ECE 526 Distance Element Examples

Transcription

1 Session 7; Page /2 ECE 526 Distance Element Examples The impedances for the system below are given in secondary ohms. The zone reach of the relay at BUS is set to cover 85% of the line. BUS BUS 2 Z S Relay Z L ZR V secll 2V Note that this is a line to line voltage. Z L 6ohme j85 deg Z S 2.5ohme j85 deg Z R j3 ohm θ L arg Z L θ L 85deg Z L2 Z L Z S2 Z S Z R2 Z R Z L 3Z L Z S 3Z S Z R 3Z R A. With the breaker at Bus 2 open calculate how much fault resistance can be present before the the distance element is unable to see the a SLG fault on phase A in Zone for faults at the following locations: (a) % of the line, (b) 5% of the line and (c) 8% of the line if the relay is () self polarized, (2) cross polarized (use V B +V C ) (3) positive sequence memory polarized (use the prefault source voltage)

2 Session 7; Page 2/2 Define some useful quantities: Z MAG Z L Z ANG arg Z L k 79 MVA kw pu a e j2deg A 2 Z L Z L k k.667 3Z L a 2 a a a 2 Sequence impedances to left of fault: Z Left ( m) Z S mz L Z Left (.5) ( i)Ω Z Left2 ( m) Z S mz L Z Left2 (.5) ( i)Ω Z Left ( m) Z S mz L Z Left (.5) ( i)Ω V f jv secll V f iV 3 Define a vector of R f values (this is just for illuatration, not the actual solution values) R f ohm.ohm.ohm

3 Session 7; Page 3/2 Zero sequence fault current as a function of fault resistance (recall that the R f vector starts at ) I mr f V mr f V f Z Left ( m) Z Left2 ( m) V f I mr f I 2 mr f V 2 mr f I mr f V mr f Z S Z S2 Z S Z Left ( m) 3R f I mr f I mr f I mr f I 2 mr f A 2 I ABC mr f I mr f I mr f I 2 mr f A 2 V ABC mr f V mr f V mr f V 2 mr f Soluton Option : Using the m-equations (calculate the effective reach to the fault) M ag = Re Re V a V a Zmag e jθ ZL I a k I R V a V amem jv secll 3

4 Session 7; Page 4/2 Self Polarized Case: M ag_self mr f Re ReV ABC mr f V ABC mr f Z MAG e jz ANG I ABC mr f k 3 I mr f V ABC mr f Memory Polarized Case: M ag_mem mr f Re ReV ABC mr f V amem Z MAG e jz ANG I ABC mr f k 3 I mr f V amem Cross polarized case Vpol_cross mr f V ABC mr f V ABC mr f 2 M ag_cross mr f Re ReV ABC mr f Vpol_cross mr f Z MAG e jz ANG I ABC mr f k 3 I mr f Vpol_crossmR f

5 Session 7; Page 5/2 R f_m ohm Given M ag_self.r f_m =.85 FindR f_m Ω Repeating the process for the other cases: R selfa R self5a R self8a 3.36ohm R crossa 7.94ohm R mema 6.327ohm 4.8ohm R cross5a 6.234ohm R mem5a 5.423ohm.7ohm R cross8a 2.26ohm R mem8a 2.ohm Check results using the resistance values: Fault at % of the line: M ag_self.r selfa.85 M ag_cross.r crossa.85 M ag_mem.r mema.85 Fault at 5% of the line: M ag_self.5r self5a.85 M ag_cross.5r cross5a.85

6 Session 7; Page 6/2 M ag_mem.5r mem5a.85 Fault at 8% of the line: M ag_self.8r self8a.85 M ag_cross.8r cross8a.85 M ag_mem.8r mem8a.85 Plot how quickly M ag increases with R f for different V pol and different location on line..85 M ag_self. R f M ag_cross. R f M ag_mem. R f R f

7 Session 7; Page 7/2.85 M ag_self.5 R f M ag_cross.5 R f M ag_mem.5 R f R f Instead of using m-equation trying using the Mho circle equations All of the cases will use the same equation for Z AG, only the mho circle changes with polarization. Z AG mr f V ABC mr f k 3 I ABC mr f I mr f The zone Mho characteristic. General form for center and offset for Mho circles (holds for any type of polarization) Radius 2 Zone V AG k pol V pol = reachz L Z PAG where: Z PAG = I A k I R Offset = 2 Zone reachz L Z PAG

