GE Multilin
T60 Transformer Protection System
9-11
9 COMMISSIONING
9.2 DIFFERENTIAL CHARACTERISTIC TEST EXAMPLES
9
9.2.5 TEST EXAMPLE 4
D/D0° TRANSFORMER WITH PHASE B TO C FAULT ON THE SECONDARY DELTA WINDING.
Transformer: D/D0°, 20 MVA, 115/12.47 kv, CT1 (200:1), CT2 (1000:1)
Figure 9–5: CURRENT DISTRIBUTION OF D/D TRANSFORMER WITH AN a TO b FAULT ON THE LV SIDE
Slope 2
A
4
∠
0°
0
∠
0°
0
∠
0°
0
∠
0°
Operate
I
d
/
I
r
= 95.7%
> Slope 2 = 95%
B
8
∠
–180°
0.6
∠
0°
12.13
∠
–180°
12.73
∠
0°
C
4
∠
0°
0.6
∠
–180°
12.13
∠
0°
12.73
∠
–180°
TEST
PHASE
INJECTED CURRENT
DISPLAYED CURRENT
STATUS
W1 CURRENT
W2 CURRENT
DIFFERENTIAL
RESTRAINT
Balanced
Condition
A
0
∠
0°
0
∠
0°
0
∠
0°
0
∠
0°
Not Applicable
B
0.435
∠
–90°
0.8
∠
–270°
0
∠
0°
0.8
∠
–270°
C
0.435
∠
–270°
0.8
∠
–90°
0
∠
0°
0.8
∠
–90°
Min Pickup
A
0
∠
0°
0
∠
0°
0
∠
0°
0
∠
0°
Block
I
d
= 0.065 < Min PKP
B
0.09
∠
–90°
0.23
∠
–270°
0.065
∠
0°
0.230
∠
–270°
C
0.09
∠
–270°
0.23
∠
–90°
0.065
∠
0°
0.230
∠
–90°
Min Pickup
A
0
∠
0°
0
∠
0°
0
∠
0°
0
∠
0°
Operate
I
d
= 0.101 > Min PKP
B
0.21
∠
–90°
0.486
∠
–270°
0.102
∠
0°
0.486
∠
–270°
C
0.21
∠
–270°
0.486
∠
–90°
0.101
∠
0°
0.486
∠
–90°
Slope 1
A
0
∠
0°
0
∠
0°
0
∠
0°
0
∠
0°
Block
I
d
/
I
r
= 14% < 15%
B
0.651
∠
–90°
1.39
∠
–270°
0.195
∠
0°
1.39
∠
–270°
C
0.651
∠
–270°
1.39
∠
–90°
0.195
∠
0°
1.39
∠
–90°
Slope 1
A
0
∠
0°
0
∠
0°
0
∠
0°
0
∠
0°
Operate
I
d
/
I
r
= 16.8% > 15%
B
0.63
∠
–90°
1.39
∠
–270°
0.233
∠
0°
1.39
∠
–270°
C
0.63
∠
–270°
1.39
∠
–90°
0.233
∠
0°
1.39
∠
–90°
Intermediate
Slope 1 & 2
A
0
∠
0°
0
∠
0°
0
∠
0°
0
∠
0°
Block
I
d
/
I
r
= 52.6%
< 60% computed
B
1.2
∠
–90°
4.63
∠
–270°
2.44
∠
–270°
4.63
∠
–270°
C
1.2
∠
–270°
4.63
∠
–90°
2.44
∠
–90°
4.63
∠
–90°
Intermediate
Slope 1 & 2
A
0
∠
0°
0
∠
0°
0
∠
0°
0
∠
0°
Operate
I
d
/
I
r
= 68.8%
> 60% computed
B
0.8
∠
–90°
4.63
∠
–270°
3.18
∠
–270°
4.63
∠
–270°
C
0.8
∠
–270°
4.63
∠
–90°
3.18
∠
–90°
4.63
∠
–90°
TEST
PHASE
INJECTED CURRENT
DISPLAYED CURRENT
STATUS
W1 CURRENT
W2 CURRENT
DIFFERENTIAL
RESTRAINT
828739A1.CDR
D
/d
0
° Transformer
F
I (f) = 0
A
I (f) = 0.866 pu
–90°
b
∠
I (f) = 0
a
I (f) = 0.866 pu
–270°
c
∠
I (f) = 0.866 pu
–90°
B
∠
I (f) = 0.866 pu
–270°.
C
∠
A
B
C
A
B
C
H
w
inding
X
w
inding
Summary of Contents for T60
Page 6: ...vi T60 Transformer Protection System GE Multilin TABLE OF CONTENTS ...
Page 14: ...xiv T60 Transformer Protection System GE Multilin TABLE OF CONTENTS ...
Page 34: ...1 20 T60 Transformer Protection System GE Multilin 1 5 USING THE RELAY 1 GETTING STARTED 1 ...
Page 490: ...5 344 T60 Transformer Protection System GE Multilin 5 10 TESTING 5 SETTINGS 5 ...
Page 522: ...6 32 T60 Transformer Protection System GE Multilin 6 5 PRODUCT INFORMATION 6 ACTUAL VALUES 6 ...
Page 536: ...7 14 T60 Transformer Protection System GE Multilin 7 1 COMMANDS 7 COMMANDS AND TARGETS 7 ...
Page 568: ...10 12 T60 Transformer Protection System GE Multilin 10 6 DISPOSAL 10 MAINTENANCE 10 ...
Page 596: ...A 28 T60 Transformer Protection System GE Multilin A 1 PARAMETER LISTS APPENDIX A A ...
Page 716: ...B 120 T60 Transformer Protection System GE Multilin B 4 MEMORY MAPPING APPENDIX B B ...
Page 762: ...E 10 T60 Transformer Protection System GE Multilin E 1 IEC 60870 5 104 PROTOCOL APPENDIX E E ...
Page 774: ...F 12 T60 Transformer Protection System GE Multilin F 2 DNP POINT LISTS APPENDIX F F ...