Over current & Earth fault relay co-ordination Over current fault relay co-ordination:
Let us consider one small section of Aditya cement plant,Sambhupura and let us see how over current and earth relay co-ordination is done on it.
Over Current Relay Co-ordination Guidelines:
1. Over current relays are co-ordinated from load end till the source. Generally the operating time of relay at load end is assumed to be at 0.4 seconds and considering this assumption operating time of back up relays are calculated. 2. Discrimination time for relays on series element is considered as 0.05 and for non-series elements discrimination time is considered as 0.3. 3. For calculation of TMS of upstream/backup relay, the downstream relay having higher operating time is considered. 4. Generally it is preferable to have the plug setting (amperes) of upstream relays to be equal or little more than plug setting (amperes) of downstream relays. 5. Instantaneous settings for upstream relays are provided so as to provide protection against the close in faults occurring at the respective buses such that respective relays trip much before the maximum fault current flows through them. Instantaneous settings should be provided such that they should have a discrimination of about 0.3 second with thermal curve of their back up relay. 6. But for transformer primary relays instantaneous setting should be set such that the relays don’t trip for charging current but protect the transformer against close in faults. 7. Time grading is an important criterion for instantaneous protection. The relays should be time graded in such a way that back up relays trip instantaneously with certain standard time delay, such that for any fault at lower stream, the upstream relays don’t undergo cascaded tripping.
8. As per the standard convention, instantaneous time delay of 0.05 seconds is provided for relays at extreme load end and at transformer primary. 9. In case of phase overcurrent protection generally an instantaneous time delay of 0.20 seconds should exist between primary and it’s backup. 10. In case where condition of critical clearing time needs to be satisfied, there the instantaneous time delay is kept as 0.15 seconds between two relays acting as primary and back up. 11. If there is a fault at one voltage level then the relay at that voltage level should trip instantaneously and it’s back up at a different voltage level should trip thermally. 12. It is always advisable to run co-ordination & simulation in Mi-power, to check whether the relays operate in co-ordination with proper time discrimination.
Considering the above guidelines let us proceed to perform the overcurrent relay co-ordination of the section considered above.
Relay.No.104:
Let the operating time of relay.No.104 be 0.4 seconds The three-phase fault current viewed by relay no. 104 is I 3phf104 = 25693.37 A Plug setting of relay.No.104 in amperes = 3600 A. Therefore TMS of relay 104 is TMS = 0.4*[(25693.37/3600) 0.02 –1]/0.14=0.11 Since the TMS range of REX 521 relay is in range of 0.05 – 1 in steps of 0.05 Hence above calculated TMS is approximated to 0.15 Now as per the approximated TMS, recalculated operating time is : TopR104’ = 0.15*0.14/[(25693.37/4000)0.02 –1] = 0.5238 sec.
Relay.No.100: Since relay.No.100 is in the same series element as relay no. 104 hence discrimination time between them is 0.05 sec. Hence remote operating time for relay no. 100 is T opR100remote = 0.5238 + 0.05 = 0.57 sec. Remote fault current viewed by relay no. 100 is I remote = 973.12 A Plug setting of relay no.100 is 140 A Therefore TMS of relay 100 = 0.5738*[(973.12/140) 0.02-1]/0.14 = 0.16 But as step value is 0.05 hence approximated TMS of relay no.100 is 0.20 Close in fault current viewed by relay no.100 is I 3phf100 = 31168.57 A Hence close in operating time of relay no. 100 is TopR100 = 0.20*0.14/[(31168.57/140)0.02 –1] = 0.4534 sec. Note: If M= [close in fault operating current/Plug setting in amperes] > 20, then M = 20 (Beyond M=20, relay is considered to be saturated). Hence in the above calculation as M = (31168.57/140)>20 hence M=20 has been considered. Exactly the calculations for relay no. 105 and Relay.No 101 are made and results are same as above.
Let the operating time of relay.No.7 be 0.965 sec (Calculated similarly as above). And the operating time of relay.No.66 be 1.99 sec. (Calculated similarly as above).
Relay.No.6: For calculation of operating time for relay no.6, operating time of relay no.100 and relay no. 101 is considered and out of them whichever’s operating time is higher that operating time is considered for calculation of TMS of the relay no.6. but as relay.No.66 has an higher operating time of 1.99 sec hence operating time of relay. No.66 is considered for TMS calculation of R6. Hence operating time of relay R6=operating time of relay.No.66+discrimination time. Operating time of relay R6, TopR6 = 1.99 + 0.3 = 2.29 sec. Close in fault current viewed by relay no.6 is I3phf6= 5264.3 A Plug setting of relay R6 in amperes =1800 A Hence TMS of relay R6 = 2.29*[(5264.3 /1800) 0.02-1]/0.14 =0.355 Approximating the above result we get TMS =0.4 So recalculating the operating time as per new TMS, Top6’=0.4*0.14/[(5264.3 /1800)0.02-1] = 2.58 sec.
