G. Ziegler. Digital Differential Protection

G. Ziegler, 10/2005 page 241

Differential Protection Symposium

Belo HorizonteNovember 2005

0204060

2

0204060

2

0204060

1

2

t

0204060

2

1

0204060

1

2

I1

I2

0204060

1

2

I

Op

= |I1+I2|

∆I=I

Op

-I

Res

k·I

Res

=k·(|I1|+|I2|)

I1

I2

I

Res

=|I1|+|I2|

Digital busbarprotection 7SS5, Measuring technique

External fault, fault current with DC offset

G. Ziegler, 10/2005 page 242

Differential Protection Symposium

Belo HorizonteNovember 2005

0204060

2

t

0204060

2

0204060

2

1

2

3

0204060

2

t

0204060

2

1

2

3

t

0204060

3

1.75

0.5

0.75

2

I1

I2

Tripping!

I

Op

= |I1+I2|

∆I=I

Op

-I

Res

k·I

Res

=k·(|I1|+|I2|)

I1

I2

I

Res

=|I1|+|I2|

Digital busbarprotection 7SS5, Measuring technique

Internal fault, fault current with DC offset

G. Ziegler, 10/2005 page 243

Differential Protection Symposium

Belo HorizonteNovember 2005

CT dimensioning for busbarprotection 7SS5, Example (1)

110kV

S

SC

‘‘= 4 GVA

G

10 kV

10 MVA

Xd‘‘= 15%

T

G

= 100 ms

110/10 kV

25 MVA

u

T

= 14%

T

= 60 ms

M

5MW

300/1

150/1

1500/1

600/1

2MW

5MW

I

St

= 6·IN

Xd‘‘= 17%

T

M

= 35 ms

M

1440A

3

1010

6

1025

TN

I =

⋅⋅

⋅

=

−

577A

3

1010

6

1010

GN

I =

⋅⋅

⋅

=

−

577A

3

1010

6

105

2

HV-MN

I =

⋅⋅

⋅

⋅=

−

Σ

kA 11.3

0.14

14401.1

TF

I =

⋅

=

−

kA 4.2

0.15

5771.1

GF

I =

⋅

=

−

kA 3.7

0.17

5771.1

I =

⋅

=

−

Σ

MF

G. Ziegler, 10/2005 page 244

Differential Protection Symposium

Belo HorizonteNovember 2005

Total fault current with a fault at the transformer HV terminals:

kA 19,23,74,211,3

MF

ΣI

GF

I

TF

IΣI =++=

−

+

−

+

−

=

F

Equivalent time constant:

ms 64

3.74.211.3

353.71004.26011.3

MF

ΣI

GF

I

TF

I

M

T

MF

ΣI

G

T

GF

I

T

TF

I

Equiv.

T =

++

⋅+⋅+⋅

=

−

+

−

+

−

⋅

−

+⋅

−

+⋅

−

=

As worst case, the CT 150/1 A in bay 1 is considered.

We consider a CT type 5P?, 30 VA, internal burden Pi= 15% (4.5 VA):

Connected burden Pa= 1 VA

CT over-dimensioning factor for 3ms saturation free time: K

TF

ca. 0.45

Corresponding to an overburdening factor of k

OB

= 1/K

TF

= 2.2

5845.0

150

200.19

' =⋅=⋅

−

Σ

=

TF

K

CTN

I

F

I

ALF

3.958

5.430

5.41

' =⋅

+

=⋅

+

= ALF

Pi

N

P

i

P

a

P

ALF

We finally choose: CT 5P10, 150/1, 30 VA, R2≤ 4.5 Ohm

Checking of the k-setting

(Stability with symmetrical fault currents):

()

5.0

12.24

2,2

14

=

−⋅

=

−⋅

>

OB

k

OB

k

(chosen: k=0.6)

CT dimensioning for busbarprotection 7SS5 and 7SS6, Example (2)

G. Ziegler, 10/2005 page 245

Differential Protection Symposium

Belo HorizonteNovember 2005

0102030405060708090

100

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

1.1

1.2

1.3

1.4

1.5

012345678910

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

1.1

1.2

1.3

1.4

1.5

K

TF

K

TF

T

N

T

N

TM

5 ms

4 ms

3 ms

TM

5 ms

4 ms

3 ms

Transient performance of iron closed CT cores (type TPX)

