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Measurenrents..\
ANALYSIS AND PROTECTION OI'POI,IER SYSTBIS COURSE
A}TALYSIS OF BALANCED FAUITS
BY
" J.. .3. RSYIEI
e €r_alvi Traiuing Course
f, ,:.. It, N,v. [o 30, Nov.n 19g9,
.'t .-J-,;c 0J,Gf C :js:r;m Sliras,r.romenrs Ltd"5r".i'., . i:\8 i ,j. : : : :, r .tr l\.t :.j, i$lPs--i ,;rucering (pvt) ttd. WI l- i.. ,..irroef$-
a_ LAHO&E.f sl;- 8t*ttig. g?f20?
1.
1.0 INTRODUCTION
An understand'ing and' vorking knowredge of systen- Analysis is very inportantto the hotection Engineer as he must know iros the systen operates under,.f1il:"::"rt eondiiione berore "h";ri;;";1";i; ;:iaj; ii=ilr"n the systen
Analysls of load and fault conditions also provides rrseful information for;-(i) Choice of power Systen Anangenent.(ii) Required Sreaking capacity of switchg:ear and Fusegear.(iif) Application of Control Equipnent,(i") Required. Load. and Short Circuit Ratinga of pLant.(") Systen Operation. Securlty of Supply. Econonics .("1) rnvestlgation of unsatisfactory pla.t perfor*ance.
2.O .IIECTORS
Z
rf z1
= R+ jX
and Z2
= lzl (coe 9 + J sin 9) = lzf "je
.)-
a=antl-clocknise
anti-clocknl.sein s3rnnetrlcal
through 90o.
th:oough 1200conponent analyel.s.
a2 ='t/z4ooia2+a+1 = o.
4.O cCINvENtION USm FoR YOLTAGE DINECTION
produces a voltage drop ln Z such that
Vp,6 ' Vor-rAGf oF A A.ove
: r.f Z.
= lzlE= lzi l9r
ir.V, = lzl lrzl /\uz
).O OPMATORS
= l'zl@and h= lelld2 lzzl
rotates a vector
rotates a vectorused extensively
in dlrection shownrespect to B.
ITt /lzoo
Current I flowingA is positive with
5.0
Per unit valuesfollovs :-
Per unit
5.o
lnpedance Z =/ P.U. @-tsase inpedance
B.A,SE QUANTTTfrS' ap euR.ubri svinnr:T l-..' j,. l'",r' l'"i,
= za'x
2
I
This is partieularry tiiefilr r{he: .qpShlsins rarge systens with severalvoltage revels' *fg* ffi-srsteia, i.i""ilti*3 ".. tal<e place the systenparaneters nust all be refeirla to cornnon uase-q,rartlties. The basequantitles are {lta on one p""t or !!" "v"t"J ana base qr:antities on
;:H ifi::r::r$::""""r "orias"" will depend-on the ratio oi-i,,t""oenins:
The base qrrantlties used aret-
Sase voltage = kVU = phase to phase voltage ln kV.
Base PfyA = MVA6 = three phase MVA.
Other base qr:antities can then be establiehed. :-Sase inpedance = Zb = glh,l, in ohms.
MVAU
Base current = f6 = in kA.MVAU
Tffiaie obtained by divid.ing 4gtr,4l values by base values as
MYA
lmP=Za
Z6
Per rrnit voltage OOn.r.
Per rrnlt *O, *nn.o.
Per rrnit crgreat In.*.
agztl
TRAI|SFERRING pF,R itNIT qUANTITIES FROrr{ONE SET OP BASE VAIUES TO A}IOTHM
= kVakvt
waaMVA6
IcI6
Pereentage valueg are co'qnonly used. for trangformer inpedances and whereper unit values €ure veqr snali. pereentage ""io", are 100 tines theequivalent per unit values.
Let ZFoU. J
.azaz
oL
PoU.l x zb!zaz
Per rrnit value on base 262 is zp.u.Z =.
3.
aootb
P.u.2 = zp.o.1 x (Evut )2. x w4qz, ,.':.llvlfr (FEI2
i- :+ iY.-
'! -.::.'l'" 9:' :'*
'I
.. f 5..r'r' '! i .
r:* -.; I 1-
o
Ar-\o
aolt
4.5)4ILstv)
.+r-co
J'1ornO.
r..'\ Ln FC\I
trlrj r45Hdt!
a
lf\@r
Itr\
, .rrJ\o
.N" .
s:n ^J\or;
dii., ll 'oll o\ ll
p& sHb xx;x3,s,RYdib.O^
lt-ll .n
.v C\J J4r ta\ t{\r|{\HHF{
ata
AcoCl\\o
ooIlor
-lst:il
HH
aa
a
atO
Arnr--
aolt
olor-{otx
aoov)dp!oEoOgo
a5o
N
t'tc.'3i, ,i'"". i;,.
