01 introduction to ransmission line faults

Transmission Line Protection - PROT 407

Section 1 - Introduction to Transmission Line Faults

Transmission Line Protection

l ntroduction to Transmission Line Faults

Copyright OSEL 2010 E!

Technical papers recommended for this section:

"Introduction to Symmetrical Components," Stanley E. Zocholl

"Symmetrical Components: Line Transposition," Stanley E. Zocholl



Simple Short Circuits

Three Single Phase Phase to Two Phase Phase to Ground Phase to Ground

Unbalanced

The most common and harmful line faults are the short circuits, or transversal faults. During these faults, two points of the line at different potential make contact. This includes the structure that supports the line.



Fault Type Distribution in 500 kV Lines

Fault Type %

Single-Line-Ground 93

Line-Line 4

Line-Line-Ground 2

Three-Phase 1

The high incidence of single-phase-to-ground faults is more noticeable in very high voltage transmission lines.



Voltage and Current Phasors Normal Operation

Va Vo < Fau It -1-1-1 ---

Normal Operation

During normal balanced operation, the currents have a relatively low magnitude and the phase-to-ground voltages are close to the system's nominal voltage.



Voltage and Current Phasors Phase-to-Phase Fault

L-L Fault *T,

For phase-to-phase fault (B-C fault in this case), the current magnitude increases in the faulted phases, and the voltage magnitude decreases in the same phases. The phase angle between the faulted phases' voltages decreases. The voltage and current in the sound phase remain close to the pre-fault values.



Decomposition of an Unbalanced System

Zero-Sequence Positive-Sequence Negative-Sequence

Single-Phase Balanced Balanced Copyrrght OSEL 2010

Recall that the symmetrical components method is used to analyze unbalances in power systems.

Any three-phase set of unbalanced currents, or voltages, can be decomposed in three sets of currents, or voltages, with the following characteristics:

One three-phase BALANCED set with POSITIVE-sequence rotation, a-b-c by convention

One three-phase BALANCED set with NEGATIVE-sequence rotation, a-c-b

One three-phase with the three currents that are equal in magnitude and in phase. This set is called ZERO sequence.

The convention rules are as follows:

1. For a system with a-b-c rotation, the positive-sequence set will have a-b-c rotation. The negative-sequence set will have the a-c-b rotation.

2. For a system with a-c-b rotation, the positive-sequence set will have a-c-b rotation. The negative-sequence set will have the a-b-c rotation.

- - - 2- - I , = KO + I,, + I,, = loo + a I,, + aI,, Drop "a" subindex & = & + a2& + aF2

- - - - - - 2 - Ic = IcO + Icl + Ic2 = IaO + aI,, + a I,,



Symmetrical Components as a Function of Phase Quantities

Solving for the sequence components, the equations give the sequence currents in terms of the phase currents.



Inverse Matrix Form

Define: r l 1 1 1 Then:

[Is] = [A]-' . [I,]

The inverse transformation is easily obtained, as shown above.


Section 1 - Introduction to Transmission Line Faults g __._. - --- - . --

Some Overhead Line Configurations

(Not to scale)

copyright asel 2010

Overhead lines and cables are among the most important elements of the power system. They transport and distribute the energy along vast geographical regions. There is a large variety of line configurations (conductor arrangement, tower design, etc.). However, the lines can be represented with an acceptable degree of accuracy with relatively simple models.



Symmetrical Transmission Line

- Magnetic Coupling I, "3J%,

c - V , V , V , c; F7' 4 4 4 --- --- 4 4 4 --- - - - --- - - -

- - - - v, = qr;, +Z,I, +Z,,I, +c q7 = z,nc7 + zsq, + ZTEC + q v C = % T ( + z & + z s C + ~

The figure shows the original phase equations for a perfectly symmetrical transmission line. Note that the value of the mutual and self impedances is represented by a single term Zm, because Z,, = Z,, = Z,,.



Positive-Sequence Impedance of a Transmission Line - - v,, = ZSIa1 + Zm&, + ZmC, = (2, -Zm)Ial

- - 5, = ZmTal + ZsI,, + Zm<, = ($ - Zm)Ibl - - - - - - v,, = z, I,, + ZmIb1 + zs I,, = (2, - Zm)Tcl

The result gives us the value of the positive-sequence impedance as a function of the self and mutual inlpedances of the line.



