EE 369POWER SYSTEM ANALYSIS

Lecture 18 Fault Analysis

Tom Overbye and Ross Baldick

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AnnouncementsRead Chapter 7. Homework 12 is 6.59, 6.61, 12.19, 12.22,

12.24, 12.26, 12.28, 7.1, 7.3, 7.4, 7.5, 7.6, 7.9, 7.12, 7.16; due Thursday, 12/3.

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Transmission Fault Analysis

The cause of electric power system faults is insulation breakdown/compromise.

This breakdown can be due to a variety of different factors:– Lightning ionizing air,– Wires blowing together in the wind,– Animals or plants coming in contact with the wires,– Salt spray or pollution on insulators.

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Transmission Fault TypesThere are two main types of faults:

– symmetric faults: system remains balanced; these faults are relatively rare, but are the easiest to analyze so we’ll consider them first.

– unsymmetric faults: system is no longer balanced; very common, but more difficult to analyze (considered in EE 368L).

The most common type of fault on a three phase system by far is the single line-to-ground (SLG), followed by the line-to-line faults (LL), double line-to-ground (DLG) faults, and balanced three phase faults.

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Lightning Strike Event Sequence1. Lighting hits line, setting up an ionized path to ground

30 million lightning strikes per year in US! a single typical stroke might have 25,000 amps, with a rise

time of 10 s, dissipated in 200 s. multiple strokes can occur in a single flash, causing the

lightning to appear to flicker, with the total event lasting up to a second.

2. Conduction path is maintained by ionized air after lightning stroke energy has dissipated, resulting in high fault currents (often > 25,000 amps!)

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Lightning Strike Sequence, cont’d3. Within one to two cycles (16 ms) relays at both ends of

line detect high currents, signaling circuit breakers to open the line: nearby locations see decreased voltages

4. Circuit breakers open to de-energize line in an additional one to two cycles: breaking tens of thousands of amps of fault current is no

small feat! with line removed voltages usually return to near normal.

5. Circuit breakers may reclose after several seconds, trying to restore faulted line to service.

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Fault AnalysisFault currents cause equipment damage due to

both thermal and mechanical processes.Goal of fault analysis is to determine the

magnitudes of the currents present during the fault:– need to determine the maximum current to ensure

devices can survive the fault,– need to determine the maximum current the circuit

breakers (CBs) need to interrupt to correctly size the CBs.

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RL Circuit AnalysisTo understand fault analysis we need to

review the behavior of an RL circuit

( )

2 cos( )

v t

V t

(Note text uses sinusoidal voltage instead of cos!)Before the switch is closed, i(t) = 0.When the switch is closed at t=0 the current willhave two components: 1) a steady-state value2) a transient value.

R L

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RL Circuit Analysis, cont’d

2 2 2 2

1

1. Steady-state current component (from standard phasor analysis)

Steady-state phasor current magnitude is ,

where ( )

and current phasor angle is , tan ( / )Corresponding in

ac

Z Z

VIZ

Z R L R X

L R

ac

stantaneous current is:

2 cos( )( ) ZV ti tZ

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RL Circuit Analysis, cont’d

1

1

ac dc 1

1

2. Exponentially decaying dc current component

( )

where is the time constant,

The value of is determined from the initialconditions:

2(0) 0 ( ) ( ) cos( )

2

tT

dc

tT

Z

i t C eLT T R

C

Vi i t i t t C eZ

VCZ

cos( ) which depends on Z 10

Time varying currenti(t)

time

Superposition of steady-state component andexponentially decaying dc offset.

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RL Circuit Analysis, cont’d

dc

Hence ( ) is a sinusoidal superimposed on a decayingdc current. The magnitude of (0) depends on whenthe switch is closed. For fault analysis we're just concerned with the worst case.

Highest DC c

i ti

Z 12urrent occurs for: = ,

( ) ( ) ( )

2 2( ) cos( )

2 ( cos( ) )

ac dc

tT

tT

VCZ

i t i t i t

V Vi t t eZ ZV t eZ

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RMS for Fault CurrentThe interrupting capability of a circuit breaker isspecified in terms of the RMS current it can interrupt.

2The function ( ) ( cos( ) ) is

not periodic, so we can't formally define an RMS value

tTVi t t e

Z

. However, if then we can approximate the current as a sinusoid plus a time-invarying dc offset. The RMS value of such a current is equal to thesquare root of the sum of the squares of theindivi

T t

dual RMS values of the two current components. 13

RMS for Fault Current2 2

RMS

22 2

I ,

2 where , 2 ,

2

This function has a maximum value of 3 .Therefore the worst case effect of the dc component is included simply by

mu

ac dc

t tT T

ac dc ac

tT

ac ac

ac

I I

V VI I e I eZ Z

I I e

I

ltiplying the ac fault currents by 3.14

Generator Modeling During FaultsDuring a fault the only devices that can contribute fault current

are those with energy storage.Thus the models of generators (and other rotating machines) are

very important since they contribute the bulk of the fault current.Generators can be approximated as a constant voltage behind a

time-varying reactance:

'aE

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Generator Modeling, cont’d

"d'd

d

The time varying reactance is typically approximatedusing three different values, each valid for a differenttime period:

X direct-axis subtransient reactance

X direct-axis transient reactanceX dire

ct-axis synchronous reactance

Can then estimate currents using circuit theory:For example, could calculate steady-state currentthat would occur after a three-phase short-circuitif no circuit breakers interrupt current. 16

Generator Modeling, cont’d

'

"

''

ac

" '

"

For a balanced three-phase fault on the generatorterminal the ac fault current is (see page 362)