, $^'1
r(\t+?N\\o:"fll\;'
ilrl
J4
4olsf
ao,o
;rrr p N l-opi ts ilEiooIaadd-oFq Fq ci
a5a
Ar(\FF
aoil
oln
atag
.-ro
> n-oJOm--(r)-ct?(na-o
v',
$R$ $
t!zla€a,do
uJ9
xo\n
:C
d0
zoi(\I
jd-Io
aFq
+td*);.1'd
cltH
oa)d
AP.
oo{H$F>dt{oLI
o.Fl{Jooo
odo
F{
+oF.rt{)0't+r=F{3h,gr
FoEfAEd{t!X .r{tr] F-{
o\r+r(\
i,'' l
trL
c{M{
atag
rao
ii-"
."1 fs t{ElsIp
H
5.
DUMPI,E 2
The base voltage on each sid.e of a transforner ruust be in the sane ratioas the voltage ratio of the transformer. . l
'
f l.gkv u.z/t+tutt t3:./trt v
DIS??,8un o^,
'Ys'€M.
fncorrect Selection of kV6
Correct Selection of kV6 1<2 x 'lll.gt4t
Alternative Correct Selection of kV6
Elu],tPtE lThe per rrnit irnped.ance of a transfo:mer istfangforner.
Consider a transformer with voltage ratio
OMvA@kV6,kV, a)n kV6= kV2
I
I
11.gkv I
I
= 11.05kV1I
I
1 1.8kV I
I
I
i
| 11kV
I
1 1kVI
I
I
141 f11 -'t1,75kv112
1r2kY
1tzkv
1 41kY
Actual'.
Actual
'n.o.,
zp.o.2
but 2"2 =
tt t 'n.o.,
impedance of the
lnped.anee of the
= 2"1 = za| x
%t
transfo:mer
transformer
MVA
F"7
the ga.ne on
uv1/ky2.
froo side
fron slde
;l
each slde of the
viewed
vieredi
: 'l'
Zal'
za2"
a-
z"p
zaz
Za1
= Z^x MVAazW
x kY22
Ez
'n.o.r'
t3zfvRv
za1 x -ff,
A
7.o crncurn,tiws
,: - .: l-"
There are'three.Uasic
(.) ohns Lar
s=r
Rorrnd any.uesh 't E" =' LiZ:ergr in nesh (1) E1 = i1 21
=L121
8.0 CIRCOIT TITEOREMS
Y=I7,
2t-o
I
lr'
Itzr-722,(rr - L) z1
The three nost comrnonly used, forTransforn and Superposition
4j.:.' lit".t.' -.'' : ;'. .
=. 4.*r .- '
,pg!.Cf g ndtwork which is not of particular
(u)
!
-
.
-
V
Kirchoffts Junction
Kirchoffts Mesh Lau
junction (or node)
+12+fr=O
Lavt
,, :
lAl ,any
i.e. f1
(o)
I'l
+
+
z3 (', @
)
fron any 2 terninals can be replaced. by a singlewith a single i.ropedance "fre".ii-
.f,o
Drivlng voltage
Inpedance
Example:
-
ZtZ = ZtO+Z2O+Z,IO.ZZO
Open circuit voltage between terninalse "
frnpedance of the network as viewed from the twote::ninals with all driving voltages. short ei.reuited..:; , .- :_i --,:. .., , ..
77z'
IlEz = t,' I +_ I
t,.
where Et = U_ . v., andm r---
(U) A/^ and. x/a Transfom
I
z'to =
ol
Theorens
zro
z1z.z71
-
-z1Z+221+211
Superposition Theoren
linear network the cu*ent in any branch due totirr" "i;"ir"""o* '
of several d.ifferent d.riving vo1laqgs.fs.,q-qXqil,to't5e rygctor,,j*1, ...,Jcurrents caused by each drivrng ,roftage "iit'iil
"rorr" with t},.d,",: -. ;".short circuited..
'. "1 ^i :i: i '9r"':j' '{le:l:r i'ir'': :' "^:'
'.-.:,,: ': ;", ;.; iil lir :- r.: ;'11;-i;;1"' i-r' ":
':::'"1
2
(")
In arlyactionof theothers
Z1
B.
E, l
IE,
I
tlEz
I
I
{
lr.
9.0
(*),:' ::;l
. : '.. . 1. 'r .i'" .', . 'ri' . '1':.; 'l ll"' i 1
USE OF PRE-FAULT VOIIAGE FOR I'AUIJT CAICUTATIONS- ; r .': l '
Consider a systen fed fron ? sourees of generation E1 and Ey..---..--.--. " " .,:.... l" ...
Zt Zs Ze 21,
i Circuit (a)
Systen inpedances are-sholtn by Zl, 22 etc. lthere 22 and. 25 representload inped.ances. Let the distribution of load current be as shown.
Let P1 be the point at which a ttrree phase fault will be consideredand let Yp.f. be the voltage at this point due to the loaded systernbefore the fault oeculs.
Ir'. l
Ilt'
rgtr)
72
I etr)
I*t')
Ve't.
v
9.
