

The positive sequence impedance to the fault point is

Z1 ¼jð0:25þ 0:18þ 0:04þ 0:24Þ � j0:37� j1:67

jð0:37þ 1:67Þjð0:25þ 0:18þ 0:04þ 0:24Þ þ j0:37� j1:67

jð0:37þ 1:67ÞThis gives Z1 ¼ j0:212:


Table 2-2 Impedance Data for Example 2.2

Equipment Description Impedance data

Per unitimpedance 100-

MVA base

G1 13.8-kV, 60-MVA, 0.85 Subtransient reactance ¼ 15% X1 ¼ 0:25power factor generator Transient reactance ¼ 20% X2 ¼ 0:28

Zero sequence reactance ¼ 8% X0 ¼ 0:133Negative sequence

reactance ¼ 16:8%T1 13.8–138 kV step-up

transformer, 50/84 MVA,delta–wye connected, wye

neutral solidly grounded

Z ¼ 9% on 50-MVA base X1 ¼ X2 ¼ X0

¼ 0:18

L1 Transmission line, 5 mileslong, 266.8 KCMIL,

ACSR

Conductors at 15 ft (4.57 m)equivalent spacing

X1 ¼ X2 ¼ 0:04X0 ¼ 0:15

T2 138–13.2 kV, 30-MVA step-down transformer, wye–

delta connected, high-voltage wye neutral solidlygrounded

Z ¼ 8% X1 ¼ X2 ¼ X0

¼ 0:24

G2 13.8-kV, 30-MVA, 0.85 Subtransient reactance ¼ 11% X1 ¼ 0:37power factor generator Transient reactance ¼ 15% X2 ¼ 0:55

Zero sequence reactance ¼ 6% X0 ¼ 0:20Negative sequence reactance

¼ 16:5%M 10,000-hp induction motor

loadLocked rotor reactance ¼ 16:7%

on motor base kVA (consider 1

hp � 1 kVAÞ

X1 ¼ 1:67X2 ¼ 1:80X0 ¼ 1

Resistances are neglected in the calculations.

KCMIL: Kilo-circular mils, same as MCM.

ACSR: Aluminum conductor steel reinforced.


Z2 ¼jð0:28þ 0:18þ 0:04þ 0:24Þ � j0:55� j1:8

jð0:55þ 1:8Þjð0:28þ 0:18þ 0:04þ 0:24Þ þ j0:55� j1:8

jð0:55þ 1:8

56 Chapter 2

Figure 2-10 Sequence network connections for single line-to-ground fault, (Example 2.2).


This gives Z2 ¼ j0:266.Z0 ¼ j0:2. Therefore,

I1 ¼E

Z1 þ Z2 þ Z0

¼ 1

j0:212þ j0:266þ j0:2¼ �j1:475pu

I2 ¼ I0 ¼ �j1:475

Ia ¼ I0 þ I1 þ I2 ¼ 3ð�j1:475Þ ¼ �j4:425pu

The fault currents in phases b and c are zero:

Ib ¼ Ic ¼ 0

The sequence voltages at a fault point can now be calculated:

V0 ¼ �I0Z0 ¼ j1:475� j0:2 ¼ �0:295

V2 ¼ �I2Z2 ¼ j1:475� j0:266 ¼ �0:392;

V1 ¼ E � I1Z1 ¼ I1ðZ0 þ Z2Þ ¼ 1� ð�j1:475� j0:212Þ ¼ 0:687

A check of the calculation can be made at this stage; the voltage of the faultedphase at fault point B ¼ 0:

