This document provides details on the design of a 500kV extra high voltage transmission line that is 600 miles long. It discusses selecting an economic conductor size, calculating line parameters such as resistance, inductance and capacitance, and ensuring safety clearances are met. The selected conductor is a bundle of 3 ACSR conductors with a cross-sectional area of 468 mm2 each. Line losses are calculated to be 51.23 MW, which is 5.123% of the 1000MW transmission capacity. Surge impedance is determined to be 276.6 ohms. Safety clearances are in accordance with National Electrical Safety Code specifications.

1. Addis Ababa Science and Technology University
Electrical and Computer Engineering Department
EHV 500KV TRANSMISSION LINE DESIGN
Prepared by: DemsewM.
JULY 2017

2. Content
• Introduction
• Economic Size of Conductor
• Transmission Line Parameter Calculation
• Transmission Circuit and Bundle Conductor Calculation
• Surge Impedance Loading (SIL) of the line
• Sag-Tension Calculation of conductor
• Insulator Sizing
• Field Effect of EHV transmission line
• Environmental Criteria

3. Electrical Design Parameters

4. Introduction
• Voltage Classes:
o Low Voltage (LV) (<1 kV)
o Medium Voltage (MV) ≥1kV and ≤72.5kV),
o High Voltage (HV) ≥72.5kV and ≤242kV),
o Extra-High Voltage (EHV) ≥242kV and <1000kV),
o Ultra-High Voltage (UHV) ≥1000kV).

5. Design Input
• 500kV EHV transmission line Design Parameters:
o Transmission distance 600mile
o Plant generation capacity 1000MW
o System power factor 0.90
o Design Voltage level 500kV and Frequency 50Hz

6. Economic Transmission Voltage
• The most economic three phase number of circuit to
transmit 1000MW power from the generation plant
through 600mile 500kV transmission line is given by:
• V=500kV, L=600mile pf=0.9 Load in
KVA=1000MW/0.9=1111.11*103kVA
𝑽 = 𝟓. 𝟓 𝑳 +
𝑳𝒐𝒂𝒅 𝒊𝒏 𝑲𝑽𝑨/𝑵 𝒄
𝟏𝟓𝟎
𝒌𝑽
𝑁𝑐=0.97 ≅ 1.

7. Cont’d…
• The most economic transmission voltage level is:
𝑉 = 5.5 600 +
1111.11 ∗ 103
150
𝑘𝑉 = 492𝑘𝑉
• 𝑉𝐸𝑐𝑜𝑛𝑜𝑚𝑖𝑐 =492KV, the nearest standard transmission
voltage is 500KV

8. Conductor Selection
• Expanded ACSR and ACCR
o Good strength weight ratio
o Better conductivity-weight ratio
o Better performance at high temperature
o Long ruling span length
o Reduces huge investment on lattice structures
o Cost…

9. Economic Size of Conductor
• Kelvin’s Law represented by the following formula used to determine the economic current
density conductor.
𝑪 = 𝟎. 𝟎𝟏𝟑
𝒂. 𝒑
𝒒
𝑨𝒎𝒑𝒆𝒓𝒆
𝒎𝒎 𝟐
Where,
o C = most economical density of current (Ampere/mm2)
o a = percent annual expense to the construction cost of conductor (13.5%)
o p = price of conductor (4.00$/kg)
o q = cost of electricity (0.03$/kWh)
𝐂 = 𝟎. 𝟎𝟏𝟑
𝟏𝟑. 𝟓 ∗ 𝟒
𝟎. 𝟎𝟑
= 𝟎. 𝟓𝟓 𝐀𝐦𝐩𝐞𝐫𝐞/𝒎𝒎 𝟐

10. Cont’d..
• The current I is calculated as follows:
I =
μp
3 ∗ V ∗ pf
Ampere
• Where,
o µ = utility factor being (0.6 )
o pf = power factor being 0.90
o V = line voltage (500kV)
o P = Maximum Power (1000MW)
• The most economic size of the conductor is, A=I/C (mm2)
𝐼 =
0.6 ∗ 106
3 ∗ 500 ∗ 0.9
Ampere = 769.8Ampere
𝐴 =
𝐼
𝐶
=
769.8Ampere
0.55 Ampere/mm2
= 1400𝑚𝑚2

11. Cont’d…
• The numbers of bundle conductors (n) if we
choose 468-mm2 expanded ACSR, 26/7
conductor of r = 14.055mm.
𝑛 =
1400𝑚𝑚2
468𝑚𝑚2
= 2.99 ≈ 3

