Transcript specialtopics_10

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
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
Lightning surge
Disturbance on a lineTraveling wave
Travel both Direction, 1/2IZ0
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


I:

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
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.)
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: r1=10 cm to r2=1 km)
 C=ε0 A/d =
=8.854x10^(-12)xΠx10^6/(10^3x4)=6.95 nF
(or: if 4 canceled ≈28 nF)
Parameters of Lightning Model
 Z0=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
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
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)
Isokeraunic Map for a Region
 ISOkeraunic
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.
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
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
Surge Impedance of Towers
 Zt (class 1)=
30ln[2(h +r )/r ]
Zt (class 2) =1/2(Zs+Zm)
Zs=60ln(h/r)+90(r/h)-60
zm=60ln(h/b)+90(b/h)-60
 Zt (class 3)=
60ln[ln(√2 2h/r)-1]
a 35-m class 1 tower base
2r=12m,Z=88.4Ω
Thevenin Eq. CCT. Of Lightning
 Tower tops connected to
a GW:
ZGW=520Ω,
Zeff=[ZTxo.5ZGW] /
[ZT+0.5ZGW]=65.97Ω
 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
Example continued
 Surge voltage:
Is. Z. Zs/ [Z + Zs]=IsZ/[1+Z/Zs]=
IsZGW/2{1/[1+(ZGW/2Zs)}
 Zs, few 1000 Ω & ZGW few 100 Ω
 Therefore surge voltageIsZGW/2
 Waves encounter discontinuities:
1-adjacent towers, 2-tower footing resist
 Low footing res. neg. ref. coef. Which
reduce tower potential
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<K<0.3
Discussion continued
 at least one ph. Opp.
Polarity of Lightning Surge
(TABLE Earth Resistivity) 
Material
Ωm
general av 100
Sea water 0.01-1.0
 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
swampy G 10-100
Dry earth
1000
Pure slate 10^7
sandstone 10^8
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”
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
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
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
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
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
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
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.
Withstand Voltage Probability
Distribution
 uncertainty physics of :
electric discharge & insulation B.D.
 Suppose “n” tests with each
VT1,VT2,…,VTr on any sample result in :
relative frequency of failure :
where:
νk/n
νk: number of failure at VTk
 Graph expressing dependence of failure prob.
P= νk/n
on
VTk
CUMULATIVE DIS. FUNC.
 F(VT)=p[VW<VT]
 f(VT)=dF(VT)/dVT its DENSITY FUNCTION
DENSITY FUNCTION of VOLTAGE
WITHSTAND
 H.V. gaps approx.
Normal Dis. ,Gaussian

1
f ( x) 
e
 2
1
F (VT ) 
 2
 μ=mean value: CFO

1
2
VT


2
(
x


)
2

e
1
2
2
(
x


)
2
dx
Discussion on Density & Cumulative
functions
F area under f(x) between x1,x2
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 Z1 to Z2 (x1 to x2)
 Reduce No. of required Normal curves to 1





Example on Application of
Transformed Normal EQ. & Table
 A string of Insulators CFO=920 kV
+ve switching Imp.s & σ=5%
 P(820<VT<880)=?
 σ= 0.05x920=46 kV
Z1=(820-920)/46=-2.174,Z2=(880-920)/46=-0.86
 P(820<VT<880)=P(-2.174<Z<-0.8696)
=p(Z<-0.8696)-P(Z<-2.174)=
0.1922-0.0148=0.1774
 Proability : 17.74%
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)
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
Test Voltage Waveforms & Transient
Ratings: 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 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.)
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
BIL & BSL Continued
◊ Margin between rated & BIL
reduce as V increase
◊ Vmax 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
Statistical Approach to Insulation
Coordination

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
Ω,
C=90.2μF
 Z0=10√1.3/9.02=3.
796Ω
 Vc(0)=11.27KV
 Ipeak=18000/3.796=
4.74 KA
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
Vp2=-(24.73+2x6.73)=-38.19 KV
Question 2
C.B. reignites during
opening&1st
 Peak voltage on L2
L2=352,L1=15mH,

C=3.2nF
So reigniting at Vp, 2 comp.:
 Ramp:Vs(0).t/[L1+L2]=
138√2x10 /[√3(352+15)x10]=0.307x10^6 t
 Oscill.of : f01=1/2Π x
{√[L1+L2]/L1L2C}
 Z0=√{L1L2/[c(L1+L2)]}
 component2:as Sw closes
Ic=[Vs(0)-Vc(0)]
/√{L1L2/[c(L1+L2)]}
≈2Vp√C/L1=104.1 A
Question 2 continued
 Eq of Reignition current
I’ t + Im sinω0t which at current zero:
sinω0t=-I’t/Im , ω0=1/√LC1=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
-0.6259 -0.3780
-0.2064 0.2005
67
-0.2409
-0.1376
66.7
-0.1987
-0.1967
66.8
-0.1959
-0.1966
Question 2
 t=66.68μs I1=0.307x66.68=20.47 A
 Vp=I1√L2/C=20.47x10.488=214.7 KV
Question 3
 69 KV, 3ph Cap. N
isolated, poles interrupt N.Seq.
 160◦ 1st reignite
 Xc=69
/30=158.7
C=20μF,CN=0.02μF
Vs-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
 Vrest=56.34+28.17+52.94=1
37.45 KV
Question 3 continued
 Z0=√L/CN=√5.3x0.2 x100=514Ω
 Ip-restrike=137.45/514=0.267KA=267A
 F0=1/[2Π√LCN]=10^6/{2Π√53x2}=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