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8/3/2019 HW 5 Praveen
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EEE 579
POWER TRANSMISSION AND DISTRIBUTION
SPRING 2011
Dr. GEORGE. G. KARADY
HOMEWORK 5
LIGHTNING PROTECTION
SHIELDING PROTECTION
BACK FLASHOVER
Praveen Ramiah Subramanian
ASU ID: 1202919089
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1. Introduction:
Lightning is a natural phenomenon occurring due to discharge from the atmosphere and is a major
cause of damage to transmission line system. This is because of the fact that most transmission lines
traverse through barren uninhabited expanses of land and are totally conducting. Hence due to the
extreme heights of the transmission lines and the fact that lightning strikes the highest conducting point
closest to it, there exists a high probability of the lightning strokes to strike transmission lines.
When a lightning strikes a transmission line, it induces currents in the range of 150 to 400 kA, voltages
of the order 106 kV and temperatures close to 50,000 F. Lightning striking transmission lines cannot be
prevented. The transmission lines are hence designed with shield conductors. The most important
application of these shield conductors is to ground the lightning induced currents and prevent the
transmission tower structure, phase conductors and insulators from damage.
The focus of this project is on designing a robust lightning protection system. It is to be noted that
lightning striking a transmission line cannot be completely avoided. However, with a systematicanalysis, the probability of successful performance of the lightning protection scheme can be gauged.
For this analysis, the following parameters need to be calculated:
Shielding performance
Back flashover rate
Expected outage per 100 miles
Based on the study, the effectiveness of the shielding and lightning protection scheme can be observed
and steps can be taken if possible to solve potential problems.
2. Transmission Line Overview:
Project name : High Plains Express Project
Point of origin : Gladstone, NM
Point of termination : Boone, COLine length : 182 miles
Voltage class : 500 kV AC
3. Conductor and Tower Data:
3.1. Phase Conductor Data:
Conductor Code = ACSR Kiwi Conductor
Number of conductors per bundle (p) = 2Distance between bundles (dbun) = 18 inches
Cross Section = 2167 kcmil
Conductor Diameter = 1.735 inches
Core Diameter = 0.347 inches
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Resistance at 60 Hz at 50C (R50) = 0.0511 /mile
Geometric Mean Radius (GMRc) = 0.0570 ft
Maximum Current Carrying Capacity = 1000 A
Conductor Span = 1000 ft
3.2. Ground Conductor Data:
Shield Conductor Code = Aluminum CladAlumoweld 7
Number of Shield Conductors = 2
Distance between Shield Conductors = 47.536 ftDiameter of Shield Conductor (Dshl) = 0.545 inches
Resistance at 75C (Rshl75) = 1.669 /mile
Geometric Mean Radius (GMRshl) = 0.00296 ftGround Resistance (Rground) = 0.095 /mile
3.3. Transmission Line Data:
Figure 1: Transmission TowerDimensions3.4. Conductor Coordinates:
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Figure 2: Conductor Coordinates
Conductor Horizontal Coordinate Vertical Coordinate
Phase A x0 = -Dc = -40 yo = H0 = 109
Phase B x1 = 0 ft y1 = H0 = 109
Phase C x2 = Dc = 40 y2 = H0 = 109
Ground 1 X3 = -Dg / 2 = -23.75 y3 = Hg = 150
Ground 2 X4 = Dg / 2 = 23.75 y4 = Hg = 150
Table 1: Conductor Coordinates
SECTION I : SHIELDING4. Ground Flashover Density:
The ground flashover equations are used to calculate the number of lightning strokes per square
kilometer per year (Ng) and can be found using either of the following:
Number of thunderstorm days (TD)
Ng = 0.04 x TD1.25
Number of thunderstorm hours(TH)
Ng = 0.054 x TH1.1
h
Hc Hg
Dc
Dg
X
Y
x0
y0
x4
y4
x1
y1
x2
y2
x3
y3
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The TD and TH are obtained from the Isokeraunic map showing the mean annual days and hours of
thunderstorm activity. The transmission line under study is to be installed from New Mexico to
Colorado. The TD and TH values for these areas are as follows:
TD = 70
Ng = 0.04 x (70)1.25
= 8.1
TH = 100
Ng = 0.054 x (100)1.1
= 8.6
5. Electric Shadow:
Average height of conductor is given by:
hcond = hat_tower - (2/3)sag
Average height of phase conductors
yc1 = yc - (2/3)sagc= 109(2/3)41.26= 81.493 ft = 24.839 m
Average height of ground conductorsyg1 = hat_g - (2/3)sagg
= 150(2/3)21.36
= 115.76 ft = 35.284 m
The shadow is given by:
W = b + (hcond)1.09
Where,
b is the breadth of the transmission line at its base in meters.
b = 40 ft = 12.192 mW = 12.192 + (33.22)1.09
= 57.73 m
6. Shielding Angle:
Shielding angle is given by:
= tan-1 [ ( xg - xc ) / ( yg1 - yc1 ) ]
Where,
xg is the horizontal distance between the shield conductors and vertical axisyg is the vertical distance between the shield conductors and ground
xc is the horizontal distance between the phase conductors and vertical axis
yg is the vertical distance between the phase conductors and ground
All the above values can be obtained from table 1.