(b) Consid.er a Three phase Fau1t at p
'ti? rJ.;r"i ta;t;'o""a 1l
One nethod' of calculating the current d,lstributlon is to consider thegenerated €.rl.f.rs at the noment of fault.--ryr"-eistribution of faurf-.',,current so obtained can be shorrn on the di-agran and rabelrea s'rrix (F).
with a sinple systen i.e. with one lnfeed this is gasiry.aghieved.However, with a d::!r." or nur.tipre-en. {"a "i"t", it is generallyd'lfficult to establish the r"errigag "oa
*gi"-Jr "rr generated s.D.f.rSat the nonent of fault and evJn if these "rJ-""r:.rabre, the calcuratlonof distributlon of fault
".,rr-.rrt is conp1ic"t"a.TLre pre-fau1t voltage at the\fault pointsimplify the calculitlons
"s toito"s :_(") The originar circrrit ean be Theve'nised'acr;i" f and. N to d.eternlnethe cunent in the fault path.
ctrcuit (b)
Circuit (c)
1Q.
(c)":':- i (Cbhtintibd). :v ,'.' " :
Apart froa the fault pathr the values of crrrrent in each branchthe Thevenised. circui.t do not equal the currents obtalned froneircuit (U) i.e. the actr:al cunents in the branches.
:-.".Consider circuit." (d) again. '
We can i.nsert a fictitious e.n.f. Yp.f. nithout affecting the loadcurrent. i -
7t
5(L)
7*7s
i'-l
Ir,
l
This can be analysed W superposition ia 2 parts :-
1.,
l--
(i).:
In currCnt:tetss (i)
and tiil
(il)
is equivalent to clrcuit (b).
is equivalent to-(circuit (c)) i.e. thecurtent 1n each branch will be reversed.fron that in circuit (").
.'. In any branch r
r, .r(D .: T(i,) ; rtnt
":rri,.ii, .,,r:,irl: i:. l,:-J.1f) , =, *I(L)' + IG)
- "-":"-"';iiiiere I11; is d.etenoined from a pre-fault load study.
If .r,) is d.eter$ined fron the Thevenised. circuit using\^/t pre-fault voltage at the fault point.
'l(
A/, I
astu)i..:
Lz
fz(r-)
' Iltr-)
--. l,
) l,,r
1
l
75
Isrr.)
11.
For the majority of applications pre-fault load cu:rents are snallconpared with cr:ments obtained fron the fheveni.ged cir.guit and, ':
can be neglected..- +' ,; ,^ l.; .. ::,
- -';l : '_ :,.
- t'1. ' ; ...-L ^'l,l
in the Thevenised, circuit isSeul,1 Po+n!.i.r..r.i . ":.i, - :.:
i.o. t(r) *
In these cases the value
the no-load open circuit
t(n). : '_i ':
of V ^ usedPoI o
voltage at the
' .':
10.0 NE$IIORK REDUCTION AND DISTRISUTION FACTORS
EXAMPIE USTNG SETERAL OF IAWS A}.ID TIIEOREIVI,S DISCUSSED
Find the crrnent distribution for a three phase faultshown. Use netrrork reductionr and back substituti-onsequence diagra^m.
Positive sequence diagran :-
2.5Jr
".j {
.' "'",,].
atAonand the
the netvorkpositive
4 singleA three phase fault is balanced and can therefore. be analysed onphase basts using only the positive sequence inpedancs.,;
Convert ,A inpedances A0, B0 and t'l10''to eguJivalent a inpedances.
Zort = zto *'*',0 * 'oo='*.,.0, :.,9r'15 + 18'8], I 941rkr@ =
zso :
LOAD
(i)
51 tt
(i)l'
F .^r
foN1"-*-t '
zts
*
"10 + Zlo'ztt1o = o'45 + 18.85
zto
12.
+ 0.45 x 18.85 - t}.Gsto.7,
0.75 x 0.4q = 1.22 tu1 B.g5
J^
= Zlo * zlo + zn.\o = o.Ti + 0.4, +
'*ro(ii) By Thevenins Theorem the network can be red.uced to :-
2.sll
I 6ltA 1.22,. g o.4jt
'"ldi-isr* fi ro..,,f.
I
I
(iii)
Vpf is the pre-fault voltage of A above.N1.
The netrvork is reduced to a singre voltage and inpedance as folrows:_':A o.szn i'r'
O.3qEtt
'' I
Vpo I
i
(i") fhe three phase fault is now appried. by connecting A to N1.
Let fault eu:erent be of unity value i.e. Ip = 1.
'., Tl" fX*t current."distribution can be obtained by back substitution.'I'he varues of cungnt obtained in the branehes with rF = 1 representthe p*ojiortion or if, in tire uianches and are refe*ed. to as thetrPositive sequence Distribution Factors'r for the circuit for afault at A.... 1. , ,
r.
- ,t.--4.i r: l
The criirifii* in the fault path can be obtained. from a knowredge ofthe pr'etriiult vottage vpp. cr:rrent in any particular branch is,,,?r.nttipgd. by. nultiplying .*.fre" fault cu*ent Uy tUe appropriate'distribution factor.
Total current in t[g lrgncb is obtained. by adding the pre-fau1t 1oad.lh l*S*.;; & {: i* i i refrten ti*ii n . ifr irt. - irr ancrr.
