Fault Analysis Using Z-bus Fault Analysis Using Z-bus 1.
1.00 InIntrtrododucuctitionon
Th
The e pprrevevioious us nnototes es oon n ZZ-b-bus us rresesuultlteed d inin
so
some me ususefeful ul knknowowleledgdge: e: didiagagononal al elelememenentt
Z
Zkk kk of the Z-bus is the Thevenin impedance of the Z-bus is the Thevenin impedance
seen looking into the network from bus k. seen looking into the network from bus k.
Combining this knowledge with equation (4 Combining this knowledge with equation (4 from notes called !"#mmetrical $aults %&' from notes called !"#mmetrical $aults %&' which was which was Thev Thev f f f f f f Z Z Z Z V V I I + + = = ′′′′ (( en
enabableles s uus s to to efeffficicieienntltl# # obobtatain in ththe e ffauaultlt
current for an# bus in the network. current for an# bus in the network. This is eas# if we
This is eas# if we have Z-bus and )have Z-bus and )f f ..
*
*oowweevveerr' ' oonne e ootthheer r tthhiinng g tthhaat t wwe e wwiillll
frequentl# need is the line currents' because frequentl# need is the line currents' because the circuit breakers are going to be in series the circuit breakers are going to be in series with the lines' not with
with the lines' not with the faults.the faults.
2.0
2.0 FauFault calt calculalculationtions uss using Zing Zbusbus
+ecall that +ecall that I I Z Z V V (%(%
"ince (% represents a set of linear equations' "ince (% represents a set of linear equations' superposition holds' and we ma# write:
superposition holds' and we ma# write:
I I Z Z V V == ∆∆ ∆ ∆ (,(,
This sa#s that the change in voltage at all This sa#s that the change in voltage at all buses
buses ) ) ma# ma# be be computed if computed if the the change inchange in
inections at all buses / are known. 0e can inections at all buses / are known. 0e can write eq. (, in
write eq. (, in e1panded form as:e1panded form as:
∆ ∆ ∆ ∆ ∆ ∆ = = ∆ ∆ ∆ ∆ ∆ ∆ N N k k NN NN Nk Nk N N kN kN kk kk k k N N k k N N k k I I I I I I Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z V V V V V V ( ( ( ( ( ( ( ( ( ( (( (( ( ( (4 (4
2ow consider a fau
2ow consider a fault at bus k' where the pre-lt at bus k' where the
pre-fault voltage at bus k is )
fault voltage at bus k is )f f . 3et the fault. 3et the fault
cu
currrrenent t be be //f f ' ' aannd d aassssuumme e ththaat t tthhe e ffaauulltt
impedance Z
impedance Zf f 56 (this is t#picall# worst-case56 (this is t#picall# worst-case
scenario. scenario.
"ince a fault is a
"ince a fault is a short circuit' then )short circuit' then )k k 5-)5-)f f ..
% %
7lso' since the fault current is out of bus k' 7lso' since the fault current is out of bus k' then /
then /k k 5- /5- /f f . "ubstituting these into eq. (4. "ubstituting these into eq. (4
results in results in ′′′′ −− == ∆∆ −− ∆∆ →→ 66 66 (( (( (( (( (( (( (( kk f f NN NN Nk Nk N N kN kN kk kk k k N N k k N N f f row row I I Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z Z V V V V V V (8 (8 2oting that the right-hand-side results in' for 2oting that the right-hand-side results in' for
each row ' onl# the Z
each row ' onl# the Z k k being multiplied b# being multiplied b#
a non-9ero current.
a non-9ero current. Therefore:Therefore:
′′′′ −− ′′′′ −− ′′′′ −− == ∆∆ −− ∆∆ →→ f f Nk Nk f f kk kk f f k k N N f f row row I I Z Z I I Z Z I I Z Z V V V V V V (( (( kk ((
0e observe from row k that: 0e observe from row k that:
f f kk kk f f Z Z I I V V ==−− ′′′′ − − (;(; "olving (; for /
"olving (; for /f f results in results in
kk kk f f f f Z Z V V I I ′′′′ == (<(< , ,
2otice
2otice that that eq. eq. (< (< is is consistent consistent with with eq. eq. ((
when Z
when Zf f 56.56.
2ow substitute eq. (< into e
2ow substitute eq. (< into eq. ( to get:q. ( to get:
( (
))
( (
))
−− −− −− == ∆∆ −− ∆∆ →→ f f kk kk Nk Nk f f f f kk kk k k N N f f row row V V Z Z Z Z V V V V Z Z Z Z V V V V V V == == (( (( kk (> (> 2ow2ow eq. eq. (> (> provides provides the the change change in in the the busbus
voltages due to the fault. Change from what? voltages due to the fault. Change from what? /t is the change from the voltage without the /t is the change from the voltage without the fault' i.e.' it
fault' i.e.' it is the pre-fault voltage.is the pre-fault voltage.
