1. TRANSMISSION & DISTRIBUTION
Protection & Control
Course PC3, Dubai
Application of Protective Relaying to
Distribution and Sub-Transmission Systems
25th - 29th March 2000
Current Transformer Requirements for
Protection
,... AlSTOM T&D Protection & Control ltd
St leonards Works
Stafford
ST17 4lX
England
Tel: +44 (0)1785 223251
p&c002 Fax: +44 (0)1785 212232
(Lecture)
Presented by
A. Varghese
AlSTOM T&D Protection & Control ltd
Registered Office:
St leonards Works
Stafford
Registered in England No. 959256
•
2. CURRENT TRANSFORMER FUNCTION-------,
1, Reduce power system current
to lower value for measurement
2, Insulate secondary circuits
from the primary.
3, Permit the use of standard current
ratings for secondary equipment
REMEMBER :
The Relay Performance DEPENDS
on the C.T which drives it ~
-
3. CURRENT TRANSFORMER FUNCTION-------.
Two Basic Groups of C.T.
1, Measurement C.Ts
- Limits well defined
2, Protection C.Ts
- Operation over wide
range of currents
Note . They have DIFFERENT
characteristics
-
4. INSTRUMENT TRANSFORMER STANDARDS
lEe lEe 185:1987 CTs
lEe 44-6:1992 CTs
lEe 186:1987 . VTs
EUROPEAN BS 7625
BS 7626
·BS 7628
BRITISH BS 3938:1973
VTs
CTs
CT+Vf
CTs
BS 3941:1975 VTs
AMERICAN ANSI C51.13.1978 CTs and'VTs
CANADIAN CSA CAN3-C13-M83 "
AUSTRALIAN AS 1675-1986 CTs
•
6. )
Insulation to stop
flash-over front HV
prinlary to core &
secondary circuit.
'Feeder' or 'Bus-bar'
forming I turn of
primary circuit.
IOOOA?
Generator,
or systenl - - I
voltage source. ..........~
)
Lruninated 'strip' wound
steel toroidal core.
)
1000 turns sec. 1
Insulation covered
wire, giving inter-turn
insulation & secondary
to core insulation.
Typical protection bar-primary current transformer.
7. $', S2
'--~-~
r-r.~(:. c.urrenC" drre,c.t:ions·
PI '> P2
S r --:;:. S '2.
POLARITY
pr
s.
Pl
FLIC~ TEST
FWb ~~c:k 0" QPpl;e4.l",~" J
REV kick 0" "e.~ov~J o~
AVO T ve ,(l.o..t/.... t6 S,
8. •
- BASIC THEORY------------,
-
R
1 Primary Turn
oN Secondary Turns
For an ideal transformer :-
PRIMARY AMPERE TURNS = SECONDARY AMPERE TURNS
9. BASIC THEORY-----------..
For Is to flow through R there must be some
potential - Es = the E.M.F.
Es is produced by an alternating flux in the core.
EsC>< d0
dt
-
14. BASIC FORMULAE----------,
Maximum Secondary Winding Voltage:
Ek = 4.44 x B A f N Volts ....1
Where :-
Ek = Secondary Induced Volts
( Knee-point Voltage )
B = Flux Density ( Tesla )
A = Core Cross-sectional Area
( Square Metres )
f = System Frequency ( Hertz )
N = Number of Turns
•
15. BASIC FORMULAE·----------..
Circuit Voltage Required:
Es = Is ( ZB + ZCT+ ZL) Volts ....2
. Where :-
Is = Secondary Current of C.T.
( Amperes )
ZB = Connected External Burden ( Ohms )
ZCT= C.T Winding Impedance ( Ohms )
ZL = Lead Loop Resistance ( Ohms )
Require Ek > Es
-
'"" .
17. BASIC FORMULAE~·- - - - - - - - - ,
Example Calculation:
C.T. Ratio = 2000/5 A
Rs= 0.31 Ohms
IMaxPrimary = 40 kA
Max Flux Density = 1.6 T
Core C.S.A = 20 cm2
Find maximum secondary burden permissible
if no saturation is to occur.
