ECE 476 Power System Analysis

Lecture 21: Unsymmetrical Faults, System Protection

Prof. Tom Overbye

Dept. of Electrical and Computer Engineering

University of Illinois at Urbana-Champaign

overbye@illinois.edu

Announcements

• Read Chapters 10 and 11• Design project due date has been extended to Tuesday,

December 8– A useful paper associated with the design project is

T. J. Overbye, "Fostering intuitive minds for power system design," IEEE Power and Energy Magazine, pp. 42-49, July-August 2003.

– You can download it from campus computers at

http://ieeexplore.ieee.org/xpl/abstractAuthors.jsp?arnumber=1213526

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Unbalanced Fault Analysis

• The first step in the analysis of unbalanced faults is to assemble the three sequence networks. For example, for the earlier single generator, single motor example let’s develop the sequence networks

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Sequence Diagrams for Example

Positive Sequence Network

Negative Sequence Network

4

Sequence Diagrams for Example

Zero Sequence Network

5

There is no zero sequenceconnection to the machinesbecause of the D-Y transformers

Create Thevenin Equivalents

• To do further analysis we first need to calculate the thevenin equivalents as seen from the fault location. In this example the fault is at the terminal of the right machine so the thevenin equivalents are:

0.2 in parallel with j0.455

0.21 in parallel with j0.475

th

th

Z j

Z j

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Single Line-to-Ground (SLG) Faults

• Unbalanced faults unbalance the network, but only at the fault location. This causes a coupling of the sequence networks. How the sequence networks are coupled depends upon the fault type. We’ll derive these relationships for several common faults.

• With a SLG fault only one phase has non-zero fault current -- we’ll assume it is phase A.

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SLG Faults, cont’d

?

0

0

fa

fb

fc

I

I

I

0

2 0

2

Then since

1 1 1 ?1 1

1 03 3

01

f

ff f f f a

f

I

I I I I I

I

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SLG Faults, cont’d

0

2

2

0

1 1 1

1

1

This means

The only way these two constraints can be satisified

is by coupling the sequence networks in series

f fa f a

ffa

ffb

fc f

fa f f f

V Z I

VV

V V

V V

V V V V

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SLG Faults, cont’d

With the sequence networks inseries we cansolve for the fault currents(assume Zf=0)

01.05 01.964

(0.1389 0.1456 0.25 3 )

5.8 (of course, 0)

f f ff

ff fs a cb

I j I Ij Z

I j I I

I A I

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Line-to-Line (LL) Faults

• The second most common fault is line-to-line, which occurs when two of the conductors come in contact with each other. With out loss of generality we'll assume phases b and c.

0

bg

Current Relationships: 0, , 0

Voltage Relationships: V

f f fa c fb

cg

I I I I

V

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LL Faults, cont'd

0

2

2

0

2 2

Using the current relationships we get

1 1 1 01

13

1

0

1 1

3 3

Hence

f

ff b

fbf

f

f ff fb b

f f

I

I I

II

I

I I I I

I I

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LL Faults, con'td

0

2

2

2

2

Using the voltage relationships we get

1 1 11

13

1

Hence

1313

ff ag

ff bg

ff cg

f ff ag bg

f ff ag f fbg

V V

V V

V V

V V V

V V V V V

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Recall duringthis fault theb and c phasevoltagesare identical

LL Faults, cont'd

To satisfy &

the positive and negative sequence networks must

be connected in parallel

f f f fI I V V

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There is nozero sequencecurrent becausethere is nopath to ground

LL Faults, cont'd

2

2

Solving the network for the currents we get

1.05 03.691 90

0.1389 0.1456

1 1 1 0 0

1 3.691 90 6.39

3.691 90 6.391

f

fa

fb

fc

Ij j

I

I

I

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LL Faults, cont'd

2

2

Solving the network for the voltages we get

1.05 0 0.1389 3.691 90 0.537 0

0.1452 3.691 90 0.537 0

1 1 1 0 1.074

1 0.537 0.537

0.537 0.5371

f

f

fa

fb

fc

V j

V j

V

V

V

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Double Line-to-Ground Faults

• With a double line-to-ground (DLG) fault two line conductors come in contact both with each other and ground. We'll assume these are phases b and c.

0 ( )f f f f fa cg f cbg bI V V Z I I

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DLG Faults, cont'd

0

2

2

0

From the current relationships we get

1 1 1

1

1

Since 0 0

Note, because of the path to ground the zero

sequence current is no longer

ffa

ffb

fc f

fa f f f

II

I I

I I

I I I I

zero.

