lecture_18 - power system analysis
TRANSCRIPT
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EE 369
POWER SYSTEM ANALYSIS
Lecture 18
Fault Analysis
Tom Overbye and Ross Baldick
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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 ("
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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"
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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"
(
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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) $'th the tota e/entast'n up to a second"
&" Conduct'on path 's ma'nta'ned . 'on':ed a' a4te'htn'n sto%e ene. has d'ss'pated) esut'n 'nh'h 4aut cuents >o4ten F &+),,, ampsD?
+
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)cont;d
3" W'th'n one to t$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"
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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"
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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$o components7 1? a stead.@state /a
&? a tans'ent /aue"
R L
*
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RL C'cu't Ana.s's) cont;d
2 2 2 2
1
1. Steady-state current component (from standard
phasor analysis)
Steady-state phasor current magnitude is ,
where ( )
and current phasor angle is , tan ( / )
Corresponding in
ac
Z Z
V I
Z
Z R L R X
L R
ω
θ θ ω −
=
= + = +
− =
ac
stantaneous current is:
2 cos( )( ) Z
V t i t
Z
ω α θ + −=
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RL C'cu't Ana.s's) cont;d
1
1
ac dc 1
1
2. !ponentially decaying dc current component
( )
where is the time constant,"he #alue of is determined from the initial
conditions:
2($) $ ( ) ( ) 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,
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T'me /a.'n cuent
i>t ?
t'me
Supepos't'on o4 stead.@state component andeGponent'a. deca.'n dc oset"
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RL C'cu't Ana.s's) cont;d
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&
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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
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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 '
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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+
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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
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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!
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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*
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eneato Shot C'cu'tCuents
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eneato Shot C'cu'tCuents
&,
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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
$.2 seconds
d d d
d d
A
X X X
T T
T
= = =
= =
=&1
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eneato S"C" EGampe)contKd
2.$
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 =&&
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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$o% 2aut Ana.s's
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Net$o% 2aut Ana.s'sS'mp'nonsp'nn'n? oads ae 'noed
&(
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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$$
line X = =
eneato has 1",+tem'na /otae supp'es 1,, MA
$'th ,"9+ a p4
&+
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Net$o% 2aut EGampe)contKd
2auted net$o% pe un't d'aam
<'
"o determine the fault current we need to first estimate
the internal #oltages for the generator and motor &or the generator 1.$6, 1.$ 1.2
1.$ 1.2$.;62 1.2 1.1$ 8.1
1.$6
T G
Gen a
V S
I E
= = ∠ °
∠ = = ∠ − ° = ∠ ° ÷&6
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Net$o% 2aut EGampe)contKd
"he motor's terminal #oltage is then
1.$6 $ - ($.;$77 - $.2;8) $. 1.$$ 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
∠ ° ∠ − °+
= ∠ − ° + ∠ − ° =
&!
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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$o oppos'n/otae souces- so/e s.stem . supepos't'on7
8
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Supepos't'on Appoach
2auted Cond't'on
EGact EBu'/aent to 2auted Cond't'on2aut 's epesented. t$o eBua and
oppos'te /otaesouces) each $'tha man'tude eBuato the pe@4aut /otae
&9
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Supepos't'on Appoach)cont;d
S'nce th's 's no$ 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,
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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
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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&
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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
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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(
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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+