Relay.No.5: Operating time of relay.No.5 for remote fault is Top5remote = 2.58+0.05 = 2.63 sec Remote fault current as viewed by relay no.5 = 5264.3 A Plug setting of relay no.5 in amperes = 1800 A TMS of relay no. 5 = 2.63*[(5264.3/1800)0.02-1]/0.14 = 0.41 Approximating the TMS as per the step value of REX 521 we get TMS=0.45 Remote operating time as per approximated TMS Top5remote’ = 0.45*0.14/[(5264.3/1800)0.02 -1] =2.90 sec
Close in fault current I3phf5 = 8816.8 A Close in operating time of relay no.5 is given as: TopR5 = 0.45*0.14/[(8816.8/1800)0.02 -1] =1.95 sec.
Instantaneous Setting : Relay.No. 104: (Feeder) Close in fault current viewed by relay.No.104 is I3phf104 = 25793.7 A CT Ratio = 4000/1 Therefore Ip>> is set such that the relay trips instantaneously before the maximum fault current and has a discrimination of 0.3 with thermal curve of it’s back up relay. Hence if relay is set at 300% then the relay trips at 12000 A which is much before the maximum fault current and discrimination of 0.3 is maintained with the thermal curve of it’s back up relay. Factor to be entered in Mi-power database = 3 (for relays at load end directly the instantaneous factor is entered in database) Instantaneous time delay tp>> = 0.05 sec
Relay.No. 100:(Transformer Primary) Rated current of relay.No. 100 , Irated = 131.2 A Remote fault current = 973.12 A Charging current = 10*Rated current = 1312 A Back up current =1.3*Rated current = 1265.06 A Since Icharg > Iback, hence Icharg is considered for instantaneous calculation Instantaneous setting, Ip>> = Icharg/CT ratio = 1312/200= 6.56
Approximating the above value to nearest whole number we get Ip>> = 7 times or 700%.The factor to be entered in database = 7 x CT ratio/Remote fault current = 1.44 Instantaneous phase time setting = 0.05 sec. Instantaneous setting for Relay no. 105 is exactly same as that of relay no.104 & instantaneous setting for Relay no. 101 is exactly same as that of relay no.100.
Relay.No.6:(Feeder) Close in fault current viewed by relay no.104 is I 3phf104 = 5264.3 A CT Ratio = 2000/1 Therefore Ip>> is set such that the relay trips instantaneously before the maximum fault current and has a discrimination of 0.3 with thermal curve of it’s back up relay. Hence if relay is set at 150% then the relay trips at 3000 A which is much before the maximum fault current and discrimination of 0.3 is maintained with the thermal curve of it’s back up relay. Factor to be entered in Mi- Power database = 3000/5264.3 = 0.57. Instantaneous phase time setting for relay.No.6 is done taking the higher instantaneous time setting of relays connected to its outgoing feeder and then adding an instantaneous time delay of 0.15 sec to it. Since the instantaneous phase time setting for relay. No 7 is 0.35 sec, hence the instantaneous phase setting of relay.No.6 is 0.50 sec(0.35 sec+ 0.15 sec).
Relay.No. 5:(Transformer Primary) Rated current of relay.No.55 is I rated = 1784.5 A Remote fault current =5264.3 A Charging current, Icharg= 10*Rated current = 17845 A Back up current, IBack =1.3* Remote fault current = 6843.59 A
Since Icharg > IBack , hence Icharg is considered for instantaneous calculation Instantaneous setting, Ip>> = Icharg/CT ratio = 17845/2000 = 8.9 Approximating the above value to nearest whole number we get Ip>> = 9 times or 900%. The factor to be entered in Mi- Power database = 9 x CT ratio/Remote fault current = 3.41 Instantaneous phase time setting = 0.05 sec The Phase co-ordination curves are given in the figure 1.1 below.
Figure 1.1. Phase Co-ordination curves
Earth Fault Relay Co-ordination :
Earth Relay Co-ordination Guidelines:
1.
In case of single line to ground fault the plug setting of the relay depends upon the single line to ground fault current viewed by the relay and the setting range of the relay used. Generally plug setting of 10% - 20% is set for the relay, as an unbalance of about 10% always exists in the system.
2.
Relay used should be sensitive enough to operate for an single line to ground fault occurring in it’s zone but should not trip for too low SLG current.
3.
The Plug setting should be provided such that the plug setting current(amperes) of upstream relays should be either equal to or greater than plug setting current (amperes) of the downstream relays.
4.
For transformers having delta-star configuration or vice versa, the relay located at transformer primary can’t be co-ordinated with relay located at transformer secondary because delta windings isolate the fault current from the healthy side and hence SLG fault current doesn’t find a path to flow to the other side. In such a case relay located on star side should be set so as to only operate for close in fault current.
5.
