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Lightning  Lightning greatest cause of outages: 1- 26% outages in 230 KV CCTs & 65% of outages in 345 KV Results of study on 42 Companies in USA & CANADA.

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Presentation on theme: "Lightning  Lightning greatest cause of outages: 1- 26% outages in 230 KV CCTs & 65% of outages in 345 KV Results of study on 42 Companies in USA & CANADA."— Presentation transcript:

1 Lightning  Lightning greatest cause of outages: 1- 26% outages in 230 KV CCTs & 65% of outages in 345 KV Results of study on 42 Companies in USA & CANADA  And 47% of 33 KV sys in UK study of faults reports Also Caused by Lightning  Clouds acquire charge& Electric fields within them and between them

2 Development  When E excessive: space Insulation Breakdown or lightning flash occur  A high current discharge  Those terminate on or near power lines  similar to: close a switch between cloud & line or adjacent earth  a direct con. Or through mutual coupling

3 Lightning surge  Disturbance on a line  Traveling wave  Travel both Direction, 1/2IZ 0 I: lightning current Z 0 =Surge Imp. Line  The earth carries a net negative charge of 5x10^5 C, downward E=0.13KV/m  An equivalent pos. charge in space  Upper Atmosph. Mean potential of 300 KV relative to earth

4 Lightning  Localized charge of thunder clouds superimposes its field on the fine weather field, freq. causing it to reverse  As charges within cloud & by induction on earth below, field sufficient Breakdown(30 KV/cm)  Photographic evidence: a stepped leader stroke, random manner &short steps from cloud to earth  Then a power return stroke moves up the ionized channel prepared by leader

5 Interaction Between Lightning & Power System  Goal: reduce service Interruption by lightning  Need : A model for lightning stroke  used with Sys Eq. CCT & study Interaction  Lightning strikes a power line: -a current injected to Power sys. -through an Impedance(i.e.tower impedance)  The voltage across insulator & flashover  To avoid it, ground wires are used and then: 1-tower imp. Parallel Gr. w. Imp.,&reduce Eq. Imp. 2-shield the phase conductors, i.e. lightning strike Gr. W. 3-total Imp. Reduced & tower top voltage is less (tower Gr. Resistance should be low.)

6 Lightning Equivalent CCT  Assuming cloud & earth forming a cap., Discharged by stroke  return CCT completed by displacement current in Elec. Field  Bewley, calculated Induct. Of path: L=2x10^(-7)∫[1-(x/r)]dx/x H/m = =2x10^(-7) ln(r2/r1)=2.18mH (integ. From: r 1 =10 cm to r 2 =1 km)  C=ε 0 A/d = =8.854x10^(-12)xΠx10^6/(10^3x4)=6.95 nF (or: if 4 canceled ≈28 nF)

7 Parameters of Lightning Model  Z 0 =540 Ω & a period of 24μs  resistance of ionized path, damp Osc. if: resistance 5000 Ω, result in approx.  a 1.5:30 μs wave  More accurate representation: Consider the leader stroke & prestrike  parallel plate capacitor not adequate

8 Griscom Eq. CCT. For Lightning considering prestike  (a) Traveling Model  (b) Lumped Model  Prestrike initiat. Switch closes& charge circulate in CCT : (a)  Simplified CCT :(b)  Distributed Rep. of Arc channel & Tower replaced by: Lumped CCT. Constants

9 Isokeraunic Map  How vulnerable is Trans. & Dis. To Lightning : 1-depends on geographic location Lightning activity varies place to place 2-depends how attractive is a line as termination for lightning  Keraunic level (T): degree of lightning activity :No. of days/yr thunder heard  GFD: a new parameter defined as Ground flash rate (number of cloud to ground flashes per square meter /year)  GFD=0.04 T ^(0.25)

10 Isokeraunic Map for a Region  ISOkeraunic

11 kraunic Level  is statistical & sometimes: vary : yr to yr & season to season  Other factors also introduce uncertainties in predicting lightning performance of lines  Taller structures being more likely to struck  According to Anderson: N0. Lightn./100 km/yr, NL=0.004 x T^1.35 x (b+4h^1.09)  Defined shadow angle as Fig in next slide h: average height of shield wires, b: spacing between S.W.

12 Electrical Shadow  h=h max -2/3 sag  ex: T=30,h=26m, b=6.7m for a 230 kV line then: N L =0.004x30^1.35(b+ 4x26^1.09)=57.67 The impact on line depends on: 1-stroke current Mag. 2-r.r.of stroke current

13 Stroke currnet Magnitude  Anderson & Erikson collected Data  Fig illustration of prob. Of a range of stroke current magnitude  P I =1/[1+(I/31)^2.6] pu P I: probability of exceeding stroke current I I: stroke current in kA  Velocity of surges on eq. line model of tower is approx. 85% speed of light  Different tower design  different Z

