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Zero Sequence Mutual Induction in DoubleCircuit Lines & Its Effect on Distance

Protection Relay Performance

PEARLet #1

Author: Pradeep Kumar Gangadharan

EEnn gg iinn ee ee rree dd ttoo IInn nn oo vvaa tt ee

Protection Engineering And Research Laboratorieswww.pearlabs.com

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About PEARLet:PEARLets are technical notes on specific topics in the area of Power System Protection. Thesenotes are designed to explain concepts that a power system protection application engineerswill encounter in their day-to-day professional careers. Through these PEARLets, PEARLattempts to de-mystify many protection application phenomena, kindling interest in fresh

electrical engineers to purse a career as Power System Protection Application Engineer. At thesame time PEARLets are designed to be interesting to advanced protection applicationengineers as a good supplementary source of knowledge (we deliberately call itsupplementary, as there is nothing to replace real practical knowledge gained on field).

To achieve this, PEARLets will include simple physical explanations with analogies, supportedby rigorous mathematical derivations and figures to support these explanations. Whereverpossible, numerical examples will be included to explain the concept.

The author of PEARLets assumes that the readers have a basic, undergraduate levelknowledge in electrical engineering. Wherever the author feels it necessary, some fundamentalsrequired to understand the topic of the document will be re-visited.

Organization of a PEARLetAll PEARLets will follow the following architecture, with some or all of the components listedbelow;

> Introducing the topic, concept or issue to state the problem at hand.> Brief overview of the problem and its significance to the given application> List of the protection terms that are used in the document and their meaning> Refresh background knowledge required to understand the concept> A detailed analysis discussing various situations, highlighting all important influencing

factors.> All known and generally adopted solutions.> Reference to a list of published material including books, articles, papers etc.

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PEARLet #1Rev.: A

12th June, 2006Zero Sequence Mutual Induction in Double CircuitLines & Its Effect on Distance Protection RelayPerformance Author : Pradeep Kumar Gangadharan

©2006 Protection Engineering And Research Laboratories 4 Of 39

1.0 IntroductionPerformance of distance relays on double circuit lines are affected by the faultcurrent flow in the parallel line. The zero sequence current flowing in the parallel line

will introduce an error in the phase to ground impedance measured by the relay onthe protected line. The error could be either positive or negative depending on therelative direction of fault current flow in the parallel line with respect to the protectedline. This results in the distance relay under-reaching or over-reaching for faultsinvolving ground.

In this document this issue is explained from the basics and various scenarios arediscussed to help in understanding the problem better. This document will help thereader to understand the causes of the problem, learn how to analyze any givensituation for effects of mutual induction and to decide what kind of corrective action(if required) needs to be taken.

2.0 Zero sequence mutual ind uction and Distance relaysIn a double circuit line configuration, two lines are strung on the same tower and runparallel, in close proximity to each other for the entire or part of the distancebetween two substations. In such a system, during faults, fault currents will flow inboth the lines. The amount and direction of fault currents in the two lines woulddepend on the location of fault, type of fault and status of the lines (in service, out of service, out of service and grounded, etc).

2.1. Cause of zero sequence mutual inductionThe current flow in the parallel line will induce voltage in the protected line due tomutual induction. The amount of voltage induced would depend on the towerconfiguration, which determines the distance between the two lines on the tower.

Symmetrical component analysis is a very useful tool to analyzeunbalanced faults. Using symmetrical components any three phaseunbalanced signal (voltage or current) can be represented as three sets of three phase signals, positive, negative and zero sequence. Positive andnegative sequence quantities are balanced, meaning their three phases willhave same magnitude and will be displaced by 120 deg. The zero sequencequantities of all three phases are equal in magnitude and in phase.

Since positive and negative sequence currents are balanced, their net resultant fluxthat links with the other line (protected line) to induce voltage would be very lessand is generally ignored. However since zero sequence currents of all the three

phases are in phase, they will have substantial net resultant flux to link with theprotected line and induce voltage in it.

In simple terms, only zero sequence current flowing in the parallel line will inducevoltage in the protected line (on the same tower). Thus the fault voltage measuredby the relay on the protected line would include this induced voltage, which causesthe impedance measured by the relay to be higher or lower (depending on thepolarity of the induced voltage) than the actual value.

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Since only the voltage induced by the zero sequence currents are substantial and cancause significant mutual induction in a double circuit line, only the unbalanced faultsinvolving ground (which is the only case when zero sequence currents will flow) willcause problems.

2.2. Effect on distance relaysHaving seen the cause of induced voltage in the protected line, let us nowunderstand what happens as a result.

