RV Problem With IEEE C37.013-1997 By homas Field Southern Company Services
Question Asked of IEEE about Standard C37.013-1997 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?
Presentation Outline Utility RV Standard Applications Standard C37.013-1997 RV Problem Similar Problem in Proposed Changes to C37.011 RV Standard Solutions to Problems
Utility Application of RV Standards Generate Ideal Breaker Capability Curve Generate System RV Response Curve Comparison of Ideal and System Curve
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
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
C37.011-1994 Curve Equations hree Phase Bus Fault Short ine Fault
C37.011-1994 hree Phase Fault
EMP Program for Ideal Response ACS HYBRID C C ENER HE BREAKER VOAGE RAING IN VOS 11VRA 13000. C ENER HE BREAKER CURREN RAING IN AMPS 11IRA 40000. C ENER HE BREAKER IME IN SECONDS 11.00060 C ENER HE BREAKER R VAUE IN VOS/SECOND C this comes from table 5 of ansi c37.06-1979 11R 1.8E9 C C ENER HE ACUA FAU CURREN IN AMPS C 11CUR 30544. 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 C37.011-1994 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 C37.011-1994 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...^...^..
EMP Output of Ideal Response
C37.011-1994 Short ine Fault
Short ine Fault Boundary Equations
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
Simple System Fault Model Pre-Opening Current Injected Model Capacitances and Inductances Fault hree Phases Monitor Voltage Across First Open Contact
Drawing of Simple Fault Circuit
Program for Circuit Simulation C <Bus M<Bus K<Bus 3<Bus 4<----R<----<----C<---R<---<---C<---R3<---3<---C3 SOURA 0.100 SOURB 0.100 SOURC 0.100 C...^...^...^...^...^...^...^...^...^...^...^...^...^ C <Bus M<Bus K<Bus 3<Bus 4<----R<----<----C<---R<---<---C<---R3<---3<---C3 SOURA.50000 SOURB.50000 SOURC.50000 C...^...^...^...^...^...^...^...^...^...^...^...^...^ C <Bus M<Bus K<Bus 3<Bus 4<----R<----<----C<---R<---<---C<---R3<---3<---C3 INEA 5.500 INEB 5.500 INEC 5.500 C...^...^...^...^...^...^...^...^...^...^...^...^...^ C <Bus M<Bus K<Bus 3<Bus 4<----R<----<----C<---R<---<---C<---R3<---3<---C3 INEA.55000 INEB.55000 INEC.55000 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 1. 1. C...^...^...^...^...^...^...^...^...^...^ C <Bus K<Bus M<---close<----open<-------Ie<---Vflash<-Special-<-Bus5<-Bus6<--O INEA INEB 0. 1. INEB INEC 0. 1. C...^...^...^...^...^...^...^...^...^...^ BANK CARD ENDING SWICHES C C SOURCE C C <--Bus<V<Amplitude<-----Freq<----Phase<-------A1<-------1<---start<----stop 14SOURA -1-157871. 60. 90. 14INEA -1 157871. 60. 90. 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
Output of System Response
Comparison of Breaker Capability Curve and System Response P3P>SOURA -INEA (ype 8) 50000 P3P>SOURA -INEA (ype 8) P3P>ACS -BRKRV(ype 9) 00000 150000 Voltage (V) 100000 50000 0 0 50 100 150 00 50 300 ime (us)
C37.013-1997 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
Minimum Range Specified Rate of Rise ime Delay Peak Voltage angent in Undefined Region op of Curve Function Described No Defining Equations
Minimum Breaker Requirements
Minimum Breaker Requirements
System Source Fault ong Delay Short arge Window for System Response 1-cos Waveshape Fits Into Window
System Source Fault
System Source Fault Simulation GEN>SOURA -INEA (ype 8) GEN>SOURA -INEA (ype 8) GEN>ACS -SOPE1(ype 9) 40000 GEN>ACS -SOPE(ype 9) 30000 Voltage (V) 0000 10000 0 0 5 10 15 0 ime (us)