12. Cont’d…
Table 3.1: Conductor Data Sheet Aluminum Conductors Steel Reinforced (ACSR)

13. Cont’d…
• The following conductor specification is selected for
economic and best environmental operation of 500kV
overhead transmission line. Please refer Table 3.1.
o 500-kV transmission line e = 288.7 kV
o 468-mm2 expanded ACSR, 26/7 conductor r = 14.055 mm
o The number of conductor bundles n=3
o 457-mm (18~in) spacing on conductor bundle S = 45.72 cm
o 13.1-m (43-ft) equilateral phase spacing D = 1310cm

14. Conductor Configuration
• An equilateral conductor configuration is chosen
for the following reason.
o To balance the magnetic field stress for each phase of a bundled
conductor,
o A balanced power loss and
o Corona discharge for each phase is maintained.

15. Line Parameter calculation
• Fundamental AC transmission line parameters:
o Line Inductance
o Line Capacitance
o AC Resistance

16. 500kV Line AC Resistance Calculation
• Determinant Parameter:
o Material Type
o Cross Sectional Area
o Line Length
o Temperature and
𝑹 𝒅𝒄 = 𝝆.
𝒍
𝑨
𝐑 𝐓 = 𝐑 𝐨[𝟏 + 𝛂(𝐓 − 𝐓𝟎)]

17. Cont’d…
• From Table 3.1 the dc resistance of the chosen
conductor at 20oc is 0.0717Ω/km.
• But when the conductor is energized, it
operates a temperature of around 50oc.
• 𝛼=0.004308/oc for aluminum conductor.
𝑹 𝑻=50oc = 𝑹 𝒐[𝟏 + 𝜶(𝑻 − 𝑻 𝟎)]= 𝟎. 𝟎𝟖𝟏Ω/𝐤𝐦

18. Cont’d…
• The dc resistance of the conductor increased by 13% only by
temperature factor.
• The per phase dc resistance (3 bundle conductor per phase) is
0.081/3=0.027 Ω/km=0.027*1.6091 Ω/mile=0.04345 Ω/mile.
• We should find out the ac resistance by accounting the skin
effect:
𝑹 𝒂𝒄 = 𝐊 . 𝑹 𝒅𝒄 𝐰𝐡𝐞𝐫𝐞, 𝐊𝐢𝐬 𝐚 𝐟𝐮𝐧𝐜𝐭𝐢𝐨𝐧 𝐨𝐟 𝐗,
𝑿 = 𝟎. 𝟎𝟔𝟑𝟓𝟗𝟖
𝝁. 𝒇
𝑹 𝒅𝒄
= 𝟎. 𝟎𝟔𝟑𝟓𝟗𝟖
𝟏 ∗ 𝟓𝟎
𝟎. 𝟎𝟒𝟑𝟒𝟓
= 𝟐. 𝟏𝟔

19. Cont’d..
Table 2.1: Skin Effect Parameter Values

20. Cont’d…
• Since K is a function of X and x1<X<x2, using linear interpolation:
𝑲 𝑿 = 𝑲 𝒙𝟏 +
𝑿 − 𝒙𝟏 [𝑲 𝒙𝟐 − 𝑲 𝒙𝟏 ]
𝒙𝟐 − 𝒙𝟏
• Referring Table 2.1, X=2.16 is between x1=2.1 (K1=1.09375) and
x2=2.2(K2=1.11126)
𝑲 𝑿 = 𝟐. 𝟏𝟔 = 𝟏. 𝟎𝟗𝟑𝟕𝟓 +
𝟐. 𝟏𝟔 − 𝟐. 𝟏 [𝟏. 𝟏𝟏𝟏𝟐𝟔 − 𝟏. 𝟎𝟗𝟑𝟕𝟓]
𝟐. 𝟐 − 𝟐. 𝟏
𝑲 𝑿 = 𝟐. 𝟏𝟔 = 𝟏. 𝟏𝟎𝟒𝟐𝟓𝟔

21. Cont’d…
• Rac=K*Rdc=1.104256*0.04345 Ω/mile=0.048 Ω/mile
• The per phase ac resistance of a 500kV transmission line
having 3 bundle conductor at 50oc is 0.048 Ω/mile.
• The 3-phase 500kV power loss having the above conductor
specification at 50oc is given by:
𝑷 𝑳𝒐𝒔𝒔 = 𝟑𝑰 𝟐
. 𝑹 𝒂𝒄
𝑷 𝑳𝒐𝒔𝒔 = 𝟑 ∗ 𝟕𝟔𝟗. 𝟖 𝟐
∗ 𝟎. 𝟎𝟒𝟖 ∗ 𝟔𝟎𝟎 𝑾 = 𝟓𝟏. 𝟐𝟑𝑴𝑾
• Which is 5.123% of the transmission capacity and well performed
system.