The magnitude and phase of the induced voltage in the protected line depends onthe magnitude and direction of zero sequence current flow in the parallel line. Thiswould in turn depend on factors like;

> Type of fault> Fault location> System configuration> System impedances> Status of the parallel line

The voltage measured by the distance relay will now include,- the voltage drop in the line between the relay location and fault due to the

current flow in the protected line- the zero sequence voltage induced by the parallel line.

Depending on the polarity of the induced voltage, the measured voltage can behigher or lower than the voltage drop in the protected line.

Since the measured impedance by the relay is the ratio of the measured voltage andfault current at the relay location, the zero sequence induced voltage can cause therelay to measure higher (when the induced voltage is of same polarity) or lower(when the induced voltage is of opposite polarity) impedance than actual. This willresult in the relay under-reaching or over-reaching.

Under-reaching can cause a particular zone element of the distance relay to notoperate for a fault well within its boundary, whereas over-reaching can cause thezone element of the relay to operate for a fault outside its boundary.

3.0 Definition of terms:Before getting started with the discussion we will define few terms, which will beused throughout the document.

Distance relay: The distance relay whose measurement accuracy is beinganalyzed, when the line it is protecting is subjected tomutual induction due to the presence of a parallel line.

Double circuit line: Two, three phase transmission lines running on the sametower. Unless specifically mentioned, both lines run parallelon the same tower for the entire length between the twostations (which are connected by them).

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Protected line: Transmission line which the distance relay whosemeasurement is being analyzed is protecting.

Parallel line: A Transmission which runs parallel to the protected line andis strung on the same tower.

Loop-in-loop-out: A line running between two sub-stations, which is broken onthe way to connect to a third sub-station.

4.0 Review of mutual induction fundamentalsWe will start with the review of electromagnetism. When current flows in a conductorit produces a magnetic field around it. The magnetic field is in a plane orthogonal tothe plane carrying the current. The magnetic field forms closed flux lines. Thedirection of the magnetic flux can be explained using the famous “Maxwell’s righthand screw rule“, illustrated in figure 4.1. If the thumb of the right hand is aligned inthe direction of the current flow in a conductor, the direction of magnetic fluxproduced by it is given by the direction of the curling fingers. When the currentflowing in the conductor is an alternating current, the magnetic field and theresultant flux lines produced are also alternating.

Fig. 4.1: Direction of current flow and resultant magnetic field

We can use the knowledge of the above basic phenomenon to understand thedirection of induced voltage in a conductor placed in a magnetic field. Figure 4.2shows a conductor carrying an alternating current I 1 , flowing from P 1 to P 2 asmarked. This current produces an alternating magnetic field around it and therelative direction of the alternating magnetic flux lines ( Φ1) is shown in figure 4.2.

When a secondary conductor is placed in this alternating magnetic field of theprimary conductor, the alternating flux lines will cut the secondary conductorinducing voltage in it. This is defined by Faraday’s law. The polarity of the inducedemf can be understood by applying the Lenz’s law.

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Fig. 4.2: Current flowing in a conductor and the magnetic flux produced around

Faraday’s law

The induced emf in a coil of N loops produced by a change in flux in acertain time interval is given by:

dt d

N eφ −=

Lenz’s law

The induced emf generates a current that sets up a magnetic field whichacts to oppose the change in magnetic flux.

Effect opposes the cause

Let us use Faraday’s and Lenz’s laws to understand the relative direction of theinduced emf in the secondary conductor. Figure 4.3 shows a secondary conductorplaced in the magnetic field of the primary conductor. When the alternating flux Φ1 ,

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cuts the secondary conductor, an emf is induced in the secondary conductor, whichwill cause a current (I 2 ) to flow in it.

Fig. 4.3: Voltage induced in a secondary conductor placed in the magnetic field of a

primary conductor

Fig. 4.4: Relative directions of current flows and flux lines in primary and secondaryconductors – Lateral view

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The direction of the magnetic field produced by the current I 2 (which is the effect)should be such as to oppose the magnetic field of the primary conductor (which isthe cause). The direction of flux Φ2 will be in opposition to Φ1 , as illustrated moreclearly in figure 4.4, which is the lateral view of the setup shown in figure 4.3.

Applying Maxwell’s right hand rule we can get the direction of current I 2 , which wouldbe from S 2 to S 1 , as marked in figure 4.3. For a current I 2 to flow to the externalcircuit in the direction marked, the polarity of the voltage E 2 should be as shown inthe figure 4.3.

Now this is a place when some confusion can creep in. Remember thatalways current flows from higher potential to lower potential through theexternal connected load. Whereas in the source, current always flow fromlower to higher potential. In the discussion above the primary conductor isthe load for the voltage E 1 and thus the current flows from the higherpotential P 1 to lower potential P 2 . On the other hand the secondaryconductor is the source E 2 (the voltage is produced in it due to induction),thus the current in the secondary conductor (which is the source) flowsfrom lower potential S 2 to the higher potential S 1 .