Solution:
N = 2000/5 = 400 Turns
Is max= 40,000/400 = 100 Amps
From equation 1 the knee point voltage is
( 4.44 xl.6 x 20 x 50 x 400 )
Vk = 10 4 = 284 Volts ~
Therefore Maximum Burden = 284/100
= 2.84 Ohms.
•
Hence Maximum CONNECTED burden .
2.84 - 0.31 = 2.53 Ohms
20. FUJX. bEJ.JS-rf (B,
-reSLA (wbf",")
I ~
QEE
rc»iT-
1-+
I
/
0-8 I'
I
J
o V
t
MA..,-e2.&Ai... : c:.R.osS,
I
~
~
~/
,,-
~
--o 0-02 o~ 0-«. 0-01 0-1 O·f~
MAGIE~FoA:e. (H)
~1lE-~R.NSItNI.
(~~~S~!M~ A/m)
•
21. ( SECDNI)Al.y
i.M. t:.)
Es = 4·44 N f A 8 = l<v 8
w~:- ~ • 4-44 N f A
Ie = H.L - Ki,. • H--N
IAJ"'.... :- k' - l../ttJ. -
L· M€AN MAG-J E1'lc PATH
H 2'
AM P . .,.."tl.N$ / I"£~
l:.e ~ AMef>S
-
•
t"lcl ~~
22. ~.UL-r,t'L~ p;y
Kv
"T"o 08rAv-J VOLl'S
8
~~
_1
I
J
~ --/
V
I
I
~
V ...... H
f'f..,'-1i PL.'-I rt1 l<.t To oITAIN AM.PS
I<~ ,c. B = Vo(..~ (e.)
<.t x. H : AMPS
UNtTS : t<:~ : ~ :. 4-4r4 NfA
~
Kt: .f!.
H
L
-= N
r:
•
23. 1-
WbI"" r-;-~li~""I"""",r--r-_-:-_!"'1-'r-"_-rt_-r'.-'.. ,-._T""'.-r-_"""T.I--'_r.ir-"J~!-"""t!_-rj -,_."-_-.!_'~'-::~--i
'8'lJ:tJtt~t1-ril-r~It_jrt~tf~r·~tl~t"j:~:t1" I I I I i I I I ! • iii!~ I :0-
r ... I! I·'i' I' I ~r- I !. ,- -I-
I : I I I ' ~~l'lj !
'111:;li:~I!"'!'i"lil
I I I' I' !! t--;·- ';'-1' ! - r _1 ,-.! _., -~" ,.,l_t-t-'t·· i--li-+~--t
1-6 I I I ! . I I ! . : , 1 I I . , I !!i 0
i! il l:l.i./'fii!i!I:!1 : III!-
! I I i i i ; I. : Yo I I 1 ! ! i I I ! i '! I ii I I; I I ! I ! LLJ 1 : ! ; : 1! I : ; , , ! I L-i i I
I 1 I i ~ , i J V ~ i t : i! i I ; ~ CH. ; ~
1-401 !. : I O.g
L I I I I !Ii. : ! I I ' I 'k!::::1: I i i , ; i ! i
1 i , I I !. I : I , ! ! Y 'i I i ~ i ! I ! !
'" i i i I V I ! . I I .~ . I ; ; i ! i ' i I I !
t:l i I '--1-1..·1-: ! -~;-~-I-t I : ' I- , I
o I.: ~ I I i ! ! : I iLl. ,; I . ' i ' i i i ' , ' 0-8
> ! i i IiI: I i Vi 1 ' i ; I 1 J I ! I : i I I I
~ ;.I! I !j i ! I iL.! ! i I I 'l : 1 I! I
~ : ' I I I , JI i ! I _ I ; i j 1 I I , ! l I ;
~ '-0 ' ~ : 1ft! ! ·V: , ; I I I ! 1 I I I , i 0-7
u ; !,lll.Y.::~j 1 !:Il,~ .! l~!