Negative-Sequence Impedance of a Transmission Line

The result gives us the value of the negative-sequence impedance as a function of the self and mutual impedances of the line. It is the same as the positive sequence.



Zero-Sequence Impedance of a Transmission Line

The result gives us the value of the zero-sequence impedance as a function of the self and mutual impedances of the line. This value is always larger than the positive-sequence because we are adding two times the mutual impedance to the self impedance, instead of subtracting the mutual impedance from the self impedance.

Even though this result is valid for symmetrical lines, the trend is the same for non- symmetrical lines. The zero-sequence impedance of a transmission line is always larger than the positive-sequence impedance of the same line.

b.

PROT407 - TransmissionLineFaults - r14



Transmission Line Fault

Faults on overhead transmission lines are the most common in power systems. The high exposure of the lines to ambient and natural elements make them the most vulnerable element of the system.



Line Fault Transients: Voltages Phase A-to-Ground Fault

x lo5 1.5

Time (s)

The graphic shows how the voltages behave in the substation during a single-phase-to- ground fault. Notice the magnitude sag in the A-phase voltage and the high-frequency components that appear following the inception of the fault.



Transmission Line Protection

l ntroduction to Transmission Line Faults

Technical papers recommended for this section:

"Introduction to Symmetrical Components," Stanley E. Zocholl

"Symmetrical Components: Line Transposition," Stanley E. Zocholl



Transmission Line Fault

13.8 kV

Faults on overhead transmission lines are the most common in power systems. The high exposure of the lines to ambient and natural elements make them the most vulnerable element of the system.



Line Fault Transients: Voltages Phase A-to-Ground Fault

x l o 5

Time (s)

The graphic shows how the voltages behave in the substation during a single-phase-to- ground fault. Notice the magnitude sag in the A-phase voltage and the high-frequency components that appear following the inception of the fault.



Simple Short Circuits

Three Phase

Single Phase Phase to Two Phase to Ground Phase to Ground

Unbalanced

The most common and harmful line faults are the short circuits, or transversal faults. During these faults, two points of the line at different potential make contact. This includes the structure that supports the line.



Fault Type Distribution in 500 kV Lines

Fault Type %

Single-Line-Ground 93

Line-Line 4

Line-Line-Ground 2

Tt-~ree-P hase 1

The high incidence of single-phase-to-ground faults is more noticeable in very high voltage transmission lines.



Voltage and Current Phasors Normal Operation

v, '5 y- Fault -1-1-1

Normal Operation

Copyr~ght OSEL 2010 M

During normal balanced operation, the currents have a relatively low magnitude and the phase-to-ground voltages are close to the system's nominal voltage.



Voltage and Current Phasors Phase-to-Phase Fault

L-L Fault i7;,

Copyright OSEL 2010 R

For phase-to-phase fault (B-C fault in this case), the current magnitude increases in the faulted phases, and the voltage magnitude decreases in the same phases. The phase angle between the faulted phases' voltages decreases. The voltage and current in the sound phase remain close to the pre-fault values.



Decomposition of an Unbalanced System -

Zero-Sequence Positive-Sequence Negative-Sequence

Single-Phase Balanced Balanced Copyright .*EL 2010 @%

Recall that the symmetrical components method is used to analyze unbalances in power systems.

Any three-phase set of unbalanced currents, or voltages, can be decomposed in three sets of currents, or voltages, with the following characteristics:

One three-phase BALANCED set with POSITIVE-sequence rotation, a-b-c by convention

One three-phase BALANCED set with NEGATIVE-sequence rotation, a-c-b

One three-phase with the three currents that are equal in magnitude and in phase. This set is called ZERO sequence.

The convention rules are as follows:

1. For a system with a-b-c rotation, the positive-sequence set will have a-b-c rotation. The negative-sequence set will have the a-c-b rotation.

2. For a system with a-c-b rotation, the positive-sequence set will have a-c-b rotation. The negative-sequence set will have the a-b-c rotation.

- - - - 7 - - & = FbO + Ibl + Ib2 = IaO + a2Ta1 + aIu, Drop "a" subindex Fb = & + a-Il + a12



Symmetrical Components as a Function of Phase Quantities

Solving for the sequence components, the equations give the sequence currents in terns of the phase currents.



Inverse Matrix Form

Define: 1 1 Then:

[A]-' = - 3 [ 1 a2 a'] [Ts ] = [A]-' . [ Ip ] l a a

m Copyright OSEL Z O l O i%i

The inverse transformation is easily obtained, as shown above.


