Lecture 7 Relay Coordination.pptx

Power system ptotection

1. PROTECTIVE RELAY COORDINATION AND GRADING NATIONAL POWER TRAINING INSTITUTE OF NIGERIA ……………………………power trainer with a difference 1

2. ..............................................power trainer with a difference 2 Contents

3. .............Power Trainer with a difference 3

4. 4 Note: A common aim of all these methods is to give correct discrimination or selectivity of operation.

5. Co-ordination Procedure ..............................................power trainer with a difference 5 Information required to achieve proper co-ordination are: A single line diagram of the power system. The impedance of transformers, feeders, motors etc. in ohms, or in p.u . or % ohms. The maximum peak load current in feeders and full load current of transformers etc , with permissible overloads. The maximum and minimum values of short circuit currents that are expected to flow. The type and rating of the protective devices and their associated protective transformers . Performance curves or characteristic curves of relays and associated protective transformers.

6. Protection Co-ordination Principles ..............................................power trainer with a difference 6

7. Criteria for setting instantaneous units ..............................................power trainer with a difference 7

8. Time Graded Systems ..............................................power trainer with a difference 8 In this method, selectivity is achieved by introducing time intervals for the relays. The operating time of the relay is increased from the farthest side towards the generating source. This is achieved with the help of definite time delay over current relays. When the number of relays in series increases, the operating time increases towards the source .

9. Current Graded Systems 9 This principle is based on the fact that the fault current varies with the position of the fault because of the difference in impedance values between the source and the fault. The relays are set to pick up at progressively higher currents towards the source. This current grading is achieved by high set over current relays and with different current tap positions in the over current relays. Since their selectivity is based solely on the magnitude of the current, there must be a substantial difference ( preferably a ratio of 3:1)

10. Challenges with Current Graded System ..............................................power trainer with a difference 10 The magnitude of the fault current cannot be accurately determined since all the circuit parameters may not be known exactly and accurately. There may be variations in the fault level depending upon the source generation , thereby necessitating the frequent change in the settings of the relay. Thus discriminating by current grading alone is not a practical proposition for exact grading. As such current grading alone is not used , but may be used to advantage along with a Time Graded System .

11. Time and Current Graded System ..............................................power trainer with a difference 11 Due to the limitations of either current or time graded protection, time and current grading is employed. Overcurrent relays with inverse time characteristics are used to achieve co-ordination. In such relays the time of operation is inversely proportional to the fault current level and the actual characteristics is a function of both time and current settings. The most widely used is the IDMT characteristic where grading is possible over a wide range of currents and the relay can be set to any value of definite minimum time required. There are other inverse relay characteristics such as very inverse and extremely inverse, which are also sometimes employed. If the fault current reduces substantially as the fault position moves away from the source, very inverse or extremely inverse type relays are used instead of IDMT relays.

12. Inverse Definite Minimum Time (IDMT) Operating Characteristic ..............................................power trainer with a difference 12 Operate current = 1.05 x setting At 2x setting operate time = 10s At 10x setting operate time = 3s At 30x setting operate time = 2s Definite minimum time Numeric IDMT relay operating algorithm:

13. Time and Current Graded System ..............................................power trainer with a difference 13

14. Factors that determine Co-ordination Time interval 14

15. Over Current Protection Grading Margin – Between Relays 15 Op time of 0.5s 0.375s margin for EM relay, oil CB 0.24s margin for static relay, vacuum CB  

16. Determination of Time dial setting 16 1. Calculate the multiple of pickup value for the secondary short circuit current corresponding to the instantaneous setting of the relay where the process starts. If the instantaneous unit is overriden , the calculation is carried out with the total secondary short circuit current at the relay location . 2. With the value calculated above, determine the operating time t1 of the relay for the given Time Dial. 3. Determine the operating time of the upstream relay with the expression . 4. Calculate the multiple of pickup value of the upstream relay using the same short circuit current used in the first relay (step 1). 5. Knowing , and having calculated the multiple of pickup value of the upstream relay, select the above nearest time dial for that relay . The process follows the same steps for the next upstream relay and is repeateduntil the settings of the farthest up relay are calculated. Operating time defined by IEC 60255 and IEEE C37.112 are  