From this discussion it is clear that voltage would be induced in a secondaryconductor placed in the magnetic field of a primary conductor and,

- the magnitude of induced voltage would depend on the rate of change of flux thatlinks with the secondary conductor and the number of turns in the secondaryconductor.

- the polarity of the voltage induced in the secondary conductor would depend onthe direction of the current flow in the primary conductor.

We will use this understanding in the following sections to analyze the effect of mutual induction on the performance of distance protection.

5.0 Analysis of the effect of mutual induction on impedancemeasurement [ 2]

Let us start with the simple single ended system shown in figure 5.1. The systemshows a local source (E x) and two parallel lines (Line 1XY & Line 2XY). The distance(R) relay is protecting Line 1XY.

Fig. 5.1: A simple single system with double circuit transmission line

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Figure 5.2 shows the equivalent positive, negative and zero sequence circuits of thesystem shown in figure 5.1.

Fig. 5.2: Sequence equivalent circuit of the system shown in figure1

where,

Expn - Phase to neutral system source voltageZSX1 - System positive sequence source impedanceZ1XY1 - Positive sequence impedance of “Line 1XY” Z2XY1 - Positive sequence impedance of “Line 2XY” ZS2 - System negative sequence source impedanceZ1XY2 - Negative sequence impedance of “Line 1XY” Z2XY2 - Negative sequence impedance of “Line 2XY” ZS0 - System zero sequence source impedanceZ1XY0 - Zero sequence impedance of “Line 1XY” Z2XY0 - Zero sequence impedance of “Line 2XY” ZM0 - Zero sequence mutual inductance between the two lines (Line 1XY

& Line 2XY)

IR1 , I R2 , I R0 - Positive, negative and zero sequence currents at the relay locationVR1 , V R2 , V R0 - Positive, negative and zero sequence voltages at the relay locationIP0 - Zero sequence current flowing in the parallel line.V1M0 - Zero sequence voltage induced in parallel “Line 2XY” due to zero

sequence current flowing in the protected line “Line 1XY”. Thevalue of this induced voltage is given by, V1M0 = IR0xZM0.

V2M0 - Zero sequence voltage induced in the protected “Line 1XY” due tozero sequence current flowing in the parallel “Line 2XY”. The valueof this induced voltage is given by, V2M0 = IP0xZM0

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Fig. 5.3: Sequence circuit connection for A-G fault at bus Y – Parallel line in service.

The impedance measured by the “AN” element of relay R (since it’s an “A” to “Ground” fault) can be calculated using the expression,

K I I

V Z

RN RA

RAN meas AN ×+

=_ - (1)

where,

ZAN – “A” phase to neutral fault impedance.VRAN – “A” phase to neutral fault voltage at the relay location.IRA – “A” phase fault current measured at the relay location.IRN – Neutral fault current measured at the relay location.K – Earth fault compensation setting given by.

11

1101

3 XY

XY XY

Z

Z Z K ×

−= - (2)

Z1XY0 – Zero sequence line impedance.Z1XY1 – Positive sequence line impedance.

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The “A” phase and “Neutral” currents at the relay location can be expressed as;

0

021

3 R RN

R R R RA

I I

I I I I

×=

++=- (3)

From the figure 5.3, we can write the expression for the “A” phase to neutral voltageat the relay location as,

00010212111

02010212111

021

M P XY R XY R XY R

M XY R XY R XY R

R R R RAN

Z I Z I Z I Z I

V Z I Z I Z I

V V V V

×+×+×+×=+×+×+×=

++=

In a transmission line the positive and negative sequence impedances are equal, i.e.,Z1XY1 = Z 1XY2 . Applying this knowledge and also adding and subtracting the term (I R0 xZ1XY1 ), we can re-write the expression for the fault voltage as,

001101002111

11011000010112111

)()( M P XY XY R R R R XY

XY R XY R M P XY R XY R XY R RAN

Z I Z Z I I I I Z

Z I Z I Z I Z I Z I Z I V ×+−×+++×=

×−×+×+×+×+×=

- (4)

Substituting the values of V RAN, I RA, I RN and K from equations 2 to 4 in equation 1, weget,

K I I

Z I Z

Z Z Z I I

Z I Z

Z Z I I Z

Z

Z Z I I

Z I Z Z

I I Z

Z

Z Z I I

Z I Z Z I I I I Z

K I I

V Z

RN RA

M P XY

XY

XY XY RN RA

M P XY

XY XY RN RA XY

XY

XY XY RN RA

M P XY XY

RN RA XY

XY

XY XY RN RA

M P XY XY R R R R XY

RN RA

RAN meas AN

×+×

+=

×−

×+

×+⎟⎟

⎠

⎞⎜⎜

⎝

⎛ ×

−×+×

=

×−

×+

×+−

×+×=

×−

×+

×+−×+++×=

×+=

0011

11

1101

0011

110111

11

1101

001101

11

11

1101

001101002111

_

3

3)(

3

3)(

3

)()(

- (5)