'o't"tCi)r''{':'': 1Fa 1 'r:
t.l
X
5A4VA
sz
GECAII Trainrrg CourscLehorc lt, Nov. lo 30, Nov., 1989,
Conductcd by,GEC Alsthom Mctsutcmcats Lt4Stafford, UK.Reprcrcntativcl,Pcarl Engineering (PYt) Ltd.ll-L, Gulberg- 2, LAHORE'
Tcl:- l?9710 - 87520?
250fiV4
i O'oZ p.u
,tal!rl t-.-zb
SnsE voryAA€ 3
I qsr HvA =
8ng rHP€DA^/C6
?t p.... = .' tt r75o
ilkv5 },IVA
= lla:
s:tl*
s i.r.r.25o '
Z-r = 8Z = O'O8p..t.
lP."-
t-t<Vp.*.
7P''.
MVAp.r
,?:l-'' , ,.'5r .''"6 ZjS fiVA
M VA p..r
r GNolFDo.og
crccrrrr sc$k€( e*nuc ussD rr 'FRurr".sft' Ai r#&r^J6 6u{4tsrvo7 kNowu
l'1vA F*)L-f
lF SOrr(c€
lP.*.
2.Le* MVA gor" (. soahV)
: o.oO2 p.u.
o.e52 x o.5lla
O'o8Z xo.Slla
O.O5 p.re .
lf'*.
5oo - 18r 3ao kVA
!$"
vlCe
: O.5
o.52%
llzo.S
s O.ool p,*,
= O,ooo31 p.u .
Oool+\9.9973
(i)
lP.r.
' ...4
-" ..\ . .i
rl,o.o?'t3
\lrlrA'bus'.,'1 . 1-.
1oinz
Ssehia ofc^ :.
o.oor*io.ooolllO.oE
I
-
o.a573,$o - q,5ro kvA
i.e. fau$ level is rd'^ced h5 ress *r,an sog due to*\e e$$ect of *lre soure i,rngedeaeq.
f.lVAg,.
-l , .::: i :
Ccbtc {€ish',,ce , g3gL' '\ i. ': 7**
Cablc reac#^^.a r C.cgLzur?
rcr4rce , SX iWi++\ bus gct,i.r clogd )-
I p.,^.0.oot+)o.o923
Meacurenents
ANALYSIS AND PROTrcTION OF POWER SYSTEMS COURSE
TUTORIAL ^. SYSTEM ANql YSTS tlt\FAl ANCFD FAULTS
1. With reference to the typical povrer system network used in Tutorial 1,consider a single phase to ground fau'lt on busbar C. Conslder al'lgenerating plant and transformers in service except T3 and T4 andassume a'l'l breakers at mesh station E to be ope.n,'
Find the per unit phase currents (neg'lecting'load) seen by re'layssituated on'ljnes Dl and D2 close to busbars A and D respectively anda'lso the per unit phase voltages at these polnts.' Assume pre-fau1tvoltage at C to be I p.u.
NOTE: Use value of Z, calculated f rom Tutoria.l, 1,1.e. 7, = Z, ='i0.194 p.u,
2. In the circuit below a'|1 lmpedances are based on
il.8kv v.s/66w
Xgd" ' O't3V;l' Xtt t O'tOP.n,
Xcz ' o.13p.u. X*.z'Q'lop.u.Xeo = o.o6Vu. Xre = o'1o7.,t.
Xu . O.3p.s..,- 11;
Xr.z = 9.?1 ?.n.
Xr.o ' l'27,*.
Eatt\ing f+o,ts{or^.r
oscrnned to hcue ne3lr9'ble
rnfeJaace t" fo .
3O r,rq-fs oF eb.kV
TS,ANsMtssloN"Ltia
75 MVA and 66kV.
Find the fau'lt current and voltage for a dqub]e_phiase to earth faultat A. Assume a pre-fault voltage at A of I p.u. Neglect loadcurrent. Express current and voltage ,in vec!..9r, form i,.'e. I/0, U/0,
Using the same system as for Ouestion,2. ca:lcu1ate. the f-au.lt currentfor a three phase to ground fault at A. Ttieihce expless the; singiephase fau'lt current as a function of the ZO / Z, ratlo of the
sequence networks and the three phase fau'lt current. ,
Find the single phase fau'lt current ,ring tne OerlveO ""f""rrion.
GECAM Training Course
Lah,.rrc ll, Nov. !o 30' Nov', 1989'
Couductcd bY'
GtC Alstboo Mersnrcocatg Ltd'Stafford, UK. !Rcprcscnlatives.Pcarl Engineering (Pvt) Ltd'
l1-L, Gutbsrg' 2, LAHORE'
Tel:- E79?10 ' 87520?
3.
4.
-2-Ca'lculate the fault current I, and the currents in each phase at the'l ine ends tXt and f Yf . I
All inpedances are based on 132kv and 100 MVA and are consideredpurely reactlve.
2g1.7522 O'o4P'u' X Ztt =|tZ = Q.Qg?.rr .
