TRV Problem With IEEE C By Thomas Field Southern Company Services

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1 RV Problem With IEEE C By homas Field Southern Company Services

2 Question Asked of IEEE about Standard C How do I use the values in ables 5, 6, 7, and 8 along with Figure 15 to draw a circuit breaker RV capability curve so that the actual system RV curve can be compared to the capability curve?

3 Presentation Outline Utility RV Standard Applications Standard C RV Problem Similar Problem in Proposed Changes to C RV Standard Solutions to Problems

4 Utility Application of RV Standards Generate Ideal Breaker Capability Curve Generate System RV Response Curve Comparison of Ideal and System Curve

5 Ideal Breaker Capability Curve Ideal Breaker Capability Curve represents the Minimum Requirements for Circuit Breakers Actual Circuit Breakers Response May be Greater than the Standards Minimum Computer Programs Used For Curve Reproduction

6 EMP Curve Generation Equations Defining Continuous Minimum Capability of Breaker Defined in Standard Curve Generated Based on Equations Curve Compared to System Response on Single Graph

7 C Curve Equations hree Phase Bus Fault Short ine Fault

8 C hree Phase Fault

11 EMP Program for Ideal Response ACS HYBRID C C ENER HE BREAKER VOAGE RAING IN VOS 11VRA C ENER HE BREAKER CURREN RAING IN AMPS 11IRA C ENER HE BREAKER IME IN SECONDS C ENER HE BREAKER R VAUE IN VOS/SECOND C this comes from table 5 of ansi c R 1.8E9 C C ENER HE ACUA FAU CURREN IN AMPS C 11CUR C C DEERMINE HE FRACION OF RAED CURREN 99EV = (VRA.E.7500) 99EV1 = (VRA.G.7500) 99F1 = EV*.5 + EV1*5. 99F = EV*1.5 + EV1*. 99F3 = EV*1.0 + EV1*3. 99F4 = EV*.5 + EV1*1. 99E = CUR/IRA 99E1 = (E.GE.0.AND. E.E..3) 99E = (E.G..3.AND. E.E..6) 99E3 = (E.G..6.AND. E.E.1.0) 99KR = (E1*0.) + (E*((E-.3)/.3)*) + (E3*(-(E-.6)/.4)) 99KR = (EV*0.) + (EV1*KR) 99K = (E1*F1) + (E*(F+F3*((.6-E)/.3))) + (E3*(F-F4*(E-.6)/.4)) 99KE = 1+((1-E)/.55)*.1 C 99E1 = 1.5*SQR(./3.)*VRA 99E = (EV*1.88*VRA) + (EV1*1.76*VRA) 99B = /K 99EB = E*KE 99RB = R*KR C C C CACUAE HE RV C C this is equation ) on page 5 of C C the form is (E/.0)*k1*(1-cos(pi*t*K/) C ke is k1 and K is 1/kt C the different equations for the value under, kt, are not presented C but are in figure 5 and able 1 99BRKRX =EB/.*(1-COS(PI/B*IMEX)) C this is equation 1) on page 5 of C C the form is E1*(1-e**(1R*Kr*t/E1)) 99BRKRZ =E1*(1.-EXP(-RB*IMEX/E1)) IF (BRKRX.G.BRKRZ) HEN 98FAG =1 ENDIF IF (FAG.EQ.1)HEN 98BRKRV =BRKRX ESE 98BRKRV =BRKRZ ENDIF C 33BRKRVBRKRXBRKRZ C C <Name-<>InitialVal<- 77FAG 0.0 C...^...^..

12 EMP Output of Ideal Response

13 C Short ine Fault

14 Short ine Fault Boundary Equations

15 System RV Response System Fault Modeled in Simulation Program Response across First Pole o Open Graphed Graph Compared to Minimum Breaker Capability Curve Capacitance added if Minimum Breaker Capability Curve Exceeded

16 Simple System Fault Model Pre-Opening Current Injected Model Capacitances and Inductances Fault hree Phases Monitor Voltage Across First Open Contact

