Lecture_18 - Power System Analysis

8/17/2019 Lecture_18 - Power System Analysis

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EE 369

POWER SYSTEM ANALYSIS

Lecture 18

Fault Analysis

Tom Overbye and Ross Baldick

1


2/35

Announcements

Read Chapte !"

#ome$o% 1& 's 6"(3) 6"(*) 6"+9)6"61) 1&"19) 1&"&&) 1&"&,) 1&"&()

1&"&6) 1&"&*) 1&"&9- due Tuesda.No/" &+"

#ome$o% 13 's 1&"&1) 1&"&+)

1&"&!) !"1) !"3) !"() !"+) !"6) !"9)!"1&) !"16- due Thusda.)0eceme ("

&


3/35

Tansm'ss'on 2aut Ana.s's

The cause o4 eect'c po$e s.stem 4auts's 'nsuat'on ea%do$n5compom'se"

Th's ea%do$n can e due to a /a'et.

o4 d'eent 4actos78 L'htn'n 'on':'n a')

8 W'es o$'n toethe 'n the $'nd)

8 An'mas o pants com'n 'n contact $'th the$'es)

8 Sat spa. o pout'on on 'nsuatos"

3


4/35

Tansm'ss'on 2aut T.pes

Thee ae t$o ma'n t.pes o4 4auts78 s.mmet'c 4auts7 s.stem ema'ns aanced-

these 4auts ae eat'/e. ae) ut ae theeas'est to ana.:e so $e; cons'de them 0L?4auts) and aanced thee phase 4auts"

(


5/35

L'htn'n St'%e E/entSeBuence

1" L'ht'n h'ts 'ne) sett'n up an 'on':ed path to

ound 3, m''on 'htn'n st'%es pe .ea 'n SD

a s'ne t.p'ca sto%e m'ht ha/e &+),,, amps) $'th a'se t'me o4 1, µs) d'ss'pated 'n &,, µs"

mut'pe sto%es can occu 'n a s'ne ash) caus'n the'htn'n to appea to 'c%e) $'th the tota e/entast'n up to a second"

&" Conduct'on path 's ma'nta'ned . 'on':ed a' a4te'htn'n sto%e ene. has d'ss'pated) esut'n 'nh'h 4aut cuents >o4ten F &+),,, ampsD?

+


6/35

)cont;d

3" W'th'n one to t$o c.ces >16 ms? ea.s at othends o4 'ne detect h'h cuents) s'na'nc'cu't ea%es to open the 'ne7 nea. ocat'ons see deceased /otaes

(" C'cu't ea%es open to de@ene':e 'ne 'n an

add't'ona one to t$o c.ces7 ea%'n tens o4 thousands o4 amps o4 4aut cuent 's

no sma 4eatD

$'th 'ne emo/ed /otaes usua. etun to neanoma"

+" C'cu't ea%es ma. ecose a4te se/easeconds) t.'n to estoe 4auted 'ne to se/'ce"

6


7/35

2aut Ana.s's

2aut cuents cause eBu'pment damae dueto oth thema and mechan'ca pocesses"

oa o4 4aut ana.s's 's to detem'ne the

man'tudes o4 the cuents pesent du'nthe 4aut7

8 need to detem'ne the maG'mum cuent toensue de/'ces can su/'/e the 4aut)

8 need to detem'ne the maG'mum cuent thec'cu't ea%es >CHs? need to 'nteupt tocoect. s':e the CHs"

!


8/35

RL C'cu't Ana.s's

To undestand 4aut ana.s's $e need toe/'e$ the eha/'o o4 an RL c'cu't

( )

2 cos( )

v t

V t ω α

=

+

>Note teGt uses s'nuso'da /otae 'nstead oHe4oe the s$'tch 's cosed) i>t ? ,"When the s$'tch 's cosed at t , the cuenha/e t$o components7 1? a stead.@state /a

&? a tans'ent /aue"

R L

*


9/35

RL C'cu't Ana.s's) 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

V I

Z

Z R L R X

L R

ω

θ θ ω −

=

= + = +

− =

ac

stantaneous current is:

2 cos( )( ) Z

V t i t

Z

ω α θ + −=

9


10/35


1

1

ac dc 1

1

2. !ponentially decaying dc current component

( )

where is the time constant,"he #alue of is determined from the initial

conditions:

2($) $ ( ) ( ) cos( )

2

t T

dc

t T Z

i t C e

LT T R

C

V i i t i t t C e Z

V C

Z

ω α θ

−

−

=

=

= = + = + − +

= − cos( ) which depends on Z α θ α −1,


11/35

T'me /a.'n cuent

i>t ?

t'me

Supepos't'on o4 stead.@state component andeGponent'a. deca.'n dc oset"

11


12/35


dc

%ence ( ) is a sinusoidal superimposed on a decaying

dc current. "he magnitude of ($) depends on when

the switch is closed. &or fault analysis we're ust

concerned with the worst case.