Va ¼ V0 þ V1 þ V2 ¼ �0:295� 0:392þ 0:687 ¼ 0

The voltages of phases b and c at the fault point are

Vb ¼ V0 þ aV1 þ a2V2

¼ V0 � 0:5ðV1 þ V2Þ � j0:866ðV1 � V2Þ¼ �0:295� 0:5ð0:687� 0:392Þ � j0:866ð0:687þ 0:392Þ¼ �0:4425� j0:9344

jVbj ¼ 1:034pu

Similarly,

Vc ¼ V0 � 0:5ðV1 þ V2Þ þ j0:866ðV1 � V2Þ¼ �0:4425þ j0:9344

jVcj ¼ 1:034pu

The distribution of the sequence currents in the network is calculated from theknown sequence impedances. The positive sequence current contributed from theright side of the fault, i.e., by G2 and motor M is

�j1:475jð0:25þ 0:18þ 0:04þ 0:24Þ

jð0:25þ 0:18þ 0:04þ 0:24Þ þ j0:37� j1:67

jð0:37þ 1:67ÞThis gives �j1:0338. This current is composed of two components, one from thegenerator G2 and the other from the motor M. The generator component is

ð�j1:0338Þ j1:67

jð0:37þ 1:67Þ ¼ �j0:8463



The motor component is similarly calculated and is equal to �j0:1875:The positive sequence current from the left side of bus B is

�j1:475

j0:37� j1:67

jð0:37þ 1:67Þjð0:25þ 0:18þ 0:04þ 0:24Þ þ j0:37� j1:67

jð0:37þ 1:67ÞThis gives �j0:441. The currents from the right side and the left side should sum to�j1:475. This checks the calculation accuracy.

The negative sequence currents are calculated likewise and are as follows:

In generator G2 ¼ �j0:7172In motor M ¼ �j0:2191From left side, Bust B ¼ �j0:5387From right side ¼ �j0:9363

The results are shown in Fig. 2-10. Again, verify that the vectorial summation at thejunctions confirms the accuracy of calculations.

Currents in generator G2

IaðG2Þ ¼ I1ðG2Þ þ I2ðG2Þ þ I0ðG2Þ¼ �j0:8463� j0:7172� j1:475

¼ �j3:0385

jIaðG2Þj ¼ 3:0385pu

IbðG2Þ ¼ I0 � 0:5ðI1 þ I2Þ � j0:866ðI1 � I2Þ¼ �j1:475� 0:5ð�j0:8463� j0:7172Þ � j0:866ð�j0:8463þ j0:7172Þ¼ �0:1118� j0:6933

jIbðG2Þj ¼ 0:7023pu

IcðG2Þ ¼ I0 � 0:5ðI1 þ I2Þ þ j0:866ðI1 � I2Þ¼ 0:1118� j0:6933

jIcðG2Þj ¼ 0:7023pu

This large unbalance is noteworthy. It gives rise to increased thermal effects due tonegative sequence currents and results in overheating of the generator rotor. Agenerator will be tripped quickly on negative sequence currents.

Currents in Motor M. The zero sequence current in the motor is zero. Thus,

IaðMÞ ¼ I1ðMÞ þ I2ðMÞ¼ �j0:1875� j0:2191

¼ �j0:4066

jIaðMÞj ¼ 0:4066pu

58 Chapter 2


IbðMÞ ¼ �0:5ð�j0:4066Þ � j0:886ð0:0316Þ ¼ 0:0274þ j0:2033

IcðMÞ ¼ �0:0274þ j0:2033

jIbðMÞj ¼ jIcðMÞj ¼ 0:2051pu

The summation of the line currents in the motor M and generator G2 are

IaðG2Þ þ IaðMÞ ¼ �j0:4066� j3:0385 ¼ �j3:4451

IbðG2Þ þ IbðMÞ ¼ �0:1118� j0:6993þ 0:0274þ j0:2033 ¼ �0:084� j0:490

IcðG2Þ þ IcðMÞ ¼ 0:1118� j0:6933� 0:0274 þ 0:2033 ¼ 0:084� j0:490

Currents from the left side of the bus B are

Ia ¼ �j0:441� j0:539

¼ �j0:98

Ib ¼ �0:5ð�j0:441� j0:5387Þ � j0:866ð�j0:441þ j0:5387Þ¼ 0:084þ j0:490

Ic ¼ �0:084þ j0:490

These results are consistent as the sum of currents in phases b and c at the faultpoint from the right and left side is zero and the summation of phase a currents givesthe total ground fault current at b ¼ �j4:425. The distribution of currents is shownin a three-line diagram (Fig. 2-11).

