protection

1. Power Transmission and Distribution
Differential Protection (7UT)
Dynamic tests, Tests

2. PTD PA13 N. M Tests 7UT 05/03 No. 2
Method of Vector Group Determination
IL1,S1
IL1,S2 = IL1 - IL2
330°
(n * 30°)
Vector group is
Y d 11
Side 2: Side 1:
1L1
1L2
1L3
2L1
2L2
2L3

3. PTD PA13 N. M Tests 7UT 05/03 No. 3
Definitions in SIPROTEC 4 Relays
How to see the vector group Yd11?
Original Siprotec 4 (Browser) Phase angles
Side 1: 0°
(reference phase)
Side 2: 210°
Please subtract 180°,
than you get 30°
(360° - 30° = 330°)
Vector definition to a node is positive
The shown vectors or phase angles are transformed in
this positive definition
The phase angle is displayed mathematics positive.
The reference phase is always phase L1 on side 1

4. PTD PA13 N. M Tests 7UT 05/03 No. 4
Web Tool (Browser) Vector group Yd11

5. PTD PA13 N. M Tests 7UT 05/03 No. 5
Vector Group Determination via Fault Record
IL1;Side 1
IL1;Side 2 lags 150° or leads 210° Star point is towards the protected object
Way 1:
IL1 Side: 150° + 180° = 330°
Way 2:
IL1 Side: 210° - 180° = 30°
leads 30° or lags 330°
Phase shift is according the vector group definition 330°  11 * 30°

6. PTD PA13 N. M Tests 7UT 05/03 No. 6
Test Configuration
7UT612
20 kV 110 kV
2500A/1A 500A/1A
500A/1A
Dynamic calculation with NETOMAC
Transient files were transplayed with a CMC-256 Test device from Omicron
80 MVA
YNd11

7. PTD PA13 N. M Tests 7UT 05/03 No. 7
7UT612 Differential Protection Settings

8. PTD PA13 N. M Tests 7UT 05/03 No. 8
Test: Inrush followed by an external fault L1-L2-L3 (LV side)
Infeed: HV side
7UT612 TQL 000009 17.06.2003,19:09:31
Inrush
External
fault

9. PTD PA13 N. M Tests 7UT 05/03 No. 9
Test: Inrush followed by an internal fault L1-L2-L3 (LV side)
Infeed: HV side
7UT612 TQL 000010 17.06.2003,19:11:37
Inrush
Internal
fault

10. PTD PA13 N. M Tests 7UT 05/03 No. 10
Test: External fault L1-L2-L3 (LV side) with CT saturation
Infeed: HV side, Saturation detector is active (Diff BL exF L1)
7UT612 TQL 000001 18.06.2003,14:16:23
Differential current
due to CT saturation

11. PTD PA13 N. M Tests 7UT 05/03 No. 11
Test: Internal fault L1-E (HV side) with CT saturation
Infeed: HV side
7UT612 TQL 000008 17.06.2003,19:02:00
IDiff>>

12. PTD PA13 N. M Tests 7UT 05/03 No. 12
Test: External Fault L1-L2-L3, evolving fault intern L1-L2-L3
Infeed: HV side
7UT612 TQL 000001 17.06.2003,18:35:45
Internal
evolving fault
1
3
2
4

13. PTD PA13 N. M Tests 7UT 05/03 No. 13
Test: Internal Fault L1-E HV side (Io Elimination)
Infeed: LV side
7UT612 TQL 000002 18.06.2003,14:53:43
2/3 Idiff
1/3 Idiff

14. PTD PA13 N. M Tests 7UT 05/03 No. 14
Test: Internal Fault L1-E HV side (Io Correction)
Infeed: LV side
7UT612 TQL 000005 18.06.2003,15:01:20
IDiff>>
3/3 Idiff

15. PTD PA13 N. M Tests 7UT 05/03 No. 15
Test: External Fault L1-E HV side (Io Correction)
Infeed: LV side
7UT612 TQL 000008 18.06.2003,16:51:06

16. PTD PA13 N. M Tests 7UT 05/03 No. 16
Test: External Fault L1-E HV side (Io Elimination)
Infeed: LV side
7UT612 TQL 000009 18.06.2003,16:52:34

17. PTD PA13 N. M Tests 7UT 05/03 No. 17
Test: External Fault L1-E HV side (without Io Elim./Correction)
Infeed: LV side
7UT612 TQL 000010 18.06.2003,16:55:08

18. PTD PA13 N. M Tests 7UT 05/03 No. 18
7UT: Test of Add On Stabilisation with sequencer (1 of 5)
for test 1
Test:
- Through flowing current 5  InTr for minimum 10-12ms
(Side 1: 18.375A , Side 2: 30.25A)
- Reducing from one current to 5A to get a differential current.
(Duration less than the number of periods under parameter 1617 )
- Increasing the current again to its pre value.
Relay: 7UT5125 (5A In)

19. PTD PA13 N. M Tests 7UT 05/03 No. 19
7UT: Test of Add On Stabilisation with sequencer (2 of 5)
Test 1: with Io-Elimination

20. PTD PA13 N. M Tests 7UT 05/03 No. 20
7UT: Test of Add On Stabilisation with sequencer (3 of 5)
Add On Stabilisation
2/3 of 10
Restraining current L1 (A) : 2/325 InTr = 6.6InTr
Restraining current L2/3 (B/C) : 1/325 InTr = 3.3InTr
Trip L1 (A) is blocked, but L2 (B) and L3 (C) will trip.
To prevent a Trip of L2/L3 , the through flowing current must be higher than 35/2
(Setting 1618=5) , means 7.5 InTr.

21. PTD PA13 N. M Tests 7UT 05/03 No. 21
7UT: Test of Add On Stabilisation with sequencer (4 of 5)
Test 2: without Io-Elimination

22. PTD PA13 N. M Tests 7UT 05/03 No. 22
7UT: Test of Add On Stabilisation with sequencer (5 of 5)
Add On Stabilisation
Restraining current L1 (A) : 25 InTr = 10InTr
Restraining current L2/3 (B/C): 0 InTr = 0
There is no Trip !!

23. PTD PA13 N. M Tests 7UT 05/03 No. 23
7UT6: Test characteristic for YNyn0 with Io-Elimination
Slope 1: 0.25
Slope 1: 0.25

24. PTD PA13 N. M Tests 7UT 05/03 No. 24
7UT6: Test characteristic for YNyn0 with Io-Elimination
Slope 1: 0.25
Slope 2: 0.5