EEE470chap7 Lecture 9 Symmetrical Faults

8/4/2019 EEE470chap7 Lecture 9 Symmetrical Faults

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1

Symmetrical Faults

Chapter 7


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2

Faults

Shunt faults:

Three phaseabc

Line to line

Line to ground

2 Line to ground

ba

c

ab

c

ab

c


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3

Faults

Series faults

One open phase:a

bc

2 open phasesa

bc

Increased phase

impedance

Z a

b

c


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4

Why Study Faults?

Determine currents and voltages in thesystem under fault conditions

Use information to set protective devices

Determine withstand capability thatsystem equipment must have:

Insulating level

Fault current capability of circuit breakers: Maximum momentary current

Interrupting current


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Symmetrical Faults

t=0

2 V

i(t)

Fault at t = 0AC

R L

)sin(2)( tVte


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Symmetrical Faults

For a short circuit at generator terminals at t=0

and generator initially open circuited:

dt

di

LRite )(

dt

diLRitVSin )(2

by using Laplace transforms i(t) can be found

(L is considered constant)


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Symmetrical Faults

]/

)()([2

)(Tt

eSintSin

Z

Vti

2222)( XRLRZ

R

XTan

R

LTan

11

Where:

R

X

R

L

T Time Constant

]/

)()([2)(Tt

eSintSinac

Iti

Where: Iac = acRMS fault current at t=0 (Examples)

Note that for a 3-

phase system will be

different for each

phase. Therefore, DC

offset will be different

for each phase


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t = 0

acI2

iac

Idc = 0

]/

)()([2)(Tt

eSintSinac

Iti

o

90

V2 e(t)

o90


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]/

)()([2)(Tt

eSintSinac

Iti

0

o90

V2 e(t)

t = 0

iac02 acI

02 acIidc


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10

iac

02 acI

02 acI id

c

0

22ac

I

t

0

o90

]/

)()([2)(Tt

eSintSinac

Iti

)(ti


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Symmetrical Faults

Iac and Idc are independent after t = 0

22

dcI

acI

RMSI

Tt

eacoIdcI

2

Substituting:

Tte

acI

Tt

eac

Iac

IRMSI

221)222((max)


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Asymmetry Factor

IRMS(max) = K() Iac

Asymmetry Factor = K()

rx

eK

4

21)(

Where:

= number of cycles

(Example 7.1)


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Example 7.1

Fault at a time to produce maximum DC offset

Circuit Breaker opens 3 cycles after fault inception

IFault at t = 0AC

R = 0.8 XL = 8

V = 20 kVLN-

+

CB

Find:

1. Iac at t = 0

2. IRMS Momentary at = 0.5 cycles

3. IRMS Interrupting Current


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Example 7.1

a. RMSACkAI 488.2

88.0

20)0(

22

b.

438.121)5.0(

)10

5.(4

eKKAImomentary 577.3)488.2)(438.1(

c.023.121)3(

)10

3(4 eKKAI ngInterrupti 545.2)488.2)(023.1(


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AC DecrementIn the previous analysis we treated the

generator as a constant voltage behind aconstant impedance for each phase. The

constant inductance is valid for steady state

conditions but for transient conditions, thegenerator inductance is not constant.

Recall that for steady state conditions, the

equivalent machine reactance is made upof 2 parts: a) Armature leakage reactance

b) Armature reaction

(See Phasor Diagram)


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AC Decrement

Steadystate model of generatorXL is leakage reactance

XAR is a fictitious reactance and XAR>> XL

XARis due to flux linkages of armature current with the fieldcircuit. Flux linkages can not change instantaneously.Therefore, if the generator is initially unloaded when afaultoccurs the effective reactance is XL which is referred to asSubtransient Reactance, x.

AC

EI

R XL XAR

Load

I I


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17IL

jILXL

jILXAR(t)

EIField

Flux

Armature Reaction

Resultant

Field

ET

XL XAR

-

+

EI

I=IL

Load

Loaded Generator

I 0


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EField

Flux

Armature Reaction = 0

Resultant

Field

ET0

t = 0 -

XL XAR=0ET0

-

+

E = E = E = ET0

I=0

Unloaded Generator

I 0


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XL XAR

-

+

E = E = E = ET0

I=0

t=0

EField

Flux


Resultant

Field

ET0 = 0

Faulted Generator

I I


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XL XAR=0

-

+

E = E = E = ET0

I = I

E = jIXL

t=0+

Field

Flux

Resultant

Field

ET = 0

I


I I


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XL XAR

-

+

E = E = E = ET0

I = I

E = jI(XL + XAR)

t 3Cyc.

Field

Flux

Resultant

Field

ET = 0

I


I I


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XL XAR

-

+

E = E = E = ET0

I = I

E = jI(XL + XAR)

t =

Field

Flux

Resultant

Field

ET = 0

I



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AC Decrement

As fault current begins to flow, armature reaction willincrease with time thereby increasing the apparent

reactance. Therefore, the ac component of the fault

current will decrease with time to a steady state

condition as shown in the figure below.