Continuing with the example, the currents and voltages in the transformer T2

windings are calculated. We should correctly apply the phase shifts for positive andnegative sequence components when passing from delta secondary to wye primary ofthe transformer. The positive and negative sequence current on the wye side oftransformer T2 are

I1ðpÞ ¼ I1 < 30� ¼ �j0:441 < 30� ¼ 0:2205� j0:382

I2ðpÞ ¼ I2 < �30� ¼ �j0:539 < �30� ¼ �0:2695� j0:4668

Also, the zero sequence current is zero. The primary currents are

IaðpÞ ¼ I0 þ I1ðpÞ þ I2ðpÞ

0:441 < 300� þ 0:5397 < 240� ¼ �0:049� j0:8487

IbðpÞ ¼ a2I1ðpÞ þ aI2ðpÞ ¼ �0:0979

IcðpÞ ¼ aI1ðpÞ þ a2I2ðpÞ ¼ �0:049� j0:8487

Currents in the lines on the delta side of the transformer T1 are similarlycalculated. The positive sequence component, which underwent a 30� positive shiftfrom delta to wye in transformer T2, undergoes a �30� phase shift; as for an ANSIconnected transformer it is the low-voltage vectors which lag the high-voltage sidevectors. Similarly, the negative sequence component undergoes a positive phase shift.The currents on the delta side of transformers T1 and T2 are identical in amplitudeand phase. Figure 2-11 shows the distribution of currents throughout the distribu-tion system.



The voltage on the primary side of transformer T2 can be calculated. Thevoltages undergo the same phase shifts as the currents. Positive sequence voltageis the base fault positive sequence voltage, phase shifted by 30� (positive) minus thevoltage drop in transformer reactance due to the positive sequence current:

V1ðpÞ ¼ 1:0 < 30� � jI1ðpÞX1t

¼ 1:0 < 30� � ðj0:441 < 30�Þð�j0:24Þ¼ 0:958þ j0:553

V2ðpÞ ¼ 0� I2ðpÞX2t

¼ �ð0:539 < �30�Þð0:24 < 270�Þ¼ 0:112� j0:0647

Thus,

VaðpÞ ¼ 0:9577 ¼ j0:553þ 0:112� j0:0647 ¼ 1:0697þ j0:4883 ¼ 1:17 < 24:5�

VbðpÞ ¼ �0:5ðV1ðpÞ þ V2ðpÞÞ � j0:866ðV1ðpÞ � V2ðpÞÞ¼ �j0:9763

VcðpÞ ¼ �0:5ðV1ðpÞ þ V2ðpÞÞ � j0:866ðV2ðpÞ � V1ðpÞÞ¼ �1:0697þ j0:4883 ¼ 1:17 < 155:5�

Note the voltage unbalance caused by the fault.

60 Chapter 2

Figure 2-11 Three-line diagram of fault current distribution, (Example 2.2).


2.7 SYSTEM GROUNDING AND SEQUENCE COMPONENTS

The alternating current power system grounding is concerned with the nature andlocation of an intentional electric connection between the electrical system phaseconductors and ground. The utility systems at high-voltage transmission level,subtransmission level, and distribution level are solidly grounded. The utilitygenerators connected through step-up transformers are invariably high-resistancegrounded. The industrial systems at medium-voltage level are low-resistancegrounded. Recent trends for low-voltage industrial systems are high-resistancegrounded systems for continuity of processes in a continuous process plant. It isnot the intention here to describe the power system grounding methods or theircharacteristics, except to point out the role of symmetrical components and faultcurrent calculations that are often required for these systems.

