03 Diff Principles

Power Transmission and Distribution

Principles of system protection technology

Transformer Differential Protection

Principles Transf. Diff 2

Basic principles : Kirchhoff’s knot rule

I1

I2

I3

I4

I1 + I2 + I3 + I4 = 0 ? ? I = 0

Basis for Differential-Protection:

Definition:Currents, which flow into the knot (protected object), are counted positive.Currents, which flow out of the knot (protected object), are counted negative.

Protection objects:Line, Transformer, Generator/Motor, Bus bar


Basic principles: current comparison

IF

?I IDiff =¦ I1 + I2¦

IP1

IS1 IS2

IP2

external fault or load

Protectedobject

Protectedobject

IDiff =¦ I1 + I2¦

IS1 IS2

IP1

IS1 IS2

IP2

internal fault

? I

I1F I2F

Assumption: CT- ratio: 1/1IP1 = I1F

IP2 = I2F

IDiff = ¦ IP1 + IP2 ¦ = ¦ I1F + I2F ¦? Trip

Assumption: CT- ratio: 1/1IP1 = IF

IP2 = -IF

IDiff = ¦ IP1 + IP2 ¦ = IF - IF = 0? no Trip

Requirements for Differential Protection:

1) Internal faults ( faults between CT-sets ) ? Trip2) External faults ? no Trip


Basic principles : restrained current comparison (1/2)

assumption: CT- ratio: 1/1IDiff = ¦ IS1 + IS2 ¦ = ¦ (1+e1)· IP1 + (1+e2)·IP2 ¦ =¦ 0.95· IP1 – 1.05· IP1¦ = 0.1·IP1

-normal operation: IP1 = IN

IDiff = 0.1·IP1 = 0.1·IN

-external fault: assumption: IP1 = 10·IN

IDiff = 0.1·IP1 = 1·IN

IF

IDiff = 0.1·IP1

IP1

IS1=0.95·IP1

IS2=1.05·IP2

IP2 = -IP1

Example: external fault with linear CT-errors

Protectedobject

?I

Ct 2:e2 = +5%

Ct 1:e1= - 5%

As the setting IDiff> for usual applicationslays below nominal current,it would cause a wrong trip in case ofexternal faults with heavy current!

Restrained characteristic necessary!


Basic principles: restrained current comparison (2/2)

Under the following assumption¦ e1 ¦ = ¦ e2 ¦ and I1 = I2

the result for a conventionalDifferential Prot. characteristicshould be:IDiff = IDiff> + e1·I1 + e2·I2= IDiff> + 2·e1 ·I1

with IDiff> = setting

IDiff>

IDiff=¦ IS1+IS2¦

IRest =¦ IS1¦ +¦ IS2¦

IN

Resultingcharacteristic

Block

Trip

Linear error due todifferentCT transformation

Setting due tomagnetising or charging currents

2 10

IRest

IF

IDiff= ¦ IS1+ IS2¦IDiff = 0.1·IP1

IP1

IS1=0.95·IP1

IS2=1.05·IP2

IP2 = -IP1

Example: external fault with linear CT- errors

Protectedobject

IDiff

CT 2:e2 = + 5%

CT 1:e1= - 5%

IRest =¦ IS1¦ +¦ IS2¦IRest = 2·IP1


Basic principles: measuring circuit for a 3-phase system

Rest. current

Diff.

ConventionalDifferential Protection

L1

L2

L3

Basic circuit for a 3- phase system:Generator / Motor / Reactor


Transformer Differential Protection special qualities

Angle shifting N·30° due to vector group (0 = N = 11)for 3-phase transformers.

Different current values of the CT- sets on the high voltage side (HV)and on the low voltage side (LV)

Zero sequence current in case of external faults will causedifferential current

Transformer-tap changer, magnetising current

Transient currentsInrushCT-saturation


3-phase Transformer: primary values

I1L1

I1L2I1L3

I2*L1 = -I1L1·ku /v3 + I1L2·ku /v3

I2L1?

I2L3?

? I1L1

Load: 100MVA ,vector group: Yd5 side 2: 20kV, 2887A side 1: 110kV, 525A

1L1

1L2

1L3

2L1

2L2

2L3? I1L2

I2L2?

? I1L3

I2*L1

I2*L3

I2*L2

750/1A3000/1A

kU = U1N/U2N = 110kV/20kV = 5.5kWinding = w1/w2 = kU/v3

I2*L3

I2*L2

5·30°

I2*L1

I1L1ku/v3

-I1L1ku /v3

I1L2ku /v3


3-phase Transformer : secondary values

0.96A3000A

1A20kV3

100MVAI2

0.7A750A1A

110kV3100MVA

I1

II

U3S

I

L1sec

L1sec

NCTlprim

NCT1sec

N

NL1sec

???

?

???

?

??

?

I2L1sec = 0.96A , -150°

I1L1sec= 0.7A , 0°

IDiff L1 = ¦ I1L1sec+ I2L1sec¦

= 0.5A


Vector group and current value adaptation in case of (1/2)conventional Transformer Differential Protection

nominal Load (no fault): 0.70A ·23Wdg = 0.555A ·29Wdg, IR = 0.555·v3 = 0.96A

Matching transformer-Vector group adaptation-Current value adaptation-Zero seq. current handling

ConventionalDifferential Prot.