= tan-1 [ ( 23.75 - 40) / ( 115.76 - 81.493) ]
= -25.369
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Bundle radius with corona is given by:
Rbundle_corona = rbund + rcorona= 0.753 m
Surge impedance of phase conductor with corona is given by:
= 348.184
9. Minimum Stroke Current:
The minimum stroke current that would be created by a lightning is given by:
IL_min = 20.467 kA
10. Strike Distance vs. Current Functions:
The strike distance for both the phase conductors and shield conductors is given by the followingequation:
The strike distance for the phase conductors to ground is:
Sph_g = 71.151 m
The distance for the shield conductors to ground is:
Sshld_g = 233.436 ft
Similarly, the strike distance to the ground is calculated by:
Sg(IL_min) = 60.68.3 m = 199.092 ft
11. Shielding Effectiveness:
The shielding effectiveness is determined by plotting the strike circles for the phase conductors and the
shield conductors.
Zsurge_corona 60 ln 4yc
rbundl_corona
ln 4 yc
rbund
IL_min
2 Vins
Zsurge_corona
Sc_g IL 10IL
kA
0.65
m
Sg IL if yc 40m 3. 6 1. 7 ln 43 ycm
IL
kA
0.65
5.5 ILkA
0.65
m
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11.1. Phase Conductor Strike Circle:
The phase conductor strike circle equation is:
Sc_g IL 2
x xc 2
y yc1 2
Lower part of the circle is represented by the equation:
Upper part of the circle is represented by the equation:
11.2. Shield Conductor Strike Circle:
The phase conductor strike circle equation is also similar to the phase conductor equation. The lower
part of the shield conductor strike circle is given by:
Upper part of the circle is represented by the equation:
Lower part of the circle is represented by the equation:
yshield_lx IL yg1 Sc_g IL 2
x xg 2
11.3. Ground Strike Circle Equation:
The ground strike circle equation is represented by the following equation:
The circles are plotted for the value of IL_min obtained from section 9 previously. These plots wouldhelp obtain the unprotected area of the transmission line conductors. The parameters given as inputs to
obtain the plots are:
Distance along the x axis, in steps of 1 ft. Range [0 to 350].
Strike circles y coordinates in ft, for phase conductors, shield conductors and ground conductors as a
function of distance x. Range [0 to 120].
ycond_l x IL Sc_g IL 2
x xc 2
ycond_u x IL yc Sc_g IL 2
x xc 2
yshield_l x IL yg Sc_g IL 2
x xg 2
yshield_u x IL yg Sc_g IL 2
x xg 2
yground x IL Sg IL
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Figure 3: Strike Circles
12. Calculation of Unprotected Area vs. Lightning Current:
From the previously obtained strike circle curves, the y axis coordinates of the point of intersection of
the shield strike circle and the ground strike circle is obtained an incorporated into the circle equation.
This is eventually solved for the roots.
Sc_g IL 2
yc1 Sg IL 2 xc xint
2
Sc_g IL 2
yc Sg IL 2 xc xint
The two roots of the equation are positive and negative, are determined by:
xint_ground_nIL xc Sc_g IL 2
yc Sg IL 2
xint_ground_pIL xc Sc_g IL 2
yc Sg IL 2
The coordinates at minimum current are:
xint_ground_n(IL) = -53.447 m
xint_ground_P(IL) = 77.831 m
The initial assumed value for the negative root is x int_cg = 30 m. This value is utilized to obtain the true
value of the negative value using the equation solver in MATHCAD.
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Negative root:
xint_cg_n(IL_min) = -54.371 m
Similarly, the positive root is also solved for in the similar method.
xint_cg_p(IL_min) = 73.808 m
Unprotected area is calculated using the following formula:
Xnot_prot_p IL xint_ground_p IL xint_cg_p IL
xnot_prot_p(IL) = 4.023 m
IL_min = 20.467 kA
The maximum lightning current when the unprotected area becomes zero is obtained by solving for theroots of xnot_prot_p(IL).