Section 1 - Introduction to Transmission Line Faultsg -._- ---- - --

Some Overhead Line Configurations

(Not to scale)

Copyright @SEL 2010

Overhead lines and cables are among the most important elements of the power system. They transport and distribute the energy along vast geographical regions. There is a large variety of line configurations (conductor arrangement, tower design, etc.). However, the lines can be represented with an acceptable degree of accuracy with relatively simple models.



Symmetrical Transmission Line

r, Magnetic Coupling

Copyright OSEL 2070 M

The figure shows the original phase equations for a perfectly symmetrical transmission line. Note that the value of the mutual and self impedances is represented by a single term Z,, because Z,, = Zbc = Z,,.



Positive-Sequence Impedance of a Transmission Line

The result gives us the value of the positive-sequence impedance as a function of the self and mutual impedances of the line.



Negative-Sequence Impedance of a Transmission Line

2, = 5, ITa2 = v,, /I, , = V,, IT,,

Copylight 0 SEL 2D7D a

The result gives us the value of the negative-sequence impedance as a function of the self and mutual impedances of the line. It is the same as the positive sequence.



Zero-Sequence Impedance of a Tra~ismission Line

The result gives us the value of the zero-sequence impedance as a function of the self and mutual impedances of the line. This value is always larger than the positive-sequence because we are adding two times the mutual impedance to the self impedance, instead of subtracting the mutual impedance from the self impedance.

Even though this result is valid for symmetrical lines, the trend is the same for non- symmetrical lines. The zero-sequence impedance of a transmission line is always larger than the positive-sequence impedance of the same line.



Typical Values of Transmission Line Sequence Impedances

Nominal Voltage

(kV)

Copyright. SEL 2010

-

138

345

500

The table shows the result obtained in the calculation of series impedances for sample transmission lines. It is out of the scope of this course to explain the methods used to calculate transmission line sequence impedances. However, it is important to note the factors that have a major influence on the values of these impedances. They are:

Conductor ACSR

(#/phase)

The type and size of the phase and ground (shield) conductors

Drake (1)

Drake (2)

Falcon (3)

The geometric position and distances of the conductors

Geometry

The average soil resistivity

Grounding &

Triangle

Horizontal

Horizontal

The number of circuits in the structure and the existence of external parallel lines

The height of the conductor over the earth (more important for capacitance calculations)

318" steel ( I )

7/16 steel (2)

#8Alumoweld (2)

An average resistivity of 100 ohm-m was considered in the calculations. The line parameters must be calculated for each specific case.

The table shown here is an example and should not be taken as a specific recommendation.

0.1 17 + j0.622

0.059 + j0.588

0.021 + j0.564

0.577 + j2.73

0.344 + j2.235

0.518 + j1.713



Simplified Generator Sequence Networks

r-- - - - - - - - - - - - - - - r - - - - - - - - - - - - - - -s- - - - - - - - - - - - - - - - - - - I I

I Positive I Negative I Zero I I

I I I ! I

Copyright B SEL 2010 @I

The three sequence networks of the simplified generator are shown in the figure. Note the following:

1. There is no magnetic coupling.

2. There is a source in the positive-sequence network. This source is the original ("natural") positive-sequence source present in the three-phase model.

3. There is no source in the negative- nor the zero-sequence networks, as expected, because a generator is supposed to produce only positive-sequence voltages.

4. The neutral grounding impedance appears in the zero-sequence network. This is due to the fact that, unlike the case of positive- and negative-sequence tests, for the zero-sequence test, there is a current flowing through this impedance.



Line Fault Analysis and Calculation Using Symmetrical Components



Equivalent System Equivalent

Source Bus S Equivalent

Bus R Source

----------------------------------------------------------------.------------.**.-------------.-.. N

J Line Under Study rn

Collyripht@SEL Z O l O

The diagram above shows the traditional two-source equivalent. This is used to explain most of the effects that the whole system's characteristics have on a particular line fault.

Strictly speaking, the equivalent must have two voltage sources (S = Sending, R = Receiving) with their respective impedances, and an additional impedance in parallel with the line under study.

In many practical cases, this impedance has a magnitude several times larger than the line's impedance magnitude, making its effect negligible. When there are additional links between the systems at both line ends, this additional impedance must be considered or, alternatively, multi-bus matrix-based methods have to be used to solve for the faults of interest.