Y t32 /gkv
I3Z KV
9us8A&S
TF
a-€ fr,*l{Zrr=Zrz=|ao -
= O.t18?.iv
-'.;
-4-
TUTORIAL 1. SOLUTIONS
t.For a faul.t at
tlRV
Gt
busbar C.
il/ r+5i<V
TI
HYoRo.
rr.8tr/ tr. E/t+sEv
Base qr:a.ntities are '100 I'fVA and lt?rY on the 1r2kT systeo.
GUCAIvI Trainiag CourseLeiorc ll, Nov. fo 30, Nov., 1989,Conductcd by,CEC Alsthom Mcesurements Lrd,stafford' uK'R ePresen tatives,
Pearl Engineering (Pvt) Ltd.1l-L, Grilberg- 2, LAHOiITel - 8197 l{) - 8752{la
-5-
GENF.RA1OR DEgDA.\CES
Generators G1 and G2 :-.t'
Ni'r o'ase voltage in generator cj.reuj.t = 1)2 x 11 = 10.0kV.145
., .. " : "'': ' , ', Nev base MIIA = 100 I'fI/t
Generators Gt, G4, G5 and G6 :-
Nen base voltage in generator cj.rcuit = 152 x {1# = 10.74kV.
. .,ft.r,-:-..^d sr' = .Iattc4 = xd'"G5 = x'd"G5 =
l
TRAr:ISFORMER n'IgEDAl{gE
iO.'111 xl1]Qx (tt.g)2 = iO.1-1 o\l
75 (10.74) \/
hr? - \re . io.125 t g t (l!!!)2 = jo.2o1 p.u.7' (t2)
Irr.-.= xrz = i0. 12J x H
r l#\' = j0.201 p.u.
hr' = &r5 =
TBATTSMIS S TOU "T,
-6-
}IEII/ORK RgNUCTICN
Equivalent inped,ance at stea.rn station = X;" + X,n =
TEqrivalent inpedance
For a fault at C, thefolJows:-
I{YDR,O
j o.ear
at hydro station = Xd," * XT =Tsysteo positive sequence diagraa
Zs3 : 4 o'385
,1 o.3tS
Zs r j o'oQ2 ZUrj o.llo
21 ' Zro. Zne r?c.
Zeo. ZrcrZco
' j: I: ! i
can be sinpli3led as
,.i0, r 7q + r0. ?9,,14
JO.L72 + j0.2Q12
; " ;:'::,r.:t .
0,.=t j0.0!! p.:.
.=' jO.137 p.u.
\tl,^." e i
1.i | "Yt'j o,oca
Zq : )o.R!, Zs2 r io.zrr
Z. r ja.oSS
lt''*
2.
-8-
Tb,e per uni.t i-ropeda.nees j.n ti:e positlve sequence dlagran for questioaI caa be converted into ohnic Lnpedances to a base of 1l2kV asfollowa :
Z actva.I
z-base
ZactuaL
fhe positlve sequenceslrailas way to I.
zrn
zt,zcr
zr+z
fault caleulatioa diagra.oas showa below::
= J100 x .00185 x 174.2
= j50 x .00185 x 174.2
= j50 x .00185 x 174.2
= 1'14.2 r j0.1935
can be reduced ia a
i67.07
jtt.5tj4o.24
Jtt.7t
L
= p.ll o
1?22100
= 174.2
xZ.0ase
= 174.2
xZ prUo
i33'73
EA{3
.'.b = #r, = 2260l
i3*t c )y
J.
-9-Tbe positive sequence fault Ciagtan is:-
A z-,.?og = io.J8g
t r; ':.
\r/.,':' r. t
io.158
E I p.,^.
J-F
ia'l3e>?er = j o"l !J , Zcz =1o,231.
io8365
= MVA base r €ggl!.sl-g)2%.V.
= 1oo = 632.9tr[vAJO.15E -#
'5 'qx 105(ir)rp = #172x1o7 = tJ-6eL
@r) ro'
.: IDl
aaa
Sault MVA
+ID' =
0. "FE0.909
Inz = 7T.4Au)
o.\16,o.6t35
x 1470
x 2768 = 14?0A
= 59.6A
4.
-t0-
A tJfpical veetor diagfara for power floin; froro A ',o , is shown belowtvrhere i, and X are ihe resi.slence and reactance of the transni-ssisrIi.ne.
Ser:di.ng end power = Received pov€r + liae l,osses.
PA = VpI coso( + I2B
Since angle O is required aad d is ruloownl a:e expresstoa oust bederived grvi::g r( iJ! terms of aL[ tbe othen ]oro'ra paraneters and e.
Ig = V1 cos (e +.r ) - vl coso.
= YA coE e co34 - YA s1n e sb'or - VD coso.