22. Discussion
• Compare the loss and the ac resistance with
your Practical exposure?

23. 500kV Line Inductance Calculation
• Consider a single-circuit, three-
phase system with multi conductor
configured phase conductors as
shown in Figure 3.1.
• And assume equal current
distribution in the phase
subconductors and complete
transposition.
• We can show that the phase
inductance for the system is the
following expression:

24. Cont’d…
• For proper phase to phase Clearance and to obtain low
magnetic stress and low corona discharge:
o 500-kV transmission line e = 288.7 kV
o 468-mm2 expanded ACSR, 26/7 conductor r = 14.055 mm
o The number of conductor bundles n=3
o 457-mm (18~in) spacing on conductor bundle S = 45.72 cm
o 13.1-m (43-ft) equilateral phase spacing D = 1310cm

25. Cont’d..
GMD =
3
13.1 ∗ 13.1 ∗ 13.1 = 13.1𝑚
r’=0.768r=0.768*14.055 mm =10.8mm
𝐴 =
𝑆
2 sin
𝜋
𝑁
= 26.4𝑐𝑚
𝐺𝑀𝑅 = [𝑁𝑟′(𝐴) 𝑁−1]
1
𝑁=13.12cm
𝐿 = 2 ∗ 10−7 ln(
13.1𝑚
0.1312𝑚
) = 0.923𝜇𝐻/𝑚
𝑋 𝐿 = 2𝜋𝑓𝐿 = 6.28 ∗ 50 ∗ .875 ∗ 10−6 =
0.2898𝑚Ω
𝑚
= 0.4643Ω/𝑚𝑖𝑙𝑒

26. Discussion
• Discuss the line inductive reactance and
inductance value computed above with your
practical exposure?

27. 500kV Line Capacitance Calculation
• Assume equal current distribution in the phase
subconductors and complete transposition, the
phase capacitance for the system has the
following expression:
𝑪 𝒑𝒉𝒂𝒔𝒆 =
𝟐𝛑𝜺 𝒐
𝐥𝐧[
𝑮𝑴𝑫
𝑮𝑴𝑹
]
, 𝜺 𝒐 =
𝟏
𝟑𝟔𝝅
∗ 𝟏𝟎−𝟗

28. Cont’d…
Cphase =
2πεo
ln[
13.1
0.1312
]
=
2𝜋 ∗
1
36𝜋
∗ 10−9
4.603645
= 12.068 ∗ 10−12
𝐹/𝑚
• The line reactance for a 50Hz system is given by the following
expression:
𝑋 𝐶 =
1
2𝜋𝑓𝐶
= 2.64 ∗ 108Ω − 𝑚𝑒𝑡𝑒𝑟 = 1.65 ∗ 105Ω − 𝑚𝑖𝑙𝑒

29. Discussion
• What do you think about the result of the
capacitive reactance,
𝑋 𝐶 = 0.165 ∗ 𝑀Ω − 𝑚𝑖𝑙𝑒
???

30. Surge Impedance of Line
• Natural impedance, or characteristic impedance of the
line.
𝒁 𝟎 =
𝑽(𝒙, 𝒕)
𝑰(𝒙, 𝒕)
𝒁 𝟎 =
𝑹 + 𝒋𝒘𝑳
𝑮 + 𝒋𝒘𝑪
𝒁 𝟎 =
𝑳
𝑪
= 276.6Ω
• Which is independent of line length!