All the equivalent impedances of the system around the line can be calculated by using either short-circuit information at both buses or the original bus impedance matrices of the system.



Pre-Fault (Normal) Condition

Bus S Bus

Sending end is exporting 210 MVA at PF = 0.8 with rated voltage at Bus S

S = 210 MVA

No Fault

In order to make the two-source equivalent as valid as possible, the voltage sources must be set to match the load flow conditions existing on the 138 kV line before the fault is simulated.

This is the equivalent system's normal operation condition. For this particular case, it is known from a load flow study that the sending end is exporting 2 10 MVA at power factor 0.8 and with 1.0 per-unit voltage (138 kV) at the sending bus. This information is enough to calculate the internal voltages of each source.



Three-Phase Fault Equivalent Equivalent

Source s Three-Phase - . Bus R source

A fault analysis begins with a three-phase, perfectly balanced fault. It is assumed that this fault occurs at some distance d from the sending end.

Our goal is to calculate the sequence and phase voltages measured by a relay located at substation S.

Note:

Most authors and protection engineers use the per-unit distance m (in per unit of the line's length, L) instead of the distance d i n miles or kilometers. The per-unit distance is defmed as:

where d = distance to the fault in miles

L = line's length in miles

Thus, the fault occurs at a distance m (in per-unit) or 100 . m (in percent) of the line's length.



Numerical Example

Calculate the currents and voltages at Substation S for a three-phase fault located 10 miles from this substation

Use a fault resistance of 2 ohms

Three-Phase Bus

rn Copyright O SEL 2010

These are the results for the currents and voltages for a fault located 10 miles fiom the sending end (m = 0.33). Note that the voltages and currents are all balanced as expected.

The voltage drops to 18% and the current reaches twice the value during normal load. These effects (voltage sag and current increase) would be larger if no fault resistance is considered).

Complete all the calculations of these voltages and currents.



Phase-to-Phase Fault Equivalent

m Copyright OSEL 2010

The second fault we will analyze is a short circuit between two phases. Again, the distance to the fault is considered to be a certain unit value m.



B-C Fault Numerical Example Calculate the currents and voltages at Substation S for a B-C fault located 10 miles from this substation


B-C Bus Fault 1

Bus R 1 n

i. The slide shows the numerical results obtained for the phase voltages and currents for a B-C fault at m = 0.33. Note the following:

a. Voltages B and C (faulted phases) have a sag of about 50%.

b. Phases B and C current magnitudes increase considerably.

c. Phase A current and voltage remain similar to the normal load conditions.




Single-Phase-to-Ground Fault Equivalent

Source s A-G r-. .IL

Equivalent R Source

The third fault we will analyze is the single-phase-to-ground short circuit. This is the most common fault in transmission lines.

Traditionally, an A-phase-to-ground fault at some distance m is considered. The fault resistance is considered to have some non-zero value.



A-to-Ground Fault Numerical

Calculate the currents and voltages at substation S for a A- phase-to-ground fault located 10 miles from this substation


Bus S A-G - .. Bus R

The result for the A-to-ground fault is shown. The voltage magnitude in the faulted phase drops to 20%, while the voltage magnitudes in the sound phases increase. The current in the faulted phase increases, while the current in the sound phases remain similar to the pre- fault current.

Complete all the calculations for these voltages and currents.



Two-Phase-to-Ground Fault Equivalent Equivalent

Source s l5-L-b R Source

- A two-phase-to-ground fault can be the result of an evolving fault. This means that what originally is as a B-to-ground fault, evolves to a B-C-G fault when a C-to-ground fault occurs following the original one.

This is likely to occur during multiple insulator flashovers.

In this case, this fault looks like a B-C-G short circuit with impedances among phase B, phase C and ground. For the sake of simplicity, the fault will be represented as shown in the figure, with a fault resistance existing only in the ground return path. The study of this situation does not jeopardize the analysis and conclusions about this type of fault. Students can find this case in the Additional Resources section at the end of this manual.



Numerical Example B-C-G Fault Calculate the currents and voltages at s~~bstation S for a B-C-to-ground fault located 10 miles from this substation


The B-C-to-ground fault at 10 miles from the sending end apparently produces similar results to the B-C faults.




One Phase Open

Only One Pole Open

In transmission systems, two events can occur that demonstrate the effect of having one phase open:

When a double dead-end tower jumper is lost.

During a single-pole-open operation.

The figure shows the latter situation.