Consider an# bus' lets sa# bus ' with a Consider an# bus' lets sa# bus ' with a pre-fault voltage of )
fault voltage of ) . Then we can compute. Then we can compute
the bus voltage under the faulted condition the bus voltage under the faulted condition as as j j j j jf jf V V V V V V == ++∆∆ (6(6 4 4
$rom eq. (>' we know that $rom eq. (>' we know that
f f kk kk jk jk j j V V Z Z Z Z V V ==−− ∆ ∆ ((
"ubstitution of ( into (6 results in "ubstitution of ( into (6 results in
f f kk kk jk jk j j jf jf V V Z Z Z Z V V V V == −− (%(% 2ow
2ow eq. eq. (% (% is is useful useful for for computing computing faultfault
currents in the circuits. Consider $ig. . currents in the circuits. Consider $ig. .
Z Z b b
bus i
bus i bus .bus .
$ig. $ig. 0
0e e ccaan n uusse e eeqq. . ((%% tto o wwrriitte e ddoowwn n tthhee
vo
voltltagages es unundder er ththe e ffauaultlted ed cconondidittioion n foforr
buses i and ' as buses i and ' as f f kk kk ik ik ii if if V V Z Z Z Z V V V V == −− (,(, f f kk kk jk jk j j jf jf V V Z Z Z Z V V V V == −− (4(4 2ow
2ow we we can can compute compute the the subtransientsubtransient
current flowing from bus i to bus under the current flowing from bus i to bus under the fault condition as fault condition as b b jf jf if if ij ij Z Z V V V V I I ′′′′ == −− (8(8 8 8
"
"ububststititututining g eqeqs. s. ((, , aand nd ((4 4 ininto to ((88 results in results in kk kk b b jk jk ik ik f f b b j j ii b b f f kk kk jk jk j j f f kk kk ik ik ii ij ij Z Z Z Z Z Z Z Z V V Z Z V V V V Z Z V V Z Z Z Z V V V V Z Z Z Z V V I I − − − − − − = = − − − − − − = = ′′′′ ( ( 0e can use eq. ( to get the fault current 0e can use eq. ( to get the fault current in the circuits. These values provide us with in the circuits. These values provide us with the appropriate information for selecting the the appropriate information for selecting the circuit breakers in the lines.
circuit breakers in the lines.
3.0
3.0 Some Some impoimportanrtant t commcommentsents
•
• ZZbbuus s sshhoouulld d bbe e ddeevveellooppeed d uussiinngg
subtransient reactances in generator=motor subtransient reactances in generator=motor models.
models.
•
• @@eeccaauusse e ffaauult lt ccuurrrreenntts s aarre e tt##ppiiccaallll##
much larger than load currents' it ma# be much larger than load currents' it ma# be assumed that there are no
assumed that there are no loads.loads.
o
o 7ll pre-faults currents are 6.7ll pre-faults currents are 6.
o
o 7ll buses have voltage (pre-fault equal7ll buses have voltage (pre-fault equal
to ) to )f f .. o
o Aquation ( becomes:Aquation ( becomes:
kk kk b b jk jk ik ik f f ij ij Z Z Z Z Z Z Z Z V V I I ′′′′ ==−− −− (;(;
•
• $rom ( and (;' we see that onl# the$rom ( and (;' we see that onl# the
k
k thth ccoolluummn n oof f tthhe e ZZ--bbuus s iis s rreeqquuiirreed d ttoo
anal#9e a fault at bus k. anal#9e a fault at bus k.
The last observation can be utili9ed in an The last observation can be utili9ed in an ef
effefectctivive e ffasashihioon n wwhehen n peperrfoformrmining g fafaulultt
aannaall##ssiiss. . 33eetts s aassssuumme e tthhaat t wwe e wwaannt t ttoo
compute the short circuit currents for a fault compute the short circuit currents for a fault at onl# one bus k. "o we ust want to get the at onl# one bus k. "o we ust want to get the k
k thth column of Z-bus' but we do not need the column of Z-bus' but we do not need the
entire Z-bus. entire Z-bus. T
Thheerre e iis s aan n eeffffiicciieennt t wwaa# # tto o ggeet t tthhe e k k thth
column of Z-bus. 3ets stud# it. column of Z-bus. 3ets stud# it.
Consider that the Z-bus and the B-bus are Consider that the Z-bus and the B-bus are inverses of each other' i.e.'
inverses of each other' i.e.'
( ( − − = =Y Y Z Z (<(< T
Thhiis s mmeeaanns s tthhaat t tthheeiir r pprroodduucct t ggiivvees s tthhee
identit# matri1. identit# matri1. I I Z Z Y Y == (>(>
where / is given b# a matri1 of 9eros e1cept where / is given b# a matri1 of 9eros e1cept the diagonal which contains all ones' i.e.'
the diagonal which contains all ones' i.e.'