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The actual fault impedance is,

11_ XY act AN Z Z = - (6)

Comparing equations 5 and 6 we can see that the error in the measured impedancedue to mutual induction is;

K I I

Z I Error

RN RA

M P

×+×

= 00

In this case the relay will under-reach by a factor of,

100

100100% 11

00

11

00

××+

×=

××+

×=×

×+×

=

K I I

K I

K I I

Z

Z I

Z

K I I

Z I

Underreach

RN RA

M PN

RN RA

XY

M P

XY

RN RA

M P

- (7)

Where,

KM - Mutual compensation factor,11

0

3 XY

M M Z

Z K ×

=

IPN - Parallel line neutral current, 03 PPN I I ×=

On analyzing equation 7 we can see that the degree of under-reaching due to mutualinduction depends on,

- The mutual compensation factor K M - The ratio of the neutral current in the parallel line (I PN) to the compensated

current flowing in the protected line (I RA+I RN * K)

It is worthwhile to mention here that the neutral current flowing in the parallel line tofeed a fault at the remote bus will be very close to the neutral current flowing in theprotected line only if the two lines are of the same type and also originate and endon the same bus at both substations. There can be instances when the parallel lineoriginates or/and ends at a bus different from that of the protected line and the bus

coupler between the buses is open. In such cases the fault current flow in theparallel line is independent of the current flow in the protected line and only dependson the strength of the source feeding the parallel line.

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zero sequence current in the parallel line was due to the fault zero sequencecurrent. Thus the zero sequence current in the parallel line flows in the oppositedirection when compared to the zero sequence current flow in the protected line(observe the current flow direction marked in figure 5.4).

5. The polarity of the zero sequence voltage induced in the protected line (due tothe zero sequence current flow in the parallel line) is opposite to its own zerosequence voltage (observe the induced voltage polarity marking in figure 5.4).

Following a similar procedure as in case 1, the “A” phase to “Ground” fault voltage atthe relay location can be derived and expressed as,

021101011 )( M XY XY R RA XY RAN V Z Z I I Z V −−×+×= - (8)

The zero sequence induced voltage in the protected line (V 2M0 ) is given by,

0002 M P M Z I V ×= - (9)

The zero sequence current flow in the parallel line (I P0 ) is given by,

02

010

XY

M P Z

V I = - (10)

The zero sequence induced voltage in the parallel line (V 1M0 ) is given by,

0001 M R M Z I V ×= - (11)

Using equations 9 to 11 in equation 8, the equation for the fault voltage becomes,

02

20

01101011

002

011101011

001101011

021101011

)(

)(

)(

)(

XY

M R XY XY R RA XY

M XY

M XY XY R RA XY

M P XY XY R RA XY

M XY XY R RA XY RAN

Z

Z I Z Z I I Z

Z Z

V Z Z I I Z

Z I Z Z I I Z

V Z Z I I Z V

×−−×+×=

×−−×+×=

×−−×+×=

−−×+×=

- (12)

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The measured “AN” impedance now becomes,

K I I

Z

Z I

Z

Z

Z Z I I

Z

Z I

Z

Z Z I I Z

Z

Z Z I I

Z

Z I Z Z I I Z

K I I

V Z

RN RA

XY

M R

XY

XY

XY XY RN RA

XY

M R

XY

XY XY RN RA XY

XY

XY XY RN RA

XY

M R XY XY R RA XY

RN RA

RAN meas AN

×+

×−=

×−

×+

×−⎟⎟

⎠

⎞⎜⎜

⎝

⎛ ×

−×+×

=

×−

×+

×−−×+×=

×+=

02

20

0

11

11

1101

02

20

011

110111

11

1101

02

20

01101011

_

3

3)(

3

)(

- (13)

In this case the measured impedance is less than the actual impedance (Z 1XY1 ). Thiswill cause the distance relay to over-reach. The percentage over-reach is given by

)I(assumiing 10013

1003

100100%

RA02

0

02

0

1102

000

11

02

20

0

RN M

XY

M

RN RA

M XY

M RN

RN RA XY XY

M M R

XY RN RA

XY

M R

I K

K Z

Z

K I I

K Z

Z I

K I I

Z Z

Z Z I

Z

K I I

Z

Z I

Overreach

=×+

××

=

××+

××

×=

××+××

×

=××+

×

=

- (14)

We can understand the factors that influence the degree of over-reaching byanalyzing equation 14. The main factors are,

> Ratio of the mutual compensation factor K M to the line earth fault compensationfactor plus 1 (K+1)

> The ratio of the zero sequence mutual impedance to three times the parallel linezero sequence impedance.