In case of earth overcurrent protection generally an instantaneous time delay of 0.20 seconds should exist between primary and it’s backup. Rest all other guidelines are similar to that of over current fault co-ordination Now let us see how earth relay co-ordination is set using the above guidelines.
Relay.No. 104:
The single line to fault current viewed by relay.No.104 is, I slgf104 = 28833.3 A The plug setting = 5% of CT ratio = 0.05*4000= 200 A Let the relay operating time be =0.4 sec Therefore TMS of relay.No.104 = 0.4*[(28833.3/200) 0.02-1] = 0.176 Here M=20 Approximating the TMS as per step value of REX 521 is = 0.20 So new operating time as per the approximated TMS TopSLG104= 0.20*0.14/[(28833.3/200)^0.02-1] =0.453 sec As relay. No 100 is on delta side of transformer hence no co-ordination is not possible with relay.No 104.
Relay.No. 100: The single line to fault current viewed by relay.No.104 is, I slgf104 = 4066.6 A The plug setting = 10% of CT ratio = 0.1*200= 20 A Let the relay operating time for close in fault be =0.05 sec Therefore TMS of relay.No.104 = 0.05*[(4066.3/20) 0.02 -1]/0.14 = 0.02 Approximating the TMS as per step value of REX 521 = 0.05 New operating time as per approximated TMS is given as TopSLG100 = 0.05*0.14/[(4066.3/20)0.02 -1] =0.113 sec The operating time of relay.No 7 is T opSLG7 = 0.79 sec (calculated similarly as above). Similarly operating time of relay.No 66 is T opSLG7 = 0.42 sec (calculated similarly as above).
Relay.No. 6:
For calculating the operating time of relay.No.6 ,the highest operating time of the relay, among the relays, connected to its outgoing feeder is considered. Here the operating time of relay.No 7 is considered for calculation of TMS of relay.No 6 is considered. TopSLG6 = 0.79 + 0.3 = 1.09 sec Single line to ground fault current at relay.No 6,I SLGf6 = 642 A Plug setting = 10% of CT ratio = 200 A Hence TMS of relay.No 6 is = 1.09*[(642/200)0.02-1] / 0.14 = 0.184 Approximating the TMS as per steps provided in REX 521,we get TMS=0.20 As per the new TMS, the operating time of relay.No.6, TopSLG6 =0.20*0.14 / [(642/200)0.02-1] = 1.186 sec,
Relay.No.5: Relay.No.5 can’t be co-ordinated with real.No.6 as relay.No 5 is on delta side. But however it should operate for any close in fault. Single line to ground fault current at relay.No.5, I SLGf5 = 100 A Plug setting = 1% of CT ratio =0.01*2000= 20 A Let the operating time for the close in fault be T opSLG5 = 0.05 sec. Hence TMS = 0.05*[(100/20)0.02 – 1]/0.14 = 0.01 Approximating the TMS as per step value we get new TMS = 0.05 Operating time as per approximated TMS = 0.05*0.14 /[(100/20) 0.02 – 1] = 0.21 sec.
Instantaneous setting : Relay.No.104:(Feeder) SLG Fault current viewed by relay.No.104, I slgf104 = 28833.3 A Instantaneous earth setting, Io>> = 300% of CT ratio = 3*4000 =12000 A Instantaneous earth time setting, To>> =0.05 sec Instantaneous Factor to be entered in database =3 (load end and delta side of transformer directly instantaneous factor is entered in Mi-power database)
Relay.No.100:(Transformer primary) SLG Fault current viewed by relay.No.104, I slgf100 = 4066.6 A Instantaneous earth setting, Io>> = 500% of CT ratio = 5*200 =1000 A Instantaneous earth time setting, To>> =0.05 sec Instantaneous Factor to be entered in database = 5 (load end and delta side of transformer directly instantaneous factor is entered in Mi-power database)
Relay.No.6:(Feeder) SLG Fault current viewed by relay.No.104, I slgf6 = 642 A Instantaneous earth setting, Io>> = 15% of CT ratio = 0.15*2000 =300 A Instantaneous phase time setting for relay.No.6 is done taking the higher instantaneous time setting of relay connected to out going feeders and then adding a time delay of 0.20 seconds tom it. Instantaneous earth time setting, To>> = Instantaneous earth time setting of relay.No.7 + 0.20 sec =0.45 + 0.20 = 0.65 sec Instantaneous Factor to be entered in database = 300/642 = 0.467
Relay.No.5:(Transformer Primary) SLG Fault current viewed by relay.No.104, Islgf5 = 100 A As per the range of instantaneous earth setting provided by REM 543 ( 0.1 – 12 x In ), relay.No.5 is not sufficient to provide instantaneous earth protection. The lowest instantaneous current that can be set is 0.1*2000 = 200 A, but the fault current is 100 A, hence relay is unable to operate instantaneously as relay would only be able to sense current above 200 A.
Figure 1.3. Earth relay co-ordination curves
Figure 1.4. Earth relay co-ordination curves