In a solidly grounded system, no intentional impedance is introduced betweenthe system neutral and ground. These systems meet the requirement of ‘‘effectivelygrounded’’ systems in which the ratio X0=X1 is positive and less than 3.0 and theratio R0=X0 is positive and less than 1, where X1, X0, and R0 are the positivesequence reactance, zero sequence reactance, and zero sequence resistance, respec-tively. The coefficient of grounding (COG) is defined as a ratio of ELG=ELL inpercentage, where ELG is the highest rms voltage on a sound phase, at a selectedlocation, during a fault affecting one or more phases to ground, and ELL is the rmsphase-to-phase power frequency voltage that is obtained at the same location withthe fault removed. Calculations in Example 2.2 show the fault voltage rises onunfaulted phases. Solidly grounded systems are characterized by a COG of 80%.By contrast, for ungrounded systems, definite values cannot be assigned to ratios X0=X1 and R0=X0. The ratio X0=X1 is negative and may vary from low to high values.The COG approaches 120%. For values of X0=X1 between 0 and �40, a possibilityof resonance with consequent generation of high voltages exists. The COG affectsthe selection of rated voltage of the surge arresters and stresses on the insulationsystems. The overvoltages based on relative values of sequence impedances areplotted in Ref. 4.

In impedance grounded systems, an intentional low or high impedance isintroduced between the neutral and ground. In a high-resistance grounded system,the resistance is calculated so that Ir is at least equal to Ic, where Ir is the currentthrough the resistor and Ic is the system capacitance current returning to ground dueto distributed phase capacitances of the power system equipment like motors, cables,and transformers. The total ground current is therefore

ffiffiffi2

pIc.

Example 2.3

Figure 2-12(a) shows a 5-MVA 34.5–2.4 kV delta–delta transformer serving indus-trial motor loads. It is required to derive an artificial neutral through a wye–deltaconnected transformer and high-resistance ground the 2.4-kV secondary systemthrough a grounding resistor. The capacitance charging current of the system is8A. Calculate the value of the resistor to limit the ground fault current throughthe resistor to 10 A. Neglect transformer resistance and source resistance.

The connection of sequence networks is shown in Fig. 2-12(b) and the givenimpedance data reduced to a common100-MVA base are shown in Table 2-3. Motorwye-connected neutrals are left ungrounded in the industrial systems, and therefore



motor zero sequence impedance is infinite. (This contrasts with grounding practicesin some European countries, where the motor neutrals are grounded.) The sourcezero sequence impedance can be calculated based on the assumption of equal posi-tive and negative sequence reactances. The motor voltage is 2.3 kV and, therefore itsper unit reactance on 100-MVA base is given by

16:7

1:64

� �2:3

2:4

� �2

¼ 9:35

Similarly, the grounding transformer per unit calculations should be adjustedfor correct voltages:

X0 ¼1:5

0:06

� �2:4

2:4=ffiffiffi3

p� �

¼ 75

The equivalent positive and negative sequence reactances are: 1.41 per unit each. Thezero sequence impedance of the grounding transformer is 50þ j75 per unit. The totalfault current should be limited to 10-j8 ¼ 12:80 amperes. Thus, the required impe-dance is

62 Chapter 2

Figure 2-12 (a) Artificially derived neutral grounding in a delta–delta system through a

wye–delta grounding transformer, (Example 2.3); (b) connection of sequence impedances forhigh-resistance fault calculations, (Example 2.3).

(a)


Zt ¼2400=

ffiffiffi3

p

12:8=3

� �¼ 324:8 ohms

The base ohms (100-MVA baseÞ ¼ 0:0576. The required Zt ¼ 324:8=baseohms ¼ 5638:9 per unit. This shows that the system positive and negative sequenceimpedances are low compared to the desired total impedance in the neutral circuit.The system positive and negative sequence impedances can, therefore, be neglected.