A point worth noting here is that there is no requirement to measure the parallel lineneutral current (as was required in case 1) to estimate the degree of over-reach.Just by knowing the values of zero sequence mutual impedance and zero sequenceimpedance of parallel line we can estimate the error and thus compensate for it. Wewill utilize this fact later when we discuss various solutions.

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5.3. Parallel line in service, with current flow in the oppositedirection to protected line

In some situations we can have a condition where current flow in the parallel line is

in the opposite direction to that of the protected line. This can happen when theparallel line is terminated on a bus different from the one on which protected line isterminated and the bus coupler between the two buses is open. If there are in-feedsat the remote bus, then during faults on one of the buses of the remote substation,current flow directions in the protected and parallel lines can be opposite. Figure 5.5shows such a system and figure 5.6 shows the sequence network connection for an

“A” phase to “Ground” fault at remote bus “Y1”.

Fig. 5.5: Two ended system with the remote bus-coupler open.

Since we have not considered any fault resistance in this case, the expression forfault voltage at the relay location will be the same as given in equation 8, which isagain listed below,

021101011 )( M XY XY R RA XY RAN V Z Z I I Z V −−×+×=

Since the zero sequence current in the parallel line now depends on the strength of the remote source connected to bus “Y2” (the value of Z 2S0 ), the zero sequenceinduced voltage on the protected line is fairly independent of its own zero sequencecurrent. This is unlike in case 2, where the zero sequence induced voltage was afunction of its own zero sequence current (see equation 14). Thus the fault voltageat the relay location for this case can be written as,

001101011 )( M P XY XY R RA XY RAN Z I Z Z I I Z V ×−−×+×= - (15)

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Again as in case 2, the measured impedance in this case also is less than the actualvalue, causing the relay to over-reach. The percentage over-reach for this case isgiven by,

100

100100% 11

00

11

00

××+

×=

××+

×=×

×+×

=

K I I

K I

K I I

Z

Z I

Z

K I I

Z I

Overreach

RN RA

M PN

RN RA

XY

M P

XY

RN RA

M P

- (17)

On analyzing equation 17 we can see that the degree of over-reaching due to mutualinduction depends on,

- The mutual compensation factor K M - The ratio of the neutral current in the parallel line (I PN) to the compensated

current flowing in the protected line (I RA+I RN * K)

Also note that the parallel line neutral current depends primarily on the strength of the source feeding it.

5.4. Parallel line is looped-in-looped-outOnce in a while a situation may occur where a line is broken in between twosubstations and taken to a third substation to form what is called a “Loop-in-loop-out” (LILO) system. The degree of influence the parallel line can have on theprotected line relay measurement would depend on the distance for which both linesrun parallel. The system being discussed here is shown in figure 5.7 below.

Fig. 5.7: Double circuit line with LILO and source on all three ends.

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In the system shown in figure 5.7, the protected “Line XY” goes from bus X to bus Y(over a distance of “d” km) The parallel line which is running on the same tower isbroken at point “T” (at a distance of “d1” km from bus X) and taken to another bus

“Z” (over a distance of “m1” km). A line from bus “Z” goes to bus “Y” to completethe “Loop-in-loop-out” (over a distance of “m2”+”d2” km). This is a very commonarrangement in a LILO system. It has to be noted that the “Line XZ” runs parallel tothe protected line from bus X to point T. Whereas from point T to bus Y, the “LineZY” becomes the parallel line. Both these segments will induce zero sequencevoltages in the protected line during ground faults. Usually in such applications thedistance d1+d2=d and m1=m2, but this is not always true and entirely depends onthe line construction.

The magnitude and polarity of the zero sequence voltage induced in the protectedline would depend on the magnitude and direction of current flowing in each of thetwo parallel line segments. As you would have guessed by now, there are manypossibilities for this depending on the relative strengths of the three sources and

fault location.

To proceed with our analysis, we will consider a fault at bus Y (as in the earliercases). This gives us three possible scenarios of fault current flow in the parallel line.They are shown in figure 5.8.

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(i). Scenario 1:In this situation, the fault current flow in the parallel line is also in the same directionas that of the protected line. This will happen when the local source is much strongerthan the two remote sources and/or the source Z is electrically much closer to theremote bus “Y”. The equivalent circuit and their connection for this situation for an

“A” phase to “Ground” fault is shown in figure 5.9.

Fig. 5.9: Sequence circuit connection for A-G fault at the remote bus of a two endedsystem – Parallel line LILO with fault current flow in same direction.