17 Drawing of Simple Fault Circuit

18 Program for Circuit Simulation C <Bus M<Bus K<Bus 3<Bus 4<----R<----<----C<---R<---<---C<---R3<---3<---C3 SOURA SOURB SOURC C...^...^...^...^...^...^...^...^...^...^...^...^...^ C <Bus M<Bus K<Bus 3<Bus 4<----R<----<----C<---R<---<---C<---R3<---3<---C3 SOURA SOURB SOURC C...^...^...^...^...^...^...^...^...^...^...^...^...^ C <Bus M<Bus K<Bus 3<Bus 4<----R<----<----C<---R<---<---C<---R3<---3<---C3 INEA INEB INEC C...^...^...^...^...^...^...^...^...^...^...^...^...^ C <Bus M<Bus K<Bus 3<Bus 4<----R<----<----C<---R<---<---C<---R3<---3<---C3 INEA INEB INEC C...^...^...^...^...^...^...^...^...^...^...^...^...^ C BANK CARD ENDING BRANCH DAA C SWICHES C C <Bus K<Bus M<---close<----open< Ie<---Vflash<-Special-<-Bus5<-Bus6<--O SOURA INEA C...^...^...^...^...^...^...^...^...^...^ C <Bus K<Bus M<---close<----open< Ie<---Vflash<-Special-<-Bus5<-Bus6<--O INEA INEB INEB INEC C...^...^...^...^...^...^...^...^...^...^ BANK CARD ENDING SWICHES C C SOURCE C C <--Bus<V<Amplitude<-----Freq<----Phase< A1< <---start<----stop 14SOURA INEA C...^.^...^...^...^...^...^...^...^ BANK CARD ENDING SOURCE CARDS C OUPU SPECIFICAION CARDS C <Bus 1<Bus <Bus 3<Bus 4<Bus 5<Bus 6<Bus 7<Bus 8<Bus 9<Bus10<Bus11<Bus1<Bus13 INEA SOURA C...^...^...^...^...^...^...^...^...^...^...^...^...^ BANK CARD C POS CARDS BANK CARD BANK CARD BEGIN NEW DAA CASE BANK CARD

19 Output of System Response

20 Comparison of Breaker Capability Curve and System Response P3P>SOURA -INEA (ype 8) P3P>SOURA -INEA (ype 8) P3P>ACS -BRKRV(ype 9) Voltage (V) ime (us)

21 C RV Problem Minimum Range Specified For Circuit Breakers Capability Curve No Boundary Defined for System Response No Equations for Minimum Circuit Breaker Capability Curve Unable to Determine System Capacitance Needed

22 Minimum Range Specified Rate of Rise ime Delay Peak Voltage angent in Undefined Region op of Curve Function Described No Defining Equations

23 Minimum Breaker Requirements

24 Minimum Breaker Requirements

25 System Source Fault ong Delay Short arge Window for System Response 1-cos Waveshape Fits Into Window

26 System Source Fault

27 System Source Fault Simulation GEN>SOURA -INEA (ype 8) GEN>SOURA -INEA (ype 8) GEN>ACS -SOPE1(ype 9) GEN>ACS -SOPE(ype 9) Voltage (V) ime (us)

28 Generator Source Fault Short Delay ong Small Window 1-cos Waveshape Does Not Fit in Window

29 Generator Source Fault

30 Generator Source Fault Simulation GEN>INEA -SOURA (ype 8) GEN>INEA -SOURA (ype 8) GEN>ACS -SOPE1(ype 9) GEN>ACS -SOPE(ype 9) Voltage (V) ime (us)

31 Problem Area In ast Graph GEN>INEA -SOURA (ype 8) GEN>INEA -SOURA (ype 8) GEN>ACS -BRKCOS(ype 9) GEN>ACS -SOPE1(ype 9) GEN>ACS -SOPE(ype 9) 6956 Voltage (V) ime (us)

32 Similar Problem With C Proposed Changes IEC Harmonization with IEEE IEC Standard 56 Waveshapes to Replace Defined Functions Parameter Function 4 Parameter Function Circuit Breaker Range No System Response Boundary Defined

33 Parameter Function of IEC 56 (Similar to IEEE C )

34 4 Parameter Function of IEC 56

35 Possible Solutions to Problem IEC Reference Waveshape Delay ine to E Delay ine to %1 and then ROR ine Delay ine to %1 and then Slope to ROR ine at % Delay ine to %1, then Slope to ROR ine at %, then Curve to E at