8 Session 7; Page 8/2 For the self polarized case: k pol_self V pol_self mr f V ABC mr f This is V AG. Z P_AG_self mr f k pol_self V pol_self mr f k 3I mr f V ABC mr f I ABC mr f Note that for the self polarized case, the numerator is always. rad Mhozone_self offset Mhozone_self mr f mr f 2.85 Z L Z P_AG_self mr f 2.85 Z L Z P_AG_self mr f As we would expect, based on Z P always equal to, the radius and offset don't change with Rf or m. offset Mhozone_self.5R f i i i i i i i i i i i Ω - So we can define the circle with a single value of m and R f rad Mhozone_self Ω.5R f

9 Session 7; Page 9/2 Zone k offset Mhozone_self.5ohm ( ) rad Mhozone_self (.5ohm) e jk.5deg Line imedance vector for diagram: LineZ Z L For the self polarized case we want to find R such that: Z AG mr f offset Mhozone R f_m Given As a check: ohm offset Mhozone_self.R f_m Z AG.R f_m R self_a_mho = rad Mhozone_self.R f_m Find R f_m R self_a_mho 3.36Ω earlier we found: R selfa 3.36Ω rad Mhozone In this case we want to be exactly on the circle offset Mhozone_self.R self_a_mho Z AG.R self_a_mho rad Mhozone_self.R self_a_mho Ω Now for the cross polarized case - now the Mho circle changes, but not the Zag equation From above we had defined k cross_pol V p_cross mr f Vpol_crossmR f V ABC mr f k cross_pol Vpol_cross mr f

10 Session 7; Page /2 Z p_cross mr f k 3 I ABC mr f V p_cross mr f I mr f Z offset_cross mr f Z p_cross (.8ohm) 4.5Ω.5.85Z L Z p_cross mr f 95 arg Z p_cross (.ohm) Z radius_cross mr f deg.5.85z L Z p_cross mr f Z offset_cross.r f Ω Z radius_cross.r f Ω Do not vary with Rf or with m (only 2 cases shown here)

11 Session 7; Page /2 Z offset_cross R f Ω Z radius_cross R f Ω Zone crossk Z offset_cross (.5ohm) Z radius_cross (.5ohm) e jk.5deg Now for the memory polarized case - now the Mho circle changes, but not the Zag equation From above we had defined V amem iV k mem_pol V p_mem mr f V ABC mr f k mem_pol V amem Z p_mem mr f k 3 I ABC mr f V p_mem mr f I mr f Z p_mem (.8ohm) 2.5Ω 95 arg Z p_mem (.8ohm) deg

12 Session 7; Page 2/2 Z offset_mem mr f.5.85z L Z p_mem mr f Z radius_mem mr f.5.85z L Z p_mem mr f Z offset_mem.r f i i i i i i i i i i i Ω Z radius_mem.r f 3.8 Ω Do not vary with Rf or with m (not shown here) Zone memk Z offset_mem (.5ohm) Z radius_mem (.5ohm) e jk.5deg Now plot the calculated effective impedances against the Mho characteristic (note these are generic values of R f )

13 Session 7; Page 3/2 5 Im( LineZ) ImZone k Im Zone crossk Im Zone memk Im Z AG. R f Im Z AG.5 R f Im Z AG.8 R f 5 5 Re( LineZ) Re Zone k Re Zone crossk 5 Re Zone memk Re Z AG. R f Re Z AG.5 R f Re Z AG.8 R f

14 Session 7; Page 4/2 Observations: The case with fault resistance falls on the line impedance vector. Note that as the fault resistant varies from to ohms, the ZAG calculation moves to the right When the cross or memory polarized case is used the mho expansion increases fault resistance coverage most dramatically for faults closer to the bus. The increase in fault resistance coverage is less for faults closer to the zone boundary (with no gain at all at the boundary Now find the solutions for all of the cases asked for: R f_m ohm Given offset Mhozone_self.5R f_m Z AG.5R f_m R self5_a_mho = rad Mhozone_self.5R f_m Find R f_m R self5_a_mho 4.8Ω earlier we found: R self5a 4.8Ω Summary: m-equations Mho circle approach R selfa 3.36Ω R self_a_mho 3.36Ω R self5a 4.8Ω R self5_a_mho 4.8Ω R self8a.7ω R self8_a_mho.7ω Both methods give the same results