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The expression for the fault voltage at the relay location for this scenario can bewritten as,

0000110101121)(

M PTY M PXT XY XY R RA XY RAN Z a I Z a I Z Z I I Z V ×+×+−×+×=

where,

a1 – Per-unit length for which the line segment XT is in parallel with theprotected line.

d d

a1

1 = , {d1<d}

a2 – Per-unit length for which the line segment TY is in parallel with theprotected line.

d d

a2

2 = , {d2<d}

If the LILO is made at the two adjacent towers, then we can write a2 as,

d d

d d d

a1

1)1(

2 −=−=

Thus,

000011010111

11

)( M PTY M PXT XY XY R RA XY RAN Z d d

I Z d d

I Z Z I I Z V ×⎟ ⎠ ⎞

⎜⎝ ⎛ −×+××+−×+×=

- (18)The expression for the measured impedance can be written as,

K I I

Z d d

I Z d d

I Z

Z

Z Z I I

Z d d

I Z d d

I Z

Z Z I I Z

Z

Z Z I I

Z d d

I Z d d

I Z Z I I Z

K I I

V Z

RN RA

M PTY M PXT

XY

XY

XY XY RN RA

M PTY M PXT XY

XY XY RN RA XY

XY

XY XY RN RA

M PTY M PXT XY XY R RA XY

RN RA

RAN meas AN

×+

×⎟ ⎠ ⎞

⎜⎝ ⎛ −×+××

+=

×−

×+

×⎟ ⎠ ⎞

⎜⎝ ⎛ −×+××+

⎟⎟

⎠

⎞⎜⎜

⎝

⎛ ×

−×+×

=

×−

×+

×⎟ ⎠ ⎞

⎜⎝ ⎛ −×+××+−×+×

=

×+=

0000

11

11

1101

000011

110111

11

1101

00001101011

_

11

13

11

13

)(

3

11

1)(

- (19)

As in case 1, in this case also the measured impedance is higher than the actualimpedance, causing the relay to under-reach. The percentage under-reach is given,

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PEARLet #1Rev.: A



As you would have realized by now, this will only work if the neutral current in theparallel line is equal to the neutral current in the protected line and also in the samedirection. This will be true only when the fault is on or beyond the remote bus withthe remote bus coupler closed.

The biggest problem with this solution is that since this is a static compensation(done once via setting), when the parallel line current is different, the correction canbe erroneous. The worst case is when the neutral current in the parallel line is in theopposite direction to neutral current in the protected line.

For this reason, though used in some application in earlier days, this is not anadvisable remedy.

6.3. Dynamic compensation by measuring parallel line neutralcurrent

In this solution, dynamic compensation is provided for the zero sequence mutualinduction. This is done by measuring the neutral current flowing in the parallel lineand using this to offset the error in measurement. The relay is set with both the self earth fault compensation (K) and mutual zero sequence compensation (K M). Therelay measured impedance for a remote bus fault is then given by,

33

33)(

33

33)(

33

)()(

11

11

0

11

1101

11

0

11

110111

11

0

11

1101

0110111

11

0

11

1101

001101002111

_

XY

XY

M PN

XY

XY XY RN RA

XY

M PN

XY

XY XY RN RA XY

XY

M PN

XY

XY XY RN RA

M PN

XY XY RN RA XY

XY

M PN

XY

XY XY RN RA

M P XY XY R R R R XY

M PN RN RA

RAN meas AN

Z

Z

Z I

Z

Z Z I I

Z

Z I

Z

Z Z I I Z

Z

Z I

Z

Z Z I I

Z I Z Z I I Z

Z

Z I

Z

Z Z I I

Z I Z Z I I I I Z

K I K I I

V Z

=×

×+×

−×+

⎟⎟

⎠

⎞⎜⎜

⎝

⎛ ×

×+×

−×+×

=

××+

×−

×+

×+−×+×=

××+

×−

×+

×+−×+++×=

×+×+=

This is the correct fault impedance. This type of compensation works effectively forall faults in the protected line and faults beyond the remote bus.

However this kind of compensation can result in the relay mal-operating for close-upfaults on the parallel line. This is particularly true when the local source is verystrong, in which case the zero sequence current in the parallel line can be very highas compared to the zero sequence fault current in the protected line. In such casesthe relay can wrongly decide the fault (which is on the parallel line, which is actually

8/7/2019 Pear Let 1


8/7/2019 Pear Let 1


PEARLet #1Rev.: A



ZMO = 0.8 ∠75 ° Ω Zero sequence mutual impedance of line

For a solid “A” phase to “Ground” fault at 80% of the protected line from Bus X, thevoltages and currents at the two relay locations can be calculated. The simplest wayis to draw the sequence network and calculate the sequence currents and voltagesand then the phase values. The values obtained for this fault are given below.