31. Surge Impedance Loading (SIL)
• Surge Impedance Loading (SIL) or the natural loading
of a transmission line is the power delivered by a line
to a purely resistive load equal to its surge impedance.
SIL(3ϕ) =
(𝑘𝑉𝐿𝐿)2
𝑍 𝑜
(MW)
SIL(3ϕ) = 904MW
|𝐼𝐿| =
𝑉𝑅 𝐿−𝐿
3 ∗ 𝑍 𝑜
(𝐴)
|𝐼𝐿|=1043.7𝐴𝑚𝑝𝑒𝑟𝑒

32. Group Discussion
• So what is the implication of the SIL result
calculated in the previous slide?

33. Safety Design Parameter

34. Sag-Tension Calculation
• The clearances required vary with the area being traversed.
• The clearances must meet the specifications outlined in the
National Electrical Safety Code (NESC) (Safety Rules for
the Installation and Maintenance of Overhead Electric
Supply and Communication Lines,).
• The conductor low point must never violate the clearance
levels specified in Table 3.3.

35. Basic Clearance Data for 500kV (NESC)

36. Cont’d…
System
Voltage, kV
ROW Width,
m
TC, m Clearance, m
Phase to
phase
Height of
mid span
Average height
at lowest phase
345 38m 45.7m 7.6m 8.5m 12m
500 55m 67m 10m 9.9m 14m
745 72m 88m 14m 14.3m 18.5m

37. Cont’d…
• Generally the minimum ground clearance requirement
for EHV transmission line of flat pedestrian area only
is given by the formula:
• 𝐻 𝑚𝑖𝑛 = The minimum clearance of the transmission line
conductor from ground
• V= Transmission line voltage maximum Value (550kV for
example for 500kV line)
𝑯 𝒎𝒊𝒏 = 𝟐𝟐 +
𝑽
𝟑
− 𝟓𝟎 ∗
𝟎. 𝟒
𝟏𝟐
(𝒇𝒆𝒆𝒕)

38. Cont’d…
• The curve shape assumed by a conductor when
suspended between towers is very close to a
catenary or a parabola.
• The equation for the conductor low point (i.e.
the conductor sag at mid span, assuming
suspension points at equal elevations) using
parabolic model is:

39. Cont’d…
𝐋 = 𝐏 +
𝟖𝑺 𝟐
𝟑𝑷
𝐒 𝒙 = 𝐖
𝑿 𝟐
𝟐𝑯
𝐒 = 𝐖
𝑷 𝟐
𝟖𝑯
𝐇 = 𝟐 ∗ 𝒂 ∗ 𝐖
𝐕 =
𝐖(𝟑𝑷 𝟐 + 𝟖𝑺 𝟐)
𝟔𝑷
𝐓 =
𝑾𝑷
𝟖𝑺
𝑷 𝟐 + 𝟏𝟔𝑺 𝟐

40. Cont’d…
• The average span length of 500kV line outside the city are
1200- 1500 ft (366-457 m)
• Therefore we can take a ruling span of, P=411.5m for 500kV
transmission line.
• Rated conductor Strength 139.7kN (Refer Table 3.1)
• W=(1627kg/km)*(10m/s2)=16.27N/m (Refer Table 3.1)
• Stringing the conductor with a maximum tension force, T=70kN
(50% 0f the material rated strength to avoid mechanical stress)

41. Cont’d…
• The maximum sag developed for a span length of 411.5m is given by:
𝐻 = 𝑊
𝑃2
8𝑆
= 16.27 ∗
411.52
8∗4.925
= 69.924𝑘𝑁
V =
W 3P2
+ 8S2
6P
=
16.27 3 ∗ 411.52
+8 ∗ 4.9252
6 ∗ 411.5
= 3.35𝑘𝑁
𝑇 = 𝐻2 + 𝑉2 = 70𝑘𝑁
𝑆 =
𝑃2
8𝑇
𝑊𝑃
2
−16
=4.925m

42. Cont’d…
Therefore the minimum clearance of the lowest
phase from ground at pole point is:
• Pedestrian: 𝐻ℎ𝑒𝑖𝑔ℎ𝑡 = 𝑆 + 𝐻 𝑚𝑖𝑛 = 9.1𝑚 + 4.925𝑚 = 14.025𝑚
• Highway: 𝐻ℎ𝑒𝑖𝑔ℎ𝑡 = 𝑆 + 𝐻 𝑚𝑖𝑛 = 12.2𝑚 + 4.925𝑚=17.125m
• Rail Road: 𝐻ℎ𝑒𝑖𝑔ℎ𝑡 = 𝑆 + 𝐻 𝑚𝑖𝑛 = 11.9𝑚 + 4.925𝑚=16.825m

43. Discussion
• Reflect your opinion on the calculated result of
Sag and Tension.
• What would happen if the stringing tension
becomes:
a) 140kN
b) 35kN