"" ; ! -r-t-.~. 'i'- ~.- j' , - , - ':'-!-f-; I' I I'
U) 'I-.J.:, I t ' i ; l ! ~ I i ! i ! - I l i I
.en ; I _ 'J ! III i i 1-i I; I! 1 I ! J I i
, I : i.J. l . : ; : t I' I I I : I I I 0.6
, I ! I~ I i i I l i i! I I! i I! I i I
~ !
r:i. 0·"
'%
< !: i I I I! ; I I Iii i l ! i! ! I I I I
o : ~ 1 I I ~~'"~€ ~ffi! i i i I i 0.S
~ Ili/ll/ I I ! ! ii I
o 0"6 ' I ~.S~t~ -'it l l' 1 I l' It- ! f(,.-.,.r.) a2 I
I I iii I 1Io... ~ ~ I. I I ! I> I J I .~T'~ -uo~-+__~~~-+~~~
:IC i.11 I " W I 1 I I I
l I! I; MAGNEilSING COMPOMENT I I I ;
~ 0-4 I I r I SERIAL HUM~!
~ I!I I i CONTRACT NUM&EP.Q..
5 i J I I TYPE
~ 1ft I I RATIO
% 0'2 I tI I I TURN RATIO
II I I I ! SECONDARY RESISTANCE
AMP
OHMAT7SC !
;
i1 I I I IA I I i I I J 1: I I
o V I I , J l I I I i I I 1 I 1 :
In
o
u
o 0'02 0'04 0'06 0'08 0'1 0-11 A.T/MM.
MUlTlPlY BY Ki TO OBTAIN R.MS. TOTAL EXCITING CURRENT IN AMPS
SO H2.
IDRA·~::t Ir- f. A-
~'r-~-I-=-----~-------~I~iVI'''.
l~f~OYED I
--
-.
25. C.T EQUIVALENT ·CIRCUIT---------,
Ip
!
p. Is
I~ +N I
5. I
I
N Ze Es
Ip = Primary rating of C.T.
N = C.T. ratio.
ZCi
Zb = Burden of relays in ohms (r+jx)
ZCT = C.T. secondary winding impedance
in ohms (r+jx)
Ze = Secondary "excitation impedance
In ohms (r+jx)
Ie = Secondary excitation current.
Is = Secondary current.
Es - Secondary excitation voltage.
Vt = Secondary terminal voltage across
the C.T. terminals
,I
I
I
I
Vt Zb
I
I
~
I
,
--."
-
26. PHASOR DIAGRAM -----------,
Es-.------~~=======-~
Ep Ic
Ep= Primary Voltage.
Es= Secondary Voltage.
m= Flux.
Ic = Iron losses (hysteresis & eddy currents)
1m = Magnetising current.
Ie - Excitation current.
Ip Primary current.
Is Secondary current.
•
29. CURRENT TRANSFORMER
FUNCTION
Two Basic Groups of CT :
1. Measurement CTs
- Lim'its well defined
2. Protection CTs
- Operation over wide range of
currents
NOTE:
They have DIFFERENT characteristics
•
-
30. Mc-..SuR Na- C:. $ ..
- REQlQ£ Coo'b AU!lIAC.:y UP -ro A-ffJtoX tz0 4 IlATEb OJ~T
- Rl:$u,ae. lJ:)W SA"tlJaAT'ON LeJ£L To ~1'EC-T ,f&STiN~S
HJS USE N1C.lC&l- ,-.oN ALLI:Jf ~ '-'l11f LOW £)(c:.(T,~<;
<:.ua«eNT NIb QIee. PaIN' AT L.£)-.J ~ ~rrf
PQ..OTECTtot-( I, (:..".IS
- It~ ~T ~ I~fo~ ~ ABoU£
- _UtIlE. Act:.1JItAcr -.lP To MIW~ TMES RA-=b c:SJ_~T.
iUS use C4DrfN oQ~TAEb Sit_teeN. STES- f&21'H
H~H SA'TUQA'Tlot-l FLUX betlS'f.