Numerical Example

Calculate the currents and voltages at substation S for a single-pole-open situation at Substation S circuit breaker

One Phase Open Bus S / Bus R

As a result of having one pole open, notice how the bus-side voltage on the affected phase increases to a value even larger than the internal voltage of the equivalent source.




One Pole Open Results One Pole Open Phase Voltages (pu)

4 1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

One Pole Open Phase Currents (A)

300 .1

One Pole Open Phase Currents (A)

The phasor diagrams show a remarkable presence of negative-sequence current.

The zero-sequence current could be even lower if the system is not solidly grounded.



Sequence Network Connections

Copyright@ SEL 2010 a

The sequence networks are connected as shown in the diagram. The sequence currents and voltages are calculated with the expression shown below. First, define the auxiliary impedances Z,,, Zk2 and ZkO:



B-C-Ground Fault Results B-C-to-Ground Fault Sequence Voltages (pu) 6-C-to-Ground Fault Sequence Currents (A)

&C-to-Ground Fault Phase Voltages [DU) B-C-to-Ground Fault Phase Currents (A)

These phasor diagrams show the similitude with the phase-to-phase fault. The difference is more evident when the sequence phasors are analyzed.

The presence of zero-sequence voltage and current changes the general pattern of the B-C fault.




The sequence networks for a B-C-G fault must be connected as shown in the diagram. Similar network techniques as previously shown are used to determine the currents and voltages in this case. The following expressions are obtained.

7 , = - (2, + 3 ~ ~ ) I (T + To + 3R,) - - - IF0 = - 4 1 - IF2 - - - - Il = IPF + c1 . IF1; - - - I, = c, . I,, ; - - - I, =co .I,,;

I, = f b + T + & 2- - z = & + a I ,+UI,

- - - I , = I, + aI, +a2<



A-Ground Fault Results A-G Fault Sequence Voltages (pu)

-0.34.2-0.1 0 0.1 0.2 0.3 0.4 0.5 0.6

A-G Fault Phase Voltages (pu)

-1 -0.8 4 . 6 -0.4 4 . 2 0 0.2 0 4 0.6 0.8

A-G Fault Sequence Currents (A)

100

A-G Fault Phase Currents (A)

Copyright O SEL

The phasor diagrams clearly demonstrate that the zero-sequence quantities are present in a similar amount as the negative-sequence quantities. Moreover, the phase angle between voltages and currents of negative- and zero-sequence are very similar.

Notice that the positive-sequence voltage leads the positive-sequence current by an angle less than 90°, while the negative-sequence current leads the negative-sequence voltage by an angle very close to 90".

In this system, the negative-sequence source impedance has a resistive part equal to zero, which means Z, =: -V2 / I2 must have an angle close to 90". This is equivalent to the angle of +V2 / I2 being close to 270" for faults on the line. For a fault right behind the relay's bus, the current phasor direction reverses, and the angle is close to 90" (270" - 180"). This forms the basis for designing ground directional elements.




For a single-phase-to-ground fault, the sequence networks are connected in series, as shown.

The Thevenin voltage will be in series with the three sequence equivalent impedances (Z,, Z,, and Z,), and three times the fault resistance.

This network connection is algebraically described by the following formulas and expressions: -

v

Transmission Line Protection - PROT 407 - - - - - - - - - -


B-C Fault Results &.C Fault Sequence Voltages (pu)

-0.4 4 2 0 0.2 0.4 0.6 0.8

B-C Fault Sequence Currents (A) 6001-. . . . . . I

I . ; . . ... -200 0 200 400 600 8W 1000 1200

B-C Fault Phase Currents (A)

Copyright Q SEL

The observations made from the previous example are evident in this graphic representation of the currents.

The phasor diagrams not only show the magnitudes but also the relative angles of voltages and currents during the fault.



Sequence Network Connections f P re------ I

Copyright 0 SEL 2010 #

According to the symmetrical components method, for this type of fault the zero-sequence currents and voltages must be zero. Therefore, the zero-sequence network is not considered.

The Thevenin voltage, Erk and the positive-sequence Thevenin impedance Z,, are calculated in the same way as for the three-phase fault. Here we use the negative-sequence values of the original impedances. All these derivations lead to the results in the following expressions: -

c7



Results Three-Phase Fault Sequence Voltages (pu)

Three-Phase Fault Phase Voltages (pu)

--------------

Three-Phase Fault Sequence Currents (A) - Three-Phase Fault Phase Currents (A)

The diagrams show the graphical representation of the voltage and currents at the sending end.

