Transformer Protection

transcript

TRANSFORMER PROTECTION

TRANSFORMERS ARE EVERY WHERE :

RANGES FROM 3 kVa to 500 MVA

GENERALLY OF 2 TYPES :

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

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

In addition to that the load current of Phase A

is still supplying

PHASE TO GROUND FAULT

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

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

TRANSFORMER DAMAGE CURVE

DOWNSTREAM FEEDER

PROTECTION

What is the backup Protection if the feederProtection fail ?

132 kV Bus

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

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.

TRANSFORMER DAMAGE

SEC. AMPS × 1000

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

RECLOSER FUSE

X-FOR.DAMAG

51 51 G

So the relay will not operate

XI1 I2

Differential Protection

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

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

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

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

Variable Restraint(%)

OperateCurrent(Iop)

Restraint Current (IR)

TRANSFORMER DIFFERENTIAL PROTECTION

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.

PHASE SHIFT IN DELTA -WYE TRANFOMERS

COMPENSATE BY CAREFUL CONNECTION OF THE SECONDARIES OF CTS

PHASING

DIFFERENTIAL RELAY

DELTA-DELTA CONNECTION WYE- CONNECTION IN CTS SECONDERIES

CT SECONDERIES are connected to the respective phases Restrained coils

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)

132 kV 33 kV

Correction of

30 degree phase Shift

DIFFERENTIAL RELAY

IA - IB

IB - IC

IC - IA

PRIMARY SIDE CT CONNECTION

IA - IB

IB - IC

IC - IA

IA - IB

IC - IA

I aI aSECONDARY SIDE CT CONNECTION

Ic-Ia Ib-Ic Ia-Ib

Ic-Ia Ib-IcIa-Ib

IA - IB

IB - IC

IC - IA

132 kV 33 kV

C.T RATIO SELECTED SO THAT PRIMARY AND SECONDARY CURRENTSMATCH AT RELAY

DIFFERENTIAL RELAY

IA - IBIB - IC

Ia -IbIb -IcIC - IA IC - Ia

132 kV 33 kV

50MVAOP

LOAD ZEROIf

INTERNAL GROUND FAULT AT PHASE C

INCREASED FAULT CURRENT AT PRIMARY PHASE B & C

NO BACK FEED BECAUSE OF RADIAL FEEDER RELAY @ B & C WILL OPERATE

132 kV 33 kV

50MVAOP

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.

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

Transformer Protection

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