44. Insulator Sizing
• The insulation capacity of the insulator is the measure of the
power frequency and lightning with stand voltage gradient to
provide proper insulation between tower and live part of the
conductor.
𝐈𝐧𝐬𝐮𝐥𝐚𝐭𝐨𝐫 𝐰𝐢𝐭𝐡𝐬𝐭𝐚𝐧𝐝 𝐯𝐨𝐥𝐭𝐚𝐠𝐞 =
𝟐 𝐕𝐬𝐲𝐬𝐭𝐞𝐦(𝐊𝐕)
𝟑
∗ [𝐒𝐚𝐟𝐞𝐭𝐲 𝐌𝐚𝐫𝐠𝐢𝐧]
• Safety margin=1.2 (Switching Pulse)-1.4(Lightning Pulse)
• 𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑤𝑖𝑡ℎ𝑠𝑡𝑎𝑛𝑑 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 =
500 2
3
∗ 1.2 = 490𝑘𝑉
• Which is equivalent to 850kV a switching transient over voltage!!

45. Cont’d
• Withstand voltage for 210 kN Suspension insulator (plastic
type) =13 kV/disc. Thus, Number of insulator discs = 490/13 =
38 discs (for 210 kN insulators).
• Withstand voltage for 300kN Suspension insulator (glass
type)=15kV/disc. Thus, Number of insulator discs = 490/15 =
33discs (for 300kN insulators).

46. Group Discussion
• Discuss the different types of insulators for EHV!
• Reflect the capacity of the different insulator
type used in EHV transmission and how you can
choose them?

47. Environmental Design Parameter

48. • Environmental impacts from transmission lines are:
o conductor surface gradient,
o Corona Discharge
o Radio interference,
o Audible noise and
o Random noise.
o Deforestation throughout the transmission ROW

49. Conductor Surface Gradients
• Mean conductor surface gradient (g), kV/cm
g=
1+
1.4055
45.72
∗288.68
(2∗1.4055) log 𝑒
1310
1.4055∗45.72
= 20.775kV/cm

50. Cont’d…
• The critical visual corona gradient is calculated
through (go), kV/cm
𝑔 𝑜 = 21.1 ∗ 0.88 ∗ 1 1 +
0.301
1.4055
= 23.3𝑘𝑉/𝑐𝑚

51. Corona Discharge
• The power loss due to corona discharge, in kW/km at 50
Hz (per phase)
𝑃𝑘 = 𝑛2
𝑟2
∗ 𝑓 ∗
𝑔
𝑔𝑜
= 32 0.014055𝑚 2 ∗ 50𝐻𝑧 ∗
20.775
23.3
= 0.0793𝑘𝑊/𝑘𝑚
𝑃𝑘 3𝜙 = 0.2379𝑘𝑊/𝑘𝑚

52. Group Discussion
• Why the power loss due to corona discharge
becomes very small?

53. Audible Noise of Corona
• The average value of AN in rainy weather for any
phase of the transmission line is given as:
𝐴𝑁𝑖 = 120 log g + 𝑘 log 𝑛 + 55 log 2𝑟 − 11.4 log 𝑅 + 𝐴𝑁𝑜
k =26.4 for n ≥3, and 0 for n <3
ANo= -128.4(In case of n≥3) and -115.4 (In case of n<3)
𝐴𝑁𝑖= 120 log 20.775 + 26.6 log(3) + 55 log 2.811 − 11.4 log 100 − 128.4
𝐴𝑁1= 𝐴𝑁2= 𝐴𝑁3= 44.3dB(A)

54. Cont’d…
• Therefore the resultant sound pressure SPL, in dB (A)
is given by:
• Where, SPL = resultant sound pressure, dB (A),
• ANi = Noise Level due to phase I, dB (A)
SPL = 10 log[ 10
44.3
10 + 10
44.3
10 + 10
44.3
10 ] = 49.07dB(A)

55. Radio Interference
• RIfair = −105.81 + 117.41 log g + 40.38log 2r + 1.54 logN −
10.22logR − 27.1 log(f)
• RIfoul = −81.89 + 119.56 log g + 43.57log 2r + 3.97 logN −
19.05logR − 25.07log(f)
𝑹𝑰 𝒇𝒂𝒊𝒓 = 𝟏𝟏. 𝟓𝐝𝐁 (𝐀)
𝑹𝑰 𝒇𝒐𝒖𝒍 = 𝟏𝟔. 𝟑𝟗𝐝𝐁 (𝐀)