1 1 1

( ) 2 sin( )1 1

where

direct-axis su

d

d

tT

d dda t

T

d d

d

eX XX

i t E t

eX X

T

'

btransient time constant ( 0.035sec)

direct-axis transient time constant ( 1sec)dT

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Generator Modeling, cont'd

'

"

''

ac

" '

'

DC "

The phasor current is then

1 1 1

1 1

The maximum DC offset is

2( )

where is the armature time constant ( 0.2 seconds)

d

d

A

tT

d dda t

T

d d

tTa

d

A

eX XX

I E

eX X

EI t eX

T

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Generator Short Circuit Currents

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Generator Short Circuit Currents

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Generator Short Circuit ExampleA 500 MVA, 20 kV, 3 is operated with an

internal voltage of 1.05 pu. Assume a solid 3 fault occurs on the generator's terminal and that the circuit breaker operates after three cycles. Determine the fault current. Assume

" '

" '

0.15, 0.24, 1.1 (all per unit)

0.035 seconds, 2.0 seconds0.2 seconds

d d d

d d

A

X X X

T TT

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Generator S.C. Example, cont'd

2.0

ac0.035

ac

6

base ac3

0.2DC

Substituting in the values1 1 1

1.1 0.24 1.1( ) 1.05

1 10.15 0.24

1.05(0) 7 p.u.0.15500 10 14,433 A (0) 101,000 A3 20 10

(0) 101 kA 2 143 k

t

t

t

eI t

e

I

I I

I e

RMSA (0) 175 kAI 22

Generator S.C. Example, cont'd

0.052.0

ac 0.050.035

ac0.05

0.2DC

RMS

Evaluating at t = 0.05 seconds for breaker opening1 1 1

1.1 0.24 1.1(0.05) 1.05

1 10.15 0.24

(0.05) 70.8 kA

(0.05) 143 kA 111 k A

(0.05

eI

e

I

I e

I

2 2) 70.8 111 132 kA

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Network Fault Analysis Simplifications To simplify analysis of fault currents in networks

we'll make several simplifications:1. Transmission lines are represented by their series

reactance2. Transformers are represented by their leakage reactances3. Synchronous machines are modeled as a constant voltage

behind direct-axis subtransient reactance4. Induction motors are ignored or treated as synchronous

machines5. Other (nonspinning) loads are ignored

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Network Fault ExampleFor the following network assume a fault on the terminal of the generator; all data is per unitexcept for the transmission line reactance

219.5Convert to per unit: 0.1 per unit

138100

lineX

generator has 1.05terminal voltage &supplies 100 MVAwith 0.95 lag pf

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Network Fault Example, cont'dFaulted network per unit diagram

*'

To determine the fault current we need to first estimatethe internal voltages for the generator and motorFor the generator 1.05, 1.0 18.2

1.0 18.2 0.952 18.2 1.103 7.11.05

T G

Gen a

V S

I E

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Network Fault Example, cont'dThe motor's terminal voltage is then 1.05 0 - (0.9044 - 0.2973) 0.3 1.00 15.8The motor's internal voltage is1.00 15.8 (0.9044 - 0.2973) 0.2

1.008 26.6We can then solve as a linear circuit:

1f

j j

j j

I

.103 7.1 1.008 26.60.15 0.5

7.353 82.9 2.016 116.6 9.09j j

j

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Fault Analysis Solution Techniques Circuit models used during the fault allow the network to be

represented as a linear circuit There are two main methods for solving for fault currents:

1. Direct method: Use prefault conditions to solve for the internal machine voltages; then apply fault and solve directly.

2. Superposition: Fault is represented by two opposing voltage sources; solve system by superposition:– first voltage just represents the prefault operating point– second system only has a single voltage source.

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Superposition ApproachFaulted Condition

Exact Equivalent to Faulted ConditionFault is representedby two equal andopposite voltage sources, each witha magnitude equalto the pre-fault voltage

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Superposition Approach, cont’dSince this is now a linear network, the faulted voltagesand currents are just the sum of the pre-fault conditions[the (1) component] and the conditions with just a singlevoltage source at the fault location [the (2) component]

Pre-fault (1) component equal to the pre-fault power flow solution

Obvious the pre-fault “fault current”is zero!

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Superposition Approach, cont’dFault (1) component due to a single voltage sourceat the fault location, with a magnitude equal to thenegative of the pre-fault voltage at the fault location.

(1) (2) (1) (2)

(1) (2) (2)0

gg g m m m

f f f f

I I I I I I

I I I I

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Two Bus Superposition Solutionf

(1) (1)

(2) f

(2) f

(2)

Before the fault we had E 1.05 0 ,

0.952 18.2 and 0.952 18.2

Solving for the (2) network we getE 1.05 0 7

j0.15 j0.15E 1.05 0 2.1j0.5 j0.5

7 2.1 9.1

0.952

g m

g

m

f

g

I I

I j

I j

I j j j

I

18.2 7 7.35 82.9j

This matcheswhat wecalculatedearlier

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Extension to Larger Systems

bus

bus

The superposition approach can be easily extendedto larger systems. Using the we have

For the second (2) system there is only one voltagesource so is all zeros except at the fault loca

Y

Y V I

I tion

0

0fI

I

However to use thisapproach we need tofirst determine If

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Determination of Fault Currentbus

1bus bus

(2)1(2)

11 1 2

(2)1 1

(2)

(1)f

Define the bus impedance matrix as

0Then

0

For a fault a bus i we get -I

bus

n

f

n nn n

n

ii f i

V

Z Z VI

Z Z V

V

Z V V

Z

Z Y V Z I

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Determination of Fault Current

(1)

(1) (2) (1)

Hence

Wheredriving point impedance

( ) transfer point imepdance

Voltages during the fault are also found by superposition

are prefault values

if

ii

ii

ij

i i i i

VIZ

Z

Z i j

V V V V

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