36 Delay ine to %1 and then ROR ine

37 Delay ine to %1 and then Slope to % ROR ine

38 Delay ine to %1, then Slope to ROR ine at %, then Curve to E at

39 otal Suggested Solution

40 Possible Generic Solution

41 ( ) ( ) t E c t E c t Rt t ae t ae be t d d t R d t + + = = + = = = = 5 1) ( 5 5 cos 1 ) ( ) ( ) ( π π

43 Without imit on Cosine PROP>ACS -INEDR(ype 9) PROP>ACS -INEDR(ype 9) PROP>ACS -INROR(ype 9) Current (A) ime (us)

44 imit on Cosine PROP>ACS -INEDR(ype 9) PROP>ACS -INEDR(ype 9) PROP>ACS -INROR(ype 9) Current (A) ime (us)

45 ACS HYBRID C C ENER VAUE (in us) 11I 7. C ENER E VAUE (in kv) 11EI 7.6 C ENER HE IME DEAY (in us) 11ID 1.0 C ENER CONSAN A IN % 11AICON 10.0 C ENER CONSAN B IN % 11BICON 30.0 C ENER CONSAN C IN % 11CICON 50.0 C 99 =I/ D =ID/ E =EI* ACON =AICON/ BCON =BICON/ CCON =CICON/ =0.85* 99ROR =E/3 991 =(ACON*E/ROR)+D 994 =BCON*E/ROR 995 =CCON*E/ROR C IF (IMEX.. D) HEN 98INEDR =0.0 ENDIF IF (IMEX.GE. D.AND.IMEX..1) HEN 98INEDR =ROR*(IMEX-D) ENDIF IF (IMEX.G. 1.AND.IMEX..4) HEN 98INEDR =((BCON-ACON)*E/(4-1))*(IMEX-1)+ACON*E ENDIF IF (IMEX.G. 4.AND.IMEX..5) HEN 98INEDR =ROR*IMEX ENDIF IF (IMEX.G.5)HEN 98EMP1 =(1-CCON)*E 98ARGCS =(PI/.0)*(IMEX-5)/(-5)+PI/.0 98EMP =EMP1*(1-COS(ARGCS)) 98INEDR =EMP+((*CCON)-1.0)*E ENDIF 98CHECK =ROR*IMEX IF (INEDR.G. CHECK) HEN 98INEDR =CHECK ENDIF IF (IMEX.. 3) HEN 98INROR =ROR*IMEX ESE 98INROR =E ENDIF C C <name1<name<name3<name4<name5<name6<name7<name8<name9<nam10<nam11<nam1<nam13< 33INEDRINROR C...^...^...^...^...^...^...^...^...^...^...^...^...^.

46 A=30%, B=50%, C=70% PROP3>ACS -INEDR(ype 9) PROP3>ACS -INEDR(ype 9) PROP3>ACS -INROR(ype 9) Current (A) ime (us)

47 Cosine Oscillation PROP4>ACS -INEDR(ype 9) PROP4>ACS -INEDR(ype 9) PROP4>ACS -INROR(ype 9) Current (A) ime (us)

48 ( ) ( ) t E c t E c t ae t ae be t d d t R d t + + = + = = = = 4 1) ( 5 5 cos 1 ) ( ) ( ) ( π π

49 4 Segment Method a=10%, b=90%, c=67.5% PROP5>ACS -INEDR(ype 9) PROP5>ACS -INEDR(ype 9) PROP5>ACS -INROR(ype 9) Current (A) ime (us)

50 ACS HYBRID C C ENER VAUE (in us) 11I 7. C ENER E VAUE (in kv) 11EI 7.6 C ENER HE IME DEAY (in us) 11ID 1.0 C ENER CONSAN A IN % 11AICON 10.0 C ENER CONSAN B IN % 11BICON 80.0 C ENER CONSAN C IN % 11CICON 67.5 C 99 =I/ D =ID/ E =EI* ACON =AICON/ BCON =BICON/ CCON =CICON/ =0.85* 99ROR =E/3 991 =(ACON*E/ROR)+D 994 =BCON*E/ROR 995 =CCON*E/ROR C IF (IMEX.. D) HEN 98INEDR =0.0 ENDIF IF (IMEX.GE. D.AND.IMEX..1) HEN 98INEDR =ROR*(IMEX-D) ENDIF IF (IMEX.G. 1.AND.IMEX..4) HEN 98INEDR =((BCON-ACON)*E/(4-1))*(IMEX-1)+ACON*E ENDIF IF (IMEX.G.4)HEN 98EMP1 =(1-CCON)*E 98ARGCS =(PI/.0)*(IMEX-5)/(-5)+PI/.0 98EMP =EMP1*(1-COS(ARGCS)) 98INEDR =EMP+((*CCON)-1.0)*E ENDIF 98CHECK =ROR*IMEX IF (INEDR.G. CHECK) HEN 98INEDR =CHECK ENDIF IF (IMEX.. 3) HEN 98INROR =ROR*IMEX ESE 98INROR =E ENDIF C C <name1<name<name3<name4<name5<name6<name7<name8<name9<nam10<nam11<nam1 <nam13< 33INEDRINROR