14 Surge Impedance of Towers  Z t ( class 1) = 30ln[2(h +r)/r] Z t ( class 2) =1/2(Z s +Z m ) Zs=60ln(h/r )+ 90(r/h )-60 z m=60ln(h/b )+ 90(b/h )-60  Z t (class 3)= 60ln[ln (√2 2h/r )-1] a 35-m class 1 tower base 2r=12m,Z=88.4Ω

15 Thevenin Eq. CCT. Of Lightning  Tower tops connected to a GW: Z GW =520Ω, Z eff =[Z T xo.5Z GW ] / [Z T +0.5Z GW ]=65.97Ω  Lightning stroke as a current source: its Thevenin eq. CCT.   Z S:Impedance of L.Channel  Z: Impedance of stricken object Z in top example=65.97 Z(stroke mid span)= 0.5 ZGW

16 Example continued  Surge voltage: I s. Z. Z s / [Z + Z s ]=I s Z/[1+Z/Z s ]= I s Z GW /2{1/[1+(Z GW /2Z s )}  Z s, few 1000 Ω & Z GW few 100 Ω  Therefore surge voltage  I s Z GW /2  Waves encounter discontinuities: 1-adjacent towers, 2-tower footing resist  Low footing res.  neg. ref. coef. Which reduce tower potential

17 Conclusions  if footing res. High, top voltage increase  Potential diff. across string insulators can cause flashover  Cross-arm potential between tower top and tower foot potentials  Wave traveling on GW induce voltage on ph. Conductors by a factor: 0.15

18 Discussion continued  at least one ph. Opp. Polarity of Lightning Surge (TABLE Earth Resistivity)   This ph. more likely to flash & called: Back flashover  Tower footing resistance very important & depend on: 1-local resistivity of earth, 2-connection between tower & ground MaterialΩmΩm general av100 Sea water swampy G Dry earth1000 Pure slate10^7 sandstone10^8

19 Insulation Coordination  Basic Ideas: overvoltages on PWR SYS 1-switching operations 2-faults & abnormal conditions 3-Lightning  How to protect PWR SYS: is an Economic 1-unrealistic to insulate against any surge. 2-unrealistic to only insulate against S.S.  A compromise is needed: A reasonable investment in 1- insulation 2-reliable protective devices; guard against uncontrollable transients  Above item called “INSULATION COORDINATION”

20 Objectives of Insulation Coordination  Design the insulation of a power system with all its components to: Minim. damage & service interruption as a consequence of: a- S.S. b- dynamic c- transient O.V. s and do so ECONOMICALLY ◊to achieve this goal need information

21 Information needed for Ins.Cor.  A-STRESS: 1- likely mag. & frequency of occur. of Lightn. And sw. surges; PWR SYS EQUIP. will be subjected to 2-how distribute between &within components  B-strength: dielectric withstand of various ins. Sys.s  C-protection devices & arrangements to eliminate or reduce their effect  D-Economics: item 1,2&3 coordinated to be effective and Economic

22 The Strength of Insulation  voltage withstand of an insulation depends on: 1-magnitude of stress 2-rate at which is applied 3-duration of the stress  dielectric strength is waveform dependent  Dielectric strength is statistic

23 Insulation Withstand Evaluation  wave form dependence& breakdown time lag can be quantified by: VOLT TIME CURVE  Gen. of V.T.C. for a string of INSUL. 1-series of surges from low to high, in step 2-waveform fixed just mag. changed  At least 3 Tests at each voltage level  Critical flashover:50% flash& 50% do not  This called CFO

24 Examples of Volt-time curve  a 20-inch rod gap: sharp turn-up  a long air gap  B.D. in open air depend on: 1-relative humidity 2-air pressure  sw surge strength for neg impulse higher ignored in : Flashover Failure Rate  B.D. liquid similar to gas up to streamer M.  Solid ins. B.D. progressive, P.D. occur in voids

25 Discussion on different INS. B.D.s  B.D. in solid ins. Is not self-healing  Insulators of T.L. flashover then : (self-restoring) 1-C.B. operate and eliminate fault 2-arc path deionizes 3-& C.B. can be reclosed in less than a second ◊ Solid ins. Of Transformer or cable 1-fault destructive 2-fault permanent 3-equipment should be removed from service & repaired  These faults should be avoided and protected against

26 Statistical properties of Voltage Withstand of an equipment  Insulation can withstand one surge appliaction & fail in second,  withstand voltage of equipment is definable in statistical term  W.Volt. has a probability Dis. With: a mean & standard deviation  W.Volt.:self-restoring INS can be det.

