G. Ziegler. Digital Differential Protection

G. Ziegler, 10/2005 page 140

Differential Protection Symposium

Belo HorizonteNovember 2005

I0-

elim.

Com-

parison

∆I

IR

IS

IT

Ir**

Is**

It**

IR*

IS*

IT*

Ir*

Is*

It*

Vector

group

adapt.

N

o

r

m

N

o

r

m

JR-sec

JS-sec

JT-sec

Jr-prim

Js-prim

Jt-prim

JR-prim

JS-prim

JT-prim

Jr-sec

Js-sec

Jt-sec

CT 1

CT 2

W1

W2

I0-

elim.

Ir

Is

It

1-N

U3

N

S

W1TransfN

⋅

=

−−

I

2-N

U3

N

S

W2TransfN

⋅

=

−−

I

sec

secR

J

1

sec

secR

J

W1-TransfN

I

1 CTPrimN

I

T

I

S

I

R

I

−

⋅

−

=

−

⋅

−

−−

=

T

J

S

J

CT

k

T

J

S

J

sect

secs2CT

sect

secs

J

Jk

J

−

−−

⋅=⋅=

secrsecr

W2-TransfN

2 CTPrimN

t

s

r

JJ

I

Digital transformer protection

Adaptation of currents for comparison (1)

G. Ziegler, 10/2005 page 141

Differential Protection Symposium

Belo HorizonteNovember 2005

I0-

elim.

Com_

parison

∆I

IR

IS

IT

Ir**

Is**

It**

IR*

IS*

IT*

Ir*

Is*

It*

Vector

group

adapt.

N

o

r

m

N

o

r

m

JR-sec

JS-sec

JT-sec

Jr-prim

Js-prim

Jt-prim

JR-prim

JS-prim

JT-prim

Jr-sec

Js-sec

Jt-sec

CT 1

CT 2

W1

W2

I0-

elim.

Ir

Is

It

()

TSR0

III

3

1

I ++⋅=

0TT

0SS

0RR

II*I

−=

−

=

()

t

I

s

I

r

I

3

1

0

I ++⋅=

0

I

t

I*

t

I

0

II*

s

I

0

I

r

I*

r

I

−=

−

=

s

*

110

011

101

3

1

**

t

I

s

I

r

I

t

I

s

I

r

I

⋅

−

=

Example Yd5:

**

*

t

s

r

T

S

R

T

S

R

I

+=

−∆

Digital transformer protection

Adaptation of currents for comparison (2)

G. Ziegler, 10/2005 page 142

Differential Protection Symposium

Belo HorizonteNovember 2005

Input data:

•n times 30

O

vector group number

(only for 2nd and 3rd winding,

1st winding is reference)

•UN (kV)Rated winding voltage

•SN (MVA)rated winding power

•INW (A)Primary rated CT current

•Line or BBdirection of CT neutral

•Elimination / I

0

-treatment

Correction /

without

•Side XXAssignment input for REF

•INW S (A)Primary rated current of neutral CT

•Neutral CT Earth side connection to relay: Q7 or Q8?

minmax

N

UU

2U

+

⋅

⋅=

At windings with tap

changer:

Winding 1 (reference) is normally:

•High voltage side

Adaptation of currents for comparison

Relay input data

G. Ziegler, 10/2005 page 143

Differential Protection Symposium

Belo HorizonteNovember 2005

Yd5

S

N

= 100 MVA

U

N2

= 110 kV

U

N1

= 20 kV

600/1 A

3000/5 A

2.400 A

7621 A

∆I

W2

W1

1 A

5 A

Digital transformer protection

Current adaptation, Example (1)