8
Pi<oTECoTtON C..' •
H
•
31. WAYS OF IMPROVING CT PERFORMANCE
Es - 4.44 BANs f-
i) Increase B
ii) Increase Ns
iii) Increase A
iv) Reduce burden on CT
Ie = H.L/Ns
i) Use mumetal, when measurement etc is required
ii) Reduce L
iii) Wound primary; allows increase Ns
•
--
32. INSTRUMENT TRANSFORMER STANDARDS
IEC IEC 185:1987 CTs
IEC 44-6:1992 CTs
IEC 186:1987 VTs
EUROPEAN BS 7625
BS 7626
BS 7628
BRITISH BS 3938:1973
VTs
CTs
CT+VT
CTs
BS 3941:1975 VTs
AMERICAN ANSI C51.13.1978 CTs and VTs
CANADIAN CSA CAN3-CI3-M83 "
AUSTRALIAN AS 1675-1986 CTs
-.
33. CURRENT TRANSFORMER RATINGS
Rated Burden
Value of Burden upon which accuracy claims are
based
Usually expressed in VA
Preferred values :-
2.5, 5, 7.5, 10, 15, 30 VA
Continuous Rated Current
Usually rated primary current
Short Time Rated Current
Usually specified for 0.5, 1, 2 or 3 secs
No harmful effects
Usually specified with the secondary shorted
Rated Secondary Current
Commonly 1, 2 or 5 Amps
Rated Dynamic Current
Ratio of:-
(IPEAK = Maximum current C.T. can withstand without
suffering any damage)
AP03916
•
--
34. '-HOICE. OF- ~A" 10
C.L..E~L'f I Tl-E PRtf'I~ f<A..,.,,-1<:
~ ~ NO~MAL c:.URR£NT IN HE c:.1Q.c:.UT
~ TH~t-Pr'- (c::.C»l-rl~UOU~) PA.,J~ 5 NoT
10 BE excee:os.~
secolJ~PrQ.'"J AATINc:. s usuPrU...'l Oil S ~
(o-S ANb 2. R-f-tp AQ.E ALso USe~)
[~S~ ~(1<1N<:.. 1l.5Ol'E: Lal~M: 15
- IA1-P ~
IF L~<:6J2. ~MAi.'i flA'~~ I'rite:. Q~ulQ.e.b
( e. ;5. ~Q. t..AilCE <:£~S<K'["oRS)~ CAN tlSE
2.0 Pd-P S~ To<:e:n-cSjt '-l C"tH It-tteRpos~
c..'.
e..c;. 5000/'2..0 2.0/'
•
37. CURRENT TRANSFORMER ERRORS -------,
Current Error Definition.
ERROR IN MAGNITUDE OF THE SECONDARY CURRENT,
EXPRESSED AS A PERCENTAGE. GIVEN BY :-
CURRENT ERROR % -
kn = Rated Transformation Ratio
Ip= Actual Primary Current
Is = Actual Secondary Current
Current Error is :-
+VE : When secondary current is HIGHER than the
rated nominal value. -
-VE When secondary current IS LOWER than the
rated nominal value.
•
38. CURRENT TRANSFORlvfER ERRORS ------,
Phase Error Definition.
THE DISPLACEMENT IN PHASE BETWEEN THE PRIMARY
.AND SECONDARY CURRENT VECTORS, THE DIRECTION OF
THE VECTORS BEING CHOSEN SO THE ANGLE IS ZERO
FOR A PERFECT TRANSFORMER.
Phase Error is·:-
+VE : When secondary current vector LEADS the
primary current vector.
-VE : When secondary current vector LAGS the
'primary current vector.
•
39. •
METElUNG CURRENT TRANSFORMERS. (6s 3 q 38) .
TABLE 1 - LIMITS OF ERROR FOR ACCURACY CLASSES 0.1 TO 1
± ~ Ratio Error : Pha.e Displacement at percentaqe of
CLASS at percentaqe of rated current shown below.
rated current shown
KINUTES CENTIRADIANS
10 - 20 20 -100 100-120 10 - 20 20 -100 100-120 10 - 20 20 -100 100-120
NOT INC NOT INC NOT INC NOT INC NOT INC NOT INC NOT INC
20 100 20 100 20 100
0.1 0.25 0.20 0.10· 10 8 5 0.30 0.24 0.15
0.2 0.50 0.35 0.20 20 15 10 0.60 0.45 0.30
0.5 1.00 0.75 0.50 60 60 30 1.80 1.35 0.90
1.0 2.00 1.50 1.00 120 90 60 3.60 2.70 1.80
TABLE 2 - LIMITS OF ERROR FOR ACCURACY CLASSES J AND 5
+/- % Ratio Error
CLASS at percentage of
rated current shown
below.
