transformer protn. areva

Transformer Protection

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Alan WixonSenior Applications Engineer

> Transformer Protection - January 20043 3

Transformer Fault Categories

1. Winding and terminal faults

2. Sustained or uncleared external faults

3. Abnormal operating conditions such as overload, overvoltage and overfluxing

4. Core faults


- “Clock Face” numbers refer to position of low voltage phase-neutral vector with respect to high voltage phase neutral vector

- Line connections made to highest numbered winding terminal available

- Line phase designation is same as windingExample 1 : Dy 11 Transformer

B1B2 b2b1

a2a1A1A2

C1C2 c2c1

Question : How to connect windings ?

High Voltage

Windings

Low VoltageWindings

A PhaseWindings

B PhaseWindings

C PhaseWindings

Transformers Connections


1. Draw Phase-Neutral Voltage Vectors

30°

A

C B

a

c

b

Line Designation



2. Draw Delta Connection

A

C B

a

c

b



3. Draw A Phase Windings

A

C B

a

c

b

a2

a1

A2

A1



4. Complete Connections (a)

A

C B

a

c

b

a2

a1

A2

A1

C1

C2

B1 B2

b2b1c1

c2



4. Complete Connections (b)

c1 c2

b1 b2

a1 a2 a

b

c

A

B

CC2 C1

B2 B1

A2 A1



Phase displacement0Group 1

Phase displacement180Group 2

Lag phase displacement30Group 3

Lead phase displacement30Group 4

Yy0Dd0Zd0Yy6Dd6Dz6Yd1Dy1Yz1

Yd11Dy11Yz11

Transformer Vector Groups


Overcurrent & Earthfault Protection


11kV Distribution Transformers Typical Fuse Ratings

Transformer rating Fuse

kVA

100

200

300

500

1000

5.25

10.5

15.8

26.2

52.5

16

25

36

50

90

3.0

3.0

10.0

20.0

30.0

Full loadcurrent (A)

Ratedcurrent (A)

Operating time at3 x rating(s)


Transformer Overcurrent Protection

Requirements

Fast operation for primary short circuits

Discrimination with downstream protections

Operation within transformer withstand

Non-operation for short or long term overloads

Non-operation for magnetising inrush


Use of Instantaneous Overcurrent Protection

Source HVLV

50

51

50 set to 1.2 - 1.3 x through fault level


Concerns relay response to offset waveforms(DC transient)

Definition

I1 - I2I2

x 100

I1 = Steady state rmspick up current

I2 = Fully offset rms pickup current

I1 I2D.C

.

Transient Overreach


Earthfault on Transformer Winding

: 3 ratio turns Effective

. 3

. current Fault : at fault a For

. x 3

current,primary Thus, values load full to current E/F limits Resistor

F.L.2

F.L.

F.L.P

χ

χχχ

χχ

=

Ι=Ι=

Ι=Ι

secondary C.T. then r,transforme of current loadfull on based is side)primary (on ratio C.T. If

3 circuit

2χ=

1 p.u. turns√3 p.u. turns

xIP

PRR IFProtective

Relay


Overcurrent Relay Setting > IF.L.

χIF

51

OvercurrentRelay

If as multipleof IF.L.1.00.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

Star Side

Delta Side

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0χ p.u..

Overcurrent Relay Setting > IF.L.


χIF

109

8

7

6

5

4

3

2

1

Star Side

Delta Side

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0χ

p.u..

If as multipleof IF.L.

51

OvercurrentRelay

Overcurrent Relay Sensitivity to Earth Faults


χ

IF

IP IN

109

8

7

654321

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0χ

p.u..

If as multipleof IF.L.

IN

IP

IF

Overcurrent Relay Sensitivity to Earth Faults


Star Winding REF

ProtectedZone

REF

Relay only operates for earthfaults within protected zone.Uses high impedance principle.Stability level : usually maximum through fault level of

transformer


Restricted E/F Protection Low Voltage Windings (1)

A B C N

LV restricted E/Fprotection tripsboth HV and LV breaker

Recommended setting : 10% rated


Restricted E/F Protection Low Voltage Windings (2)

A B C N

LV restricted E/F protection trips both HV and LV breakerRecommended setting : 10% rated


Delta Winding Restricted Earth Fault

Delta winding cannot supply zero sequence current to system

Stability : Consider max LV fault level Recommended setting : less than 30% minimum

earth fault level

Protected zoneREF

Source


High Impedance Principle (2)