IL_max root Xnot_prot_p IL IL
IL_max = 29.542 kA
This variation of the lightning current can be plotted as follows:
Figure 4: Variation of lightning current
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13. Probability of Line Flashover:
Ibase IL if IL 20kA 61.1kA 33.3kA
IL if IL 20kA 1.33 0.605
f IL 1
2 IL IL
kA
e
lnIL
Ibase IL
2 IL 2
F(IL_min) = 6.264%
P IL 1
1IL
Ifirst
2.6
Ifirst 25kA P(IL_min) = 62.719%
Number of lightning strokes hitting the ground per year per 100 km2
Ng_day = 8.099
Number of flashovers per 100 km per year is:
Xnot_prot_p(IL_min) = 4.023 m
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SECTION II : BACK FLASHOVER
14. Flashover Voltage across the gap and insulator:
14.1 Flashover Voltage at 2 s:
= 2 s
Flashover voltage across the gap is given by:
Vgap_2 s 0.4Lins
m
0.71dmin
m
2s
s
0.75
M V
= 5.88 MV
Flashover voltage across the insulator is given by:
Vins_2s 0.4Lins
m
0.71Lins
m
2s
s
0.75
M V
= 3.601 MV
14.2 Flashover Voltage at 6 s:
= 6 s
Flashover voltage across the gap is given by:
Vgap_6 s 0.4Lins
m
0.71dmin
m
6s
s
0.75
M V
= 3.563 MV
Flashover voltage across the insulator is given by:
Vins_6s 0.4Lins
m
0.71 Linsm
6s
s
0.75
M V
= 2.563 MV
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15. Surge Impedance Calculation:
Vtop_2 Emax rcorona lnyc1
rcorona
Initial value of Emax is assumed
Emax2 15 kVcm
The rcorona2 value is found by solving for the roots of the above equation.
rcorona2= 1.172 m
The self impedance of the phase conductors with corona:
Zshild_self 60 ln2.yg
rCorona2
ln2 yg
rshild
= 376.52
Average distances:ds1_s2 = 2 xg = 14.489 m = 47.536 ft
ds1_S2 xg2
2 yg 2
= 79.578 m
Mutual surge impedance:
Zshild_mutual 60 lnds1_S2
dS1_S2
= 102.201
Surge impedance of shield conductors:
Zshild
Zshild_self Zshild_mutual
2
= 239.361
16. Coupling Factor to each Phase:
Zphase_self 60 ln2 yc
rbund
= 389.717
da_1 xc xg 2
yc yg 2
= 8.09 m
da_2 xc xg 2
yc yg 2
= 20.463 m
Da_1 xc xg 2
yc yg 2
= 73.015 m
Da_2 xc xg 2
yc yg 2
= 75.396 m
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Za_1 60lnDa_1
da_1
= 132.022
Za_2 60lnDa_2
da_2
= 78.247
Ka_12
Z
a_1
Z
a_2
Zshild_self Zshild_mutual
= 43.919 %
17. Tower Surge Impedance:
rtower = 15 ft htower = 150 ft
Ztower 60 ln 2 2.htower
rtower
1
= 131.952
Travel time is computed by:
T
htower
0.85 300m
s
= 0.155 s
top_crossarm 17.6ft
0.85 300m
s
= 0.021 s
span span
300m
s
= 1.016 s
Tower ground resistance:
Rfoot_min 30 Rfoot_max 50
Rground Rfoot_max
23
Rfoot_max
= 23
Intrinsin Circuit Impedance:
Zto p
Ztower
Zshild
2
Ztower
Zshild
2
= 62.759
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19. Shield ConductorDamping Factor:
r
Rground Ztower
Ztower Rground
= -0.703
s
Zshild
2Ztower
Ztower
Zshild
2
= -0.049
st
Ztower
Zshild
2
Ztower
Zshild
2
= 0.049
ts
Zshild
2Ztower
Ztower
Zshild
2
= -0.049
s
2Zshild
2
Zshild
2Ztower
= 0.951
20. Current Wave Plot:
Tfront 2s Thalf 50s
a 1kA
Tfront
= 500 A/s
b1kA 0.5kA
50s
= 10 A/s
The current wave is represented as:
ILi t( ) if t Tfront a t a Tfront b t
At 50 s, the current is found to be:ILi(50 s) = 500 A
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Figure 5: Current Wave
21. Tower Top Voltage:
t 2s Vtop_0 t( ) ILi t( ) Ztop
Vtop_0 t( ) 61.503 k
1 r 0.297 T 0.155 s
r s r2
s 0.01 3 T 0.466 s
r2 s2 r3 s2 3.491 10 4 5 T 0.777 s
k 1 3
rk 1
sk 1
rk
sk 1
0.257
0.025
-32.49410
1 r Vtop_0 t 1 T 16.039 k
Vground t k( ) if t k T 0kV Vtop_0 t k T rk 1 sk 1 rk sk 1
Vground t k( )
16.039
1.448
0.13
kV
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Vground_T t( )
k
Vground t k ( )V ground_T(t) = 17.617 kV
The voltage equations obtained are plotted for the required time period.
