Transcript Slide 1

Short Line Fault
 Skeats Described:
 L:sourc.Inductance
L1:line Induct. To F
E :Instant. emf
VPW=1/2 L1/[L1+L] E
 2 travel B and F
 At F, sh. CCT.
 At C.B., open CCT.
Wedge-shaped Wave Formation
 Initial Distribution
 After Wave
Separation
 Each reflected
 α C.B. Ter.=1
 α F Ter. = -1
Voltage Distribution
At different points
 C.B.
Ter.,middle,3/4to F
Wave Reflections
 At T and 2T
 How 1 and 2 reflect
 At Each end V found =
V1(t)+V2(t)
 At C.B. start:
t=0,V=Vp
t=T,V=0
t=2T,V=-Vp
 T; travel time to F
Wave Distribution
 At C.B.
Swing rate :2Vp/2T
T:Very Small,at CB Fig1 
Fast Rise of TRV
Risk for C.B.
 at Middle ----Fig2
 at ¾ to Fault--Fig3
Solution of Sh. Line Fault
application of Transmission Line
Response
 Vs= VR Coshjω√LC+IR Z0 sinhjω√LC
 Is= IR coshjω√LC+ VR/Z0 sinhjω√LC
 Zs=
=
VR cosh j LC  I R Z 0 sinh j LC
I R cosh j LC  VR / Z 0 sinh j LC
VR / I R  Z 0 tanh j LC
1  (VR / I R Z 0 ) tanh j LC
Solution of Sh. Line Fault …
ZR=VR/IR
if a Sh. CCt. At Rec.ZR=0
|Zsc|=Z0 tanh jω√LC
Substituting s for jω :
Zsc(s)=Z0 tanh√LC s
T=√LCtanhTs=sinhTs/coshTs=
[1-exp(-2Ts)]/[1+exp(-2Ts)]
 exp(-2Ts)=α

tanh Ts=(1-2α+2α^2- 2α^3+..)
 Z(s)=Z0 [1-2exp(-2Ts)+2exp(-4Ts)-2…]
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Recovery Voltage Determination
Injecting minus fault current into CCT
I=-E/[L+L1] . T
i(s)=-E/[L+L1] . 1/s
response of Sh. Trans. Line :
i(s)Z(s)=-E/[L+L1] Z0 1/s [1-2e^(-2Ts)
+2e^(-4Ts)-…]
 Inverse Transform results in TRV
 1st neg. ramp slope Z0 X current
 Next terms ramp 2xslope of 1st delayed:
2T,4T,6T,… and alternate in Sign
Sum a Saw tooth wave ∆t=4T, Vpp=E/[L+L1] Z0 2T
Z0T=L1