= coso( (V1 cos e - Vl) - sin... (YA si-a e)
11 = V1 sln (e +'r) - VO s13{
= Vg si.a e coEor + Vg sin.a cos e - VD si.n'r
= coEe. (V1 si-n 9) + sin.t (Vl cos e - VO)
rhrtiply (r) uy (v1 cos e - vD) ana (z) by Vl sin I
In (vE cos e - vo) = cos * (vl cos 9 - b)2s!r{ (v1 sin e)(vl cos I - vn)
fX (V.q, sirr c) = cos.. (VA sb e)2
+ stn* (vl sb e)(v1 cos e - vp)
.ad,dirxs (f) aaa (+)
rg (vs coE g - vp) + rx (Yl sj.a e) = cost{tv1 cos e - vl)2
+(vn sin e)2]= cos.. (tzz2)
... coecr = n(vl cos e_-dg)_+_lr(v s:a e)
Substihrtlag this i.a the erpresslon for P1
PA = vo rB'(vl cos s - vo) + vp I! (vl "fae) + t2n
= VoR (of cos e - Vl) + YII (Vl sl.rx e) + t4.22
by cosd.:re ::ar T2z2 = vA2 * vo2 - 2V1 vp cos I
1.
2i
2).
4.
)a
-il
.'. P6 - vDR (v4 cos I - vo) + vnX (vo sia 9) + R (va2 * vo2 - zVl vn cos 3)
= a (vo vl ccs e - vo2 + vl2 * vD2 - 2ve vD cos O) * ffA VD si.:e 9
=R(Vn2-VOVOcosO)+XVlVtsi:eg G.
In tbe above expression for P4,:-.' (i) pU can be ir 16 poreel aad. V4 pbase-aeutral voltage.
| "; " or'6i) ?g can be i:x 3b powet and V1 phase-pbase voltage.
Impedance of 1jne DJ between busbar A and busbar D is 0.146 + j0.J85 p.u.
..'i:..::i.Si.:aCe tbe base qrrantii;i-es are 100[fVA and 152trY, t]re oluric irTedenees ar€3-
B = 0.145 x 1222 = 25.44 oh:ns.100
,t, c'. 12O = 25.44 (tJJz - 1zl x 121 cos e) + 57.€x 1qJ r 1<1 siJl e|
=647 + 45oa
= 450000 - 44jOo0 cos e +-116bOO sl.r' e5117
tl
12O = 87.4 - 86.1 cos g + 227 s1n 9.
. 12"6 + 85.1 cos I = 227 sin e.
1 + 2.64 cos e = 5.95 sin e.
1 + 5.28 cos g + 5.9? .o"2 g = 4g.4 - 48.4 cos2 e.
55.17 "o"2 e + 5.8 cos e - 47.4
.o"2 e + 0.095 cos o - 0.855
cosO ' -0.095+ 0.009 + 1.4242
= a.ogl t {ffi =2
= -1 .948 or +1"7q822
0.
0.
Taid-ag the positlve value cos g = 0.8?9.
-O.095 A 1.85J
2
= 4.9?4 or +0.879
.'. lhase dieDfacageAt q-=?g.5o
V,A
vc
(
_1?-
= 2.7 sj:a w't + O.l sj-a Jw! + 0.1 sjn (5rrt + GOo)
= 2.7 si-a ("-u * 12C) + O.i sia t (rvt + taOo) + 0.1 sir-
(f (on + 1200) + 6o0).
(.) VCI = VC - Vl
= ,.7 lsi.n (w-t + 12oo) - sin vnr) + 0.J t "irr: (,rt + 120)
- si:r ,*t ) + 0.1i "it, (l(-t + 12Oo) + Ooo) - sin (i*t * Goo))
sinA-sin3=Z"i:rf cos 4!1
.'. vcA =2.7 {e "u 50o cos (*t * Eoo)} + 0.J{ e ":.rr
reoo
eos (rwt + 1B0o[* o.t t z s+. J0oo cos (ir* * fooo)]
= 2.7 JT "os (*u * 5Oo) - O.i.J3 cos 5nt
i.e. phase to pbase voltages do not contai.:: thid harncrricc@poaents.
rrs
In3.rI'6 +fcc = 3fRr
From ttre expressions for v1 aacl vg it can be seen tbat tb,e values1f Ylf and VC, (tUe tirirt h,a.monic componeats) are equal.Iberefsre the-thirt haroonlc voltages ire each pbase are eqrral andln pbase as ate the thtd, harogaic cutreats It , I;gl ora 43.bf 7a5 = load lnped,ance at third, barrcuic frequeacy.
fhea I11 = IBJ = I,Cr.
Cu.:reat 1n eatth path = tl, + fg, + LC, - tI,Ar.
tru gS a YA, - lt .ZL, = JI15.B
.'. I15 (N * zoS - Vt5TAq
,R + Zq,trAt
- t3 -
a22Z\ = 0.5 + jO.E p.u. = ffi= 0.871
= 0.1 p.n. = C.E71 x 0.,
Zl, = Q.i23 +jJx0.597
3R + ZO, = 0.'183 + 0.527 +
{ o.s * jo.a} oiurs.
1tt 0.6 + j0.6 ) = 0.)23 + j0.697 ohns.
= 0.261 oi1trts.
