Download presentation

Presentation is loading. Please wait.

Published byKarina Needle Modified over 2 years ago

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 50000 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 water0.01-1.0 swampy G10-100 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=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

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=- 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

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 =-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

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 =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

Similar presentations

Presentation is loading. Please wait....

OK

Calculating Separation Distance and Surge Current

Calculating Separation Distance and Surge Current

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on e commerce business models Types of window display ppt online Ppt on condition monitoring maintenance Ppt on art and craft movement furniture Ppt on pollution prevention day Ppt on world war 1 and 2 Ppt on peak load pricing electricity Ppt on different types of dance forms list Ppt on environmental pollution Ppt on regional trade agreements definition