Vpp=2EL1/[L+L1]
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CCT TRV
 Response of shorted line Calculated
 Response of source side:
If source : by parallel L & C
Rise a TRV of 1–cosine form T=2Π√LC &
Amplitude EL/[L+L1] ,i.e.:
Vsource =E L/[L+L1] {1-cos t/√LC}
 VC.B.=Vsource-Vshort fault as in Figure
Transformer Transient Model
 Have the greatest exposure to electrical
Transients (after Transmission line)
 Different Model for specific studies
 Transformer Model for:
Switching on OPEN CCT
 Im & Ih+e so a parallel RL CCT,
& Cg capacitance of winding with length l
R0=V /W0
 Cg/l per unit length & small fraction
(Cg/l)∆x
Transformer Equivalent CCT
 As a load on Auto
Transformer
 Its Imp.:l/[ωCg∆x]
looking to A seen
As : (l/x) {l/[ωCg∆x]}
 Related Cap.:
Cg x ∆x /l
 Ceff=Cg/l ∫x dx
 Integrating 0 to l
Ceff=Cg/l [x /3]=Cg/3
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Other Models
 Terminal Model & Detailed Model
1- Based on Name plate Information
A paper, not discussed here
2-Software for Transformer Transient
Study , not discussed here
 Internal Model for a Transformer
for special studies
Internal Model
 for our purpose Detailed one not desirable
 Divide the time after surge into:
1- extremely short (about fraction of μs)
2- Next in range of m S and seconds
3- the steady state
 In the fraction of μs
no current penetrate to inductance
only displacements currents to Capacitances
Initial Distribution
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Eq. CCT.
 Cg:Cap to Gr
Cs:Cap from end to
end of winding
 Gr Cap./unit length=Cg/l
Series Cap./length=Cs l
E:voltage at any point
on winding
Ig=total current in Gr Cap
Is=current in series Cap
Elementary length of Winding
 The Gr current:
∆Ig=Cg ∆x ωE/l (1)
 Also :
∆Ig=dIs/dx ∆x (2)
 With (1) & (2):
dIs/dx=CgωE/l (3)
 Series Cap. Of Elem.
Length:
l Cs/∆x
Deriving Eeuations
Voltage across Element=dE/dx ∆x
series Cap. current=C dE/dt
Is=l Cs ω dE/dx
(4)
Differentiating above Eq. :
dIs/dx=lCsω d E/dx
(5)
Equate (3) & (5):
d E/dx -1/l Cg/CsE=0 (6)
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Fast Response of winding to
switching
 The General solution of Eq.(6):
E=A1 e^px + A2 e^-px
 where p=1/l √Cg/Cs
 A1& A2 from Boundary Conditions
 NEUTRAL GROUNDED:
x=0, E=0 and x=l, E=V
 V amplitude of step surge
Coefficients for Neutral Gr
 A1+A2=0, A1e^pl+A2e^-pl=V
A1=-A2=V/[ e^pl- e^-pl ]
 If pl=√Cg/Cs=α
 A1=-A2=V/[2sinhα]
 E=V/[2sinhα] . {e^px – e^-px}=
= V sinh(αx/l)/sinhα
(7)
 If Neutral Isolated:
x=l , E=V & x=0 , Is=0 or dE/dx=0
Coefficient of Neutral Isolated
 Two Boundary Eqs:
p(A1-A2)=0
A1e^pl + A2e^-pl=V
 Coefficients are :
A1=A2=V/[2cosh pl]=V/[2coshα]
 The response :
 E= V/2coshα [e^px + e^-px]=
V cosh(αx/l) / coshα
(8)
Eqs (7) & (8) are Initial Voltage Distribution
Discussion on responses
 both depend on α=√Cg/Cs
 More Nonuniform as α increases
 α=10% to 60% voltage across the 1st 10% of
winding at Line end
 The Gradient at line terminal by differentiating:
 a- Gr Neutral
dE/dx=α/l V cosh(αx/l)/ sinhα;
x=l  dE/dx=αV/l cothα
 B-Neutral Isolated
dE/dx=α/l V sinh(αx/l)/coshα
x=l  dE/dx= αV/l tanhα
Discussion Continued
 For large values of α,
cothα=tanhα=1 ; dE/dx=αV/l
or is α times mean Gradient
 Under Surge; Turns Insulation at Line
End Severly Stressed
 Slower fronted surges, capacitance
not governing Voltage distribution
 Winding Cap.:Geometery Dependent
Discussion Continued
 Different Winding Styles :
Vary in α
 Source of steep fronted surges:
Flashover of an insulator, closing of a
switch or C.B.,reignition in a switch,Fast
Trans. in opening GISswitches,Lightning
 Remedial Methods:
strengthen End turn insulation; Failed
placing metallic shields, Design interleaved
Winding
Initial and Approximate limit of
Excursion
 Ceff of a winding, excited by V at ω
 I=ωCeffV , Is=ωCsl dE/dx
[Is]l=ωCsl (dE/dx)l;
(dE/dx)l=α(V/l) cothα
 Thus: ωCeffV=ωCsαV cothα
or : Ceff=αVscothα=√Cg/Cs Cs cothα
 Typical α, cothα=1,
therefore:Ceff=√CgCs
A sample Transformer
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Cg=3000 PF, Cs=30 PF
α=√3000/30=10 Ceff=√90000=300PF
For a Transmission Line Z0=400Ω
charging Time const.=400x3x
10^-10=0.12μs
 Points with transients:
starts at initials, and finally Osc. About some Final value:
Swing almost as far above steady value as starts below it
In winding Final Dis. Uniform or it is the α=0 line
Therefore excursions of any point lie within the envelope
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 negatve
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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I:
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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
Interaction Between Lightning &
Power System

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,
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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.:
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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