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Statistical Approach to Insulation
Coordination
 INS. COORDINATION Outcomes:
1-selection or specification of the Elec.
Strength
2-ph. to Gr and Ph. to ph. Clearances &
leakage or creepage distance
3-selection of surge arresters
Suggested Steps (air insulated
Substations)
1-rating of surge arrester
2-creepage length of porcelain insulator
(contamination & Equivalent. BIL & BSL)
3-Arrester close to transformer & its BIL, BSL
4-BIL of other Equipments &
ph. to Gr clearance, lightning O.V.s
(if BIL & clearance excessive, New Arr. & the BIL again.)
5-BSL of other equipment, ph. to Gr. & ph. to ph. Clearances
(required by switching OV.s)
6-Protection of opened CCT. Breaker (gap type or arrester)
Insulation coordination approach
 Ins. Design of substation equipments
◊Transformer as an Example:
1-highest surges enter substation/generate within it
2-stress on coil-to-coil & turn-to-turn INS.
3-INS. Sys. need capability of 20% above calculated
4-Surge arrester rating coordinated with Transformer
INS.
Safety factors employed: to cover lack of knowledge
Compensating the Over Design
◊ However many INS. Designs are overdesigned
◊ certain risk of failure is acceptable
 In Self Healing Insulation Systems:
This compensate the over design
Probabilistic Nature of voltage
stress & Insulation strength
 Insulation of O/H line :
1-reducing No. of insulators:
save money; cost risk; reduce cross arm
length, width of right of way, weight of
tower; affect foundation
2-to justify this change, risk should be
quantified
3-INS. Strength Determined by
R.O.F.
Parameters for Risk Assessment
◊stress has a f(Va) of Normal Dis.
◊ F(va) prob. of Voltage less than Va
◊ Q(Va) prob. of Voltage exceed Va
Breakdown Strength of Ins. Also
Statistic
Risk of Insulation Failure
 Matching probabilities of Insulation
stress & strength, risk of failure
determined
 Prob. of flashover g(Vw) & its cum.
G(Vw)
 then prob. of V1 occur : f(V1)
prob. Flashover at less than V1, G(V1)
◊ prob. of V1 & flash Over: f(V1)xG(V1)
Risk Assessment
 Normal Dis. For stress:
f(Va)=1/[σs√2Π] exp{-1/2[(Va-S)/σs]}
where:
σs=standard dev.
Surge with crest of S2=S+2σs,
S: mean
for sw. surge INS. Design
 Possibility of a surge exceed S2  ≈ 2%
S2 is referred to as :
”statistical
switching overvoltage”
Eq (1)
Probability of Failure
◊ Surge known by: f(Va) applied an Ins. Of
CFO & σ
 The prob. of flashover can calculated :
from
Eq (1) & Eq of Normal Dis. As follows:

1
F (VT ) 
 2
VT
e

1 x CFO 2
 (
)
2

dx
Risk Assessment continued
 resultant probability:
Y
1
z2
P(V ) 
exp  dz  f (Y )