At relay R X1 location:VRXA = 53. 782 ∠-3.45 ° kV ; I RXA = 877.55 ∠-80.82 °AVRXB = 140.65 ∠ -129.12 ° kV ; I RXB = I RXC = 0 AVRXC = 142.03 ∠ 128.67 ° kV ; I RXN = I RXA + I RXB + I RXC = 877.55 ∠-80.82 °A

At relay R Y1 location:VRYA = 23.079 ∠-3.76 ° kV ; I RYA = 1864.3 ∠-80.82 °AVRYB = 151 ∠-134.71 ° kV ; I RYB = I RYC = 0 A

VRYC = 154.87 ∠133.31 ° kV ; I RYN = I RYA + I RYB + I RYC = 1864.3 ∠-80.82 °A

The A phase to neutral loop impedance measured by relay R X1 will be,

_ K I I V

Z RXN RXA

RXA AN RX ×+

=

Where,

°−∠=×−

= 49.76686.03 1

10

Z

Z Z K

8204.798052.36

49.76686.082.8055.87782.8055.87745.353782

_

Ω°∠=

°−∠×°−∠+°−∠°−∠= AN RX Z

Since the actual fault is at 80%, the measured impedance should have been

ZRX_AN act = 0.4 ∠80 ° x 100 x 0.8 = 32 ∠80 ° Ω

Thus the relay R X1 will under-reach and see the fault well outside zone 1.

Similarly we can calculate the impedance seen by the remote relay R Y1. The valueobtained is,

80.0674345.749.76686.082.803.186482.803.1864

76.323079_

Ω°∠=°−∠×°−∠+°−∠

°−∠= AN RY Z

8/7/2019 Pear Let 1


8/7/2019 Pear Let 1


PEARLet #1Rev.: A



These measured impedances are exactly equal to the actual fault impedances. Thusit is clear that relays provided with dynamic zero sequence mutual compensation (bymeasuring the parallel line current) will measure the correct fault impedance.

Example 2:In this example I will illustrate how a zero sequence mutual compensated relay canpotentially mal-operate for close-up faults on the parallel line. I will use the samesystem given in figure 7.1 and consider a “A” phase to ground fault at 5% of theparallel line from Bus X.

The fault voltage and currents measured by relay R X1 is given as,

VRXA = 10.898 ∠ -4.91 ° kVIRXA = 546.71 ∠ 99.89 °A Leading fault current angle indicates fault is in

the reverse directionIRXN = 546.71 ∠99.89 °A

The neutral current on the parallel line (which is given to the relay R X1 for mutualcompensation) is calculated as,

IPXN =2337 ∠-81.34 ° A

The impedance seen by the “A” phase to neutral element of relay R X1 (with zerosequence mutual compensation) can be calculated as,

85.9574.16

34.81233756667.049.76686.081.9971.54681.9971.546

91.410898

_

Ω°∠=

°−∠×°−∠+°−∠×°∠+°∠

°−∠=

×+×+=

M PXN RXN RXA

RXA AN RX K I K I I

V Z

From the value of impedance measured we can see that relay R X1 will operate inzone 1. Figure 7.4 shows the actual and measured impedance seen by the relay R X1

for this case.

This illustrates the point we discussed in section 6.3 of mal-operation with un-monitored zero sequence mutual compensation. As we discussed in that section if wecontrol the compensation based on the ratio of the parallel to own zero sequence

current then in this case zero sequence compensation would not have been done(since I PXN /R RXN = 2337/546.71 = 4.27 > 1.5).

Without zero sequence mutual compensation, the relay R X1 would have seen the faultas a reverse fault as is shown in the calculation below.

8/7/2019 Pear Let 1


PEARLet #1Rev.: A



101.80-971.1149.76686.081.9971.54681.9971.546

91.410898

_

Ω°∠=°−∠×°∠+°∠

°−∠=

×+×+=

M PXN RXN RXA

RXA AN RX K I K I I

V Z

-20 -15 -10 -5 0 5 10 15 20 25

0

5

10

15

20

25

30

35

X End Relay

Measured ImpedanceActual Impedance

R

jX

Fig. 7.4: Relay seeing a reverse fault on the parallel line as a forward fault

Example 3:This example is used to illustrate the effect of zero sequence mutual induction and itscompensation when the parallel line runs only for a part of the total line length. Wewill use the system shown in figure 7.5 to illustrate this case. The values of systemparameters are same as that given in example 1. The only difference is that theparallel line runs only to a distance d1 on the same tower.

The impedance between bus Y and Z will decide the amount of fault current flowingin the parallel line.

We will consider a single phase to ground fault at bus Y with d=100km, d1= 50km.Also we will consider that the sum of impedances Z XZ+Z ZY = 5*Z XY.