56. Cont’d…
System
Voltage, kV
Acceptable
Audible Noise ,
SPL, dB (A)
Radio
Interference ,
RI, dB (A)
Corona
Discharge
345 55 55
500 55 55
745 55 55

57. Mechanical Design Parameters

59. CHAPTER FIVE
LINE LIGHTNING PROTECTION

60. THEORIES OF THUNDER CLOUD
“When atmosphere near the earth surface or ocean surface containing a
large amount of water vapor warms up by the heat from the sun and other
sources, it expands and ascends. When it reaches a high altitude, it is
cooled down, resulting the vapor in the atmosphere becomes water
droplets and then a cloud. When it ascends even higher, water droplets in
the cloud becomes ice grains and some of them concentrate and grow to
hailstones. At that time, these ice grains and hailstones are decomposed by
electric current. Then, ice grains are charged positively and hailstones are
charged negatively. Ice grains ascend even higher by riding updraft and
hailstones grow bigger and fall by the gravity. Charge separation continues
by the coulomb force. Before long, top of the cloud is positively charged
and the bottom negatively. A cloud accumulates electrical energy, which is
when a thundercloud emerges”.

61. Cont’d…
• If the space charge densities, which happen to be
present in a thundercloud, produce local field strengths
of several100 kV/m, leader discharges are formed
which initiate a lightning discharge.
• Cloud-to-cloud flashes result in charge neutralization
between positive and negative cloud charge centers and
do not directly strike objects on the ground in the
process.

62. Cont’d…

63. Downward Flash

64. Upward Flash

65. Cont’d….
• On very high, exposed
objects (e.g. wind
turbines, radio masts,
telecommunication
towers, steeples, High
voltage towers) or on
the tops of mountains,
upward flashes (earth-
to-cloud flashes) can
occur.

66. Surge voltages
• Various types of surge voltages can occur in electrical and
electronic systems.
• They differ mainly with their duration and amplitude.
• Depending on the cause, a surge voltage can last a few hundred
microseconds, hours or even days.
• The amplitude can range from a few millivolts to ten thousand volts.
• Lightning strikes are a special cause of surge voltages. Direct and
indirect strikes can result not only in high surge voltage amplitudes,
but also high and sometimes long current flows, which then have
very serious effects.

67. Surge Characteristics

68. Group Discussion
Why Lightning flash over during cold season or
rain time?

69. Cont’d…
• The arrester providing a conducting path of
relatively low surge impedance between the line and
the ground to the arriving surge.
• The discharge current to the ground through the
surge impedance limits the residual voltage across
the arrester hence the equipment and the system
connected to it.
• During normal service this impedance is high
enough to provide a near-open circuit.

70. Cont’d…
• Arresters or diverters are generally of the following types
and the choice between them will depend upon the power
frequency system voltage, and characteristics of the
voltage surges, i.e.
(i) Gapped or conventional type, and
(ii) Gapless or metal oxide type.

71. Gapped Surge Arresters
• These are generally of the following types:
I. Expulsion Arrester
II. Spark Gap Arrester
III. Valve or non Linear Resistor Arrester

72. Expulsion Arrester
• These interrupt the flow of current by an expulsion action and
limit the amplitude of the surge voltages to the required level.
• They have low residual safe or discharge voltages (Vres).
• The arrester gap is housed in a gas-ejecting chamber that
expels gases during spark-over.
• The arc across the gap is reduced and blown-off by the force
of the gases thus produced.

73. Cont’d…

74. Cont’d…
• The enclosure is so designed that after blowing off the arc it
forcefully expels the gases into the atmosphere.
• The discharge of gases affects the surroundings, particularly
nearby equipment.
• The gas ejecting enclosure deteriorates with every operation and,
therefore, has only a limited operating life.
• Moreover, these types of arresters are for low system voltage
and of specific ratings and an excessive surge than the rated may
result in its failure.

75. Spark Gap Arrester
• These have a pair of conducting rods with an
adjustable gap, depending upon the spark over-
voltage of the arrester.
• Precise protection is not possible, as the spark-over-
voltage varies with polarity, steepness and the shape
of the wave protection becomes uncertain.
• These arresters are also now obsolete for the same
reasons as the previous one.

76. Valve or non Linear Resistor Arrester
• A non-linear SiC resistor-type gapped surge arrester may generally
consist of three non-linear resistors (NR) in series with the three spark
gap assemblies.