27 Withstand Voltage Probability Distribution  uncertainty  physics of : electric discharge & insulation B.D.  Suppose “n” tests with each V T1,V T2,…, V Tr on any sample result in : relative frequency of failure : ν k /n where: ν k: number of failure at V Tk  Graph expressing dependence of failure prob. P= ν k /n on V Tk CUMULATIVE DIS. FUNC.  F(V T )=p[V W

28 DENSITY FUNCTION of VOLTAGE WITHSTAND  H.V. gaps approx. Normal Dis., Gaussian   μ=mean value: CFO

29 Discussion on Density & Cumulative functions  F area under f(x) between x 1,x 2  CFO crest of Impulse cause FOV. 50%  CFO is polarity sensitive  Disposition about CFO given by σ  In Integral EQ. ;can substitute r.h.s. 1/(2σ).(x-μ)=1/2.[(x-CFO)/σ]  Normalize EQ. by defining Z=(x-μ)/σ  Therefore integrating Z 1 to Z 2 (x 1 to x 2)  Reduce No. of required Normal curves to 1

30 Example on Application of Transformed Normal EQ. & Table  A string of Insulators CFO=920 kV +ve switching Imp.s & σ=5%  P(820

31 INSULATION COORDINATION STRATEGY  power sys components act as antenna picking up surges  surges should be prevented reaching equipments  This done by INS. COORD. 1-line ins. Flashover before solid 2-volt-time of Line Ins. lies below that of Terminal Components  Fig © coordinated with (a) not (b)

32 Coordination of Insulation Strength & expected Overvoltages  Fig: superposition of air volt-time & envelope of Sys O/Vs,lacking Co.(L.,sw)  Surge protective fitted to coordination Strategy  Su.Prot.D. operate to restrict voltage within Dielectric capability of device INS. & 1-Transf. Bushing flashover before wind 2-C.B. in open position, flashover to ground before spark over between its contacts

33 Test Voltage Waveforms & Transient Ratings: BIL - BSL  Representative surges: 1-Pwr Freq. 2-Sw surge 3-Impulse wave  t f : 1.6 x time between 30% to 90% on wave front, t t : time from origin to 1/2 value point on the back of wave  The IEC standard Imp. 1.2/50 wave  The IEC standard SW. 250/2500 wave V.W.S. in terms BIL( basic lightning impulse ins. L.) V.W.S. in terms BSL( basic sw. impulse ins. L.)

34 Examples of BIL & BSL  INS. with special BIL or BSL: lack disruptive discharges up to the Level  Different Meaning: 1-for self-restoring INS.:90% prob. of Withstand 2- for non-self-restoring INS. : No disruptive discharge  For a 13.8 kV Transf. BIL is 95 kV also 75 & 50 kV available less expensive, more vulnerable full BSL for this Transformer : 75 kV

35 BIL & BSL Continued ◊ Margin between rated & BIL reduce as V increase ◊ V max design voltage =362 kV, BIL=1300kV ◊ corresponding reduced levels 1175,1050kV ◊Rotating Machines lower BIL: According to ANSI : if E; line to line voltage in kV BIL=1.25(√2x2E+1) ◊for 23 kV generator, BIL is 83 kV

36 Statistical Approach to Insulation Coordination 

37 Assignment N0.4 (Solution)  Question 1  13.8 KV, 3ph Bus L =0.4/314=1.3 mH X c =13.8/5.4=35.27 Ω, C=90.2μF  Z 0 =10√1.3/9.02=3. 796Ω  V c (0)=11.27KV  I peak =18000/3.796= 4.74 KA

38 Question 1  1- Vp=2x =24.73 KV Trap  2- Assuming no damping, reaches Again the same neg. peak and 11.27KV trap  3- 1/2 cycle later –( )=-6.73 V p2 =-( x6.73)= KV

39 Question 2  C.B. reignites during opening&1 st  Peak voltage on L2 L2=352,L 1 =15mH, C=3.2nF So reigniting at V p, 2 comp.:  Ramp:V s (0).t/[L 1 +L 2 ]= 138√2x10/[√3(352+15)x10- ]=0.307x10^6 t  Oscill.of : f 01 =1/2Π x {√[L 1 +L2]/L 1 L 2 C}  Z0=√{L1L2/[c(L1+L2)]}  component2:as Sw closes Ic=[Vs(0)-Vc(0)] /√{L1L2/[c(L1+L2)]} ≈2V p √C/L 1 =104.1 A

40 Question 2 continued  Eq of Reignition current I’ t + I m sinω 0 t which at current zero: sinω 0 t=-I’t/I m, ω 0 =1/√LC 1 =1.443x10^5  Sin 1.443x10^5t=-0.307x10^6t/104.1= x10^3t  Sin 1.443x10^5t =-2.949x10^3t t(μs):

41 Question 2  t=66.68μs I 1 =0.307x66.68=20.47 A  V p =I 1 √L 2 /C=20.47x10.488=214.7 KV

42 Question 3  69 KV, 3ph Cap. N isolated, poles interrupt N.Seq.  160◦ 1 st reignite  X c =69/30=158.7 C=20μF,C N =0.02μF V s-at-reig =69√2/3cos160 = KV  Trap Vol. :  V’A(0)=56.34KV V’B(0)=20.62KV,V’C(0)= KV,VCN(0)=28.17KV  V rest = = KV

43 Question 3 continued  Z 0 =√L/C N =√5.3x0.2 x100=514Ω  I p-restrike =137.45/514=0.267KA=267A  F 0=1/[2Π√LC N ]=10^6/{2Π√53x2}=15.45 KHz  Voltage swing N=2x137.45=274.9 KV  V N = = KV  V B’ = = KV  V C’ = = KV


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