G. Ziegler. Digital Differential Protection
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G. Ziegler, 10/2005 page75
Differential Protection Symposium
Belo HorizonteNovember 2005
W
1
=
21 turns
1 A
W
2
=
14 turns
1,5 A
W
b
=32
turns
1 A
W
a
=16
turns
1 A
0,5 A
R
B
R
B
1,5 A
600/1 A
600/1 A
Interposing CTs, Example
G. Ziegler, 10/2005 page76
Differential Protection Symposium
Belo HorizonteNovember 2005
Yd5
I
1b
I
1a
I
1c
I
2a-c
I
2b-a
I
2c-b
I
2a
I
2b
I
2c
-I
1c
I
2a
I
2b
I
2c
I
1a
I
1b
I
1c
-I
1b
-I
1a
150
O
w
1
= w
2
w
1
w
2
Interposing CTsin Y-∆-connection
)30(
1
2
21
3
1
o
nj
e
w
w
II
⋅⋅
⋅⋅⋅=
G. Ziegler, 10/2005 page77
Differential Protection Symposium
Belo HorizonteNovember 2005
∆
I
∆I
∆I
L1
L2
L3
∆I>0without delta winding!
False operation during external fault
of transformer differential protection without zero-sequence current filter
G. Ziegler, 10/2005 page78
Differential Protection Symposium
Belo HorizonteNovember 2005
Relay
I
2b
I
1b
I
1c
I
2c
I
1a
I
2a
I
3a
I
3b
I
3c
0
222
333
0
3
3
2
2
1
1
0
3
3
2
2
1
1
0
3
3
2
2
1
1
=++
==
=⋅+⋅+⋅
=⋅+⋅+⋅
=⋅+⋅+⋅
c
I
b
I
a
I
c
I
b
I
a
I
w
c
I
w
w
c
Iw
c
I
w
b
I
w
w
b
Iw
b
I
w
a
I
w
w
a
Iw
a
I
++
−=
++
−=
++
−=
=
++
⋅===
3
111
1
2
1
2
3
111
1
2
1
2
3
111
1
2
1
2
33
111
3
1
333
c
I
b
I
a
I
c
I
w
w
c
I
c
I
b
I
a
I
b
I
w
w
b
I
c
I
b
I
a
I
a
I
w
w
a
I
E
I
c
I
b
I
a
I
w
w
c
I
b
I
a
I
Zero-sequence current filter
G. Ziegler, 10/2005 page79
Differential Protection Symposium
Belo HorizonteNovember 2005
∆
I
∆I
L1
L2
L3
∆I
With delta winding: ∆I=0!
Biasing of transformer differential protection during external earth fault
with zero-sequence current filter (closed delta winding)
G. Ziegler, 10/2005 page80
Differential Protection Symposium
Belo HorizonteNovember 2005
∆I
∆I
L1
L2
L3
∆I
Y d 5
110 kV ±16%
6.3 kV
75 / 1 A
1200 / 5 A
In-Relay = 5A
53.9A
915A
3.81A
0.719A
Current matching with interposing CTs(1)
Calculation example
20,2
3
81,3
=
G. Ziegler, 10/2005 page81
Differential Protection Symposium
Belo HorizonteNovember 2005
Current transformation ratio:
We take through flowing rated current as reference:
:
110kV-side:
Mean current value of upper and lower tap changer position:
6 kV-side:
The corresponding secondary currents are:
and
.
The current in the star connected winding of the interposing transformer is .
The current in the delta connected winding is: .
The ratio of the interposing CT must be :
A2.45
%16kV110(3
kVA000,10
I
1
=
+⋅
=
A5.62
%16kV110(3
kVA000,10
'I
1
=
−⋅
=
A9.53
2
5.622.45
I
mean1
=
+
=
−
A915
kV3.63
kVA000.10
I
2
=
⋅
=
A719.0
75
1
9.53i
1
=⋅=
813.3
1200
5
915i
2
=⋅=
1
i
3/
2
i
06.3
719.0
202.2
719.0
3/813.3
i
3/i
w
w
1
2
2
1
====
Current matching with interposing CTs(2)
Calculation example
G. Ziegler, 10/2005 page82
Differential Protection Symposium
Belo HorizonteNovember 2005
BACDEFGHILKMNOPQ
1
0,013
2
5
2
0,025
4
5
7
0,08
14
5
16
0,75
32
1
1
0,013
2
5
2
0,025
4
5
7
0,08
14
5
16
0,75
32
1
Windings
R in Ohm
U-max. in V
Rated current in A
Link selectable interposing CT 4AM5170-7AA
06.3
719,0
202,2
719,0
3/813,33/
1
2
2
1
====
i
i
w
w
09,3
11
34
1127
2 16 16
2
w
1
w
:chosen ==
+++
++
=
S1P2
P1
S2
Connection and links as shown.
i
1
i
2
G. Ziegler, 10/2005 page83
Differential Protection Symposium
Belo HorizonteNovember 2005
6
0
9
3
10
11
7
8
4
5
1
2
12
Designation of transformer or CT vector groups (1)
Clock-wise notation according to IEC 60076-1
0
4
8
6
1
0
2
4
5
8
0
6
4
1
0
8
2
9
0
1 (13)
G. Ziegler, 10/2005 page84
Differential Protection Symposium
Belo HorizonteNovember 2005
11
12
12
I
IIIII
I
II
III
I
II
III
III
III
Dyn11
III
III
III
III
I
IIIII
I
II
III
I
II
III
I
II
III
12
12
12
5
Yny0d5
HS
NS
Designation of transformer or CT vector groups (2)
Clock-wise notation according to IEC 60076-1, examples