IR0 ¼ 10=3 ¼ 3:33 A. Therefore, ZR0 ¼ ð2400= ffiffiffi3

p Þ=3:33 ¼ 416:09 ohms ¼416:09=base ohms ¼ 7223:9 per unit. The additional resistor to be inserted:

RR0 ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiZ2

R0 � X2t0

q� Rt0

¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi7223:92 � 752

p� 50

¼ 7173:5 per unit

Multiplying by base ohms, the required resistance ¼ 413:2 ohms:


Figure 2-12 (continued)


These values are in symmetrical component equivalents. In actual values, referred to120 V secondary, the resistance value is

RR ¼ 120

2400

� �2

413:2� 3 ¼ 3:1 ohms

If we had ignored all the sequence impedances, including that of the ground-ing transformer, the calculated value is 3.12 ohms. This is often done in thecalculations for grounding resistance for high-resistance grounded systems, andall impedances including that of the grounding transformer can be ignored withoutappreciable error in the final results. The grounding transformer should be rated topermit continuous operation, with a ground fault on the system. The per phasegrounding transformer kVA requirement is 4:16� 3:33A ¼ 13:8 kVA, i.e., a totalof 13:8� 3 ¼ 41:5 kVA. The grounding transformer of the example is, therefore,adequately rated.

2.8 OPEN CONDUCTOR FAULTS

Symmetrical components can also be applied to the study of open conductor faults.These faults are in series with the line and are called series faults. One or twoconductors may be opened, due to mechanical damage or by operation of fuseson unsymmetrical faults.

2.8.1 Two-Conductor Open Fault

Consider that conductors of phases b and c are open-circuited. The currents in theseconductors then go to zero.

Ib ¼ Ic ¼ 0 ð2:23ÞThe voltage across the unbroken phase conductor is zero, at the point of break, Fig.2-13(a).

64 Chapter 2

Table 2-3 Impedance Data for Example 2.3—High-Resistance Grounding

Equipment Given dataPer unit impedance on

100-MVA base

34.5-kV source Three-phase fault ¼ 1500MVA Xs1 ¼ Xs2 ¼ 0:067Line-to-ground fault ¼ 20 kA sym Xs0 ¼ 0:116

34.5–2.4 kV, 5-MVAtransformer, delta–deltaconnected

X1 ¼ X2 ¼ X0 ¼ 8% Xt1 ¼ Xt2 ¼ Xt0 ¼ 1:60

2.3 kV 1800-hp (1640 kVA)

induction motor load

Locked rotor reactance ¼ 16:7%(on motor base kVA)

Xm1 ¼ Xm2 ¼ 9:35

Grounding transformer, X0 ¼ 1:5% X0 ¼ 7560-kVA, wye–delta

connected 2400:120 V

R0 ¼ 1:0% R0 ¼ 50


Va0 ¼ Vao1 þ Vao2 þ Va0 ¼ 0

Ia1 ¼ Ia2 ¼ Ia0 ¼1

3Ia

ð2:24Þ

This suggests that sequence networks can be connected in series as shown in Fig.2-13(b).

2.8.2 One Conductor Open

Now consider that phase a conductor is broken, Fig. 2-14(a)

Ia ¼ 0 Vb0 � Vc0 ¼ 0 ð2:25ÞThus,

Vao1 ¼ Vao2 ¼ Vao0 ¼1

3Vao

Iai þ Ia2 þ Ia0 ¼ 0

ð2:26Þ


Figure 2-13 (a) Two-conductor open series fault; (b) connection of sequence networks.


This suggests that sequence networks are connected in parallel, Fig. 2-14(b).

Example 2.4

Consider that one conductor is broken on the high-voltage side at the point markedO in Fig. 2-9. The equivalent circuit is shown in Fig. 2-15.

