# 03 Diff Principles

September 23, 2017 | Author: Name Is | Category: Electrical Engineering, Electromagnetism, Electric Power, Electricity, Electronic Engineering

#### Short Description

Differential Protection...

#### Description

Power Transmission and Distribution

Principles of system protection technology Transformer Differential Protection

Basic principles : Kirchhoff’s knot rule Basis for Differential-Protection:

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

Principles Transf. Diff 2

Basic principles: current comparison IP1 IS1

I1F

Protected object IS1 ?I

I2F

IP1

IP2 IS2

IS1

IP2 Protected object

IF

IS2

IS2 ?I

IDiff =¦ I1 + I2¦

internal fault

Assumption: CT- ratio: 1/1 IP1 = I1F IP2 = I2F IDiff = ¦ IP1 + IP2 ¦ = ¦ I1F + I2F ¦ ? Trip

IDiff =¦ I1 + I2¦

Assumption: CT- ratio: 1/1 IP1 = IF IP2 = -IF IDiff = ¦ IP1 + IP2 ¦ = IF - IF = 0 ? no Trip

Requirements for Differential Protection: 1) Internal faults ( faults between CT-sets ) ? Trip 2) External faults ? no Trip Principles Transf. Diff 3

Basic principles : restrained current comparison

(1/2)

Example: external fault with linear CT-errors IP1

Ct 1: e1= - 5%

IS1= 0.95·IP1

Protected object

Ct 2: e2 = +5%

IDiff = 0.1·IP1 ?I

IP2 = -IP1

IF

IS2= 1.05·IP2

assumption: CT- ratio: 1/1 IDiff = ¦ 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·I N -external fault: assumption: IP1 = 10·IN IDiff = 0.1·IP1 = 1·IN

As the setting I Diff> for usual applications lays below nominal current, it would cause a wrong trip in case of external faults with heavy current!

Restrained characteristic necessary!

Principles Transf. Diff 4

Basic principles: restrained current comparison

(2/2)

Example: external fault with linear CT- errors CT 1: e1= - 5%

I P1

CT 2: e2 = + 5%

Protected object IRest

I P2 = -I P1

IF

I Rest =¦ I S1¦ +¦ I S2¦ I Rest = 2·I P1 I S2= 1.05·I P2

I S1= 0.95·I P1

IDiff

Setting due to magnetising or charging currents I Diff= ¦ I S1+I S2¦

I Diff= ¦ I S1+ I S2¦ I Diff = 0.1·I P1

Linear error due to different CT transformation Resulting characteristic

IN

Trip

Block I Diff> 2

10

I Rest =¦ I S1¦ +¦ I S2¦

Under the following assumption ¦ e1 ¦ = ¦ e2 ¦ and I1 = I2 the result for a conventional Differential Prot. characteristic should be: IDiff = IDiff> + e1·I1 + e2·I2 = IDiff> + 2·e1 ·I1 with IDiff> = setting

Principles Transf. Diff 5

Basic principles: measuring circuit for a 3-phase system Basic circuit for a 3- phase system: Generator / Motor / Reactor L1 L2 L3

Diff.

Rest. current

Conventional Differential Protection

Principles Transf. Diff 6

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 cause differential current Transformer-tap changer, magnetising current Transient currents Inrush CT-saturation

Principles Transf. Diff 7

3-phase Transformer: primary values Load: 100MVA ,vector group: Yd5 side 2: 20kV, 2887A side 1: 110kV, 525A

3000/1A 2L1 2L2 2L3

I2L1? I2L2? I2L3?

750/1A

I2*L1

?I1L1

I2*L2

?I1L2

I2*L3

?I1L3

1L1 1L2 1L3

kU = U1N /U2N = 110kV/20kV = 5.5 kWinding = w1/w2 = kU/v3 I2*L1 = -I1L1·ku /v3 + I1L2·ku /v3 I2*L3 I1L1k u/v3

I1L1

5·30° I2*L2 I1L2k u /v3

I1L3

I1L2

-I1L1k u /v3

I2*L1 Principles Transf. Diff 8

3-phase Transformer : secondary values

I L1sec ?

SN I ? NCT1sec 3 ?U N I NCTlprim

I1L1sec ?

100MVA 1A ? ? 0.7A 3 ?110kV 750A

I2 L1sec ?

I1L1sec= 0.7A , 0°

100MVA 1A ? ? 0.96A 3 ?20kV 3000A IDiff L1 = ¦ I1L1sec+ I2L1sec¦ = 0.5A

I2L1sec = 0.96A , -150°

Principles Transf. Diff 9

Vector group and current value adaptation in case of conventional Transformer Differential Protection 3000/1A

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

(1/2)

750/1A

L1

2887A L2 L3

0.7A

0.96A

Diff.