2
2 
 Where:
y=[S-CFO]/√(σ +σs )
◊ Transmission Lines N elements
Prob. At least one flash over:
P(N,V)=1- [1 - P(V)]^N
Example of Risk Assessment
 A line with 500 towers which CFO is
2.3 pu, & σ=4.5% and stress characterized
by: s=1.6 and σs=10%
 σ= 2.3x0.045=0.1035, σs=0.16 pu
Y=[1.6-2.3]/√(0.1035 +0.16 )=-3.67
 P(V)=1.22x10^-4 ignoring neg. surges
P(V)=6.1x10^-5
 This is for 1 Tower, for the whole line:
 P(N,V)={1-[1-0.000061]^500}=3.0x10^-2
 means 30 switch op.s per 1000 applications
Protection of Systems and
Equipment against Transient OVs
 Surges in PWR SYS can be Very destructive
 Important to protect against them
1- stop Lightning surge generated outside to enter
2- minimize effects of those entered
3- sw. surges & surges gen. by faults are
inevitable, however reduced by proper design
 How to protect PWR SYS Equip.s and
control CCTs or relays
 The surge protection devices
Surge Protection Devices & their
Application
 Ground Wires
1- 1 or 2 are bounded to tower above ph. Conductors
2- Ground wires are at ground potential under normal
3- lightning strokes stopped to be terminated on ph.
 Strike Distance: S (Love’s Formula)
1- as a leader stroke approaches earth’s surface, it is
attracted towards tall objects
2- s: strike distance ;
stroke tip reaches within distance “S” of Gr object
prob. of terminating on that >
prob. striking another object in a distance more than s
 S=10x I ^ 0.65 (I in kA, S in m) EQ.1
Strike Distance Application
If I=10 kA, S=44.7 m
if I=50 kA, S=127 m
 Anderson approach
α:shielding angle
βs: h of horizontal line
strokes within βs of Earth
terminate on it, rather
than on Gr or ph
 β=0.8 for EHV
 β=0.67 for UHV
 This line & 2 arcs
Define 3 regions, where:
◊ PQ; unshielded section
Discussion of unshielded section
 PQ reduces for greater S related to higher I
 at some I, Imax points P & Q will coincide
or no shielding failure at I>Imax
 As S reduces, arc PQ & X increases
There is a Imin below it stroke to ph., generate
insufficient voltage to cause flashover
 Imin=2VCFO/Z0
EQ(2)
VCFO=Ins. Critical flashover voltage
Z0=surge Impedance of Line
Shielding Failure Rate
 Xs: is X corresponding to Imin
Shield. Fail.: Imin<I<Imax & strokes within Xs
 P (Imin<I<Imax)=Pmin - Pmax
 No. of strokes to earth/sq. km/yr  N=kT
T: keraunic level
K: a constant 1<k<0.19
 NSF=kT/10 Xs/2 (Pmin-Pmax)/100 km/yr
 Procedure should be performed for each ph cond.
w.r.t. its most protective Gr wire to obtain overall
Shielding Failure Rate
Shielding Failure …continued
 Combining (1) & (2)
Smin=10(2VCFO/Z0)^0.65
 Q is located; being Smin from most exposed
phase conductor and βSmin above the ground
Where : Smin from most exposed ph. Cond. and βsmin,
above GR
 Shield wire now a distance Smin from Q
a circular arc centered on Q of radius Smin
 This assures P & Q coincide and leaving no
unshielded region
Lightning Shielding of Substation
 Gr wires above Substation
 Frequently equipped by
1-lightning rods above structural steel work
2-O/H ground cage solidly bounded to Gr mat
to provide a low resistance ground
economically.
Surge Suppresors & Lightning
Arreters
 To protect against possible S.F.s:
1-family of devices developed known:
(a) Surge diverter,
(b) Surge suppressor,
(c) Lightning arrester
2-these placed in parallel & close
3-permanently connected or sw. in by
Spark over of a series gap
Performance of Protective Devices
 Normal operation;
open gap represents a high impedance
 When gap flash over;
switch over to low impedance mode
 Arc voltage few hundred or a few
thousands volts for long gap
 Surge voltage divided between sys imp. &
the protector Imp.
Surge Voltage divided at protector
terminal

If:
Z1 surge Imp. of sys
generated surge
Zp surge Imp. of protector




Z2 surge Imp. Of load
Current in S when closes:
I=V/{Zp+[Z1Z2/(Z1+Z2)]}=
V[Z1+Z2]/[ZpZ1+ZpZ2+Z1Z2]
V1= V Zp(Z1+Z2)/
[ZpZ1+ZpZ2+Z1Z2]
Z2>>Z1 surge doubled
Dissipating energy pot.
Nonlinear Resistor Protectors
 Rod gap disadvantage:
1-flashover through a fault on CCT & need
CCT interruption
2- do not protect fast rising surges
 A device limit voltage without creating a fault
more attractive
 Nonlinear Resistor is
Such a device
 These resistors resistance diminish as voltage
increase
Nonlinear Surge Diverters and The
Analysis of their surge reponse
 Characteristic : I=kV^α
 SiC type 2<α<6 Fig 16.4 (V-I)
 SiC versus ZnO(20<α<50) Fig 16.13
 Employed at all voltage levels
1-small elements to protect relays,
2-in large junks,under oil, across windings of
pwr Transformer
 Determination of over voltage protection
Examples of characteristic SiC
material
 Fig gives instantaneous VI characteristic of
SiC nonlinear resistors

Operation of Nonlinear Protector
 CCT Diagram:
Thevenin eq CCT
 Zs parallel of Z1 & Z2
 V/I characteristic of
Zp
shown in Fig (d).