G. Ziegler, 10/2005 page 144

Differential Protection Symposium

Belo HorizonteNovember 2005

525A

110kV3

100MVA

I

W1TrafoN

=

⋅

=

−−

4,57A4

525

600

Norm

I

A 4,02400

600

1

sek-TS,R,

J

=⋅=

357,4

357,42

357,4

0

4,57

0

211

121

112

3

1

*

t

I

*

s

I

*

r

I

⋅−=−⋅

−−

⋅=

2887A

20kV3

100MVA

I

W2TrafoN

=

⋅

=

−−

A34,5734,4

2887

3000

Norm

I

A 34,4313200

3000

1

sek-tr,s,

J

=⋅=

3/57,4

3/57,42

3/57,4

0

357,4

110

011

101

3

1

**

−

⋅

−

=−⋅

−

=

t

I

s

I

r

I

0

34,5734,57

34,57234,572

34,5734,57

**I*I

I

tT

sS

rR

TA

S

A

RA

=

−

⋅+⋅−

−

=+

=

−

110-kV-side

20-kV-side

I0-elimination:

Vector group adaptation: Yd5

Digital transformer protection

Current adaptation, Example (2)

G. Ziegler, 10/2005 page 145

Differential Protection Symposium

Belo HorizonteNovember 2005

7UT513 (5A Relay)

Matching CTsrecommended, if

k_Wan_n > 4 or k_Wan_n < 1/4: *)

To keep the specified measuring accuracy of 7UT51. For protectio

n only

necessary if 8 <k_Wan_n <1/8

Current adaptation (1)

Practical example for the need of matching CTs

*)

500 kV,

1050 MVA

345 kV, 1050 MVA

13,8 kV,

30 MVA

1500/5 A

2000/5 A

1000/1 A

43.930 A

(1050 MVA)

1213 A

(1050

MVA)

43,43 A

4,04 A

1/0.2A

8,786

!476,1

5

786,8

k

*

3_CT

<==

G. Ziegler, 10/2005 page 146

Differential Protection Symposium

Belo HorizonteNovember 2005

809,0

5

04.4

1

04,4

5/1500

1213

sec1

1213

5003

3

101050

1

==

−

==−−−

=

⋅

=

−−

CT

k

A

WTrn

I

A

WTrn

I

!4786,8

5

93,43

3

93,43

1/1000

930.43

_3

930.43

8,133

3

101050

3

>==

−

==−−

=

⋅

=

−−

CT

k

A

sekWTrn

I

A

WTrn

I

878,0

5

39,4

2

39,4

5/2000

1757

2

1757

3453

3

101050

2

==

−

==−−−

=

⋅

=

−−

CT

k

A

sekWTrn

I

A

WTrn

I

0,122%In0,099%1,072%

1

32.768

15

Relay

40·In= 15 Bit +sign

= 2

15

= 32.768

40·In32.36351,44

Transformer

winding 1

Transformer

winding 3

Current adaptation (2)

Practical example for the need of matching CTs

G. Ziegler, 10/2005 page 147

Differential Protection Symposium

Belo HorizonteNovember 2005

7UT6

Operating characteristic

1

2

3

4

1234567891011

Stable operation

S

l

o

p

e

1

S

l

o

p

e

2

I DIFF >>

5

6

7

8

I

stab

/ I

n

I

DIFF

=

I

STAB

0

9

I

1

I

2

I

DIFF

= |I

1

+ I

2

|

I

Stab

= |I

1

| + |I

2

|

7UT6

Additional stabilisation

for high I

s.c.

I

Diff/

I

n

I DIFF >

Tripping area

G. Ziegler, 10/2005 page 148

Differential Protection Symposium

Belo HorizonteNovember 2005

è I

0

-elimination necessary at all windings with earthed neutral or

with grounding transformer in the protection range

Earth fault sensitivity reduced to 2/3 !

Incorrect fault type indication!

è I

0

-correction provides full earth current sensitivity and correct

phase selective fault type indication, however requires CT in the

neutral-to-earth connection of the transformer.

è As an alternative, earth differential protection can be used to

enhance earth fault sensitivity.

I

0

-elimination / correction: Summary

G. Ziegler, 10/2005 page 149

Differential Protection Symposium

Belo HorizonteNovember 2005

Source: P.M. Anderson: Power System Protection, McGraw-Hill and IEEE Press (Book)

Transformer winding to earth fault

Solid earthed neutral

h⋅U

R

U

R

I

F

I

K

Infeedside

I

F

I

K

200

4060

80

100

2

4

6

8

10

I

F

per unit

Short-circuited winding part h in %

G. Ziegler. Digital Differential Protection

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