Maximum voltage across relay circuit, Vs = If (RCT + 2RL)

To limit current through relay to < Is the relay impedance Rrelay > Vs/Is

ProtectedCircuit

ZM RCT

RL

ZMRCT

RLIF

VS

RL RL

IS

R


Non-Linear Resistors (Metrosils)

During internal faults the high impedance relay circuit constitutes an excessive burden to the CT’s.A very high voltage develops across the relay circuit and the CT’s which can damage the insulation of CT’s, secondary wiring and relay.Magnitude of peak voltage VP is given by an approximate formula (based on experimental results)

VP = 2 √2VK (VF - VK)Where VF = If (RCT + 2RL + RSTAB + Rrelay)Metrosil required if VP > 3kV


High Impedance Circuit Arrangement

Metrosil Characteristic

V = C I β

Suitable values of C & β chosen based on :(1) Max secondary current under fault conditions(2) Relay setting voltage

I

RST

VSVM

RR


Unrestricted Earthfault Protection (1)

51 515151N

- Provides back-up protection for system- Time delay required for co-ordination


Unrestricted Earthfault Protection (2)

- Can provide better sensitivity(C.T. ratio not related to full load current)(Improved “effective” setting)

- Provides back up protection for transformer and system

51 515151N51N


ANT1

Bus Section

T2

415 VoltSwitchboard

CB

Parallel Transformers (LV)


ANT1

Bus SectionOpen

T2

415 Volt Switchboard

CB

51N

51N

Parallel Transformers - CT In Earth


ANT1

Bus SectionOpen

T2

415 Volt Switchboard

CB

51N

51N

Parallel Transformers - CT In Earth & Neutral


ZG1289C

Bus section

415 volt switchboard

T1 N A B C

F2

F1

Will maloperate if bus section is openfor fault at F1

No maloperation for fault at F2 (but setting must begreater than load neutral current)

T2

Parallel Transformers - Residual Connection


Differential Protection



Overall differential protection may be justified for larger transformers (generally > 5MVA).

Provides fast operation on any windingMeasuring principle :

Based on the same circulating current principle as the restricted earth fault protectionHowever, it employs the biasing technique, to maintain stability for heavy thro’ fault current

Biasing allows mismatch between CT outputs.It is essential for transformers with tap changing facility.Another important requirement of transformer differential protection is immunity to magnetising in rush current.


Biased Differential Scheme

Bias = Differential (or Spill) CurrentMean Through Current

BIASBIAS I2I1

OPERATEI1 - I2

DifferentialCurrent

OPERATE

RESTRAIN

Mean Thro Current

I1 + I22

I1 - I2



LV

R

HV

PROTECTED ZONE

Correct application of differential protection requires CT ratio and winding connections to match those of transformer.CT secondary circuit should be a “replica” of primary system.Consider :

(1) Difference in current magnitude(2) Phase shift(3) Zero sequence currents


Earthfault on Transformer Winding

I

DifferentialRelaySetting = IS

χ

protected not is ng windiof 59% Thus59% i.e.

operation for 20% 3

then 20%, If e.g.

operation,relay For

3

2S

S

2

>

>=Ι

Ι>Ι

=Ι

χ

χ

χ

Differential Relay Setting % of Star Winding Protected

10% 58%20% 41%30% 28%40% 17%50% 7%


Differential Connections

P1 P2 A2 A1 a1 a2 P2 P1

A B C


Use of Interposing CT

RR

R

P1 P2 A2 A1 a1 a2 P2 P1

S2 S1 P1 P2

Interposing CT provides : Vector correction Ratio correction Zero sequence compensation

S1 S2 S1 S2


S2S1

P1P2A1A2 a2a1P2

15MVA66kV / 11kV150/5 800/5

P1

Dy1

S1S2

Given above : Need to consider -(1) Winding full load current(2) Effect of tap changer (if any)(3) C.T. polarities

Assuming no tap changerFull load currents :- 66kV : 131 Amp = 4.37 Amps secondary

11kV : 787 Amp = 4.92 Amps secondaryHowever, require 11kV C.T.’s to be connected in ∆Thus, secondary current = √3 x 4.92 = 8.52A