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DLG Faults, cont'd

0

2

2

0 2

2 2

0

From the voltage relationships we get

1 1 11

13

1

Since

Then ( )

But since 1 0 1

ff ag

ff bg

ff bg

f fcg f fbg

ff fbg

ff fbg

V V

V V

V V

V V V V

V V V

V V V

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DLG Faults, cont'd

0

0 2

0 2

2

0 0

0 0

( )

Also, since

Adding these together (with -1)

(2 ) with

3

ff fbg

f ff cb

ff f fb

fc f f f

ff f f f f f fbg

f f f f

V V V

Z I I

I I I I

I I I I

V Z I I I I I I

V V Z I

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DLG Faults, cont'd

• The three sequence networks are joined as follows

0

1.05 00.1389 0.092( 3 )

4.547 0

ff

VI

j jZ Z Z Z

Assuming Zf=0, then

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DLG Faults, cont'd

0

1.05 4.547 90 0.1389 0.4184

0.4184/ 0.1456 2.874

4.547 2.874 1.673

Converting to phase: 1.04 6.82

1.04 6.82

f

f

f f f

fb

fc

V j

I j j

I I I j j j

I j

I j

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Unbalanced Fault Summary

• SLG: Sequence networks are connected in series, parallel to three times the fault impedance

• LL: Positive and negative sequence networks are connected in parallel; zero sequence network is not included since there is no path to ground

• DLG: Positive, negative and zero sequence networks are connected in parallel, with the zero sequence network including three times the fault impedance

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Power System Protection

• Main idea is to remove faults as quickly as possible while leaving as much of the system intact as possible

• Fault sequence of events– Fault occurs somewhere on the system, changing the

system currents and voltages – Current transformers (CTs) and potential transformers

(PTs) sensors detect the change in currents/voltages– Relays use sensor input to determine whether a fault has

occurred– If fault occurs relays open circuit breakers to isolate fault

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Power System Protection

• Protection systems must be designed with both primary protection and backup protection in case primary protection devices fail

• In designing power system protection systems there are two main types of systems that need to be considered:

1. Radial: there is a single source of power, so power always flows in a single direction; this is the easiest from a protection point of view

2. Network: power can flow in either direction: protection is much more involved

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Radial Power System Protection

• Radial systems are primarily used in the lower voltage distribution systems. Protection actions usually result in loss of customer load, but the outages are usually quite local.

The figure showspotential protectionschemes for a radial system. Thebottom scheme ispreferred since itresults in less lost load.

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Radial Power System Protection

• In radial power systems the amount of fault current is limited by the fault distance from the power source: faults further down the feeder have less fault current since the current is limited by feeder impedance

• Radial power system protection systems usually use inverse-time overcurrent relays.

• Coordination of relay current settings is needed toopen the correct breakers

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Inverse Time Overcurrent Relays

• Inverse time overcurrent relays respond instan-taneously to a current above their maximum setting

• They respond slower to currents below this value but above the pickup current value

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Inverse Time Relays, cont'd

• The inverse time characteristic provides backup protection since relays further upstream (closer to power source) should eventually trip if relays closer to the fault fail

• Challenge is to make sure the minimum pickup current is set low enough to pick up all likely faults, but high enough not to trip on load current

• When outaged feeders are returned to service there can be a large in-rush current as all the motors try to simultaneously start; this in-rush current may re-trip the feeder

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Inverse Time Overcurrent Relays

Relays havetraditionally beenelectromechanicaldevices, but aregradually beingreplaced by digital relays

Current and timesettings had been adjusted using dialson the relay

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Protection of Network Systems

• In a networked system there are a number of difference sources of power. Power flows are bidirectional

• Networked system offer greater reliability, since the failure of a single device does not result in a loss of load

• Networked systems are usually used with the transmission system, and are sometimes used with the distribution systems, particularly in urban areas

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Network System Protection

• Removing networked elements require the opening of circuit breakers at both ends of the device

• There are several common protection schemes; multiple overlapping schemes are usually used

1. Directional relays with communication between the device terminals

2. Impedance (distance) relays.

3. Differential protection

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Directional Relays

• Directional relays are commonly used to protect high voltage transmission lines

• Voltage and current measurements are used to determine direction of current flow (into or out of line)

• Relays on both ends of line communicate and will only trip the line if excessive current is flowing into the line from both ends– line carrier communication is popular in which a high

frequency signal (30 kHz to 300 kHz) is used– microwave communication is sometimes used; radio can also

be used– Security is a concern

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Line Carrier Communication, Line Traps

• Line traps are connection in series with lines to allow the power frequency (50 or 60 Hz) to pass with low impedance but to show high impedance at the line carrier frequency

34Image source: qualitypower.com/line-trap.html

Impedance Relays

• Impedance (distance) relays measure ratio of voltage to current to determine if a fault exists on a particular line

1 1

12 12

Assume Z is the line impedance and x is the

normalized fault location (x 0 at bus 1, x 1 at bus 2)

V VNormally is high; during fault

I IxZ

Impedance Relays Protection Zones

• To avoid inadvertent tripping for faults on other transmission lines, impedance relays usually have several zones of protection:– zone 1 may be 80% of line for a 3f fault; trip is

instantaneous– zone 2 may cover 120% of line but with a delay to prevent

tripping for faults on adjacent lines– zone 3 went further; most removed due to 8/14/03 events

• The key problem is that different fault types will present the relays with different apparent impedances; adequate protection for a 3f fault gives very limited protection for LL faults

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