Observe that only positive-sequence voltage and positive-sequence current are present. Therefore, the phase voltages and currents are perfectly balanced.

Transmission Line Protection - PROT 407 -- - - --

Section I - Introduction to Transmission Line Faults

Sequence Network Connections r - - - - - - - E r-------I

C F 4 + Positive Negative

4 b

Zero

As required by the symmetrical component method, only the positive-sequence network is considered for a perfectly balanced fault.

There are many ways to calculate the currents and voltages in these networks. A Thkvenin equivalent is used to calculate current IF,. The Thkenin voltage source Eth is in series with the positive-sequence Thevenin equivalent impedance 2,. The following derivation serves to calculate the positive-sequence and phase currents at terminal S.

-

Notice the importance of the positive-sequence current divider factor C,.

The positive-sequence voltage V, can be calculated following the familiar circuit laws.

Transmission Line Protection - PROT 407 - - - - - - - -


Pre-Fault Condition

Bus S Bus R

No Fault

138 kV System

Es = 1.494157.79" PU

E, = 0.71610" pu

q,, = l.OO(37.04' pu

I,, = 878.5810.17" A

6, = 57.79"

These are the results for the pre-fault condition. The expressions used to determine these values are:

I,, = 2.1 pu



Example Equivalent System Equivalent

Source Bus S

Equivalent R Source

Example: 138 kV line, 30 miles

Line impedances: z,,,qO in ohms/mile or ohmikrn

Z,, , q0 in ohms

Bus S equivalent: Zs,,Zs2, Go in ohms

Bus R equivalent: ZR,,ZR2, q, in ohms

This is an example system for a 138 kV line.

Line and equivalent system data:

- z,, = 0.1 2 + J0.82 Q /mile - z,, = 0.65 + j2.47 S Z / mile



Transmission Line Fault 13.8 kV

138 kV 1 Bus S I

Analyzing line faults requires using the symmetrical components method.

For this purpose, the system external to the faulted line will be represented by a two-source equivalent.



Sequence Networks of a Simple Power System

Equivalent S R Equivalent

ZLI

Positive L ~ 2

Negative

Copyright OSEL 2010

The example shows the classical two-source equivalent over an isolated two-terminal line. Both sources are assumed to have solidly grounded neutrals.



Simplified Generator Model Bus

t t t

The figure shows the representation of a simplified generator, which is widely used, not only to model real generators, but also as an equivalent source of interconnected power systems.

The procedure described before can be applied to find the three sequence networks for this equipment.



Practical Definitions and Concepts Used in Line Protection



Importance of the Equivalent Source Strength

Source strength depends upon relative impedance magnitude:

- Source S is relatively weak if

I ZSl 1>>1 Z,, I = I T l<<l T I - Source S is relatively strong if

Izsl I<<lzRI I = I & l>>ITl

As mentioned previously, when the lines belong to a non-radial system, the two-source equivalent can be used to draw some conclusions.

When one of the equivalent sources has a relatively small equivalent impedance, its current contribution to the fault is large. Moreover, the voltage drop at the substation bus is pronounced. This source is called a "strong source."

Conversely, if the equivalent impedance is large, the contribution will be small and the voltage variation during a fault could be undetectable in some critical cases. In this case, the source is a "weak source."

In most cases, the relative value of one source with respect to the other is a useful parameter, because some types of protection elements respond better when the sources at both sides have similar relative magnitudes.



Effect of the Pre-Fault Load Flow (A-G Fault)

210 MVA Power Flow A-G Fault Sequence Currents (A)

100

Zero Power Flow A-G Fault Sequence Currents (A)

-I -ira -2 Y) W

-250 I - I

-1% -100 4 0 0 so loo 1% 200

A-G Fault Phase Currents (A)

600

4W

200

Copyright D SEL

Results are shown for a single-phase-to-ground fault in the example system.

The quantities at the left show the sequence and phase currents previously calculated with a pre-fault condition with 210 W A . The phasor diagrams at the right show the results assuming there is no power flow through the line before the fault occurs.

As expected, the relative magnitude of the positive-sequence current is larger for the case with large pre-fault power. A similar effect takes place with the positive-sequence voltage. This results in different magnitudes and angles for the phase quantities.

This effect must be considered when designing and setting protective relays and schemes.

Transmission Line Protection - PROT 407 - -


Symmetrical Line Impedance Matrix

These equations for the symmetrical line were described previously. They can be also written in matrix form.