Fig© shows variation of
surge without arrester
Lines on Fig (d) with a
slope
of tan− Zs intersects Zp
I2 flowing in Zp

I2Zs=V2-V’2


Surge Arrester
Response…continued
 As surge reaches V1,V2,V3,…
voltage on protected object passes
through V’1,V’2,V’3,…with a
considerable reduction
 It found following the surge variation
with combination of Eq CCT &
Arrester V/I Characteristic
Traditional Lightning Arresters
 Trad.L. Arr. Use nonlinear resistance
 They have gap or gaps in series
 It is possible to design the resistor
element to satisfy the energy
dissipation and voltage-limiting under
surge conditions
 Preferred material ZnO &
traditionally SiC
SiC Arresters
 When suppressor operates an arc in gap
 This arc must quenched as surge pass, or
resistor will be destroyed
 In other cases the gap is not required
 Arresters vary depending on their voltage
class & duty however has:
Gaps, coils, valve elements (nonlinear res.)
 They are stacked in series & hermetically
sealed in a porcelain housing(6 kV element)
 A station-type arrester for 96 kV shown in
next slide
station-type arrester
 In which similar
elements are
connected in series
Operation of Gapped Arresters
 little different from plain nonl. Resis.
 Initially behave like a gap with a
volt/time curve
turn-up relatively slight, less than that
of a rod gap
 Once sparkover occur(in front, peak,
or on tail)nonlinear resistor inserted
 Sequence shown in Fig. 16.8
Metal Oxide Arresters
 Metal Oxide Varistors
1- introduced for O/V protection(1960)
2-larger α than SiC
3- like SiC is crystalline
4-90% ZnO & metal oxides
5-material is ground, mixed, pressed,& sintered and shape
disk blocks
6-the nonlinear property depond on boundary
layers between crystals
7-Fig16-12,VI characteristic, Dyna Var 209kV
Traditional Lightning Arresters
 Traditional lightning arresters uses
nonlinear resistance elements as before
 however have a gap or gaps series with
them
 So resistor is isolated from cct under
normal conditions & is introduced when a
surge appears by sparkover of gap
 It is possible to design resistor element
from energy dissipation & voltage-limiting
under surge conditions
Preferred Type of Arrester
 Preferred material for application is Zinc
Oxide (ZnO)
 however traditionally SiC used
 traditional type still in a vast number are in
service
 A different approach relates to a type of
surge suppressor, in which when suppressor
operates and an arc is established in gap
this arc must be quenched when surge
passed or resistor will be destroyed by
current that flow
Arresters Assembly
 Arresters vary in sophistication upon the
voltage class & duty
 generally comprise: gap units, coil units,
valve elements of nonlinear resistance
material
 These are stacked in series & hermetically
sealed in porcelain housing
 Principle is shown in next fig.
Valve type Arrester
 path of :
a- surge current
b-follow current
 Components
Shown comprise
Requirement for
a 6 kV arrester
Operation of Valve type Arresters
 Magnetic field created by coil follow current
in coil reacts upon this current in arcs of
gap assemblies
 causing them to be driven into arc
quenching chambers
 arc extinction occur at first current zero by
elongating & cooling arc
 Operation of a gapped arrester is little
different from plain nonlinear resistor at
least up to point of gap spark-over
 The sequence illustrated in fig.
Operation of gapped type surge
Arrester
 Surge impinges on
arrester
 Voltage follows surge
voltage to point of
sparkover P
 Upon voltage drop to
Q, defined by Q’
determined by load
line P’Q’
 Then surge climbs to
R but protected
object sees only S
Examination of Efficient Performance of
this type(application of repeated 1.5/40
μs impulses)
 Done according to IEC and USA standards
 For a 12 kV maximum system voltage arrester
 average time to spark-over (a)0.4μs (b)1.8μs
Spark-over Curves
 Volt/time spark-over curves for arresters
 Surges of: 1.2/50 μs (turn up is evident)
Performance of a 36 kV arrester of
this type
 Discharging a 5000 A surge current of 8x20 μs
Note: peaks (voltage & current) do not coincide
Metal Oxide Arresters
 Metal Oxide Varistors 1st for O/V protection in
1960 (safe guard electronic components)
 Years passed until technology advanced to where
large disks of consistent quality & stability made
& applied in PWR SYS
 impact on PWR INDUS. Since then is profound
 metal oxide material different from silicon carbide
in exponent α which typically 20 rather than 4 for
SiC
 It is about 90% ZnO & rest of other metal oxides
Metal Oxide Arrester …
 Material is ground, mixed, pressed, &
sintered to form disk-shaped blocks with
a dense, fine structure
 Property of SiC derives from bulk material
itself, while in ZnO it resides in boundary
layers between crystals
 Grain size & number of boundaries is
dependent on sintering process , so VI
controlled by sintering as well as
composition as shown in next fig.
Metal Oxide Arrester ….
 Influence of ZnO
grain size upon
Varistor Voltage
 However VI
characteristic of a
real sample named
“Dyna Var” 209 kV
metal Oxide arrester
shown in fig of next
slide
Comparison of SiC & ZnO
 Fig 16.13: (for application in 345 kV)
1- a ZnO Arrester
2- a SiC Arrester
3- a Linear Resistor




