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TRANSFORMER PROTECTION
TRANSFORMERS ARE EVERY WHERE :
RANGES FROM 3 kVa to 500 MVA
GENERALLY OF 2 TYPES :
A
B
C
a
b
c
According to the IEEE standard : High side Leads low side by 30◦For Simplified Connection .
TRANSFORMER MUST BE PROTECTED
AGAINST INTERNAL FAULTS
THROUGH FAULTS
The Reasons For internal faults :
• insulation failure which creates a short circuit path between phasesor grounded causing heavy fault current flowing can serious damage to the winding and even burn the core itself .
What is the action ?Transformer must be immediately switched out of service
WHAT IS THROUGH FAULT ?
132 kV Bus
132 kV Bus
132 kV Bus
132 kV Bus
132 kV Bus
132 kV Bus
132 kV Bus
132 kV Bus
132 kV Bus
132 kV Bus
132 kV Bus
132 kV Bus
132 kV Bus
132 kV Bus
132 kV Bus
How the protection are applied in Through Fault ?
132 kV Bus
CLEARED BY THE PROTECTIVE SCHEME IN
THE FEEDER PROTECTION
If the fault is failed to clear
by the feeder protection
Then what happen ?
132 kV Bus
Heavy fault current in the
secondary winding
132 kV Bus
Heavy fault current flows through the
primary corresponds to secondary winding
THREE PHASE FAULT
THE PHASE FAULT
TWOPHASE FAULT
A
B
C
a
b
c
TWOPHASE FAULT
A
B
C
a
b
c
TWOPHASE FAULT
A
B
C
a
b
c
TWOPHASE FAULT
A
B
C
a
b
c
TWOPHASE FAULT
A
B
C
a
b
c
TWOPHASE FAULT
A
B
C
a
b
c
TWOPHASE FAULT
A
B
C
a
b
c
TWOPHASE FAULT
A
B
C
a
b
c
TWOPHASE FAULT
A
B
C
a
b
c
TWOPHASE FAULT
A
B
C
a
b
c
TWOPHASE FAULT
A
B
C
a
b
c
In addition to that the load current of Phase A
is still supplying
PHASE TO GROUND FAULT
A
B
C
a
b
c
PHASE TO GROUND FAULT
A
B
C
a
b
c
PHASE TO GROUND FAULT
A
B
C
a
b
c
PHASE TO GROUND FAULT
A
B
C
a
b
c
PHASE TO GROUND FAULT
A
B
C
a
b
c
PHASE TO GROUND FAULT
A
B
C
a
b
c
PHASE TO GROUND FAULT
A
B
C
a
b
c
PHASE TO GROUND FAULT
A
B
C
a
b
c
PHASE TO GROUND FAULT
A
B
C
a
b
c
PHASE TO GROUND FAULT
A
B
C
a
b
c
PHASE TO GROUND FAULT
A
B
C
a
b
c
This causes ,
Thermal & Mechanical Damages
Thermal damages due to the high temperature resulting from high over current
Mechanical damages causes by physical movement of the winding
Both cause rapid deterioration or breakdown of the insulation .
20001000 500 200 100 50 20 10 5 2 1 0.5 0.2 0.1 1 2 5 10 50
TRANSFORMER THROUGH – FAULT
PROTECTION CURVE TI
ME
IN S
ECON
D
MULTIPLES OF BASE CURRENT
How can we prevent it ?
132 kV Bus
20001000 500 200 100 50 20 10 5 2 1 0.5 0.2 0.1 1 2 5 10 50
TIM
E IN
SEC
OND
MULTIPLES OF BASE CURRENT
TRANSFORMER DAMAGE CURVE
DOWNSTREAM FEEDER
PROTECTION
What is the backup Protection if the feederProtection fail ?