77. Cont’d…
• The resistance has an extremely low value on surge voltages and
a very high one during normal operations to cause a near-open
circuit. It is now easier to interrupt the flow currents.
• Across the spark gaps, known as current limiting gaps, are
provided high-value resistors (HR) backed up with HRC fuses.
• The non-linear resistors have a very flat V-I curve, i.e. they
maintain a near-constant voltage at different discharge currents.
• The flatness of the curve provides a small residual voltage and a
low current.

78. Cont’d…

79. Cont’d…
• When the switching or lightning surge voltage exceeds the
breakdown voltage of the spark gap, a spark-over takes place and
permits the current to flow through the NR.
• Due to the nonlinear nature of the resistor, the voltage across the
line is limited to approximately the discharge commencing voltage
(Vres), which is below the 3–5 p.u. level for a line.
• It may be noted that the use of resistor across the spark gap
stabilizes the breakdown of the spark gap by distributing the surge
voltage between the gap and the non-linear resistor.

80. Gapless Surge Arresters
• The high resistive component of the previous system
results high power loss which generates heat is the
limitation.
• The alternative was found in ZnO.
• ZnO is a semiconductor device and is a ceramic resistor
material constituting ZnO and oxides of other metals,
such as bismuth, cobalt, antimony and manganese.

81. Cont’d…
• These ingredients in different proportions are mixed in
powdered form, ZnO being the main ingredient.
• It is then pressed to form into discs and fired at high
temperatures to result in a dense polycrystalline
ceramic.
• Surge arresters made of these elements have no
conventional spark gap and possess excellent energy
absorption capability.

82. Cont’d…
• Under rated system conditions, its feature of
high non-linearity raises its impedance
substantially and diminishes the discharge
current to a trickle.
• Under rated conditions, it conducts in mA
while during transient conditions it offers a
very low impedance to the impending surges
and thus rises the discharge current and the
discharge voltage.
• However, it conducts only that discharge
current which is essential to limit the
amplitude of the prospective surge to the
required protective level of the arrester.

83. Cont’d

84. ZnO Arrester for various Rating

85. Group Discussion
Discuss the arrester type and operation in your
area?

86. Electrical Characteristics of a ZnO Surge Arrester
• ZnO blocks have extremely non-linear, current-voltage
characteristics, typically represented by:
• K, represents its geometrical configuration, cross-sectional area
and length, and is a measure of its current-carrying capacity.
• ∞ is a measure of non-linearity between V and I, and depends upon
the composition of the oxides used. Typical values are:
 In SiC it is 2 to 6
 In ZnO – it can be varied from 20 to 50.
𝐼 = 𝐾 · 𝑉∞

87. Cont’d…
• By altering ∞ and K, the arrester can be designed for any
conducting voltage (Vres) and nominal current discharge (In).
• Vres and In define the basic parameters of a surge arrester.

88. Maximum continuous operating voltage
(MCOV), Vc
• This is the maximum power frequency operating r.m.s. voltage that
can be applied continuously (≥ 2 hours) across the arrester
terminals without a discharge (point 1 on the curve).
• It continuously draws an extremely low leakage current, IZnO,
capacitive in nature, due to ground capacitance.
• Where Vm is 5% above the system line-line nominal voltage.
𝑽 𝒄 =
𝑽 𝒎
𝟑
(Phase to phase)

89. Rated Voltage, Vr
• This is the maximum permissible r.m.s. voltage for
which the arrester is designed (point 2 on the curve).
• The arrester can withstand this voltage without a
discharge for minimum 10s under continuously rated
conditions (when the arrester has reached its thermal
stability),
• Indirectly indicating an in-built TOV (transient over
voltage) capability of 10s.

90. Cont’d…
• Now it also draws a current resistive in nature, in the
range of a few mA.
• The lower this current, lower will be the loss and the heat
generated during an over-voltage and hence better energy
absorption capability.

91. Discharge or Residual Voltage, Vres
• It is the voltage that appears across the arrester
during the passage of discharge current – that
flows through the arrester due to a surge.
• Vres is the conducting voltage of an arrester during an
over-voltage or transient condition and defines its
protective level.