12-
1&2 a protective level of 2 pu in 10kA
The 296 kV intersect MOV in < 1 mA
line intersect the SiC’s in 200-500 A
for MOV :
could be operated without a gap
If gap employed protected level can be
reduced
Comparison of a ZnO & SiC
 fig shows comparison of these two &
a linear resistor for application 345 kV

Gapless Arresters
 Must support Normal Voltage continuously
 Therefore the L.H.S. of characteristic
 Shown in Fig 16-14 is sensitive to Temp. (in




given Voltage increase with Temp.)
Working at elevated Temp. increase dissipation &
increase Temp. further
same situation for MOV operation continuously at
too high a voltage
Each device a Max. Con. Op. Volt:
MCOV
Normally close to max. Line to N rms voltage
Temperature Effect
 On V/I curve of metal Oxide Varistor
MOV Parameters & gap type
 Dashed line in Fig shows cap. &res. Current
components
 At this voltage level act as a capacitor with mild
loss
 Its dielectric constant about 1000
 During quiescent voltage Dis. Between gap and MOV
based on capacitance:
C1:across Gap ; C3 :MOV capacitance
 C2 disturb balance between gap 1 & 2,when fast rising
surge applied
 1st
, No. one spark over then
controlled by MOV
2 & then voltage
Schematic of Gapped Metal Oxide
arrester
 Under quiescent condition, voltage
distribution between gap section &
MOV determined by capacitance:
C1 across gaps &
C3, inherent capacitance of MOV
MOV’s share assures thermal
stability
even though it typically has 10%
fewer disks than a gapless type
C2 becomes influential with advent
of a fast rising surge
disturb balance between gap 1 &2
 first 1 spark over and then 2
follow quickly as voltage appear
across it
TIME TO SPARK OVER for MOV with
gap
 voltage is controlled
by MOV characteristics
rather than the initial
spark over voltage of
gaps which
determines max.
voltage experienced
by protected object
 Spark-over occurs at a
lower voltage for a
somewhat slower
rising surges
 Is Different from a SiC
type
 For extra HV, H. E.
applications 2 columns
of ZnO disks used
Surge Arresters Characteristics
 MCOV :Max Con operating Voltage
 Rating is 15-30% more than that
 & is the highest voltage at which duty cycle
test can be performed
 Test ANSI/IEEE C62.11-1987;
should
be subjected to 20, 8x20 μs
current surges at special intervals
followed by a test to show thermal stability
 Energy capability: kilojoules/kV
Maximum Energy Capability for
MOV arresrter
 TABLE
Parameters Continued
 energy limited to 85% of table & repeatable a
minute after some cooling
 Table 16.3 SiC
1st column rating in kV,
2nd “front of wave” spark over voltage with very fast surges,
3rd spark over voltage with standard 1.5x40 μs wave,
4th Max. switching spark over,
5th Max 60 Hz spark over voltage
 Similar Data on ZnO some gapped(VS,VX)
and ungapped
TYPICAL PROTECTIVE
CHARACTERISATICS of Silicon
Carbide station type Arresters
Typical Protective Characteristics of
MOV station type Arresters
Application of Surge Arresters
 Objective:
1- protect insulation of other equipment
2- without putting itself at risk
◊ Highest protective margin or protection ratio
desirable; as margin increase energy demand
increase
◊proper application need: a compromise
◊contingencies: T.O.V., Lightning , Sw. Surge
◊min protective ratios:
1-chop.wave withstand/Front-of-wave prot.level≥1.20