17. Over Current Protection Time Multiplier Setting 17 Used to adjust the operating time of an inverse characteristic Not a time setting but a multiplier Calculate TMS to give desired operating time in accordance with the grading margin. Calculate required relay operating time, considering grading margin & fault level Calculate op time of inverse characteristic with TMS = 1, T 1 TMS = T req /T 1   Current (Multiples of Is) 0.1 1 10 100 1 10

18. Coordination across Dy transformers 18 Calculate required operating current Calculate required grading margin Calculate required operating time Select characteristic Calculate required TMS Draw characteristic, check grading over whole curve Grading curves should be drawn to a common voltage base to aid comparison

19. Over Current Protection Time Multiplier Setting 19 I s = 5 Amp; TMS = 0.05, SI I FMAX = 1400 Amp B A 200/5 100/5 Relay B is set to 200A primary, 5A secondary Relay A set to 100A  If (1400A) = PSM of 14 relay A OP time = t = 0.14 x TMS = 0.14 x 0.05 = 0.13 ( I 0.02 -1) (14 0.02 -1) Relay B Op time = 0.13 + grading margin = 0.13 + 0.4 = 0.53s Relay A uses SI curve so relay B should also use SI curve

20. Over Current Protection Time Multiplier Setting 20 I FMAX = 1400 Amp B A 200/5 100/5 I s = 5 Amp Relay B set to 200A  If (1400A) = PSM of 7 relay B OP time TMS = 1 = 0.14 x TMS = 0.14 = 3.52s ( I 0.02 -1) (7 0.02 -1) Required TMS = Required Op time = 0.53 = 0.15 Op time TMS=1 3.52 Set relay B to 200A, TMS = 0.15, SI

21. System Study for IDMT Protection ..............................................power trainer with a difference 21

22. EXERCISE: Co-ordination between Time Graded Relays ..............................................power trainer with a difference 22

23. Coordination across Dy transformers ..............................................power trainer with a difference 23 In the case of overcurrent relay coordination for Dy transformers, the distribution of currents in these transformers should be checked for three-phase, phase-to-phase, and single-phase faults on the secondary winding . A phase-phase fault on one side of transformer produces 2-1-1 distribution on other side Use an over current element in each phase (cover the 2x phase) 2  & EF relays can be used provided fault current > 4x setting I line 0.866 I f3  I delta Turns Ratio 3 :1

24. Coordination across Dy transformers ..............................................power trainer with a difference 24 Turns Ratio 3 :1 I line 0.866 I f3  I delta I star = E - /2Xt = 3 E -n /2Xt I star = 0.866 E -n /Xt I star = 0.866 I f3 I delta = I star /3 = I f3 /2 I line = I f3 

25. Coordination across Dy transformers ..............................................power trainer with a difference 25 51 HV Ø/Ø 51 LV Grade HV relay with respect to 2-1-1 for - fault Not only at max fault level I f3 86.6% I f3

26. High Set Overcurrent Protection ..............................................power trainer with a difference 26

27. High Set Overcurrent Protection ..............................................power trainer with a difference 27

28. Exercise ..............................................power trainer with a difference 28 Consider the distribution network at Dogon gari N/ bussa which supplies PHCN camp through a 5km cable of 11KV ACSR 3 lines 240mm 2 conductor and Kadariko as shown. 132kV Busbar . Transfromer conductor Determine the PSM, TMS and Op time for relays at CB 31 ,CB 1 ,AG 1 &AG 2.  

29. High Set Overcurrent Protection ..............................................power trainer with a difference 29

30. ..............................................power trainer with a difference 30