8/7/2019 Pear Let 1


PEARLet #1Rev.: A



Fig. 7.5: Two ended double circuit line with parallel line only for a fraction of theprotected line length

The fault current and voltage at relay R location can be calculated as,

VRXA = 60.102 ∠ -2.80 ° kVIRXA = 867.636 ∠ -79.73 °AIRXN = 867.636 ∠ -79.73 °A

The neutral current flowing in the parallel line is,

IPXN = 173.527 ∠-79.73 ° A

The impedance seen by the “A” phase to neutral element of relay R X1 (without zerosequence mutual compensation) can be calculated as,

92.9760.41

49.76686.073.79636.86773.79636.8678.260102

_

Ω°∠=

°−∠×°−∠+°−∠°−∠=

×+=

K I I

V Z

RXN RXA

RXA AN RX

The fault is at bus Y, which means the actual impedance should have been 40 Ω. Thusthe relay will under-reach in this case.

Providing zero sequence mutual compensation would help in this case, however themutual compensation factor K M has to be set considering the fact that the parallelline only runs on the same tower for a part of the protected line. Thus in this casethe K M setting should be,

8/7/2019 Pear Let 1


PEARLet #1Rev.: A



°−∠=×

×= 53334.03

1

1

0

Z

Z

d d

K M M

The measured and actual impedance in the impedance plane and the relays zone 1characteristics is shown in figure 7.6.

-20 -10 0 10 20 30 40 500

10

20

30

40

50

60

X End Relay


R

jX

Fig. 7.6: Relay under reaching due to mutual induction from parallel line which runs

on the same tower for 50% of the line.

Now if we calculate the impedance measured by the relay provided with zerosequence mutual compensation with the above K M we get,

Ω°∠=

°−∠×°−∠+°−∠×°−∠+°−∠

°−∠=

×+×+=

8040

33.79527.17353334.049.76686.073.79636.86773.79636.867

8.260102

_ M PXN RXN RXA

RXA AN RX K I K I I

V Z

Figure 7.7 shows the measured and actual impedance when the relay is providedwith the zero sequence mutual compensation with the corrected K M.

8/7/2019 Pear Let 1


PEARLet #1Rev.: A



-20 -10 0 10 20 30 40 500

10

20

30

40

50

60

X End Relay


R

jX

Fig. 7.7: Correctly compensated relay measuring correct impedance.

This example illustrates the point that even if the parallel line runs only for a part of the total line on the same tower, still it can induce voltage to cause error in themeasurement of the protected line relay. The amount of error would depend on themagnitude of neutral current and the distance for which the lines run parallel. Alsowhen the compensation is provided, K M should be set to a fraction corresponding tothe distance for which the lines form a double circuit.

8.0 Summarizing the analysisThe analysis and discussions in this PEARLet elucidates the effect of zero sequencemutual induction on the performance of distance relays on a double circuit line.Different system conditions are discussed and their effects are analyzed. Thefollowing can be summarized from this discussion.

a) Zero sequence mutual induction cause the distance relays to under-reach orover-reach.

b) Under-reaching or over-reaching would depend on the direction of zerosequence current flow in the parallel line.

c) The amount of error would depend on the zero sequence mutual impedance,the amount of neutral current in the parallel line and the distance for which thelines run parallel on the same tower

d) Different solutions are available with their own limitations as tabulated below

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PEARLet #1Rev.: A



Solution CautionDo the correction for the mutualinduction by changing the earth faultcompensation.

This assumes that the zero sequencecurrents in the protected and parallellines are equal. This will only be truewhen the two lines are identical andrun parallel for the entire length andwith the faults at the remote bus orbeyond. Can cause severe problemswhen the neutral current in theparallel line is in the oppositedirection.

Not a recommended solution andvery rarely used.

Allowing the POR transfer trip schemeto take care

Can overreach for faults on theremote bus. The worst case is when

the parallel line is out of service andgrounded at both ends

Can be taken care by reducing thezone 1 setting by an error factor(refer equation 14 in section 5.2).Multiple setting groups of numericalrelays can be used to set this in adifferent group and enable it usingthe parallel line earth switch

Dynamic compensation by measuringthe parallel line current

Works well for almost all applicationseven when the parallel line runs onlyfor a part of the protected line. Buthas a potential to mal-operate forclose-up faults on the parallel line.

Overcome by controlling themutual compensation based on theratio of parallel line neutral current tothe protected line neutral current.Typically when the ratio is greaterthan 1.5, compensation is disabled.

The most widely used solution for this globally, is to not provide zero sequencemutual compensation for distance protection and to allow the POR scheme to provide

instantaneous protection to the entire line length. To take care of the over-reachingproblem when the parallel line is grounded, the zone 1 setting is reduced in adifferent setting group and switched by the parallel line earth switch auxiliarycontact.

However for fault location where accuracy of measured impedance is important forthe entire line length (unlike the distance protection where the concern of accuracy isat the boundary of the zones), it is recommended to provide a dynamic zero

8/7/2019 Pear Let 1


pear let 1

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