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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 And 47% of 33 KV sys in UK study of 50000 faults reports Also Caused by Lightning Clouds acquire charge& Electric fields within them and between them

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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

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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

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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

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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.)

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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)

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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

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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

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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)

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Isokeraunic Map for a Region ISOkeraunic

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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.

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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

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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

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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

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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

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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

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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

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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 water0.01-1.0 swampy G10-100 Dry earth1000 Pure slate10^7 sandstone10^8

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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”

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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

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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

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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

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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

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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

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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

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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.

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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

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

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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

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

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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)

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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

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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.)

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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

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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

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Statistical Approach to Insulation Coordination

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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

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Question 1 1- Vp=2x18-11.27=24.73 KV Trap 2- Assuming no damping, reaches Again the same neg. peak and 11.27KV trap 3- 1/2 cycle later –(18-11.27)=-6.73 V p2 =-(24.73+2x6.73)=-38.19 KV

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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

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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=- 2.949x10^3t Sin 1.443x10^5t =-2.949x10^3t t(μs): 70 68 67 66.7 66.8 -0.6259 -0.3780 -0.2409 -0.1987 -0.1959 -0.2064 0.2005 -0.1376 -0.1967 -0.1966

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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

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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 =-52.94 KV Trap Vol. : V’A(0)=56.34KV V’B(0)=20.62KV,V’C(0)= -76.96KV,VCN(0)=28.17KV V rest =56.34+28.17+52.94=1 37.45 KV

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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 =28.7-274.9=-246.73 KV V B’ =-246.73+20.6=-226.13 KV V C’ =-246.73+-76.96=-323.69 KV

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