Over-dimensioning factor K

TF

for short time to saturation

G. Ziegler, 10/2005 page 246

Differential Protection Symposium

Belo HorizonteNovember 2005

R

L

R

RV

∆I

Varistor

R

L

R

L

R

CT

R

CT

R

CT

R

CT

R

L

R

RS

R

CT

: Resistance of CT secondary

winding

R

L

: Connection cable resistance

R

RV

: Relay series resistance

R

RS

: Relay shunt resistance

I

V

I

S

I

R

High impedance busbarprotection

G. Ziegler, 10/2005 page 247

Differential Protection Symposium

Belo HorizonteNovember 2005

HI busbarprotection, calculation example

Given: :n = 8 feeders

r

CT

= 600/1 A

U

KN

= 500 V

R

CT

= 4 Ohm

I

mR

= 30 mA (at relay pick-up voltage)

Primary pick-up current:

(

)

mR

In

V

I

S

I

R

I

minF

I ⋅+++⋅=

− CT

r

()

0.0380.050.890.02

1

600

minF

I ⋅+++⋅=

−

(

)

CTN

111%I666A

minF

I

−

⋅=

−

R

L

= 3 Ohm (max.)

I

R

=20 mA(fixed value)

R

RV

= 10 kOhm

R

RS

= 250 Ohm

I

v

= 50 mA (at relay pick-up voltage)

Stability with external faults:

R

I

CT

R

L

R

maxthroughF

I ⋅

+

⋅<

−− CT

r

0.02

43

10.000

1

600

maxthroughF

I ⋅

+

⋅<

−−

n

I28kA 17

maxthroughF

I ⋅=<

−−

G. Ziegler, 10/2005 page 248

Differential Protection Symposium

Belo HorizonteNovember 2005

21

32

1

3

IM=2·I

L1

+ 1·I

L3

+ 3 ·I

E

=2 ·(–I

F

) + 1 ·(–I

F

)+ 3·0= –3·I

F

I

M

=2·I

L1

+ 1·I

L3

+ 3 ·I

E

=2·I

F

+ 1 ·I

F

+ 3·3I

F

= +12 ·I

F

I

Op

=| I

M1

+ I

M2

| = 9 ·I

F

I

Res

=|I

M1

| +|I

M2

| = 15 ·I

F

k=9/15 = 0.6

Fault L2-E

I

Res

I

Op

k=0.5

Faults in other phases:

Fault L1-E: I

Op

= (3+12) ·I

F

= 15 ·I

F

,

I

Res

= (3+12) ·I

F

= 15 ·I

F

,

Fault L3-E: I

Op

= (0+12) ·I

F

= 12 ·I

F

,

I

Res

= (0+12) ·I

F

= 12 ·I

F

,

Busbar

Busbarprotection, Composite current type (7SS600):

Performance under unfavourable system earthingconditions

k=1

G. Ziegler, 10/2005 page 249

Differential Protection Symposium

Belo HorizonteNovember 2005

110/10 kV

40 MVA

35 MW

R

E

= 6 Ohm (I

E

= 1 kA)

kA 1kV/6Ω310/

F

I

kA 2

310kV

35MVA

L

I

≈==

=

⋅

=

E

I

Worst case: Fault in phase L2

Restraint current:

kA 3II

KE

Op

=⋅=3

kA 33I

Res

⋅=

57,0

33

3

I

k

sRe

Op

=

⋅

==

Operating current:

I

Res

I

E

I

L3

2·I

L1

I

L1

I

L3

I

L2

I

M-L

|I

M-L

|

Load

Fault L2-E

I

Res

I

Op

k=0.5

Ph-E

Busbarprotection, Composite current type (7SS600):

Performance in networks with earth current limitation

G. Ziegler, 5/2005 page 248

Differential Protection Symposium

Belo HorizonteNovember 2005

Differential Protection 7UT6

Remarkable Features

G. Ziegler. Digital Differential Protection

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