50 120
J J J
5 5 5
40. ACCURACY LIMIT FACTOR - A.L.F.
(or SATURATION FACTOR)
Ratio of:-
IpRIMARY : IRATED
up to which the C.T. rated accuracy is
maintained.
e.g. 200/ 1A
A.L.F. = 5
will maintain its accuracy for
IpRIMARY < 5 x 200 = 1000 Amps
AP03917
-.
41. BS 3938
1
CLASSES :- 5P 10P 'X', ,
DESIGNATION (CLASSES 5P, 10P)
(RATED VA) (CLASS) (ALF)
AP03548
-MULTI·PLE OF RATED CURRENT (IN) UP TO
WHICH DECLARED ACCURACY WILL BE
MAINTAINED WITH RATED BURDEN
CONNECTED.
---5P OR 10P.
I..--VALUE OF BURDEN IN VA ON WHICH
ACCURACY CLAIMS ARE BASED.
(PREFERRED VALUES:- 2.5, 5, 7.5, 10, 15,
30 VA)
Zs =RATED BURDEN IN OHMS
•
42. •
PROTECTION CURRENT TRANSFORMERS.
TABLE J - LIMITS OF ERROR FOR ACCURACY CLASS SP AND lOP
ACCURACY current Error Phase Displacement at Composite Error
CLASS at rated primary rated primary current (%) at rated
current (%) accuracy limit
MINUTES CENTIRADIANS primary current
SP ± 1 ± 60 ± 1.8 5
lOP ± J 10
43. CURRENT TRANSFORMER BURDEN
Relay burdens are usually quoted in VA (volt - amperes)
A burden of 12.5VA at SA would have an ohmic value of :-
12.5/52 = 0.50
If a relay has a 1VA burden at its 20% setting, then the burden at
nominal current (1A) will be:- 1~x 1/0.22 = 25VA
Conversion from VA and ALF into volts:-
If Ret is known. generally. vK ., ALF(~~+ Rcr x IN )
Example:- 400/1 ct, rated at 15VA 10 P 20 with a Ret of 1.50
Vk ~ 20 (15/1 + 1.5 x 1) = 330v
If Ret is not known, then Vout = 15 / 1 x 20 - 300v
-
44. WHAT t)oES C.'T.. Cl.A5SII=CA,'ON T£'-L us ?
RA"t"e.~ guCt)EN = 2A'£b VA
{ftA"'te.b c:.utUlE)r)2.
= 5 :. 5 C,",MS
-)Z,
MAX c.ud&l (WITH 51. A-c.cu~PCf) =to -, : (Q AI-PS
loA
Vo/p S OHMS
VCIP =50 Vat...S
'1o" = lRATe.~ VA) )( (A.LF. x. AA'"tEb c:u••etrT)
(O"t'Eb c.ulAetrr'f
ALSO, eoMSI~' VA' IIIIl -a.e.LGAb
VA -150. 10 = 'Sao VA.
01 VA =(1lA1'9"A) "R-I-I=: X •.'-~.(~~)
(M1"CL cu~
lltUS, Cel: CIW bEU"ER a ~ VA) ~ (A.L.F.f vA.
(1140 RA,1EO ~JRDEt-:l)
•
45. CURRENT TRANSFORMER DESIGNATION-----,
Consider 5VA 5P10 (with lA secondary)
equivalent circuit - with rated burden :-
•I
I
I
RCT lOA
•I
I
I
Emax Vo/p 5 Ohms ( Rated Burden )
I I
I I
I I
'--_ _~,_ _ _.....;':.._J Vo/p= 50 Volts
Emax= 50 + 10 Re.T
Now consider same C.T with a 2 Ohm burden
RCT I max
Emu
I
1
I
,
ReT = 1 Ohm
ReT = 2 Ohm
ReT = 5 Ohm
vo/ p
I
I
I
,
Imax = Emax
2 Ohms
50 + lOR C.T
Imax
Re.T+ 2
IMAX = 20 Amps
IMAX = 17.5 Amps
IMAX = 14.2 Amps
•
47. •
CURRENT TRANSFORMER DESIGNATION------,
Class "x'
Specified in terms of :-
( i) RATED PRIMARY CURRENT
( ii ) TURNS RATIO (Max. Error =O·25~)
( iii ) KNEE POINT VOLTAGE
( iv ) MAG CURRENT
( AT SPlCIFiED VOLTAGE )
( V ) SECONDARY RESISTANCE (At 75°C)
48. CHOICE OF CURRENT TRANSFORMER-------,
( 1 ) Instantaneous Overcurrent Relays.