Generator Source Fault Short Delay ong Small Window 1-cos Waveshape Does Not Fit in Window
Generator Source Fault
Generator Source Fault Simulation GEN>INEA -SOURA (ype 8) GEN>INEA -SOURA (ype 8) GEN>ACS -SOPE1(ype 9) 40000 GEN>ACS -SOPE(ype 9) 30000 Voltage (V) 0000 10000 0 0 10 0 30 40 50 ime (us)
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) 16956 9 ime (us)
Similar Problem With C37.011 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
Parameter Function of IEC 56 (Similar to IEEE C37.013-1997)
4 Parameter Function of IEC 56
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
Delay ine to %1 and then ROR ine
Delay ine to %1 and then Slope to % ROR ine
Delay ine to %1, then Slope to ROR ine at %, then Curve to E at
otal Suggested Solution
Possible Generic Solution
( ) ( ) 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 ) (1 4 5 4 3 4 1 1) ( 1 4 1 ) ( 1 0 0 0 4 3 1 0 π π
Without imit on Cosine PROP>ACS -INEDR(ype 9) 30000 PROP>ACS -INEDR(ype 9) PROP>ACS -INROR(ype 9) 5000 0000 Current (A) 15000 10000 5000 0 0 4 6 8 10 ime (us)
imit on Cosine PROP>ACS -INEDR(ype 9) 30000 PROP>ACS -INEDR(ype 9) PROP>ACS -INROR(ype 9) 5000 0000 Current (A) 15000 10000 5000 0 0 4 6 8 10 ime (us)
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/1000000.0 99D =ID/1000000.0 99E =EI*1000.0 99ACON =AICON/100.0 99BCON =BICON/100.0 99CCON =CICON/100.0 993 =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...^...^...^...^...^...^...^...^...^...^...^...^...^.
A=30%, B=50%, C=70% PROP3>ACS -INEDR(ype 9) 30000 PROP3>ACS -INEDR(ype 9) PROP3>ACS -INROR(ype 9) 5000 0000 Current (A) 15000 10000 5000 0 0 4 6 8 10 ime (us)
Cosine Oscillation PROP4>ACS -INEDR(ype 9) 30000 PROP4>ACS -INEDR(ype 9) PROP4>ACS -INROR(ype 9) 5000 0000 Current (A) 15000 10000 5000 0 0 5 10 15 0 ime (us)
( ) ( ) t E c t E c t ae t ae be t d d t R d t + + = + = = = + + + + = 4 1) ( 5 5 cos 1 ) (1 4 4 1 1) ( 1 4 1 ) ( 1 0 0 0 4 3 1 0 π π
4 Segment Method a=10%, b=90%, c=67.5% PROP5>ACS -INEDR(ype 9) 30000 PROP5>ACS -INEDR(ype 9) PROP5>ACS -INROR(ype 9) 5000 0000 Current (A) 15000 10000 5000 0 0 5 10 15 0 ime (us)
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/1000000.0 99D =ID/1000000.0 99E =EI*1000.0 99ACON =AICON/100.0 99BCON =BICON/100.0 99CCON =CICON/100.0 993 =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
Actual Breaker RV
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
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
( ) ( ) 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 ) (1 4 5 4 3 4 1 1) ( 1 4 1 ) ( 1 0 0 0 4 3 1 0 π π d c d d c d x = 1 cos 5 1 cos 1 1 1 π π
Curve FromC37.013 a=10, d=90, c=90, d=67.5 FIN>ACS -INEDR(ype 9) 30000 FIN>ACS -INEDR(ype 9) FIN>ACS -INROR(ype 9) 5000 0000 Current (A) 15000 10000 5000 0 0 5 10 15 0 ime (us)
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 67.5 99 =I/1000000.0 99D =ID/1000000.0 99E =EI*1000.0 99ACON =AICON/100.0 99BCON =BICON/100.0 99CCON =CICON/100.0 99DCON =DICON/100.0 993 =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...^...^..
Conclusions Minimum RV Curves Should Be Specified For C37.013 5 Segment Curve Suggested Exact Values/Equations Must Be Determined From est Data C37.011 Use of IEC RV Standard May Have Similar Problem as C37.013 if Minimum RV Curve Not Specified