IR

29 Wdg.

23 Wdg.

Rest. current

Conventional Differential Prot.

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

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

IR = 0.555·v3 = 0.96A Principles Transf. Diff 10

Vector group and current value adaptation in case of conventional Transformer Differential Protection 3000/1A

~ ~

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

(2/2)

750/1A L1

13655A L2

~

L3

3I0 4.55A

5.73A

Diff.

IR

29 Wdg.

IP= 4300A

23 Wdg.

Rest. current

Conventional Differential Prot.

single pole fault HV -side:

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

5.73A ·23Wdg = 4.550A ·29Wdg , IR = 4.55A Principles Transf. Diff 11

Vector group and current value adaptation in case of numerical Transformer Differential Protection CT 2 3000/1A 2L1 2L2

Load: 100MVA ,vector group: YNd5 side 2: 20kV side 1: 110kV I2 L1P ?

?I1 L1P

I2 L2P ?

?I1 L2P

I2 L3P ?

2L3

I2 L1S I2 L2S I2 L3S

(1/2)

CT 1 750/1A 1L1 1L2

?I1 L3P

I2 A Current Vector Io – value group handling I2 B adaptation adaptation CT 2 I2 C

I1 A

comparison

?I

I1 B

Io – handling

I1 C

1L3

I1 L1S I1 L2S I1 L3S

Numerical Transformer Differential Protection

Principles Transf. Diff 12

Vector group and current value adaptation in case of numerical Transformer Differential Protection

(2/2)

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

Principles Transf. Diff 13

Tripping characteristic of Transformer Differential Protection CT-errors , Tap changer , Magnetising current I Diff InO 3.0 2.5 Trip

slope 2 Total error

2.0 Block

45° 1.5

CT- error 1.0 slope 1

Tap changer error

0.5 IDiff> 0

Transf. magnetising current 0

1.0

Characteristic:

2.0

3.0

4.0

5.0

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

6.0

7.0

8.0

9.0

I Rest InO

InO = nominal current of the protected object

Principles Transf. Diff 14

Transient currents (with Harmonics) - Inrush of Transformers even 2. Harm.

Inrush

(1 of 2) iDiff = i1

i1

Y Y t=0 i1

i2 = 0 t

even 2. Harm.

Inrush

iDiff = i1

Y

?

i1

t=0 i1

I2 = 0 t

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

i1

t=0 Inrush -T2

-T2

i1

-T1: iDiff = i1

I2 = 0

t=0

t Principles Transf. Diff 15

Inrush, cross block, over excitation [V/Hz]

(2 of 2)

filter window 1 cycle iRUSH = iDiff

Cross-block = No (phase separate blocking) Inrush current in one phase

1P

I2har IDiff

2P

IDiff, L1 > trip blocking

L2-block

IDiff, L2 > trip blocking

L3-block

IDiff, L3 > trip blocking

3P t

Cross-block = Yes (blocking of all phases)

block Setting value

15 %

L1-block

L1-block L2-block

OR =1

IDiff > trip blocking for an adjustable time

L3-block

no block 0

0

t

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 Principles Transf. Diff 16

Demonstration of Inrush with evolving fault

Internal fault

IDiff>>

Inrush

IDiff>

3 cycles Cross Blocking

Principles Transf. Diff 17

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

(1/2) iDiff = i1 + i2

Over excitation (U/f) UTr > UN

uneven 5. Harm. i2

i1

External fault with CT-saturation at the Low voltage side HV

even and uneven

LV

i1

i2

Internal fault with CT-saturation at the High voltage side HV

i1

iDiff = i1 + i2

iDiff = i1

even and uneven

LV

I2 ˜ 0

Principles Transf. Diff 18

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

(2/2)

Principle of Add-on stabilisation for external faults

Tripping characteristic 7UT6

I Diff InO IDiff>>

7 6

Trip

45°

5

Block

D C

4 3

2 1

Begin of saturation

B

IDiff>

0 0 A

2

4

6

8

10

12

14

16

I Rest InO

Principles Transf. Diff 19

Block

45° Trip

Principles Transf. Diff 20

Applications for Transformer Differential Protection three winding transformer 1 or 3 phases

two winding transformer 1 or 3 phases

1 ½ CB method on one side

7UT613 7UT633 7UT612 7UM62

7UT613 7UT633

Unit Protection

1 ½ CB method on two sides

Y 7UT635

?

7UT635

G 3~

Principles Transf. Diff 21