; ;
= = ( ( 6 6 6 6 6 6 ( ( 6 6 6 6 6 6 ( ( I I (%6(%6
ur approach will depend on two ideas: ur approach will depend on two ideas:
.. CCoolluummn n oof f Z Z : : 00e e ccaan n uusst t ccoonnssiiddeer r aa
single column of Z' instead of the entire single column of Z' instead of the entire matri1. Call it Z
matri1. Call it Zk k D it is a column vector.D it is a column vector.
The right hand side of (> will ust be a The right hand side of (> will ust be a col
columumn n of /. of /. CaCall it ll it //k k . /t will also be ust. /t will also be ust
a vector and will contain 9eros in ever# a vector and will contain 9eros in ever#
rroow w ee11cceeppt t ffoor r rroow w kk. . TThhe e rreessuullttiinngg
relation is: relation is: k k k k I I Z Z Y Y == (%(% %.
%. 3E3E-F-Fececomompopossititioion n : : /f /f #o#ou u hahavve e ttakakenen
A
AA A 4488 oor r a a lliinneeaar r aallggeebbrra a ccoouurrsse e iinn
m
maatthh' ' tthheen n ##oou u aarre e ffaammiilliiaar r wwiitth h 33EE
d
deeccoommppoossiittiioonn. . 3E 3E ddeeccoommppoossiittiioonn
provides
provides a a wa# wa# to to solve solve for for the the vector vector 11
in the matri1 relation in the matri1 relation
b b x x A A == (%%(%%
where 7 is a nGn square matri1' 1 is an where 7 is a nGn square matri1' 1 is an unknown nG vector' and b is an known unknown nG vector' and b is an known
< <
n
nGG vveeccttoorr. . TThhe e aaddvvaannttaagge e tto o 33EE
d
deeccoommppoossiittiioon n iis s iit t ddooees s nnoot t rreeqquuiirree
inverting the matri1 7. The basis of inverting the matri1 7. The basis of 3E-decomposition is that we ma# factor 7 decomposition is that we ma# factor 7 into a matri1 product 3E' i.e.'
into a matri1 product 3E' i.e.'
U U L L A A== (%,(%, whe
where re 3 3 is is a a lowlower er diadiagongonal al nGnGn n mamatritri11
of the form of the form = = ,, ,, ,% ,% ,( ,( %% %% %( %( (( (( 6 6 6 6 6 6 l l l l l l l l l l l l L L (%4(%4
and E is an upper diagonal nGn matri1 of and E is an upper diagonal nGn matri1 of the form the form = = ( ( 6 6 6 6 ( ( 6 6 ( ( %, %, (, (, (% (% u u u u u u U U (%8(%8
"ubstitution of eq. (%, in (%% #ields "ubstitution of eq. (%, in (%% #ields
b b x x U U L L == (%(% Fefining Fefining w w x x U U == (%;(%;
provides that eq. (%8
provides that eq. (%8 becomes: becomes:
b b w w L L == (%<(%< /f
/f we havwe have e 3 3 anand d E' E' ththen (%<en (%< is is eaeasisil#l# sol
solved ved fofor r w w (wi(withothout ut invinverterting ing 3 3 usiusingng
forward substitution' and then (%; can forward substitution' and then (%; can
> >
be
be easil# easil# solved solved for for 1 1 (without (without invertinginverting
E
E ususining g babackckwawardrds s susubsbstittitututioion. n. HoHorere
d
deettaaiils ls oon n 33E E ddeeccoommppoossiittiioon n mmaa# # bbee
ffoouunnd d iin n tthhe e nnoottees s ccaalllleedd
!3EFecomposition.doc.& !3EFecomposition.doc.& 2ow observe that eq
2ow observe that eq. (% and (%% are in th. (% and (%% are in thee
same form. Therefore we want to solve the same form. Therefore we want to solve the fo
follllowowining g eqequauatitionons s in in ththe e orordeder r ththe# e# araree
given: given: k k I I w w L L == (%>(%> w w Z Z U U k k == (,6(,6
*omework I,: Fue $rida#' Januar# %;. *omework I,: Fue $rida#' Januar# %;. Co
Consnsidider er ththe e 4-4-bubus s s#s#ststem em shshowown n bebelolow.w.
@oth machines have subtransient reactances @oth machines have subtransient reactances of 6.%6 pu (#ou can combine the machine of 6.%6 pu (#ou can combine the machine subtransient reactance with the transformer subtransient reactance with the transformer
iimmppeeddaanncce e tto o ggeet t a a ssiinngglle e rreeaaccttaannccee
co
connnneectctining g tthe he mmacachhinine e iintnterernnal al vvololtataggee
with the network. with the network.
6 6
@us %
@us , @us ( @us 4
6.%8 6.(%8 6.%6
6.46 6.%8
6.(6 6.(6
a. Construct the B-bus for this network (should be a 4G4 matri1.
b. Consider that there is a three-phase (s#mmetrical fault at bus %.
• Ese 3E decomposition to obtain the
%nd column of the Z-bus.
• Compute the subtransient fault
current.
• Ese eq. (% to find the voltages
during the fault.
• Ese eq. (; to find the subtransient
currents in lines ,-%' -%' and 4-%.