%ighest C c

i t

i

* 1

2urrent occurs for: + ,

( ) ( ) ( )

2 2( ) cos( )

2

( cos( ) )

ac dc

t T

t T

V C

Z

i t i t i t

V V i t t e

Z Z

V

t e Z

α θ π

ω

ω

−

−

− =

= +

= − +

= − + 1&


13/35

RMS 4o 2aut Cuent

"he interrupting capaility of a circuit reaer is

specified in terms of the S current it can interrupt.

2

"he function ( ) ( cos( ) ) is

not periodic, so we can't formally define an S #alue

t T

V

i t t e Z ω

−

= − +.

%owe#er, if then we can appro!imate the current

as a sinusoid plus a time-in#arying dc offset."he S #alue of such a current is e0ual to the

s0uare root of the sum of the s0uares of the

indi#i

T t J

dual S #alues of the two current components.13


14/35

RMS 4o 2aut Cuent

2 2S

22 2

,

2 where , 2 ,

2

"his function has a ma!imum #alue of .

"herefore the worst case effect of the dc

component is included simply y

mu

ac dc

t t T T

ac dc ac

t T

ac ac

ac

I I

V V I I e I e

Z Z

I I e

I

− −

−

= +

= = =

= +

ltiplying the ac fault currents y .1(

t M d ' 0 '


15/35

eneato Mode'n 0u'n2auts0u'n a 4aut the on. de/'ces that can cont'ute 4aut

cuent ae those $'th ene. stoae"

Thus the modes o4 eneatos >and othe otat'n mach'nes?

ae /e. 'mpotant s'nce the. cont'ute the u% o4 the 4autcuent"

eneatos can e appoG'mated as a constant /otae

eh'nd a t'me@/a.'n eactance7

'

a E

1+


16/35

eneato Mode'n) cont;d

3

d

'd

d

"he time #arying reactance is typically appro!imatedusing three different #alues, each #alid for a different

time period:

4 direct-a!is sutransient reactance

4 direct-a!is transient reactance

4 dire

==

= ct-a!is synchronous reactance

Can then estimate currents using circuit theory:

&or e!ample, could calculate steady-state current

that would occur after a three-phase short-circuit

if no circuit reaers interrupt current. 16


17/35

eneato Mode'n) cont;d

'

3

''ac

3 '

3

&or a alanced three-phase fault on the generator terminal the ac fault current is (see page 52)

1 1 1

( ) 2 sin( )

1 1

where

direct-a!is su

d

d

t

T

d d d a t

T

d d

d

e X X X

i t E t

e X X

T

ω α

−

−

+ − +

÷ = + − ÷

='

transient time constant ( $.$6sec)

direct-a!is transient time constant ( 1sec)d T

≈

= ≈ 1!


18/35

eneato Mode'n) contKd

'

3

''

ac

3 '

'

C 3

"he phasor current is then

1 1 1

1 1

"he ma!imum C offset is

2( )

where is the armature time constant ( $.2 seconds)

d

d

A

t

T

d d d

a t T

d d

t T a

d

A

e X X X

I E

e X X

E I t e

X

T

−

−

−

+ − + ÷

= − ÷

=

≈ 1*


19/35

eneato Shot C'cu'tCuents

19


20/35

eneato Shot C'cu'tCuents

&,


21/35

eneato Shot C'cu'tEGampe

A +,, MA) &, %) 3φ 's opeated $'th an'ntena /otae o4 1",+ pu" Assume a so'd 3φ 4aut occus on the eneatoKs tem'na andthat the c'cu't ea%e opeates a4te thee

c.ces" 0etem'ne the 4aut cuent" Assume

3 '

3 '

$.16, $.27, 1.1 (all per unit)

$.$6 seconds, 2.$ seconds

$.2 seconds

d d d

d d

A

X X X

T T

T

= = =

= =

=&1


22/35

eneato S"C" EGampe)contKd

2.$

ac

$.$6

ac

5

ase ac

$.2C

Sustituting in the #alues1 1 1

1.1 $.27 1.1( ) 1.$6

1 1$.16 $.27

1.$6($) 8 p.u.$.16

6$$ 1$ 17,7 9 ($) 1$1,$$$ 9 2$ 1$

($) 1$1 9 2 17

t

t

t

e

I t

e

I

I I

I e

−

−

+ − + ÷ =

− ÷

= =

×= = =×

= × = S9 ($) 186 9 I =&&


23/35

eneato S"C" EGampe)contKd

$.$62.$

ac $.$6$.$6

ac

$.$6

$.2C

S

#aluating at t + $.$6 seconds for reaer opening1 1 1

1.1 $.27 1.1($.$6) 1.$6

1 1$.16 $.27

($.$6) 8$. 9

($.$6) 17 9 111 9

($.$6

e

I

e

I

I e

I

−

−

−

+ − + ÷ =

− ÷ =

= × =2 2) 8$. 111 12 9= + =

&3

Net$o% 2aut Ana.s's


24/35

Net$o% 2aut Ana.s'sS'mp'nonsp'nn'n? oads ae 'noed

&(


25/35

Net$o% 2aut EGampe

2o the 4oo$'n net$o% assume a 4aut on ttem'na o4 the eneato- a data 's pe un'teGcept 4o the tansm'ss'on 'ne eactance