- Class P Specification
- A.L.F = 5 Usually sufficient
- For High settings ( 5 -15 times C.T Rating )
A.L.F = Relay Setting
-.
(
. .
) IDNIT Overcurrent Relays.11
- Generally Class 10P
- Class 5P where grading is critical
Note : A.L.F x V.A < 150
( iii ) Differential Protection.
- Class X Specification
- Protection relies on balanced C.T output.
50. OV£Rc..UI2RENT -RELA'f VIe. CHECK.
AsSUME V"UAES: If~..- 722..6 A.
c :T .. ,OCX) Is A
7.5VA tOP20
Rcr" O.Zt;,.JL
Rr - O.~../Z.. (MCGG)
RL - O.1~ ..tL
~ To SEE IF Vie. IS L.AR6e ENou(O.H:
R~I'lED VO&.l"AGE :& Vs = I~ (Rc:r + Rr • R~ )
• 722.fiJ ,..5 ( O.2.G + 0.02. + 0.13)
.CXX>
= 36. f,!, )( O. 4~ - 15.54 V'OI,Zs.
CUIZRENT Vie. APPRO'k11MTES 1t) :-
Vk. ~ VA )C ALF + Rcr Jt I".. ALF
I"
.. ?:P w 20 + 0 .~ K.5)t 20 ~ .EG VOL."1'S.
15
Vk. > Vs -r.HE~~E C:T {k. 15 ..De.~u~.
•
51. ~ore F=OR.. E-AJCr+l FAUL7 APPUCATlON!S ReGlUIRIi
~ &E ... E 10 ~ 10" 'lELA~ SETTING.
OfecK 1C see. IF V" IS l..AR.Ge. ENOU;'-H:
Tar~ lND CONWECTEO = 2Rl + Ra ... ZRr
- a If O. 1!S + 0 .2tD + 2. v. O. 02.
::.EG .~!.. ~ A
0."
-r'lPICAL e~ F"AUL-T seiliNG .:; 3:)-z..'I~
= '..51..
~E~£FoRe C.T cAN- ____-----______----______-'-A__
-
-
52. CHOICE OF CURRENT TRANSFORMER------.
CONSTANT V.A. VARIABLE OPERATING
CURRENT RELAYS.
Consider a constant V.A relay as a secondary burden
( Neglect C.T. secondary impedance & lead burden )
Let Is = C.T. secondary current.
ZR = relay impedance at setting.
I R = relay current setting.
VR = voltage which has to be applied to the relay
when set at I to cause operation.
I M = magnetising current of C.T. corresponding to V
Y = magnetising admittance of C.T.
VA = Volt - Amperes required operate the relay.
•
53. CHOICE OF CURRENT TRANSFORMER-------,
1M = VRY = VA . Y
IR
Also 1s = IR + IM = IR + VA . Y ................ 1
IR
Differentiating with respect to 1R
dIs = 1 - VA . Y
dI R IR2
Now, for a minimum value of I R
1 - VA Y = 0
J2R
IR2= VA Y
Substituting IR in equation 1
Is(min) = .JVA . yO + VA Y - 2 .JVA
-v':::;V=A=Y;:;o
Also 1M = lR =.JVA . yO
Y
Since Is(min) = 2 .JVA Y and IR = .JVA yO
and 1M = 1 = VR· Y = ZR . Y
1R VR/ZR
:. ZRY = 1
Minimum primary current for relay operation as
setting is varied occurs when the shunt mag
impedance and relay impedance are matched.
This coincides with the min VA output to the relay.
-
-
54. •
CHOICE OF CURRENT TRANSFORMER--------.
