specialtopics_10

8/12/2019 specialtopics_10

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Lightning

Lightning greatest cause of outages:

1- 26% outages in 230 KV CCTs & 65% ofoutages 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


2/43

Development

When E excessive: space InsulationBreakdown or lightning flash occur

A high current discharge

Those terminate on or near powerlines

similar to: close a switch between

cloud & line or adjacent earth a direct con. Or through mutual

coupling


3/43

Lightning surge

Disturbance on a lineTraveling wave

Travel both Direction, 1/2IZ0

I: lightning current Z0=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


4/43

Lightning

Localized charge of thunder cloudssuperimposes its field on the fine weatherfield, freq. causing it to reverse

As charges within cloud & by induction onearth below, field sufficient Breakdown(30KV/cm)

Photographic evidence: a stepped leaderstroke, random manner &short steps fromcloud to earth

Then a power return stroke moves up theionized channel prepared by leader


5/43

Interaction Between Lightning &Power System

Goal: reduce service Interruption bylightning

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.)


6/43

Lightning Equivalent CCT

Assuming cloud & earth forming a cap.,Discharged by stroke

return CCT completed by displacementcurrent 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:r1=10 cm tor2=1 km)

C=0A/d ==8.854x10^(-12)xx10^6/(10^3x4)=6.95 nF

(or: if 4 canceled 28 nF)


7/43

Parameters of Lightning Model

Z0=540 & a period of 24s

resistance of ionized path, damp Osc.

if: resistance 5000 , result inapprox. a 1.5:30 s wave

More accurate representation:

Consider the leader stroke & prestrike parallel plate capacitor not adequate


8/43

GriscomEq. CCT. For Lightningconsidering prestike

(a) Traveling Model

(b) Lumped Model

Prestrike initiat.

Switch closes& chargecirculate in CCT : (a)

Simplified CCT :(b)

Distributed Rep. of Arcchannel & Tower

replaced by: LumpedCCT. Constants


9/43

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 forlightning

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 persquare meter /year)

GFD=0.04 T ^(0.25)


10/43

Isokeraunic Map for a Region

ISOkeraunic


11/43

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.


12/43

Electrical Shadow

h=hmax-2/3 sag ex: T=30,h=26m,

b=6.7m for a 230 kV

line then:NL=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


13/43

Stroke currnet Magnitude

Anderson & Erikson collected Data

Fig illustration of prob. Of a range of stroke

current magnitude PI=1/[1+(I/31)^2.6] pu

PI: 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


14/43

Surge Impedance of Towers

Zt (class1)=

30ln[2(h +r )/r ]

Zt (class2) =1/2(Zs+Zm)Zs=60ln(h/r)+90(r/h)-60

zm=60ln(h/b)+90(b/h)-60

Zt (class3)=

60ln[ln(2 2h/r)-1]a 35-m class 1 tower base

2r=12m,Z=88.4


15/43

Thevenin Eq. CCT. Of Lightning

Tower tops connected toa GW:

ZGW=520,Zeff=[ZTxo.5ZGW]/

[ZT+0.5ZGW]=65.97 Lightning stroke as a

current source:

its Thevenin eq. CCT.

ZS:Impedance of L.Channel Z: Impedance of

stricken object

Z in top example=65.97

Z(stroke mid span)=0.5 ZGW


16/43

Example continued

Surge voltage:

Is. Z. Zs/ [Z + Zs]=IsZ/[1+Z/Zs]=

IsZGW/2{1/[1+(ZGW/2Zs)}

Zs, few 1000 & ZGWfew 100

Therefore surge voltageIsZGW/2

Waves encounter discontinuities:

1-adjacent towers, 2-tower footing resist Low footing res.neg. ref. coef. Which

reduce tower potential


17/43

Conclusions

if footing res. High, top voltageincrease

Potential diff. across string insulatorscan cause flashover

Cross-arm potential between towertop and tower foot potentials

Wave traveling on GW induce voltage

on ph. Conductors by a factor: 0.15


18/43

Discussion continued

at least one ph. Opp.Polarity of Lightning Surge

(TABLE Earth Resistivity)

This ph. more likely toflash & called:

Back flashover

Tower footing resistancevery important& depend on:

1-local resistivity of earth,2-connection between

tower & ground

Material m

general av 100

Sea water 0.01-1.0

swampy G 10-100

Dry earth 1000

Pure slate 10^7

sandstone 10^8


19/43

Insulation Coordination

Basic Ideas: overvoltages on PWR SYS1-switching operations2-faults & abnormal conditions3-Lightning

How to protect PWR SYS: is an Economic1-unrealistic to insulate against any surge.2-unrealistic to only insulate against S.S.

A compromise is needed: A reasonable investment in

1- insulation2-reliable protective devices; guard against

uncontrollable transients Above item called INSULATION COORDINATION


20/43

Objectives of InsulationCoordination

Design the insulation of a powersystem with all its components to:

Minim. damage & service interruption asa consequence of:

a-S.S. b-dynamic c-transient O.V. s

and do so ECONOMICALLYto achieve this goal need information


21/43

Information needed for Ins.Cor.

A-STRESS:1- likely mag. & frequency of occur. of Lightn.

And sw. surges; PWRSYSEQUIP. willbesubjectedto

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


22/43

The Strength of Insulation

voltage withstand of an insulation

depends on:

1-magnitude of stress2-rate at which is applied

3-duration of the stress

dielectricstrengthiswaveformdependent Dielectricstrengthisstatistic


23/43

Insulation Withstand Evaluation

wave form dependence& breakdown timelag 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, instep

2-waveform fixed just mag. changed

At least 3 Tests at each voltage level Critical flashover:50% flash&50% do not

This called CFO


24/43

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 humidity2-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 invoids


25/43

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 cable1-fault destructive

2-fault permanent3-equipment should be removed from service & repaired

These faults should be avoided and protected against


26/43

Statistical properties of VoltageWithstand of an equipment

Insulation can withstand one surgeappliaction & fail in second,

withstand voltage of equipment isdefinable in statistical term

W.Volt. has a probability Dis. With:

a mean &

standard deviation

W.Volt.:self-restoring INS can be det.