2

1;.6Con#ert to per unit: $.1 per unit

1

1$$

line X = =

eneato has 1",+tem'na /otae supp'es 1,, MA

$'th ,"9+ a p4

&+


26/35

Net$o% 2aut EGampe)contKd

2auted net$o% pe un't d'aam

<'

"o determine the fault current we need to first estimate

the internal #oltages for the generator and motor &or the generator 1.$6, 1.$ 1.2

1.$ 1.2$.;62 1.2 1.1$ 8.1

1.$6

T G

Gen a

V S

I E

= = ∠ °

∠ = = ∠ − ° = ∠ ° ÷&6


27/35

Net$o% 2aut EGampe)contKd

"he motor's terminal #oltage is then

1.$6 $ - ($.;$77 - $.2;8) $. 1.$$ 16.

"he motor's internal #oltage is

1.$$ 16. ($.;$77 - $.2;8) $.2

1.$$ 25.5

=e can then sol#e as a linear circuit:

1 f

j j

j j

I

∠ × = ∠ − °

∠ − ° − ×= ∠ − °

= .1$ 8.1 1.$$ 25.5$.16 $.6

8.6 2.; 2.$15 115.5 ;.$;

j j

j

∠ ° ∠ − °+

= ∠ − ° + ∠ − ° =

&!


28/35

2aut Ana.s's Sout'on Techn'Bues

C'cu't modes used du'n the 4aut ao$ the net$o%to e epesented as a 'nea c'cu't

Thee ae t$o ma'n methods 4o so/'n 4o 4autcuents7

1" 0'ect method7 se pe4aut cond't'ons to so/e 4o the'ntena mach'ne /otaes- then app. 4aut and so/e d'ect."

&" Supepos't'on7 2aut 's epesented . t$o oppos'n/otae souces- so/e s.stem . supepos't'on7

8


29/35

Supepos't'on Appoach

2auted Cond't'on

EGact EBu'/aent to 2auted Cond't'on2aut 's epesented. t$o eBua and

oppos'te /otaesouces) each $'tha man'tude eBuato the pe@4aut /otae

&9


30/35

Supepos't'on Appoach)cont;d

S'nce th's 's no$ a 'nea net$o%) the 4autedand cuents ae ust the sum o4 the pe@4autthe >1? component and the cond't'ons $'th /otae souce at the 4aut ocat'on the >&? co

Pe@4aut >1? component eBua to the pe@4autpo$e o$ sout'on

O/'ous the

pe@4autQ4aut cuent's :eoD

3,


31/35

Supepos't'on Appoach)cont;d

2aut >1? component due to a s'ne /otae sat the 4aut ocat'on) $'th a man'tude eBuaneat'/e o4 the pe@4aut /otae at the 4aut

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

(1) (2) (2)$

g g g m m m

f f f f

I I I I I I

I I I I

= + = +

= + = +31


32/35

T$o Hus Supepos't'onSout'on

f

(1) (1)

(2) f

(2) f

(2)

>efore the fault we had 1.$6 $ ,

$.;62 1.2 and $.;62 1.2

Sol#ing for the (2) networ we get

1.$6 $8

$.16 $.16

1.$6 $

2.1 $.6 $.6

8 2.1 ;.1

$.;62

g m

g

m

f

g

I I

I j

I j

I j j j

I

= ∠ °= ∠ − ° = − ∠ − °

∠ °= = = −

∠ °

= = = −= − − = −

= ∠ 1.2 8 8.6 2.; j− ° − = ∠ − °

Th's matches$hat $ecacuatedea'e

3&


33/35

EGtens'on to Lae S.stems

us

us

"he superposition approach can e easily e!tendedto larger systems. ?sing the we ha#e

&or the second (2) system there is only one #oltagesource so is all @eros e!cept at the fault loca

=Y

Y V I

I tion

$

$

f I

− =

I

M

M

#o$e/e to use th'sappoach $e need to


34/35

0etem'nat'on o4 2autCuent

us1

us us

(2)1

(2)11 1 2

(2)1 1

(2)

(1)f

efine the us impedance matri! as

$

"hen

$

&or a fault a us i we get -

bus

n

f

n nn n

n

ii f i

V

Z Z V

I

Z Z V

V

Z V V

−

−

=

− =

= − = −

Z

Z Y V Z I

M

L

M O M M

L

M

3(


35/35

0etem'nat'on o4 2autCuent

(1)

(1) (2) (1)

%ence

=here

dri#ing point impedance

( ) transfer point imepdance

Aoltages during the fault are also found y superposition

are prefault #alues

i f

ii

ii

ij

i i i i

V I

Z

Z

Z i j

V V V V

=

≠

= +

3+

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Lecture_18 - Power System Analysis