Ampere-Turns IT IT IT IT
Operating Current I 1/2 1/3 1/4
Turns on Relay T 2T 3T 4T
Reactance of relay X 4X 9X 16X
prop to turns sq
Operating Voltage IX 2IX 3IX 4IX
VA r2
x r2
X r2
X r2
X
Is=Ip/N
•• ,I
Vr
I,
a:
:>
4rX
d
(I.)
3IX
c
0.0
Excitation~
..,.J
0
2IX
Characteristic.
:>
~
::1
0...
..,.J
IX
:::l
0
abc d
magnetisng current -1m
55. CHOICE OF CURRENT TRANSFORMER-------,
.,J
~
cu
s..
s..
::t
U
Jp/N
________________~•• JP/N
-Is --------I~~ ----I.~ J.
~~----------~~ Is~----~e~====~------~p
[.
(b)
Es
z..-.
:::z
-+
>..,.E!
s.. ---«3
S II
s.. 0..
a.. -
Relay Operating Current IR
Le. Optimum relay performance is obtained at an intermediate
tap setting and not at the minimum current setting.
•
59. ME"~oSIL.$
-
* NON LI NEAl tlESIS-rANC€
~ bEPOIf)S o#tJ SI,E ANb
St-4APE.
vocrNrC Re"~I~ -ro PASS A +h~,l,
-neE' blS e. VAilE'S' ~lo~ 00 Fo~ A -rt4,IJ blSc:.
'-0 'Sooc:> Fo~ A THiele blSC
It exAMPLE SPEe)Flc.A.,..,o ~
bOO A Iss / PIS 17..7..2
600 - t:>sc: I),AMETEl ,.tItS.)("co.
3 - N0f16eL ot:- ~'Sc~
p _ AL.L.. ~IS:S- 'N PAIl.ALl.El...
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V-I Curve of 600AIS11Spec 256 Metros"!
10mA 100mA 1A 10A
DC or PEAK CURRENT
100A
V-I Curve of 600AIS11 Spec 1088 Metrosi'
_ ... _ f--
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•
61. METROSIL FOR CAG TYPE RELAYS
Introduction
The 'Metrosii' non-linear resistor units listed below have been specially designed for use with
GEC ALSTHOM T & D Protection & Control type CAG 14 and similar relays and are intended
to be connected across the relay circuit (relay and stabilising resistor).
Metrosil Units for Relays with a 1 Amp CT
The Metrosil units used with 1 Amp CT's
have been designed to comply with the
following requirements:-
1) At the relay voltage setting, the
Metrosil current should be less than
30mA rms.
2) At the maximum secondary internal
fault current the Metrosil unit should limit
the voltage to 1500V rms if possible.
The Metrosil units normally recommended for use with 1Amp CT's are as follows:-
Nominal Characteristic Recommended Metrosil Type
Relay Voltage Setting
C (3 Single Pole Relay Triple Pole Relay
Up to 125V rms 450 0.25 600AlS1/S256 600AlS3111SS02
125 - 300V rms 900 0.25 600AlS1/S10SS 600AlS3/1/S1195
Note: Single pole. Metrosil units are normally supplied without mounting brackets unless otherwise specified by the customer.
Metrosil Units for Relays with a
5 Amp CT
These Metrosil units have been designed
to comply with the following
requirements:-
100mA rms (the actual maximum
currents passed by the units is shown
below their type description).
At the higher relay voltage settings, it is
not possible to limit the fault voltage to
1500V rms hence higher fault voltages
have to be tolerated (indicated by', **,
***).
1) At the relay voltage setting, the
Metrosil current should be less than
2) At the maximum secondary internal
fault current the Metrosil unit should limit
the voltage to 1500V rms for 0.25secs.