51 Actual Breaker RV

52 Final Proposed Solution Specify a, b, and c for breakpoints in 5 segment curve Specify the point that the 1-cosine curve should be centered around Calculate the start of last segment based on time in curve instead of start of zero cosine point

53 Modified Equations 0, 1,, and 3 remain the same Parameters a, b, and c remain the same Add parameter d for percent of E that the 1-cosine curve is centered around Modify 4 by replacing 5 in cosine argument with parameter x Calculate x based on point in wave that 1-cosine centered around de equals ce

54 ( ) ( ) t E d x x t E d t Rt t ae t ae be t d d t R d t + + = = + = = = = 5 1) ( cos 1 ) ( ) ( ) ( π π d c d d c d x = 1 cos 5 1 cos π π

56 Curve FromC a=10, d=90, c=90, d=67.5 FIN>ACS -INEDR(ype 9) FIN>ACS -INEDR(ype 9) FIN>ACS -INROR(ype 9) Current (A) ime (us)

57 ACS HYBRID C C ENER VAUE (in us) 11I 7. C ENER E VAUE (in kv) 11EI 7.6 C ENER HE IME DEAY (in us) 11ID 1.0 C ENER CONSAN A IN % 11AICON 10.0 C ENER CONSAN B IN % 11BICON 90.0 C ENER CONSAN C IN % (place to start 1-cos wave) 11CICON 90.0 C ENER CONSAN D IN % (center of 1-cos wave) 11DICON =I/ D =ID/ E =EI* ACON =AICON/ BCON =BICON/ CCON =CICON/ DCON =DICON/ =0.85* 99ROR =E/3 991 =(ACON*E/ROR)+D 994 =BCON*E/ROR 995 =CCON*E/ROR 99DUMV1 =(.0/PI)*(ACOS(DCON-CCON)/(1-DCON)) 99X =(1.0/DUMV1)*(*(DUMV1-1.0)-5) IF (IMEX.. D) HEN 98INEDR =0.0 ENDIF IF (IMEX.GE. D.AND.IMEX..1) HEN 98INEDR =ROR*(IMEX-D) ENDIF IF (IMEX.G. 1.AND.IMEX..4) HEN 98INEDR =((BCON-ACON)*E/(4-1))*(IMEX-1)+ACON*E ENDIF IF (IMEX.G. 4.AND.IMEX..5) HEN 98INEDR =ROR*IMEX ENDIF IF (IMEX.G.5)HEN 98EMP1 =(1-DCON)*E 98ARGCS =(PI/.0)*(IMEX-X)/(-X)+PI/.0 98EMP =EMP1*(1-COS(ARGCS)) 98INEDR =EMP+((*DCON)-1.0)*E ENDIF 98CHECK =ROR*IMEX IF (INEDR.G. CHECK) HEN 98INEDR =CHECK ENDIF IF (IMEX.. 3) HEN 98INROR =ROR*IMEX ESE 98INROR =E ENDIF C <name1<name<name3<name4<name5<name6<name7<name8<name9<nam10<nam11<nam1<nam13< 33INEDRINROR 5 DCON CCON X C...^...^...^...^...^...^...^...^...^...^...^...^...^. C <Name-<>InitialVal<- 77INEDR 0.0 C...^...^..

58 Conclusions Minimum RV Curves Should Be Specified For C Segment Curve Suggested Exact Values/Equations Must Be Determined From est Data C Use of IEC RV Standard May Have Similar Problem as C if Minimum RV Curve Not Specified

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