∴ RATIO CORRECTION IS REQUIRED

Use of Interposing CT - Example


It is usual to connect 11kV C.T.’s in and utilise a / interposting C.T. (this method reduces lead VA burden on the line C.T.’s).Current from 66kV side = 4.37 AmpThus, current required from winding of int. C.T. = 4.37 Amp.Current input to winding of int. C.T. = 4.92 Amp.∴ Required int. C.T. ratio = 4.92 / 4.37 = 4.92 / 2.52

√3May also be expressed as : 5 / 2.56

RR

R

P1 P2 A2 A1 a1 a2 P2 P1150/5

S1 S2 S2 S1

800/5

4.37A 4.92AS1 S2 P1 P2

(5)(2.56)

Use of Interposing CT - Example


25MVA11KV

63MVA132KV

1600/5300/5

50MVA33KV

1000/5

4.59

2.88

10.33

2.88

5.51

5

5

All interposing C.T. ratio’s refer to common MVA base (63MVA)

Three Winding Transformer - Example


Combined Differential and Restricted Earthfault Protection

To differential relayS2

S1P1P2

REFS2

S1P1

P2

S2S1P1 P2a2a1A1A2


Integral Vectorial and Ratio Compensation

Power transformer

Differentialelement

KBCH Relay

Virtual interposing CT

Vectorialcorrection

Ratiocorrection

Virtual interposing CT


Vector Group Correction

HV VectorCorLV1 VectorCorLV2 VectorCor

Yy0, Yd1, Yd5, Yy6, Yd7, Yd11, Ydy00 , -30 , -150 , 180 , +150 , +30 , 0

87

KBCH

Yy0 Yd110 +30

Dy1 (-30 )


In Zone Earthing Transformer

P2P1a2a1

A2A1 P1P2

S1S2

P2P1S1S2 T1T2


Transformer Magnetising Characteristic

TwiceNormal

Flux

NormalFlux

NormalNo Load Current

No Load Current at Twice Normal

Flux


Typical Magnetising Inrush Waveforms

A

C

B


Transformer Differential Protection

Appears on one side of transformer only

Seen as fault by differential relay

Normal steady state magnetising current is less than relay setting

Transient magnetising inrush could cause relay to operate

Effect of magnetising current


Transformer Differential Protection (2)

Solution 1 : 2nd (and 5th) harmonic restraint

Makes relay immune to magnetising inrush

Slow operation may result for genuine transformer faults if CT saturation occurs


Transformer Differential Protection (3)

Inhibits relay operation during magnetising inrush

Operate speed for genuine transformer faults unaffected by significant CT saturation

Solution 2 : Gap measurement technique

Effect of magnetising inrush


Protection of Auto-Transformer by High Impedance Differential Relays (1)

(a) Earth Fault Scheme

A

B

C

High impedancerelay

87


Protection of Auto-Transformer by High Impedance Differential Relays (2)

(b) Phase and Earth Fault Scheme

A

B

Cabc

87 87 87

n


Additional Protection


Inter-Turn Fault

Nominal turns ratioFault turns ratioCurrent ratio

- 11,000 / 240- 11,000 / 1- 1 / 11,000

Requires Buchholz relay

CTE

Shortedturn

Load


100

80

60

40

20

05 10 15 20 25

10

8

6

4

2

Fault current inshort circuited turns

Primary input currentFault current(multiples ofrated current)

Primary current(multiples ofrated current)

Turn short-circuited (percentage of winding)

Interturn Fault Current / Number of Turns Shorted


Buchholz Relay Installation

5 x internal pipediameter (minimum)

3 x internal pipediameter (minimum)

Transformer

3 minimum

Oil conservator

Conservator


Buchholz Relay

PetcockCounter balanceweight

Oil level

From transformer

Aperture adjuster

Deflector plateDrain plug

Trip bucket

To oilconservator

Mercury switch

Alarm bucket


Overfluxing

- Generator transformers- Grid transformers

Usually only a problem during run-up or shut down, but can be caused by loss of load / load shedding etc.