92. Temporary Over-Voltage (TOV)
• It is determined by its low current region (d) that is
usually less than 1 A and for prospective transient
voltages it is determined by its high current region (e)
(2.5–20 kA, 8/20𝜇s current impulse).
• It is beyond the knee point and relatively long duration
voltage transient.
• Major sources are short cct fault and load rejection

93. Transient voltages (Vt)
• Depending upon the magnitude of Vt the
operating point may shift to near point 4 or
beyond and conduct a current 2.5–20 kA and
more. It is point 4 on the curve.
• The maximum surge voltage exist at this point
and the arrester should effectively clear the surge.

94. Energy Capability (J)
• Energy capability of an arrester defines its capability to
absorb the surge energy of an impending surge, usually
the long duration switching surge without any thermal
damage or heat generation.
• Energy capability values are provided as standard by the
manufacturers in their data sheets.
• For a series of consecutive discharge the ZnO discs must
attain thermal equilibrium.

95. Basic insulation level (BIL)
• BIL is the basic insulation level of equipment. When the system
TOVs or voltage surges exceed this level, the equipment may fails.
• In the latest international and national standards it is defined as
follows:
a. For systems 1 kV < Vm < 245 kV.
i. Rated lightning impulse withstand level (LIWL)
ii. Rated short time power frequency dielectric strength.
b. For systems Vm > 300 kV to 765 kV;
i. Rated lightning impulse withstand level (LIWL)
ii. Rated short time power frequency dielectric strength.
iii.Rated switching impulses withstand level (SIWL).

96. Protective Margins
• On the BIL discussed above a suitable protective margin is considered
to provide sufficient safety to the protected equipment against
unforeseen contingencies. ANSI/IEEEC62.22 has recommended
certain values to account for these and they are given in Table 5.1.
𝐏𝐫𝐨𝐭𝐞𝐜𝐭𝐢𝐯𝐞 𝐌𝐚𝐫𝐠𝐢𝐧 =
𝐁𝐈𝐋 𝐨𝐟 𝐭𝐡𝐞 𝑬𝒒𝒖𝒊𝒑𝒎𝒆𝒏𝒕
𝐈𝐦𝐩𝐮𝐥𝐬𝐞 𝐏𝐫𝐨𝐭𝐞𝐜𝐭𝐢𝐨𝐧 𝐋𝐞𝐯𝐞𝐥 𝐨𝐟 𝐭𝐡𝐞 𝐀𝐫𝐫𝐞𝐬𝐭𝐞𝐫 (𝑽 𝒓𝒆𝒔)

97. Selection of a ZnO Surge Arrester
• Service conditions:
• Mechanical soundness:
• Maximum continuous operating voltage (MCOV) Vc
(rms):
• The BIL of the equipment being protected
• The arrester’s nominal discharge current (In):

98. • For each kind of TOV and its duration, a
corresponding factor (K) is obtained and with this is
determined the required rating, Vr of the arrester.
• The most crucial TOV may be selected as the rating
of the arrester.
• If it is not a standard rating as in the manufacturer’s
catalogue one may select the next higher rating
available.
Vc= K · Vr

99. Cont’d….
Example:
Determine the rating of a surge arrester to protect a solid ground fault stay for 3 second and
load rejection of 1 second simultaneously for a 400 kV system.

100. 400kV Arrester Rating at Different TOV

101. Protective Distance of the Arrester
• If the arrester and the equipment to be
protected at different location:
Vs = Vres + 2.S.T 𝑻 =
𝒍 ∗ 𝟏𝟎−𝟑
𝟎. 𝟑
𝝁𝒔 𝒍 = 𝟏𝟓𝟎
𝑽 𝒔 − 𝑽 𝒓𝒆𝒔
𝑺
(𝒎)
o Vs= actual surge voltage at the instant of strike in the equipment
o S = steepness of the incoming wave in kV/𝜇s
o T = travelling time of the surge to reach the equipment from the arrester
terminals.
o L= line length between arrester andequipment

102. Example
• For the arrester of the previous example, Vm = 420 kV, Vr = 336 kV and Vres = 844
kV for a lightning surge protective margin at 20 kA discharge current, from the
manufacturer.
• If we consider the lightning surge with a steepness of 2000 kV/𝜇s, then for a
total distance of, say, 8 m from the arrester to the equipment please determine the
actual surge voltage at the equipment and BIL of the equipment If we maintain a
protective margin of 20%.
𝑉𝑠 = 844𝑘𝑉 + 2 ∗ 2000 ∗
8
0.3
∗ 10−3 𝑘𝑉 = 951𝑘𝑉
• Then the minimum BIL that the equipment under protection must have
= 1.20 *951kV = 1141.2 kV