An induction motor load of 10,000 hp was considered in the calculations for asingle-line-to-ground fault in Example 2.2. All other static loads, i.e., lighting andresistance heating loads, were ignored, as these do not contribute to short-circuitcurrents. Also, all drive system loads, connected through converters, are ignored,unless the drives are in a regenerative mode. If there are no loads and a brokenconductor fault occurs, no load currents flow.

Therefore, for broken conductor faults all loads, irrespective of their types,should be modeled. For simplicity of calculations, again consider that a 10,000-hpinduction motor is the only load. Its positive and negative sequence impedances forload modeling will be entirely different from the impedances used in short-circuitcalculations.

66 Chapter 2

Figure 2-14 (a) One-conductor open series fault; (b) connection of sequence networks.


Chapter 12 discusses induction motor equivalent circuits for positive and nega-tive sequence currents. The range of induction motor impedances per unit (basedupon motor kVA base) are

Xlr ¼ 0:14� 0:2 < 83� � 75�

Xþload ¼ 0:9� 0:95 < 20� � 26�

X�load � Xlr

ð2:27Þ

where Xlr ¼ induction motor locked rotor reactance at its rated voltage, Xþload ¼

positive sequence load reactance, and X�load ¼ negative sequence load reactance.

The load impedances for the motor are as shown in Fig. 2-15. For an openconductor fault as shown in this figure, the load is not interrupted. Under normaloperating conditions, the motor load is served by generator G2, and in the system ofFig. 2-15 no current flows in the transmission line L. If an open conductor faultoccurs, generator G2, operating in synchronism, will trip on operation of negativesequence current relays. To proceed with the calculation assume that G2 is out ofservice, when the open conductor fault occurs.

The equivalent impedance across an open conductor is


Figure 2-15 Equivalent circuit of an open conductor fault, (Example 2.4).


ðj0:25þ j0:18þ j0:08þ j0:24þ 9:9þ j4:79Þposþ j0:28þ j0:18þ j0:04þ j0:24þ 0:20þ j1:65ð Þnegkhj0:18þ j0:15þ j0:24ð Þzero

�¼ 9:918þ j6:771 ¼ 12:0 < 34:32�

The motor load current is

0:089 < �25:84� pu (at 0:9 power factor ½PF�ÞThe load voltage is assumed as the reference voltage, thus the generated voltage is:

Vg ¼ 1 < 0� þ ð0:089 < �25:84�Þðj0:25þ j0:18þ j0:04þ j0:24Þ¼ 1:0275þ j0:0569 ¼ 1:0291 < 3:17�

The positive sequence current is

I1g ¼ VG=Zt ¼1:0291 < 3:17�

12:00 < 34:32�

¼ 0:0857 < 31:15� ¼ 0:0733þ j0:0443

The negative sequence and zero sequence currents are

I2g ¼ �I1gZ0

Z2 þ Z0

¼ ð0:0857 < 31:15�Þ 0:53 < 90�

2:897 < 86:04�

0:0157 < 215:44�

I0g ¼ �I1gZ2

Z2 þ Z0

¼ 0:071 < 210:33�

Calculate line currents:

Iag ¼ I1g þ I2g þ I0g ¼ 0

Ibg ¼ a2I1g þ aI2g þ I0g ¼ 0:1357 < 250:61�

Icg ¼ 0:1391 < �8:78�

The line currents in the two phases are increased by 52%, indicating seriousoverheating. A fully loaded motor will stall. The effect of negative sequence currentsin the rotor is simulated by the equation:

I2 ¼ I21 þ kI22 ð2:28Þwhere k can be as high as 6. The motors are disconnected from service by anti singlephasing devices and protective relays.

The ‘‘long way’’ of calculation using symmetrical components, illustrated bythe examples, shows that, even for simple systems, the calculations are tedious andlengthy. For large networks consisting of thousands of branches and nodes these are

68 Chapter 2


impractical. There is an advantage in the hand calculations, in the sense that ver-ification is possible at each step and the results can be correlated with the expectedfinal results. For large systems, matrix methods and digital simulation of the systemsare invariable. This gives rise to an entirely new challenge for analyzing the crypticalvolumes of analytical data, which can easily mask errors in system modeling andoutput results.