2-Full wave withstand (BIL)/Imp. Prot. Level
≥ 1.20
3-Sw. surge withstand/Sw. surge prot. Level ≥ 1.15
Ratios Requirements and Protection
against Switching surge
 Ratios met, if possible exceeded
(since insulation deteriorate with time)
Energy loadings not exceeded
 Protection against Sw. surge
An example of protection of a
Transformer switched in through a
transmission line, where the line
energized from the other end
Example of SW. Surge Protection
 Arrangement:
where;
L=13 mH,Z0=350Ω,
l=200 miles
◊345 KV sys,
362 kV max design
voltage
 Arrester, with:
MCOV=362/√3=209 kV
Example continued
 Surge traveling down the line:
V(t)=(1-e^-αt)V ,α = Z0/L
V=voltage across the switch at closing
 1/α=37 μs short compared to:
travel time ≈ 1.075 ms
 When reach far end approach 2V
after 3 to 4 time constants
Discussion on Arrester Response
 Closing at peak;
Voltage at transformer : 591 kV
Transformer BIL 900 kV &
SIL=0.83x900=747kV
However, Arrester conduct,
What is the energy absorbed?
Example continued…..
 Neglecting corona, &
other dampings
1-surge at Arr. rise to 591KV
remain constant
until reflection return from
source (after 2.15ms)
2-arrester restrict
voltage at transformer
by:
its characteristic & load line
Example Continued…Normal case
◊ Q1, dissipated power
OP1Q1R1:423x480=203040 kW
◊ energy in 2.15 ms
=436.5 kJ
◊ In term of arrester:
436.5/209=2.09 kJ/kV
Example continued …with Trapped
charge
 Breaker recloses at pos peak voltage
when neg peak voltage trapped on line
 voltage at arrester attempt swing–vp to3vp
 Reach 887 kV which exceeds the transformer SIL
 Shifting load line to right at Q2, 530 kV & 1020 A
 Voltage within SIL limit of transformer
 Energy dissipation:530x1020x2.15ms=1162 kJ
 In term of arrester 1162/209=5.56 kJ/kV
Protection Against Temporary Over
Voltage
 Temporary power frequency over voltage
(due to single to ground fault, on unfaulted phases, Ferranti rise due to load
rejection and ferro-resonance)
 TOVs caused problem for gapped SiC
arresters if a surge cause spark over while
TOV present
 ZnO arresters have some tolerances for
TOVs however is limited
 Power dissipated and temperature increase
Rapidly with increase in voltage
Protection Against Temporary Over
Voltage
 A 209 kV MCOV arrester can withstand a TOV of 304
kV for one second & maintain stability
 Arresters can withstand lower TOVs for longer periods
and vice versa
 Ambient temperature and any dissipation immediately
prior to TOV will affect its tolerance
 i.e. voltage as per unit of TOV capability versus
duration
 Liklihood of TOVs should be studied where applying
arresters
 ANSI IEEE C62.2 (12) can be consulted w.r.t. magnitude
 Duration should be taken into account considering the
operating time of back-up breakers
voltage as per unit of TOV
capability
 Figure
Arrester/ Equipment Insulation
 Information on shielding against:
lightning, surge arresters and application of
surge arresters for a 230 kV system
summarized in one figure
 Equipment at this voltage normally has a
BIL of 750 kV, however in this figure
reduced to 600 kV, recognizing its aging
Metal Oxide arrester insulation
Coordination for 140 kV MCOV
arrester suitable for 230 kV system
Arrester for Line Protection
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=137.45 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