TxProtection 280904 PART 3

Institut Latihan TNB - ILSAS

Unit Perlindungan, Seksyen Latihan Penghantaran 2007 Zulkarnain Ishak

KURSUS LATIHAN KEKOMPETENAN

Unit PerlindunganSeksyen Latihan Penghantaran

Institut Latihan Sultan Ahmad Shah (ILSAS)

PERLINDUNGAN PENGHANTARAN(SETAKAT 132kV)



SKIM PERLINDUNGAN ALATUBAH.

TRANSFORMER DIFFERENTIAL



Institut Latihan Sultan Ahmad Shah(ILSAS)

ObjektifPada akhir pengajaran, pelatih-pelatih akan dapat 1. Mengenal pasti dan melukis litar skimatik

perlindungan geganti Tx. Differential.2. Mengira tatahan perlindungan geganti Tx.

Differential 3. Mengaturkan prosidur pengujian skim perlindungan

geganti Tx. Differential.4. Melukis litar pengujian “secondary”, “primary” dan

“stability” skim perlindungan geganti Tx. Differential.dengan betul apabila skim perlindungan Tx. Differential

digunakan pada alatubah.



TRANSFORMER DIFFERENTIAL PROTECTION

1. Overall differential protection may be justify for large transformers (generally >5MVA)

2. This scheme provides fast operation.

3. Measuring principles

a. Based on the same circulating current principle as the restricted earth fault protection.




b. However, it employs the biasing technique to maintain stability for heavy trough fault current.

- Biasing allows mismatch between CT outputs.

- It is essential for transformer with tap changing facility

- Another important requirement of transformer differential protection is immunity to magnetizing inrush current




Correct application of differential protection requires CT ratio and wining connection to match those of Txmer

CT secondary circuit should be a replica of the primary system.

Consider

1. Difference in current magnitude

2. Phase shift

3. Zero sequence current.



TRANSFORMER DIFFERENTIAL CONNECTION

e.g.



TRANSFORMER DIFFERENTIAL CONNECTION

Use of Interposing CT

Provides

1. Ratio Correction

2. Vector Correction

3. Zero Seq. Compensation



TRANSFORMER DIFFERENTIAL PROTECTION APPLICATION (EXAMPLE)

A1

B1B2

C1C2

a1b1

c1

a2

b2

c2

Yd 190MVA 132/33kV

A2

Given above:

Need to consider –

1. Winding full load current

2. Vector correction

3. Effect of tap changer

4. CT polarities




A1

B1B2

C1C2

a1b1

c1

a2

b2

c2

Yd 190MVA 132/33kV

A2

First consider no tap changer or tap changer must at nominal ratio.Full load current at nominal voltage ratio:-

132kV = 393.65Aprimarychoose CT ratio = 400/1 -> secondary current = 0.984A

33kV = 1574.59primarychoose CT ratio = 1600/1 -> secondary current = 0.984A

Because of the vector transformation on primary LV side of transformer then compensation of vector group is required on the secondary circuit.




Vector group compensation on secondary circuit can be done using

i. Line/Phase CT (refer Fig. ii. Interposing CT

A1

B1B2

C1C2

A2 a1

b1

c1

a2

b2

c2

a1

b1

c1

Yd1

Our normal practice, vector group compensation is done using Interposing CTs which were arranged such as the replica of the Power Transformer. This will involve the Delta connection on the LV side of ICT




Thus, sec. current after vector correction = √3 * 0.984 = 1.704A. (For 1:1 winding ratio).There will mismatch current enter the Transformer Differential Relay

= 1.704 - 0.984= 0.72 this will cause relay to operate at normal rated load condition.

Suitable ratio should be selected to eliminate the mismatch current. For the above example

Let secondary current of ICT (delta side) ILine = 1 AIphase = 1/√3 = 0.5774So turn ratio required = 1 : 0.5774

Interposing CTA

CB

N,

Vphase = VLine

VLine = 3 Vphase

ILine = 3 Iphase

VLineVphase

Iphase = ILine




Interposing CTA

CB

N,

Vphase = VLine

VLine = 3 Vphase

ILine = 3 Iphase

VLineVphase

Iphase = ILine

YΔ interposing CT connected on HV side of Transformer Differential scheme will eliminate the effect of Zero Sequence / Earth Fault current on the primary network (out zone fault) which could cause instability of the scheme.