= 0.521 + j2.0!1 ob.tros.
j2.091=1.506+j2.091
aaa
aaa
(")VA
= Jr.zo5 " i.lrr&. = {7, F'& = 2.n /epealc value of Jrd harnoaic yoltage = 500 volts.
.'. lorrrso Jrd barnonic cu:reat in 10ad = 6rfu =,
= 85.9A
Ib.e fundamentat r lrd and 5th harnonic coryoneots of voltege atecoasid.ezed separate3.y
(i) Since the load is balanced, the fund,anerrtal voLtages at Ng
ancl l{3 are equal and, tbeir ls no potentlal betweea N5 ead,
eartlt
(ii) s:-see the pbase to pbase voltages contai.a no 3:d harooalccoryonentr ed tberc ls ao retuns path for ]rd harnmlccu:reatsl there caa be Bo ,td batoooj.c ctr:rent.
.'. Vgtr gg5 = vyt - lpt (ZLt) = Y13.
Gii) :tt, ba:sonic cunents can be obtabed by using lheveairaf ssrd the Superposltion TbeoseEgr
i.€. I15 = I.U5 + Ilg5 * IAC5
l3:i.,!l
"
where Ig5 is the cr::rent !r A phase aiue to Y15 witJr Vg5
and V6q--s'6orbed out ad I1g5 ls the cuneat in A p!:ase -
due to-V35 rith V15 anct Vg5-shorted out etc.
GgeAM Trainiag Courso
lohors ttrn Nov' go 30, Nov"o
Coa$qstad bY"
SEC ntcthor& f,fisa$urern'eotg
Staff"d, CK"3,nPr**eniaiiveli,'Bpsll Sr,giueering iPvt) Ltd'
X1"1., Siilb:rg- 2, LAHORE'
Tcli- E?9?10 - 8?5207
1989,
Ltd"
- 14 -
v.- 2v ^--Jii_.g'i'[5 Za5 + Za5 32T,5
'=rT 1 -vr<.Ats5=ffizz'62
t -Vnc^Ac5 3 #).,\,
.?1(.'. r.t' = E tzvnl-vr5-vc:l
.'. 5tb batroaic voltage bertween ilg and. earth is glven by
VUS m5 = -Yl5 * lt1 Zt,
= -v45+!{zvu-vBr-rc'}= -I{vA5*YB5+vc:}
YL' .=
0.1 sln' (5rtt * 5Oo).
Ya5 . 0.1 Eia ({'it+ e4oo) + 6o0)
Yc5 = 0.1 sia (l(*t + 1200) + 50o)
Ye, * Vg5 = 0.2 s1n (:(.t + 't2fr + 5Oo) eos (+ " 12oo)
s -O.1 s$r (:(nt + 12Oo) + 50o)
.'. VA5 * Vg5 + Vg5 = O.
and tberc ls ao 5th barooaia voltagc between tgs ad eartb.
A slryler nctbod of sbo''vlag t&et tberc 1s ao 5tA hertonlc voltegebetrrea Ng ad ls1 ls to sbow tbat Vg5 leads VIS by 2{O- aad vg5leade YAS by 12W. Sbcc load^ls balaaced, a-d-voltage vectoraare aqual ad illaplaced by 12O', by a"ol6g dtb ngdaocatalcoailltlonsr ther: catr bc no voltage bette€n l{g and l{1.
"' Yol'iagc -T"il' : ;:' ::iffi::l'"'" uo'
CEC Alvi Trainirg Coursc
Lahore ll, Nov. to 30, Nov., 1989'
Conductcd bt'GEC Alsthoa Mclrstco€ltr Ltd.
Stafford, UK. ,;,hitReprescntrtiror ''FPcarl Engineeria; (lvt) Lt& '
ll-L, Gulberg' 2, LAHORB'
Tct:- t?9?10'E?52S?
t.
TUTO(IA L ... tsALAT.JCES FAULTS
It KV , 25of"1VA
Calc,^lete {he fa,^t+ MVA
fot a 3{ fault on *hr5l'4VA8x
s{ a*t+
sccry.do*,l srdt 4 #re SMVA)*+onsforrnrf'
tit *ol.a ti*o aec*^n* {ic
source rlEdancs "r| *he llhv b,ubr+s
o;r ,3"ooc *he €+fcd o{ .o,.ree ,-faA.c,.
2.
,t
Ix
3 corc, uadogund eathR,r o.z52lX r O.OBI?&
CalcuhfG *re fa.^lt MVA
fa a 3f f*"t+ d F.
r;t tril|. brs scet,in e.b. closed .
ti;t r.r'r*{ bus segftb,n e.b. aper.
GECAH Traioitg Courec
Lghort ll, NoY. to 30, Nov., l9t9'Conductcd bt,GEC AlstbotD Me$slcacats tt4Sbfford, UK.RcArcscorrdvcr.Portl Engioecit3 (!vt) Ltd'
ll'L, Gulbcr3'2, LAHORE'
Tel:. E?9?10.87520?
,t
X
{
llkV, 25ot"lVA
Measurernents
1.