Figure 6: Voltage Plots
Vtop_k t2 k Vtop_0 t2 2 k T rk
sk 1 r
k s
k
Vtop_k(t2,k) = -36.913 kV
-2.97
-0.227
at the time of
at the time of
at the time of
r r s 0.645 2 T 0.311 s4 T 0.622 s r
2 s r
2 s
2 0.064
6 T 0.932 s r3
s2 r
3 s
3 6.269 103
k 1 3 r
ks
k 1
r
ks
k
-0.645
-0.064
-3-6.26910
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t3 0s 0.01s 10s
Figure 7: Voltage at top of tower
22. Reflected Voltage in Adjacent Towers:
VTop_kt2 k if t2 2 k T 0kV Vtop_kt2 k
VTop_k t2 k -36.913
-2.97
-0.227
kV
VT op t( ) Vtop_0 t( )
k
VTop_kt k ( )
VT op t( ) 26.312 kV
t 6s
2 span 2.032 s
Vref_1 t if t 2 span 0kV Vtop_0t 2 span st ts Vref_1 t 1.145 k
VRef_6 s t VTop t Vref_1 t
VRef_6s 6s 16.867 k
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t. 0s 0.1s 5s
Figure 8: Reflected voltage from adjacent towers
23. Cross-arm Voltage:
24. Insulator Voltage at Unit Current:
Insulator voltage at 2 s:
Vinsulator_2s Vcrossarm 2s Ka_12 VT op 2s = 13.579 kV
Top Voltage at 6s:
Vref_6s = 16.687 kV
t4 2s
Vcrossarm
t4
Vground_T
t4
T top_crossarm
TV
T opt4
Vground_T
t4
VT op t4 Vground_T t4 8.695 k
Vcrossarm t4 25.135 k
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Insulator voltage at 6 s:
Vinsulator_6s VRef_6s 6s 1 Ka_12 = 9.459 kV
25. Critical Stroke Current:
Flashover causing stroke current at 2 s:
Istrok_2s
Vgap_2s
Vinsulator_2skA
= 433.064 A
Flashover causing stroke current at 6 s:
Istrok_6s
Vgap_6s
Vinsulator_6skA
= 376.88 A
60 Hz stroke current component at 2 s:
Effect of 60Hz voltage
Istrok_2s_60Hz Istrok_2s
Vgap_2s2 500 kV
3sin 90 deg
Vgap_2s
= 463.129 kA
60 Hz stroke current component at 6 s:
Istrok_6 s_60Hz Istrok_6 s
Vgap_6 s2 500 kV
3sin 90 deg
Vgap_6 s
= 419.838 kA
26. Probability of Line Flashover:
Probability Function:
P IL 1
1
IL
Ifirst
2.6
Probability of flashover after 2 s:
P(Istrok_2s_60Hz) = 0.051 %
Probability of flashover after 6 s:
P(Istrok_6s_60Hz) = 0.065 %
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27. Number of Lightning Strokes:
Average tower height yg1 = 35.284 m = 115.76 ft
Estimated tower width b = 12.192 m = 40 ft
Length = 100 km
Widthb
m
yg1
m
1.09
= 60.186
Nline_d Ng_day Width = 492.554
28. Number of Flashoers:
Number of flashovers caused by shielding failure:
Nshield = 4
Nline = Nline_d - Nshield = 488.554
29. Number of Back Flashovers per 100 km per year:
Nback_2 s Nline P Istrok_2 s_60Hz = 0.247
Nback_6 s Nline P Istrok_6 s_60Hz = 0.319
Total number of flashovers per year per 100 km:
Ntotal_line Nshild Nback_2 s Nback_6 s = 4.566
30. Conclusion:
The lightning protection of the proposed transmission line was designed in this module. The shieldingperformance was analyzed and is found to be a robust design. The shield conductors provide
substantial protection to the transmission lines. The back flashover rate was also calculated and is
found to be well within the limits. The expected outage rate is also computed and turns out to anacceptable value.