The Metrosil units normally recommended for use with 5 Amp CT's and single pole relays are as follows:-
Secondary internal fault RECOMMENDED METROSIL TYPE
current
Relay Voltage Setting
Amps rms Up to 200V rms 250V rms 275V rms
600A/S1/S1213 600AlSlIS1214 600AlSlIS1214
50A C = 5401640 C = 670/S00 C = 670/800
35mA rms 40mA rms 50mA rms
600AlS2/P/S1217 600AlS2/P/S1215 600AlS2/P/S1215
100A C =4701540 C = 570/670 C = 570/670
70mA rms 75mA rms 100mA rms
600A/S3/P/S1219 600AlS3/PIS1220 600AlS3/P/S1221
150A C = 4301500 C = 520/620 C = 570/670 **
100mA rms 100mA rms 100mA rms
300V rms
600A/S1/S1223
C = 740/870' -
50mA rms
600A/S2/P/S1196
C = 6201740 *
100mA rms
600A/S3/P/S1222
C = 6201740 *..
100mA rms
In some situations single disc assemblies may be acceptable, 2400V peak •• 2200V peak ••• 2600V peak
contact GEC ALSTHOM T & D Protection & Control for detailed applications.
Notes:
1) The Metrosil units recommended for
LIse with 5 Amp CT's can also be
applied for use with triple pole relays
and consist of three single pole units
mounted on the same central stud but
electrically insulated from each other. To
order these units please specify "Triple
pole Metrosil type" followed by the single
pole type reference.
2) Metrosil units for higher relay voltage
settings and fault currents can be
supplied if required.
-
62. METROSIL FOR MFAC TYPE RELAYS
IN SOME APPLICATIONS OF
VOLTAGE OPERATED HIGH
IMPEDANCE RELAYS A NON-LINEAR
RESISTOR IS REQUIRED TO LIMIT
THE CURRENT TRANSFORMER
SECONDARY VOLTAGE TO A SAFE
LEVEL DURING MAXIMUM INTERNAL
FAULT CONDITION.
Introduction
The GEC ALSTHOM T & D Protection
& Control type MFAC High Impedance
Differential Relay has three setting
ranges 25V to 175V in 25V steps,
25V to 325V in 50V steps and 15V to
185V in 5V steps.
Single element or three element
'Metrosil' units are provided for use
with single and triple pole relays
,..-espectively. The Metrosil unit chosen
3pends on the setting range used and
the required short time rating.
Recommended Metrosil Units for MFAC Relay
Relay setting range
/(fIJt) - +c:>t:">
25V - 175V
15V - 185V
25V - 325V
Continuous Ratings
I 'C' Characteristic
450
900
Short Time Ratings
Metrosil Assembly Type
Standard 600A/S1/S256
Standard 600A/S1/S1088
Special 600A/S2/S
comprises two standard
C = 450 discs in series
Special600A/S2/P/S6198
II.. comprises two special
discs in parallel
I
Nominal
C
)/C510
450
450
900
Characteristic
B
,.,-2S
0.25
0.25
0.25
Photo Courtesy of GEC ALSTHOM T & D Protection & Control
1 phase relay
IdCiV~/J//Sh3's
600ArS-l/S256
600AlS1/S256
600AlS1/S1088
Max. Continuous Rating rms
200 V
350 V
3 phase relay
6ooAIS311IS~37.4
600A/S;J/1/SHU2
600A/S3/1/S1195
Unit 'C' Value Short Time Rating
450 22A for 3 seconds
30A for 2 seconds
45A for 1 second
900 17A for 3 seconds
30A for 1.5 seconds
39A for 1 second
900 22A for 3 seconds
30A for 2 seconds
45A for 1 second
900 30A for 3 seconds
SOA for 2 seconds
90A for 1 second
Where higher ratings are required, special Metrosil units can be provided with more discs in parallel per elementto suit a particular application.
•
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No.tMAL. sE'tv,e~ ~- -reST WINO' NCf- ole.·
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:t:.s = ~~ + I<c:.~ • IT
::'"~ RC.T~=
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64. FAULT CURRENT IN POWER SYSTEM
(PRMARY CURRENT IN CT)
-i,t/L,
'-I = ... v,.. stn (",,1-..-9-¢) - '!.!:J SIn (9-s6).e
r, %,
1 " _f.t/I.,
= + II sin (wt-+-e-cp) - I, s.i le-c;).e.
i = YM
t ~,
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65. REPRESENTATON OF CURRENT TRANSFORMER
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68. SIMPLIFIED CoT E~JIJALENT C'RCUIT
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low +eQ<::hlnee .
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