ANALYSIS AND PROTECTTON OF POWFR SYSTEMS COURSE
TUTORIAL 1 . SYSTEI,I ANALYSTS. BAI ANCFN FAIILTS.
Refer to the attched diagram of a typica'l power system netrork.
Consider a'l'l 'lines and generating plant in serv'ice but no grid infeedi.e. a1'l clrcuit breakers at meSh substation E open.
Using network reduction methods and assLming a pre-fau'lt vo1tage ofI p.u. at busbar C, reduce the positive sequence d'iagram of thenetrork and f lnd the in iti al va1ue of the symmetrlcal fault currentfor a three phase fau'lt on busbar C.
(i) Work with base quantities of 100 MVA and 132kV on the 132kVsystem.
(ii) Neglect resistive components of irnpedance,
For the same system condltions as Ouest'lon l but using irnpedanceva1ues in ohrs referred to the 132kV system vo'ltage re-ca1cu'late theinltla'l va1ue of the symnetrical fault current for a three phasefault on busbar C. Assr.rne a pre-fau'lt voltage of l32kV at C. Neglectresistive corponents of impedance whene appropriate.
0n the typ'ica'l power system network consider a three phase fau'lt onthe busbar A with a'l'l 'l ines and generating plant in service but nogrid infeed. Assuning a pre-fau'lt vo'ltage of I p.u. at busbar A
reduce the network and ca]culate the initla'l va]ue of fau'lt MVA
without first calculating the fau'lt current. Then ca1culate theinltial symretrlca'l fault current and the correspond'ing currentinfeeds from lines Dl and D3.
(i) l{ork with base quantities of 100 MVA and l32kv on the L32kVsystem.
( i i) Neglect reslstance.
With reference to the typical power system nefoork suppose that l2Ot'lWof power ls flowing from busbar A to busbar D and that onlytranmisslon I ine D3 is in operation. If busbar A is at 133kV andbusbar D ls at 13lkv determlne the phase displacement between thebusbar vo'ltages. Assr.rne that 120tW is the sending end power.
A three phase 50Hz' 12.5 I'IVA' 3.3kV generator with an earthed starpoint generates phase voltages which are ba'lanced but containharnpnics.The terminal voltages in kV for phases A and C are :-
+ 0.I sjn (5 wt + 600)
sin3(wt + t20o) + 0.I sin(5(wt +
2.
3.
4.
5.
VA = 2.7 sin
YC =. 2,7 sin
1 20o ) +600 )
wt + 0.3 sin 3 wt
(wt+1200)+0,3
a
The a'lternatorr the impedance of wh jch can be neglectedr suppl ies anearthed star conneted 'load of 0.6 + j0.8 p.u. inpedance (basequantities 12.5 MVA and 3.3kV). The resistance of the earth pathbetween the two sta r po'ints i s 0.3 p. u .
Determlne (a) Voltage Vr^.
(b) The R.M.S. 3rd harnonic current in the load.
If the generator neutral point is disconnected from earth what is i'uspotential with respect to earth.
---+
_?_
HYOf,O POWER STATICN
I
I--\!'.]
(j
Hydro gowa? 3tation
G1 . 62 66'6 MVA 11 kvXa'0'95 f,,i'0'325 Xi - O'Za
Ir, lZ 75 MVA l1/145kV t l0* X '0'125
Stom powrr ltation
6r, 6. , G3, G5 75 MvA ll 8 kv- Xa't83 Xr'q155 xl'0'tlt\s, \t, \r,h 75 MvA 11 8/14skv !1096
X - 0'125
\r . lrz, 4r , Izo , I:r 60 MVA 132/33 kvx.0.125
ft, lt 30MVA 132/33kv X-0'th , ts 45 MVA 132/11 kv X ' 0'125re, 40 t20MVA 2751132/11 kv-- Xttt- 0'15 X,rr- 0'35 xtt' 0 25 ON 120 MvA
Grid rupplvSHORT CIRCUIY FAULT LEVEL T5OOMVA ON 275KV
Ovrrhod linaALL 132 kV LINES ARE 0'175 sq. in. EXCEPT D5 , D6WHICH ARE 0'4 sq. in.
0.t75 !c- in./l-0'0ola6 P€F MILE ON t0OMvA EASEr-0'00385 P€R MILE ON 100MvA EASE
0.1r+ in.ll .0.000627x - 0'00356
lg - t..Zt
trngtirOr 5O MILESO2 60 MILESO3 tOO MTLESDtt. Dtt 30 MTLES
PER MILE OTI IOOMVAPER MILE ON TOOMVA
O: 20 MILESOc 30 MILESo1t, Dta 15 MILES
ALL IMPEDANCES ARE PER UNIT.
13:l lv
LI
I
r32rv$-r*.IX
r32rvJ7r*"xxtt
'o-$n r.$ntt
rgrvj-r I s
'+"^6-'XTllkv-J-+c.
EASESASE
g-E4l|lgw-n:If-19!r-6.6^i
-sO O€+ i
-$,*'A* i
GRIO SUPPLY
lvpical power sysrcm netwotk.
