Resistance Coverage of Memory Polarized Mho Distance Elements

Size: px

Start display at page:

Download "Resistance Coverage of Memory Polarized Mho Distance Elements"

Noah Norris
5 years ago
Views:

1 Resistance Coverage of Memory Polarized Mho Distance Elements Pratap Mysore, P.E. Pratap Consulting Services Minnesota Power Systems Conference November 8, 2016

2 Arcing Faults AR. Van C. Warrington, Reactance Relays Negligibly Affected by Arc Impedance, Electrical World, September, 1931, pp , Published results are from field tests done HV systems of New England and Tennessee Electric Power Company ( A) The voltage across the arc is in phase with the current It is Resistive 2

3 Arc Resistance Empirical formulae used in the industry Warrington s Formula Rarc = &'() + I 1.4 L Length of the arc and I- current through the arc. Voltage drop across the arc, V arc = &'() + I 0.4 Voltage Decreases with the increase in current &'() +(-./01) Rarc = I 1.4 U: Cross wind speed, mph, t:time in seconds 3

4 Westinghouse Experiments A.P. Strom, Long 60 Cycle Arc in Air, Transactions of AIEE, Vol. 65, PP , Used 1/8 48 inch arc lengths with currents from 68A- 22kA Average Voltage Gradient: 31-38V/inch Arc voltage varies considerable from one cycle to another (21.5 V-50V/in with an average of 34V/in or 408V/ft) 4

5 Arc Resistance Empirical formula used in the industry (USA) Formula suggested in Westinghouse Relay book Rarc = 33) 4 through the arc. 5,L Length of the arc and I- current Voltage drop across the arc, V arc = 440L Voltage drop is dependent only on the length of the arc. 5

6 More Recent Research Prof. Terzija s research from 2000 onwards According to Prof. Vladimir Terzija -majority of faults (over 90%) are arcing faults. Questioned accuracy of Warrington s formula. Test Current range 2kA-12kA Arc Length 0.17m -2m (0.56ft-6.56ft) Rectangular Voltage waveforms with THD of 30% Two approaches to derive Equation for voltage Gradient I R = 8 8 : ;4 < from Arc modeling = (A+B/I)*L from spectrum domain analysis E a (1200 to 1500V/m or 365V - 457V/ft) A= ( if L is in ft) B= (1372 if L is in ft) 6

7 Voltage Across the arc Voltage drop across the arc either decreases with infeed or remains constant. It is proportional to the length of the arc. Voltage waveform has THD of 30% (Distorted Waveform) 7

8 Mho Distance element Compensated voltage(iz-v), where I and V are voltage and current at the relay location Polarizing voltage, V P This could be the fault voltage or pre-fault or Voltage from healthy phases(s). 8

9 Mho Characteristics Memory voltage allows detection of close-in faults when V=~0 9

10 Dynamic Mho Characteristics 10

11 Mho Distance Element Mho Distance Element Fault Voltage, V Fault Current, I Polarizing voltage (Pre-fault or positive sequence memory ) A-G V A I A +K 0 I N V AP B-G V B I B +K 0 I N V BP C-G V C I C +K 0 I N V CP A-B V AB I A -I B V ABP B-C V BC I B -I C V BCP C-A V CA I C -I A V CAP 11

12 Resistance Coverage Phase to Phase faults: Voltage at the relay: V AB = I A Z I B Z + V ARC Z Calculated = V AB /(I A -I B ) =Z +V ARC /(I A -I B ) =Z + V ARC /(2*I)=Z+R ARC /2; where R ARC =V ARC /I A 12

13 Resistance Coverage Phase to Ground Faults Voltage at the relay: V A = (I A + K 0 I N ) Z +V ARC Z Measured = V A /(I A +K 0 I N ) +R*I A /(I A +K 0 I N ) For a radially fed fault, I A =I N Z calculated = Z + R/(1+K 0 ) Where, R = Arc resistance + parallel combination of Tower footing resistance and ground wire return path. 13

14 Resistance Coverage Memory Polarized Mho Element O- Center of the Circle; OA = (Z S +Z R )/2 (1-K )Z R = (Z S -Z R )/2 + KZ R ; Radius, OB = (Z S +Z R )/2 jx Z R A R K B A R K α B kz R Z S O (Z S +Z R )/2 R (Z S + Z R )/2 O 14

15 Resistance Coverage At location K R K = [((Z S +Z R )/2) 2 - ((Z S -Z R )/2 + KZ R ) 2 ]; Using (A 2 B 2 ) = (A-B)(A+B); R K = (Z S + KZ R )(1-K)Z R Defining Z S /Z R as SIR ( assuming the relay reach=z L ) R K =Z R (SIR+K)(1-K) For a fault at relay location, K=0 R close-in = Z R SIR Z R = m.z L ; m=0.85 for zone 1 and 1.25 for Zone 2 R close-in = m.z L (Z S /mz L ) R close-in = Z L (m*sir) 15

16 Effect of considering the resistance of source and line A θ L R K α B OB = Radius of the circle, (Z S +Z R )/2; AB= resistance coverage, R K, to be determined OA = (Z S +Z R )/2 (1-K)Z R = 0.5(Z S -Z R ) + KZ R Angle OAB = (180-θ L ) where θ L is the line angle in degrees. β (Z S + Z R )/2 Using the law of sines: (OA/sinα) = (OB/sinOAB) = R K /sin β Angle α is determined from Sin α = (OA/OB) sin (OAB) =(OA/OB)*sin(180- θ L ) = (OA/OB)*sinθ L = ).( >?@> A.B> A ).( >?.> A sinθ L O 16

17 Resistance Coverage At location K R K = ).( C D.C E FGHI J sin(θ 4 α) R K = ).( C D.C E sin[θ QGHI 4 ).( C D@C E.SC E J ).( C D.C E sinθ 4 ] If the relay is set to Z L, Z S =SIR*Z L R K = [ ).( F5U.-)C J QGHI J ]sin[θ 4 ).( F5U@- C J.SC J ).( F5U.- C J sinθ 4 ] 17

18 Resistance coverage -R K For Z R =m.z L R K = [ 0.5 SIR.m)Z L ]sin[θ sinθ L L 0.5 SIR@m Z L.K mz L 0.5 SIR.m Z L sinθ L R K = [ 0.5 SIR.m)Z 0.5 SIR@m.K m ]sin[θ sinθ L sin L 0.5 SIR.m sinθ L For SIR=0 and Relay set to cover 100% (m=1) R K = ).(> k Sin[θ lmno + 1 k sinθ + Sin(2θ L )/Sinθ L R K = Z L Cos θ L ] =0.5Z L 18

19 Resistance coverage for close-in faults at various SIR values Line with Z L ohms at 80 degrees considered. SIR Resistance Coverage at location K (m=1) in PU (Z L =1.0 PU) K=0 K=0.25 K=0.5 K=

20 Conclusions Voltage Drop across the arc either reduces with increase in current or remains constant for constant arc length Memory polarized Mho elements resistance coverage depends on the Source impedance and the type of fault. The Arc coverage with time depends on the memory voltage- whether it is held constant or whether it decreases with time. Resistance coverage at any point within the characteristics can be determined knowing the SIR. Modern fault program developers are planning to provide a macro/script to plot arcing faults using either Warrington/Westinghouse/Terzija s arc resistance equation. 20

21 Questions? 21

Math 104 Final Exam Review

Math 104 Final Exam Review Math 04 Final Exam Review. Find all six trigonometric functions of θ if (, 7) is on the terminal side of θ.. Find cosθ and sinθ if the terminal side of θ lies along the line y = x in quadrant IV.. Find