For the symmetrical line, all the elements on the diagonal are equal. The off-diagonal elements are also equal.

- --



Non-Symmetrical Line Impedance Matrix

[ ~ ~ h ] = ['ph] ' ['ph] + [';A]

These are the line equations when asymmetry is considered.

Note that all the elements of the matrix can be different. This asymmetry changes the behavior of the line currents when compared with the symmetric case.



Typical 400 kV Transmission Line Tower Configuration

1-1 3.3 meters -+-+ 3.58 meters

8.53 meters

C-phase I 20 meters

Copyright aSEL 2010 M

In this example, a typical tower configuration for a 400 kV transmission line consists of the following:

Horizontal configuration

Two bundled conductors

Two ground wires

The modeled line length is 1 10 Ism.



Sequence Currents for a Three-Phase Fault at Line End

Equivalent Source s Bus R

Balanced I,, = 74.211-19.14" A Three-Phase -

I,, = 5484.741-85.840 A Fault

I,, = 486.26136.82" A

Copyright O SEL ZOlO

Assuming zero source impedance and a radial system, these are the sequence currents for a three-phase fault at the end of the non-symmetrical line.

As shown, there is enough negative-sequence current to possibly deceive a device whose operation is based on negative-sequence measurement.

In this case, the ratio of negative-sequence current to positive-sequence current is slightly less than 10%. For phase-to-phase and phase-to-ground faults, the same ratio approaches 100%.

As you will see later, this ratio can be used to supervise a negative-sequence directional element to prevent it from operating for three-phase faults.



Sequence Mutual Coupling Creates Unbalanced Voltages From Balanced

Currents and Vice-Versa

Factors used to evaluate current deviation:

m Copyright OSEL 2040

The two factors (a, and q) defined in the equations serve to give an idea of current deviation with respect to the perfectly balanced condition.

The larger a, is, the larger the contents of negative-sequence current will be. The larger a, is, the larger the content of zero-sequence currents.



Tower Configuration and Phasing Affect a, and a,

Line 1 Line 2 i Line 1 Line 2

a, = 0.0477 a, = 0.0489 j a, = 0.1021 a, = 0.1023

a, = 0.0828 a, = 0.0799 j a, = 0.0123 a, = 0.0159

ABC ABC Phasing ABC CAB Phasing m

Copyright OSEL 2010

L-- Next, you will study a sampling of horizontal tower configurations. Which phasing is best for protection is not as clear cut as that for the vertical tower configuration.

Here, the circuit on the left has the smaller a, factors, but the higher q, factors.



Fault analysis techniques

Symmetrical components method

Influence of the pre-fault condi,l:ion and the fault resistance

Influence of the source strength

The SIR ratio

Influence of line asymmetry

Large systems and computer programs

Copyright o S E L 2010 fl

Summary

Engineering techniques can be used to calculate and analyze the behavior of voltages and currents during line faults. With these analyses, a series of conclusions and d e f ~ t i o n s can be established to use in detecting faults and minimizing damage.

The Symmetrical Components Method serves to simplify the calculations and provides an analysis tool used to improve protection methods.

The pre-fault condition and the fault resistance magnitude are important factors that affect voltages and currents during faults.

The source strength and the source impedance ratio (SIR) are useful indicators to predict system behavior during line faults.

Line asymmetry must be considered in some fault analysis cases.

Computer programs are used to calculate faults in large systems and complex situations.



Asymmetrical Lines Create Coupling Between the Sequence Networks

Another Example

Exarr~ple Phase Exarr~ple Sequence Impedance Matrix: Impedance Matrix:

To illustrate this coupling, the example shows a phase-impedance matrix for a non- transposed line. This top row is Zaa, the self-impedance, Zab and Z,,, the mutual impedances. Notice that the mutual impedances are not equal.

Looking to the right, we see the sequence impedance matrix where the top row corresponds to the zero-sequence, the second row to the positive-sequence, and the last row corresponds to the negative-sequence. Notice that, unlike the case of a symmetrical line, the off-diagonal terms of the sequence matrix are not zero.

Transmission Line Protection - PROT 407 - - -


Transmission Line lmpedances

c Z,ZB,Zc-- Self lmpedances

Z,,,Z,,:,Z,--- Mutual Impedances Copyrtght O SEL 2010 !@I

Untransposed transmission lines have different self-impedances and mutual impedances.