Problems

1. A double line-to-ground fault occurs on the high-voltage terminals oftransformer T2 in Fig. 2-9. Calculate the fault current distribution andthe fault voltages throughout the system, similar to Example 2.2.

2. Repeat Problem 1 for a line-to-line fault and then for a line-to-groundfault.

3. Calculate the percentage reactance of a 60/100 MVA, 13.8–138 kV trans-former in Fig. 2-P1 to limit the three-phase fault current at bus A to 28kA symmetrical. for a three-phase symmetrical fault at the bus. Assumeonly nondecaying ac component of the short-circuit current and neglectresistances.

4. In Problem 3, another similar generator is to be added to bus A. What isthe new short-circuit current? What can be done to limit the three-phaseshort-circuit level at this bus to 36 kA?

5. Calculate the three-phase and single line-to-ground fault current for afault at bus C, for the system shown in Fig. 1-P1, Chapter 1. As all thegenerators are connected through delta–wye transformers, and deltawindings block the zero-sequence currents, does the presence of genera-tors (1) increase, (2) decrease, or (3) have no effect on the single line-to-ground fault at bus C.

6. In Problem 5 list the fault types in the order of severity, i.e., the magni-tude of the fault current.


Figure 2-P1 Distribution system for Problem 3.


7. Calculate the three-phase symmetrical fault current at bus 3 in the systemconfiguration of Fig. 2-P2. Neglect resistances.

8. Figure 2-P3 shows an industrial system motor load being served from a115-kV utility’s system through a step-down transformer. A single line-to-

70 Chapter 2

Figure 2-P2 System configuration for Problem 7.

Figure 2-P3 System configuration for Problem 8.


ground fault occurs at the secondary of the transformer terminals. Specifythe sequence network considering the motor load. Consider a load oper-ating power factor of 0.85 and an overall efficiency of 94%. Calculate thefault current.

9. A wye–wye connected transformer, with neutral isolated and a tertiaryclose-circuited delta winding serves a single phase load between twophases of the secondary windings. A 15-ohm resistance is connectedbetween two lines. For a three-phase balanced supply voltage of 480Vbetween the primary windings, calculate the distribution of currents in allthe windings. Assume a unity transformation ratio.

10. Calculate the COG factor in Example 2.2 for all points where the faultvoltages have been calculated.

11. Why is it permissible to ignore all the sequence impedances of the systemcomponents and base the fault current calculations only on the systemvoltage and resistance to be inserted in the neutral circuit when designinga high-resistance grounded system?

BIBLIOGRAPHY

1. WD Stevenson. Elements of Power System Analysis. 4th ed. New York: McGraw-Hill,

1982.2. CA Gross. Power System Analysis. New York: John Wiley, 1979.3. GO Calabrase. Symmetrical Components Applied to Electric Power Networks. New

York: Ronald Press Group, 1959.4. JL Blackburn. Symmetrical Components for Power Systems Engineering. New York:

Marcel Dekker, 1993.5. DR Smith. Digital Simulation of Simultaneous Unbalances Involving Open and Faulted

Conductors. IEEE Trans PAS, Vol. 89, No. 8. 1970, pp 1826–1835.

REFERENCES

1. Transformer Connections (Including Auto-transformer Connections). General Electric,Publication no. GET-2H, Pittsfield, MA, 1967.

2. ANSI/IEEE. General Requirements of Liquid Immersed Distribution, Power and

Regulating Transformers. Standard C57.12.00-1987.3. ANSI. Terminal Markings and Connections for Distribution and Power Transformers.

Standard C57.12.70-1978.

4. Electrical Transmission and Distribution Reference Book. 4th ed. Westinghouse ElectricCorp., East Pittsburgh, PA, 1964.



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