15 Session 7; Page 5/2 R crossa 7.94Ω R cross_a_mho 7.94Ω R cross5a 6.234Ω R cross5_a_mho 6.234Ω R cross8a 2.26Ω R cross8_a_mho 2.26Ω R mema 6.327Ω R mem_a_mho 6.327Ω R mem5a 5.423Ω R mem5_a_mho 5.423Ω R mem8a 2.Ω R mem8_a_mho 2.Ω B. Repeat part A with the breaker at Bus 2 closed. Comment on your results for both part A and part B. Soluton Option : Using the m-equations (calculate the effective reach to the fault) Self Polarized Case: M ag_selfb mr f Re ReV ABC_B mr f V ABC_B mr f Z MAG e jz ANG I ABC_B mr f k 3 I B mr f V ABC_B mr f Memory Polarized Case: M ag_memb mr f Re ReV ABC_B mr f V amem Z MAG e jz ANG I ABC_B mr f k 3 I B mr f V amem

16 Session 7; Page 6/2 R fb Vpol_crossB mr f V ABC_B mr f V ABC_B mr f 2 M ag_crossb mr f ohm Given M ag_selfb.r fb FindR fb 2.96Ω Re =.85 ReV ABC_B mr f Vpol_crossB mr f Z MAG e jz ANG I ABC_B mr f Results of applying the solve block for all of the cases... R selfb R self5b k 3 I B mr f 2.96ohm R crossb 5.83ohm R memb 4.64ohm 2.56ohm R cross5b 3.298ohm R mem5b 2.858ohm Vpol_crossBmR f R self8b.668ohm R cross8b.894ohm R mem8b.85ohm Notice that these are smaller than in part A Check results using the resistance values: Fault at % of the line: If use R f answers for part A instead: M ag_selfb.r selfb.85 M ag_crossb.r crossb.85 M ag_memb.r memb.85 M ag_selfb.r selfa.426 M ag_crossb.r crossa.44 M ag_memb.r mema.443

17 Session 7; Page 7/2 Fault at 5% of the line: M ag_selfb.5r self5b.85 M ag_crossb.5r cross5b.85 M ag_memb.5r mem5b.85 M ag_selfb.5r self5a.676 M ag_crossb.5r cross5a.649 M ag_memb.5r mem5a.66 Fault at 8% of the line: M ag_selfb.8r self8a.85 M ag_crossb.8r cross8a.59 M ag_memb.8r mem8a.69 M ag_selfb.8r self8a.85 M ag_crossb.8r cross8a.59 M ag_memb.8r mem8a.69 Note that these appear as if the fault is farther away due to remote infeed. Instead of using m-equation trying using the Mho circle equations (largely same as earlier Now plot the calculated effective impedances against the Mho characteristic (note these are generic values of R f )

18 Session 7; Page 8/2 Im( LineZ) ImZoneB k Im ZoneB crossk Im ZoneB memk Im Z AG_B. R f Im Z AG_B.5 R f Im Z AG_B.8 R f Re( LineZ) Re ZoneB k Re ZoneB crossk Re ZoneB memk Re Z AG_B. R f Re Z AG_B.5 R f Re Z AG_B.8 R f Observations: The case with fault resistance falls on the line impedance vector. Note that as the fault resistant varies from to ohms, the ZAG calculation moves to the right and that it moves much faster to the right than it did in par Note that it no longer moves horizontally. This is due to the difference in phase angles between the local and remote source impedances When the cross or memory polarized case is used the mho expansion increases fault resistance coverage most dramatically for faults closer to the bus. The increase in fault resistance coverage is less for faults closer to the zone boundary (with no gain at all at the boundary

19 Session 7; Page 9/2 Summary: m-equations R selfb 2.96Ω R self5b 2.56Ω R self8b.668ω Mho circle approach R self_b_mho 2.96Ω R self5_b_mho 2.56Ω R self8_b_mho.668ω R crossb 5.83Ω R cross_b_mho 5.83Ω R cross5b 3.298Ω R cross5_b_mho 3.298Ω R cross8b.894ω R cross8_b_mho.894ω R memb 4.64Ω R mem_b_mho 4.64Ω R mem5b 2.858Ω R mem5_b_mho 2.858Ω R mem8b.85ω R mem8_b_mho.85ω Both methods give the same reesults

20 Session 7; Page 2/2 C. How do the results of part B to change if the source impedance behind Bus increases by a factor of 5? Solution: There will be a couple of different effects. First, the effective impedance, ZAG for the mho circle, will move to the right, since the infeed effect will be much more pronounced. At the same time, for the cross and memory polarized cases the amount of Mho expansion will also increase significantly since the expanded circle roughly extends back by the source impedance. R selfc.92 Ω R crossc 5.29 Ω R memc 4.76 Ω R self5c.926 Ω Big decrease R cross5c Ω R mem5c 2.4 Ω small decrease Rself8C.35 Ω R cross8c.72 Ω R mem8c.575 Ω