They are: 2, # 2, # Zc.

As a result, negative- and zero-sequence currents will be generated during a three-phase fault.



Sequence Networks of a Non-Symmetrical Line

If the equations of the sequence networks belonging to a non-symmetrical line are stated in matrix form, the off-diagonal elements are not zero.

This can be interpreted as mutual coupling among the three sequence networks.



Sequence Networks of a Symmetrical Line

If the equations of the sequence networks for a perfectly symmetrical line are stated in matrix form, the off-diagonal elements result in "zero." This is because there is no mutual coupling among the three sequence networks.



Effect of Line Asymmetry



Importance of the Pre-Fault Condition Review of the Superposition Principle 1. Faulted System 2. Prefault System 3. Superimposed System

The diagram shows how a faulted system (1) can be represented in two systems:

a. System 2 represents the system during normal load conditions @re-fault), and

b. System 3 voltages and currents represent the increment (or decrement) required for the pre-fault quantities to reach the fault quantities.

Thus, the faulted system voltages and currents can be calculated by adding the superimposed. (or pure-fault) values to the pre-fault values. Therefore, the fault values are affected by the pre-fault condition.

The effect of the pre-fault condition is very evident for a perfectly balanced three-phase fault. For unbalances, the pre-fault condition affects only the positive-sequence quantities during the fault (provided that the pre-fault system is balanced).

Note: In the figure, E,, is the pre-fault voltage at the fault point.



Source Impedance Ratio (SIR)

Equivalent Source s Bus R

Radial t ine

ZL 1

I %I I SIR = ----r r'

I zf-1 I Fault

The larger the SIR is, the smaller the current and voltage magnitudes at the sending end bus

A large SIR indicates a weak source

There is a well-known apparent incongruity in protection. On one hand, we do not want currents to increase and voltages to decrease too drastically, to minimize damage to the equipment and the system. On the other hand, protective relays work better, in a more secure way, when the changes in the currents and voltage are more severe, and therefore more dangerous.

For example, an overcurrent relay will do its job better if the short circuit current magnitude is considerably larger than the load current. In one way or another, this is true in all types of relays.

Rule of Protection engineers should know how large those changes can be in order to predict the protective devices' behavior or, better yet, to properly design and set the relays. One of the parameters used for this purpose is the source impedance ratio (SIR = Z, I Z,), defined for radial lines. The effect of this number can be seen by calculating the fault current at some point on the line located at a per-unit distance, m, from the substation location.

Assuming a bolted three-phase fault and using the same notation as in the other cases, the A-phase current and the voltage seen by a device at the substation will be:

The dependence on the SIR is evident. The larger the SIR, the smaller the voltage and the current will be.


Section 1 -Introduction to Transmission Line Faults

Use of Computer Programs

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The use of computer programs is the most accurate way to determine the power system's behavior during faults. Modern fault programs include:

Simulation of any type of fault in multi-bus, multi-source systems. The maximum number of buses that can be simulated is, in some cases, very large. It is normal to see systems with thousands of buses modeled with these programs. The simulation can be done for different pre-fault load flow conditions.

Simulation of switching operations to determine the variation of the source impedances during fault clearing and reclosing sequences.

Simulation of relay operation. This is a particular feature of some programs that specialize in protection analysis.

The most sophisticated programs have complete models of the system that can include electromechanical (slow) and electromagnetic (fast) transients, as well as the effect of instrument transformers.

Transmission Line Protection - PROT 407 - - - - - - - - -


Tower Configuration and Phasing Affect a, and a, in Parallel Lines

TI = 0.31/86.49" SZ = 0.28)86.37"SZ

A * Line 1 A .

a,, = 0.0256 - 4 t A a, = 0.01 16 C.

ABC ABC Phasing ABC CBA Phasing

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The figure shows the power configuration and phasing affect the magnitude of % and ao. Certain tower configurations and phasings have lower % and cx, factors than others.

The EPRI 345 kV Transmission Line Reference Book also addresses unbalance due to line asymmetry. The example shows two phasing options for a vertical double-circuit tower.

The tower on the left has much higher % and q, factors than does the tower on the right for the same conductors and conductor spacing. The difference is the phasing. The tower on the left has an ABC CBA phasing counting counter clockwise, while that on the right has an ABC ABC phasing. It is interesting to note that the tower on the right has the lower positive-sequence impedance, which is also better for power flow.

01 introduction to ransmission line faults

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