HIGH IMPEDENCE BUSBAR PROTECTION High Impedance Bus bar Pr!ec!in Two type of relays 1. Voltage Operated 2. Current Operated Basic circuit of High Impedance Bus bar rotection CURRENT OPERATED
"O#TAGE OPERATED
S!abi$i%ing &$!age The !oltage de!eloped across the CT secondary during ma"imum fault condition. Vs # If $ % &ct ' %2 $ & ()) I* &ct & (
+
,a"imum fault current
+ CT secondary &esistance
+ (ead resistance of cable
The ma'imum (au$! curren! )I * + The ma"imum fault current %I * ) or ,a"imum short circuit current is the base !alue for design the substation system. The -uipment specifications and relay settings are calculated based on the ,a"imum fault current !alue of the system. The ,a"imum fault current is specified in the standard for each system !oltage *or -"ample / In 0-C standard T-0 + +11.2 C(340- 5.2.6 070T-, VO(T38-
0HO&T CI&C4IT C4&&-9T
:5 ;V 2: ;V 11= ;V :6.= ;V 1:.5 ;V
<:;3 <:;3 6 ;3 2= ;3 21 ;3
Consider :5;V system I* + <;3 I( + ,a"imum fault current in CT secondary > 21 3 for CT :?1 Tap selection Rc! , CT secndar- Resis!ance &ct can get from the CT specification Consider :?1 Tap used for Bus bar protection and &ct > 6.<= ohms R # , #ead resis!ance ( cab$e E'amp$e. < 0mm cable &( > .:5 ohm?meter at 2 dec C &( at @= dec C > &( ) 1' A 2 %T2+T1)+
Consider length > =m &( at @= dec C >1.55 ohms S!abi$i%ing &$!age Vs # I* $ % &ct ' %2 $ & ()) Vs > 21 % 6.<= ' % 2 1.55 )) > 1@@ !olts Consider 2 ,argin > 1@@ " 1.2 > 211.:2 !olts Hence se$ec! "s / 012 &$!s Therefore in ma"imum fault current the !oltage de!eloped across the CT secondary at the relay end is 2@= V. "O#TAGE OPERATE RE#A3 %,*3C + 3&-V3) Calculate !oltage actually need to operate the relay or total !oltage drop in the secondary circuit during the ,a"imum fault condition Is > Total secondary current drop Is > CT ,agnetiDing current ' CT super!ision relay coil current % ,VT) ' rotection relay coil current for set !alue %,*C3 0-TTI98 &398- 2=+:2= V) ' ,etrosil lea;age current > Im ' I m!tp 'Ir ' Imetrosil Consider 5 CTs are connected parallel > %.= " 5) ' .: ' .: ' .15 Is > .1 3 This is the secondary current drop during the fault condition. In other words this is the current enough to operate the protection relay. Con!ert this current to rimary !alue in :?1 ratio CT Ip > Is " : > 2@: 3mps Voltage for <:3 is 2@=V
Voltage for 2@:3 is E <: ? 2@: >2@=? Vsec Vsec > 1.2 !olts This !oltage is enough to operate the relay. This !oltage and current is too small compared to ,a"imum fault current and corresponding !oltage. This secondary operating current can achie!e e!en lea;age or through fault. 0o we ha!e to decide the fault current at which the relay should operate Shun! Resis!r 0ay The primary fault current @623 at which the relay should operate. This is called bias !alue and up to this !alue relay should not operate. To achie!e this condition add some resistance and increase the Is secondary lea;age current !alue. This -"ternal resistor should perllaly connect with &elay. Ip > %Is ' %Vs?&sh)) " : @62 > %Is ' 2@=?&sh) " : &sh > 1@=.@ ohms 0o select the 2@ ohms Variable resister and adFust the !alue. 3dFust the resistance !alue according to our primary operating fault current reuirement . E'amp$e Mani(a *ina$ re$a- se!!ing (r M*AC re$a- Vs > 1@= V Ir > .23 &sr > 5= Omhs Therefore Ip > %Is 'Vs?&sh) " : Ip > %%%.= " 5) ' .: ' .: ' .15) ' %1@=?5=) ) " :) >555 3mps should be de!elop in primary and 45066 3mps should be flow through the 0econdary circuit to operate the relay with this setting.
CURRENT OPERATE RE#A3 %,C38 + 3&-V3) 3ll other calculations are same e"cept 0tabilising &esistor &elay setting Voltage Vs > 2@= Volts &elay setting current Ir > .2 3mps &elay burden at 2@=V setting is consider as 1 V3 0tabilising resistor reuired &sr > % Vs?Ir) + % &elay burden + IrG2) &sr > 1:= Omhs 0elect the resistor range +1= Ohms CT SUPER"ISION RE#A3 This relay parallely connected with ,ain protection relay
Consider & phase wire open in CT +2. The CT+1 load current will start to flow through both relays. This is not a fault current but it causes unnecessary trip. To a!oid this trip CT super!ision relay is used. The !oltage setting of this relay is !ery low with time delay.
0o this relay will operated for lea;age current and its contact short the CT. ,ain relay setting is 8rater then this !alue and instantaneous. 0o for hea!y internal busbar fault ,ain protection will operate immediately. *inal relay setting Vs>16 ! Time > : sec Open & phase CT will sa!e by metrosil mounted near the CT marshaling bo". METROSI# This is non liner resistor uring Hea!y fault condition rotection relay will immediately clear the fault . If fault not clear and secondary fault !oltage will damage the relay. This ;ind of situation ,etrosil will short the CT secondary circuit and sa!e the relay coil. 0ome times circuit is open due to relay coil damage. In this situation if fault occur hea!y !oltage will de!elop and CT get damage. To a!oid this metosil will short the circuit and sa!e the CT. ,etrosil should withstand the for !oltage de!elop due to ma"imum shout circuit stability !oltage Vs. Busbar S!abi$i!- Tes! 75 Curren! pera!ed Re$a- 1. Calculate the rimary operating current for relay setting !alue current -"/ &elay setting 2 CT ratio > :?1 2 of secondary current .23mps rimary Current < 3mps 2. 0hort the 0tabilising resistor with small shorting wire :. rown out the CT super!ision &elay to reduce the load for rimary inFection ;it. 6. rown out the ,ain protection relay and circuit will short in side and get closed path. =. Open the metrosil one side wire % ont short the ,etrosil) =. 0lowly inFect the primary current and reach .2 3mpls secondary <. 9ow inset the rotection relay let it operate and measure the current.
@. rimary current !alue will if ,etrosil J &esistor J CT super!ision relay are include in the circuit. 05 "$!age pera!ed Re$a- 1. Calculate the rimary operating current for relay setting !alue !oltage -"/ &elay setting 1@=! &sh > 1= Ohms CT ratio > :?1 Ip > %Is 'Vs?&sh) " : > %.1' .11<<) " : > <2: 3mps 0econdary current .23mps :. rown out the CT super!ision &elay to reduce the load for rimary inFection ;it. 6. rown out the ,ain protection relay and circuit will short in side and get closed path. =. Open the metrosil one side wire % ont short the ,etrosil) =. 0lowly inFect the primary current and reach .2 3mpls secondary <. 9ow inset the rotection relay let it operate and measure the Voltage.
Dn8! in9ec! !he curren! cn!inuus$- ! !he re$a-5 Because re$a- :i$$ ge! damage5