27/43

Withstand Voltage ProbabilityDistribution

uncertaintyphysics of :

electric discharge & insulation B.D. Suppose n tests with each

VT1,VT2,,VTr on any sample result in :relative frequency of failure : k/n

where: k:number of failure at VTk Graph expressing dependence of failure prob.

P= k/n on VTkCUMULATIVE DIS. FUNC.

F(VT)=p[VW


28/43

DENSITY FUNCTION of VOLTAGEWITHSTAND

H.V. gaps approx.

Normal Dis. ,Gaussian

=mean value: CFO

2

2

1( )

2

1( )

2

x

f x e

2

2

1( )

21

( )2

TVx

TF V e dx


29/43

Discussion on Density & Cumulativefunctions

F area under f(x) between x1,x2

CFO crest of Impulse cause FOV. 50%

CFO is polarity sensitive

Disposition about CFO given by In Integral EQ. ;can substitute r.h.s.

1/(2 ).(x-) =1/2.[(x-CFO)/] Normalize EQ. by defining Z=(x-)/ Therefore integrating Z1to Z2 (x1 to x2)

Reduce No. of required Normal curves to 1


30/43

Example on Application ofTransformed Normal EQ. & Table

A string of Insulators CFO=920 kV+ve switching Imp.s & =5%

P(820


31/43

INSULATION COORDINATIONSTRATEGY

power sys components act asantenna 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 thatof Terminal Components

Fig coordinated with (a) not (b)


32/43

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 wind2-C.B. in open position, flashover to ground

before spark over between its contacts


33/43

Test Voltage Waveforms & TransientRatings: BIL - BSL

Representative surges:1-Pwr Freq. 2-Sw surge 3-Impulse wave tf: 1.6 x time between 30% to 90%

on wave front,tt : time from origin to 1/2 value point on the backof wave

The IEC standard Imp. 1.2/50 wave The IEC standard SW. 250/2500 waveV.W.S. in terms BIL(basic lightning impulse ins. L.)V.W.S. in terms BSL(basic sw. impulse ins. L.)


34/43

Examples of BIL & BSL

INS. with special BIL or BSL: lack disruptivedischarges up to the Level

Different Meaning:

1-for self-restoring INS.:90% prob. ofWithstand

2- for non-self-restoring INS. :

No disruptive discharge

For a 13.8 kV Transf. BIL is 95 kV also 75 &50 kV available lessexpensive, morevulnerable

full BSL for this Transformer : 75 kV


35/43

BIL & BSL Continued

Margin between rated & BILreduce as V increase

Vmax design voltage=362 kV, BIL=1300kVcorrespondingreducedlevels1175,1050kVRotating Machines lower BIL:

According to ANSI : if

E; line to line voltage in kVBIL=1.25(2x2E+1)

for 23 kV generator, BIL is 83 kV


36/43

Statistical Approach to InsulationCoordination


37/43

Assignment N0.4 (Solution)

Question 1

13.8 KV, 3ph Bus

L=0.4/314=1.3 mH

Xc=13.8 /5.4=35.27, C=90.2F

Z0=101.3/9.02=3.796

Vc(0)=11.27KV Ipeak=18000/3.796=

4.74 KA


38/43

Question 1

1- Vp=2x18-11.27=24.73 KV Trap

2- Assuming no damping, reaches

Again the same neg. peak and11.27KV trap

3- 1/2 cycle later (18-11.27)=-6.73

Vp2=-(24.73+2x6.73)=-38.19 KV


39/43

Question 2

C.B. reignites duringopening&1st

Peak voltage on L2

L2=352,L1=15mH,C=3.2nF

So reigniting at Vp,

2 comp.: Ramp:Vs(0).t/[L1+L2]=1382x10 /[3(352+15)x10-

]=0.307x10^6 t Oscill.of : f01=1/2 x

{[L1+L2]/L1L2C} Z0={L1L2/[c(L1+L2)]} component2:as Sw closesIc=[Vs(0)-Vc(0)]

/{L1L2/[c(L1+L2)]}2VpC/L1=104.1 A


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Question 2 continued

Eq of Reignition current

I t + Imsin0t which at current zero:

sin0t=-It/Im, 0=1/LC1=1.443x10^5

Sin 1.443x10^5t=-0.307x10^6t/104.1=-2.949x10^3t

Sin 1.443x10^5t =-2.949x10^3tt(s): 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=66.68s I1=0.307x66.68=20.47 A

Vp=I1L2/C=20.47x10.488=214.7 KV


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Question 3

69 KV, 3ph Cap. N

isolated, poles interrupt N.Seq.

1601st reignite

Xc=69 /30=158.7

C=20F,CN=0.02FVs-at-reig=692/3cos160

=-52.94 KV

Trap Vol.:

VA(0)=56.34KV

VB(0)=20.62KV,VC(0)=

-76.96KV,VCN(0)=28.17KV

Vrest=56.34+28.17+52.94=137.45 KV


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Question 3 continued

Z0=L/CN=5.3x0.2 x100=514

Ip-restrike=137.45/514=0.267KA=267A

F0=1/[2LCN]=10^6/{253x2}=15.45 KHz

Voltage swing N=2x137.45=274.9 KV

VN=28.7-274.9=-246.73 KV

VB=-246.73+20.6=-226.13 KV

VC=-246.73+-76.96=-323.69 KV

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