"2I '2II2

"2I


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AC Decrement

For a round rotor machine we only need toconsider the direct axis reactance.

dX

EI

"

"2"2 Subtransient

dX

EI

'

'2'2

dX

EI 22

Transient

Synchronous(steadystate)


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AC Decrement

Can write the ac decrement equation )])'()'"(2)( '" tSinIeIIeIItaci dT

tdTt

For an unloaded generator

(special case):TEEEE '"

Td: Subtransient time constant

(function of amortisseur winding X/R)Td: Transient time constant

(function of field winding X/R)

Look at equation for t=0 and t=infinity


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AC Decrement

For t = 0

)])'()'"(2)( '"

tSinIeIIeIItaci dT

tdTt

For t =

IIiac 2]00[2(max)

"2])'()'"[(2(max) IIIIIIiac


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ac and dc Decrement

Transform ac decrement equation to phasor form

/')'(")'"(

_

IdT

t

eIIdT

t

eIIacI

dc decrement equation:

A

T

t

eSinIdc

I

)("2

Where TA = Armature circuit time constant

(Example 7.2)


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Network Equivalent circuit


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Lecture 9 Short circuit calculation

Electrical netw ork can be represented by a Thvenin equivalent circuit, voltage source and

series connected inductive reactance and a small resistance. The resistance is neglected

most of the time.

The voltage is the rated netw ork line to neutral voltage. The reactance is calculated f rom the

short circuit current. The pow er company calculates the short circuit current at every

substation.

Example: A 500 kV netw ork short circuit current is 30kA. Calculate the Thvenin equivalent

circuit and the short circuit current time function.

Netw ork Thvenin Equivalent circuit is:

AC VlnZnet


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Vnet 500kV Ishort 30kA 2 60 Hz

The network line to line voltage and the equivalent netw ork reactance and estimated small

resistance

Vln

Vnet

3288.675kV X

net

Vln

Ishort

9.623

Rnet

Xnet

20 Estimated value Lnet

Xnet

25.524mH

The supply voltage is:

Vsup t ( ) 2 Vln sin t ( )

Laplace transformation of the supply voltage is:

2 Vln cos ( ) s sin ( )( )

2

s2

2 Vln sin t ( ) has Laplace transform


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Network Equivalent circuitLaplace transformation of the supply voltage is:

2 Vln cos ( ) s sin ( )( )

2

s2

2 Vln sin t ( ) has Laplace transform

The system equation w hen the load current is neglected

2 Vln cos ( ) s sin ( )( )

2

s2

s Lnet Rnet inet

Short c ircuit current in s domain is: inet s ( )2 Vln cos ( ) s sin ( )( )

2

s2

s Lnet Rnet

Inverse Laplace transform is:

2 Vln cos ( )( )

2

s2

s Lnet Rnet has inverse Laplace transform

2 Vln cos ( ) Rnet sin t 2

Lnet e

Rnet t

Lnet

2

Lnet cos t 2

2

2 Lnet2 2 Rnet2


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Ish1 t ( )2 Vln cos ( ) Rnet sin t

2 Lnet e

R

net

t

Lnet

2

Lnet cos t 2

2

2

Lnet2

2

Rnet2

2 Vln s sin ( )( )

2 s2 s Lnet Rnet has inverse Laplace transform

2 Vln sin ( ) Rnet cos t 2

Rnet e

Rnet t

Lnet

Lnet sin t 2

2

Lnet

2

2 R

net

2

Ish2 t ( )2 Vln sin ( ) Rnet cos t

2 Rnet e

Rnet t

Lnet

Lnet sin t 2

2

Lnet2

2

Rnet2


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The short circuit current time function

IShort t ( ) Ish1 t ( ) Ish2 t ( )

DC component of the current is

IDC t ( )

2 Vln cos ( ) Lnet e

Rnet t

Lnet

Lnet2

2

Rnet2

2 Vln sin ( ) Rnet e

Rnet t

Lnet

Lnet2

2

Rnet2

IDC t ( )2 Vln cos ( ) Lnet e

Rnet t

Lnet

Lnet

2

2 Rnet

2

2 Vln sin ( ) Rnet e

Rnet t

Lnet

Lnet

2

2 Rnet

2

IDC 0s 0deg ( ) 42.321kA IDC 0s 90deg ( ) 2.116 kA


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t 0s 0.01ms 150ms Ishort_max 78.67kA IDC_max 42.321kA

0 50 100 15050

0

50

100

IShort t 0.0deg ( )

kA

IDC t 0deg( )

kA

t

ms

Current and DC component maximum at 0deg


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t 0s 0.01ms 150ms IShort_max 42.215kA

0 50 100 150

60

40

20

020

40

60

IShort t 90.0deg ( )

kA

t

ms

No DC component at 90deg

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EEE470chap7 Lecture 9 Symmetrical Faults