132 kV Bus
132 kV Bus
51
132 kV Bus
51
132 kV Bus
FUSE
HOW CAN WE CO-ORDINATE THE O/C RELAY OR FUSE
20001000 500 200 100 50 20 10 5 2 1 0.5 0.2 0.1 1 2 5 10 50
TIM
E IN
SEC
OND
MULTIPLES OF BASE CURRENT
TRANSFORMER DAMAGE CURVE
DOWNSTREAM FEEDER
PROTECTION
HV FUSE
1000 500 200 100 50 20 10 5 2 1 0.5 0.2 0.1 .05.C
.02 .01 .1 .5 2 10 50 20 100 . 2 1 50 20 100 500
FEEDER BKR
MAIN BKR.
FUSE
TRANSFORMER DAMAGE
SEC. AMPS × 1000
TIM
E IN
SEC
ONDS
RECLOSER RECLOSER
1000 500 200 100 50 20 10 5 2 1 0.5 0.2 0.1 .05.C
.02 .01 .1 .5 2 10 50 20 100 . 2 1 50 20 100 500 SEC. AMPS × 1000
TIM
E IN
SEC
ONDS
RECLOSER FUSE
X-FOR.DAMAG
E
51
51N
51 51 G
So the relay will not operate
XI1 I2
i1i2
TC
i1
i2
Differential Protection
i1 i2
Diff. Relay
CT1 CT2
X protected Equipment CT1 and CT2 same transformation ratio
Current i1 and i2 are equal in magnitude and opposite in direction.So, the net current in the differential coil is zero at load condition (without fault),
Differential Protection XI1 I2
i1i2
TC
i1
i2
i1 i2
Diff. Relay
CT1 CT2
External Fault happens ,
I1 and I2
i1 and i2 But i1 = i2
Current in TC= 0 The Relay will not operate
XI1 I2
i1i2
TC
i1
i2
i1 i2
Diff. Relay
CT1 CT2
Differential Protection
INTERNAL FAULT HAPPES
Current in TC= i1+i2
Which Is
very High
TRIPS THE DIFFERENTIAL REALY
Biased differential relay
XI1 I2
i1i2
Op
i1
i2
i1 i2
Biased Diff. Relay
CT1 CT2
Res. Res.
Large external fault cause false operation To make more stable
Two Restraining ( Biasing ) coil
One Operating coil is introduced
What is the function of two Restraining ( Biasing ) Coils ?
Restraining coils will oppose the operation of operating coil
The relay will operate only when the operating force > the restraining force
50%
40%
25%
10%
Variable Restraint(%)
OperateCurrent(Iop)
Restraint Current (IR)
TRANSFORMER DIFFERENTIAL PROTECTION
87T
MAGNETIZING INRUSH CURRENT .
PHASE SHIFT IN WYE-DELTA TRANFOMERS C.T CONNECTIONS (PHASENG)
EFFECT OF TRANSFORMER TAPS
FACTORS TO BE CONSIDERED :
SOURCE INRUSH LOAD
6 TO 10 TIMES OF FULL LOAD CURRENT
MAGNETIZING INRUSH CURRENT .
High in Second Harmonics
INRUSH CURRENT
VOLTAGE
Remedy
Harmonic restraint was added to make the transformer differential relay less likely to operate on transformer inrush current.
87
PHASE SHIFT IN DELTA -WYE TRANFOMERS
COMPENSATE BY CAREFUL CONNECTION OF THE SECONDARIES OF CTS
PHASING
A
B
C
a
b
c
OP
OP
OP
R
R
R
R
R
R
DIFFERENTIAL RELAY
DELTA-DELTA CONNECTION WYE- CONNECTION IN CTS SECONDERIES
CT SECONDERIES are connected to the respective phases Restrained coils
A
B
C
a
b
c
OP
OP
OP
R
R
R
R
R
R
DIFFERENTIAL RELAY
WYE – WYE CONNECTION
CT SECONDARIES ARE CONNECTED IN DELTA-DELTA WHY ?