L1

Rest. current

Diff.

L2

L3

Load: 100MVA ,vector group: YNd5 side 2: 20kV side1: 110kV

750/1A3000/1A ILoad=525A

0.7A0.96A

29 Wdg.

23Wdg.

2887A

IR


Vector group and current value adaptation in case of (2/2)conventional Transformer Differential Protection

single pole fault HV -side: 5.73A ·23Wdg = 4.550A ·29Wdg , IR = 4.55A

Matching transformer-Vector group adaptation-Current value adaptation-Zero seq. current handling

ConventionalDifferential Prot.

L1

Rest. current

Diff.

L2

L3

~

~

~

Load: 100MVA ,vector group: YNd5 side 2: 20kV side1: 110kV

750/1A3000/1A

IP=4300A

3I0

5.73A4.55A

29 Wdg.

23Wdg.

13655A

IR


Vector group and current value adaptation in case of (1/2)numerical Transformer Differential Protection

I1B

I1C

I1A

I2L1P ?

I2L2P ?

? I1L1P1L1

1L2

1L3

2L1

2L2

2L3

comparison?I

Io –handlingI2L2S

I2L1S

I2L3S

I2B

I2C

I2AVectorgroup

adaptation

Current value

adaptationCT 1

Io –handling I1L2S

I1L1S

I1L3S

I2L3P ?

Current value

adaptationCT 2

? I1L2P

? I1L3P

Load: 100MVA ,vector group: YNd5 side 2: 20kV side 1: 110kV

Numerical Transformer Differential Protection

CT 1750/1A

CT 23000/1A


Vector group and current value adaptation in case of (2/2)numerical Transformer Differential Protection

Parameterisation of transformer and CT- datain a 7UT6 Differential Protection Device


Tripping characteristic of Transformer Differential Protection

IDiff = f (IRest)IRest = |I1| + |I2|

2.0 8.0 9.0

3.0

InOIRest

InOI Diff

7.0

2.5

slope 1

slope 2

IDiff>

01.0 4.0 5.03.0 6.00

1.0

0.5

2.0

1.5

Trip

Block

Total error

CT- error

Tap changer error

Transf. magnetising current

45°

CT-errors , Tap changer , Magnetising current

Characteristic: InO = nominal current of the protected object


Transient currents (with Harmonics) - Inrush of Transformers (1 of 2)

even2. Harm.Inrush

i1 I2 = 0

Y?

t = 0

i1 I2 = 0

-T1

-T2

Connecting -T2 in parallel with -T1(Sympathetic Inrush –T1)

t = 0Inrush -T2 t = 0

i1 iDiff = i1

t

iDiff = i1

-T1: iDiff = i1

i1

t

even2. Harm.Inrush

i1 i2 = 0

Y Y

t = 0

i1

t


Inrush, cross block, over excitation [V/Hz] (2 of 2)

recognise inrush condition by evaluating the ratio 2nd harmonic I2har to basic wave IDiff.

Time limit for cross-block. Reliable reaction to the inrush condition with cross-block.Trip of a short circuit after the set time delay.

recognise over excitation [V/Hz] by evaluating the ratio 3rd or 5th harmonic to basic wave

Cross-block = Yes (blocking of all phases)

O R=1

L1-block

L2-block

L3-block

IDiff > trip blockingfor an adjustable time

filter window1 cycle

t1P 2P 3P

iRUSH= iDiff Inrush current

in one phase

Settingvalue

t

15 %

no block

blockDiff

2har

II

00

L1-block

L2-block

L3-block

Cross-block = No (phase separate blocking)

IDiff, L1 > trip blocking




3 cyclesCross Blocking

Inrush

Internalfault

IDiff>>

IDiff>

Demonstration of Inrush with evolving fault


Transient currents (with harmonics) (1/2)- Over excitation and CT- saturation

uneven5. Harm.

evenanduneven

Over excitation (U/f)UTr > UN

i1 i2

Internal fault withCT-saturation at theHigh voltage side

i1 I2 ˜ 0

LVHV

External fault withCT-saturation at theLow voltage side

i1 i2

LVHV

iDiff = i1 + i2

iDiff = i1 + i2

iDiff = i1

evenanduneven


Transient currents (with harmonics) (2/2)- Over excitation and CT- saturation

0A 2 4 8 106 120

2

1

4

3

Tripping characteristic 7UT6

Trip6

7

16

IDiff>>

14

5

IDiff>

Add-onStabilisation

45°

InOI Rest

InOI Diff

Block

Begin of saturationB

DC

Principle of Add-on stabilisation for external faults


Demonstration of add-on stabilisation

Add-Stabilisation

45°

Block

Trip


Applications for Transformer Differential Protection

1 ½ CB methodon

one side

7UT6137UT633

three windingtransformer

1 or 3 phases

7UT6137UT633

two windingtransformer

1 or 3 phases

7UT6127UM62

1 ½ CB methodon two sides

7UT635

Unit Protection

Y

?

G3~

7UT635

Documents

03 Diff Principles