Flux O ∝ Vf

Effects of overfluxing :- Increase in mag. Current- increase in winding temperature- increase in noise and vibration- overheating of laminations and metal parts (caused by stray flux)

Protective relay responds to V/f ratioStage 1 - lower A.V.R.Stage 2 - Trip field


ΦV f α K

Trip and alarm outputs for clearing prolonged overfluxing

Alarm : Definite time characteristic to initiate corrective action

Trip : IDMT or DT characteristic to clear overfluxing condition

Settings

Pick-up 1.5 to 3.0i.e. 110V x 1.05 = 2.3150Hz

DT setting range 0.1 to 60 seconds

Volts / Hertz Protection


Enables co-ordination with plant withstand characteristics

1000

100

10

11 1.1 1.2 1.3 1.4 1.5 1.6

M =

Operatingtime (s)

V HzSetting

K = 63K = 40K = 20K = 5K = 1

t = 0.8 + 0.18 x K (M - 1)2

Volts / Hertz Characteristics


Effect of Overload on Transformer Insulation Life

90 100 110 120 130 140Hot spot temp C

0.1

1.0

10

100

Relativerateof usinglife

With ambient of 20 C.Hot spot rise of 78 isdesign normal.A further rise of 6 Cdoubles rate ofusing life.

98

80


Typical Schemes


Up to 1MVA

51

3.3kV

200/5 P1215051

1MVA3.3/0.44kV1500/5

1500/564

P120

MCAG14

N50N

51N


1 - 5MVA

11kV

P1215051

5MVA11/3.3kV1000/5

1000/564

P120

MCAG14

51N

MCAG1464

3.3kV


Above 5MVA

33KV

P1415051200/5

P12010MVA33/11KV 51

N 87600/5 P631

MCAG1464

600/55/5A


Worked Examples


REF Calculations


Restricted Earth Fault Protection

1MVA(5%)

11000V 415V

1600/1RCT = 4.8Ω

Vk =330V

80MVA

RS1600/1

RCT = 4.8Ω MCAG14IS = 0.1 Amp

2 Core 7/0.67mm (7.41Ω/km)100m Long

Calculate :1) Setting voltage (VS)2) Value of stabilising

resistor required3) Peak voltage

developed by CT’s for internal fault


Solution :

1 P.U.1 4

I1

4

I2

4

I0

Sequence Diagram

Earth fault calculation :-Using 80MVA base

Source impedance = 1 p.u.Transformer impedance = 0.05 x 80 = 4 p.u.

1Total impedance = 14 p.u.∴ I1 = 1 = 0.0714 p.u.

14Base current = 80 x 106

√3 x 415= 111296 Amps

∴ IF = 3 x 0.0714 x 111296= 23840 Amps (primary)= 14.9 Amps (secondary)


Solution :(1) Setting voltage

VS = IF (RCT + 2RL)Assuming “earth” CT saturates,

RCT = 4.8 ohms2RL = 2 x 100 x 7.41 x 10-3 = 1.482 ohms∴ Setting voltage = 14.9 (4.8 + 1.482)

= 93.6 Volts(2) Stabilising Resistor (RS)

RS = VS - 1IS IS

2 Where IS = relay current setting∴ RS = 93.6 - 1 = 836 ohms

0.1 0.12


Solution :

3) Peak voltage = 2√2 √VK (VF - VK)VF = 14.9 x VS = 14.9 x 936 = 13946 Volts

IS

For ‘Earth’ CT, VK = 330V∴ VPEAK = 2√2 √330 (13946 - 330)

= 6kVThus, metrosil voltage limiter will be required.


ICT’s


CT Ratio Mismatch Correction - Modern Approach

87

200/1

0.875A 1.31 Amps

I = 175A I = 525A 400/1L L33kV : 11kV

10 MVA

KBCH

1.14 0.76

HV Ratio CorLV1 Ratio CorLV2 Ratio Cor

0.05 to 2 in steps of 0.01

1A 1A


Stability for External Earth Faults - Modern Approach

Selection of suitable vector correction factor

87

Dy11 (+30 )

Yy00

Yd1-30 87

Dy11

Yy00

Yy00


Current Distributions



RR

R

P1 P2 A2 A1 a1 a2 P2 P1

S2 S1 P1 P2

Connections check : Arbitary current distribution

S1 S2 S1 S2



RR

R

P1 P2 A2 A1 a1 a2 P2 P1

S2 S1 P1 P2

Add winding current :-

S1 S2 S1 S2



RR

R

P1 P2 A2 A1 a1 a2 P2 P1

S2 S1 P1 P2

Complete primary distribution :-

S1 S2 S1 S2



RR

R

P1 P2 A2 A1 a1 a2 P2 P1

S2 S1 P1 P2

Complete secondary distribution :-

S1 S2 S1 S2

transformer protn. areva

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