IF WYE-WYE CONNECTED THEN ZERO SEQUENCE CURRENT MAY FLOW THROUGH GROUND AND CAUSE UNWANTED TRIPPING FOR THROUGH FAULT (EXTENAL FAULT)
A
B
C
a
b
c
132 kV 33 kV
50MVA
Correction of
30 degree phase Shift
OP
OP
OP
R
R
R
R
R
R
DIFFERENTIAL RELAY
A A
A
B B
B
C C
C
IA - IB
IA - IB
IA
IA
IB - IC
IB - IC
IB
IB
IC - IA
IC - IA
IC
IC
PRIMARY SIDE CT CONNECTION
IA
IB
IC
IA - IB
IB - IC
A
B
C
A
B
C
I A -
I B
IB - IC
I C -
I A
IC - IA
I B -
I C
IA - IB
IC - IA
I aI aSECONDARY SIDE CT CONNECTION
Ia
Ib
Ic
Ia
Ib
Ic
Ia
Ic
Ic-Ia Ib-Ic Ia-Ib
OP
OP
OP
R
R
R
R
R
R
Ia
Ib
Ic
Ia
Ib
Ic
Ia
Ic
Ic-Ia Ib-IcIa-Ib
IA
IB
IC
IA - IB
IB - IC
A
B
C
A
B
C
I A -
I B
I C -
I A
IC - IA
I B -
I C
A
B
C
a
b
c
132 kV 33 kV
50MVA
C.T RATIO SELECTED SO THAT PRIMARY AND SECONDARY CURRENTSMATCH AT RELAY
OP
OP
OP
R
R
R
R
R
R
DIFFERENTIAL RELAY
IA - IBIB - IC
Ia -IbIb -IcIC - IA IC - Ia
A
B
C
a
b
c
132 kV 33 kV
50MVAOP
OP
OP
R
R
R
R
R
R
LOAD ZEROIf
INTERNAL GROUND FAULT AT PHASE C
If
If
If
INCREASED FAULT CURRENT AT PRIMARY PHASE B & C
I=0
I=0
I=0
NO BACK FEED BECAUSE OF RADIAL FEEDER RELAY @ B & C WILL OPERATE
A
B
C
a
b
c
132 kV 33 kV
50MVAOP
OP
OP
R
R
R
R
R
RFOR OTHER EXTERNAL FAULT
CURRENT BALANCE ON BOTH SIDE OF THE RELAY SO THE RELAY WILL NOT OPERATE
TRANSFORMER OVER-ALL
PROTECTION
TYPICAL ARRANGMENTOF X-FOMERPROTECTIONFOR 10 MVAOR ABOVE
BOTH PRIMARY AND SECONDARY CKT BRK. TRIPPED BY ANY INTERNAL FAULT RELAY 49 – THARMAL REALY RELAY 63 – BUCHHOLZ RELAY
DIFFERENTAL PROTECTION IS FOR Ø – Ø OR Ø – GRND. FAULT- I RY & II RY BRK.OF X-FOR.
BACK UP OC –GRND. RELAY FOR PRI. CKT. BKR.52P AND BACK OF GRND. FAULT II RY BUS & FEEDERS
Grounded Neutral of tr.
Sec.
OC –GRND. RELAY FOR PRI. CKT. BKR.52S AOC –GRND. ARE CONNNTED TO IRY & IIRY CKT. FOR BACKUP PROTE. BUS & FEEDERS
BACK-UPGRND.
PROTECTION
FOR IRY CKT.
THARMAL REALY
DETECTS OVER HEATING OF THE OIL
CAUSESE : 1) CONTINIOUS OVER LOAD 2) INTERNAL FAULT IN THE WINDING. AND SO O.
IT ANNOUCES ALARMTO THE CONTROL ROOM
OR START THE FORCECOOLING SYSTEM OF THETRANSFORMER
VERY SENSITIVEOPERATES BEFORE DIFFERENTIAL RELAY