Elimination of zero seq. current on LV side. This can be achieved by using YΔY connection ICT.

Delta connection of YΔY LV ICT should be left open. Delta winding provide a path for the flow of zero sequence current to balance the LV current transformer under earth fault condition.

It is important to note that the star point of the relay must not connected to the star point of the LV ICT as this would create an alternative zero seq. path and current may flow through the relay under external fault condition.

Since effectively the relay is connected to the CT through YY then there will be no ratio different.




It is important to note that the star point of the relay must not connected to the star point of the LV ICT as this would create an alternative zero seq. path and current may flow through the relay under external fault condition.

Since effectively the relay is connected to the CT through YY then there will be no ratio different.

So turn ratio required for YΔY LV ICT normally 1: 1: 1.



400/1 A1

B1B2

C1C2

a1

b1

c1

a2

b2

c2

Yd190MVA 132/33kV

400/1

1600/1

1600/1

LV ICT1: 1: 1

HV ICT 1:0.5774

6464



TRANSFORMER DIFFERENTIAL PROTECTION APPLICATION

EFFECT OF TAP CHANGER

Example: Transformer tap changer with +5% and –15% of turn ratio / nominal voltage.

Effect of tap changer on voltageAt tap where –15% turn ratio = 112.2kV this mean that the LV terminal voltageLV = (33kV * 132kV) /112.2kV

= 38.8kV

Effect of tap changer on currentTurn ratio N1/N2 = 112.2/33Where transformation formula

N1/N2 = I2/I1At full load where LV current I2primary = 1574.6 A

I2sec. = 0.984A

Than I1primary = I2 * N2/N1= 463.12 A

I1sec. = 1.1578A

Mismatch current I1sec.- I2sec. = (1.1578A - 0.984A)= 0.1738 A

The operation current for most of differential relay is about 20% / 0.2A.With above condition relay might operate for external fault. Due to this effect then transformer differential is equipped with bias element for bias characteristic.




BIAS DIFFERENTIAL SCHEME

I1 I2

Io = I1 -I2Operate

BiasBias

Assume that external fault current = 10 * Full load current. Io = I1 – I2

= (11.578A - 9.84A)= 1.738 A

IBias = (I1 + I2) / 2= (11.578A + 9.84A) / 2= 10.709 A

Io < IBias

Io < IBias cause relay to restrain.Io > IBias cause relay to operate

0.2 A

1.74A

10.76AIBias

Io

20%slope

80% slope




EFFECT MAGNETISING INRUSH CURRENT

1. Appears on one side of transformer only2. Seen as fault by differential relay.3. Normal steady state magnetizing current is less than

relay setting4. Transient magnetizing inrush could cause relay to

operate





Solution 1: Time delay

•Allow magnetizing current to die away before relay can operate.

•Slow operation for genuine Txmer faults.

Solution 2: 2nd and 5th harmonic restrain

•Makes relay immune to magnetizing inrush.

•Slow operation may result for genuine Txmer faults if CT saturation occurs.





Solution 3: Gap measurement technique

•Inhibits relay operation during magnetizing inrush

•Use in Alstom relay (MBCH & KBCH)

•Operate speed for genuine transformer faults unaffected by significant CT saturation




TRANSFORMER GUARD



TRANSFORMER DIFFERENTIAL PROTECTION APPLICATIONEFFECT ON OVERLOADTRANSFORMER INSULATION LIFE

OVERHEATING PROTECTION



STANDBY EARTH FAULTStandby Earth fault Relay

– CT located at Neutral Cable thus sensing any earth fault through the transformer

– Stage 1 trips the bus section and/or bus coupler– Stage 2 trips HV and LV transformer feeders– To act as backup to feeder earth fault relays by

sectionalising busbar



STANDBY EARTH FAULTStandby Earth fault Relay (Continued)

– To act as backup to feeder earth fault relays by sectionalising busbar

– To act as backup to transformer incomer feeder for busbar and downstream feeder fault

– Maximum delay time based on rating of NER– Relays usually located at bus section panel



earth

STAND BY EARTH FAULT

CT

SBEF OPERATIONAT TWO STAGESFOR ANY FAULTDOWNSTREAM



earth

STAND BY EARTH FAULT

CT

SBEF WILL NOTOPERATEFOR PH-PH FAULT



THANK YOU

TxProtection 280904 PART 3

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