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Chapter 3 · 8 hours

Overhead Line Insulator Design

IOE past exam questions

Past questions and answers

25 questions set from this chapter, 5 of them more than once. Most asked first.

  • Asked 4 times
  • 2074 Magh · 1+3 marks
  • 2072 Asoj · 3 marks
  • 2071 Bhadra · 1+3 marks
  • 2067 Mangsir (old course) · 1+3 marks

State whether the following statement is TRUE or FALSE. Justify your answer with a brief explanation: The value of Switching Surge Ratio (SSR) increasing the system voltage for insulation coordination designing purpose.

Answer

Read as: "The value of switching surge ratio (SSR) chosen for insulation coordination increases with increasing system voltage." FALSE.

  • SSR = peak switching over-voltage / peak phase voltage (in pu).
  • At low voltages (up to 220 kV) insulation is governed by lightning, so switching surges up to about 3–3.5 pu can be accepted without extra cost.
  • At EHV/UHV, switching surges govern the insulation (string length, air clearance, tower size), and insulation cost rises steeply with voltage. So SSR is deliberately reduced by closing resistors, controlled switching and surge arresters.
  • Typical design values: about 3.0–3.5 pu up to 220 kV, 2.5–2.8 pu at 400 kV, about 2.0 pu at 765 kV and 1.5–1.8 pu at UHV.

So the SSR used for design decreases as system voltage increases.

  • Asked 3 times
  • 2078 Kartik · 4 marks
  • 2074 Bhadra · 1+3 marks
  • 2071 Magh · 1+3 marks

State whether the following statement is TRUE or FALSE. Justify your answer with a brief explanation: The value of SSR should not be chosen below 1.8.

Answer

TRUE (as a practical design rule).

  • SSR is the ratio of peak switching surge voltage to peak normal phase voltage. Switching a line always produces some transient over-voltage due to trapped charge, travelling-wave reflection and the Ferranti rise.
  • With closing resistors, synchronised (point-on-wave) switching and metal-oxide arresters the SSR can be reduced to about 1.8–2.0 pu; reducing it further needs very costly equipment and gives little saving in insulation.
  • Temporary over-voltages and arrester protective level also set a lower bound: the insulation must anyway withstand arrester residual voltage plus margin.

Hence in design an SSR below about 1.8 is not chosen; values of 1.8–2.0 are used for UHV and higher values for lower voltages.

  • Asked 3 times
  • 2077 Chaitra · 10 marks
  • 2073 Magh · 6 marks
  • 2070 Bhadra · 8 marks

Discuss the role of system (neutral) earthing in the selection of overhead line insulation level.

Answer

System (neutral) earthing is the connection of the star point of generators/transformers to earth, solidly or through resistance/reactance. It decides how high the voltage of healthy phases rises during an earth fault, and so it decides the rating of surge arresters and the insulation level (BIL) of the line and equipment.

Earthing coefficient

Ce=highest rms phase-to-earth voltage of healthy phase during earth faultrms line-to-line voltageC_e = \frac{\text{highest rms phase-to-earth voltage of healthy phase during earth fault}}{\text{rms line-to-line voltage}}
  • Isolated (ungrounded) neutral: healthy phases rise to full line voltage, Ce≈1.0C_e \approx 1.0 (100%). Arcing grounds can produce even 5–6 pu transients.
  • Effectively (solidly) earthed: X0/X1≤3X_0/X_1 \le 3 and R0/X1≤1R_0/X_1 \le 1; healthy phase voltage rises to about 0.8 of line voltage, Ce≤0.8C_e \le 0.8 (80%).
  • Resistance/reactance earthed: in between.

Effect on surge arrester rating

Arrester rated voltage =Ce×Vmax= C_e \times V_{max} (times a small margin):

  • 220 kV, Vmax=245V_{max} = 245 kV: effectively earthed → 0.8×245=1960.8 \times 245 = 196 kV arrester (≈"80% arrester"); isolated neutral → 245 kV arrester ("100% arrester").
  • A lower rated arrester has a lower protective (residual) level.

Effect on insulation level (BIL)

  • BIL == protection ratio × arrester protective level (protection ratio about 1.2–1.4).
  • With effective earthing the protective level is about 20% lower, so a reduced BIL can be chosen. Example for 220 kV: full insulation 1050 kV, reduced 900 kV (and 825 kV in some practice).
  • Power-frequency withstand voltages and switching surge over-voltages (which also depend on phase-to-earth voltage) are lower too.

Effect on overhead line insulation

  • Fewer discs per string, shorter cross-arms, smaller clearances, shorter towers, narrower right-of-way. Saving can be 15–20% in insulation cost at 132 kV and above.
  • With isolated neutral, the insulation must be designed for full line voltage to earth and for arcing-ground surges; this is acceptable only at low voltage (≤ 33 kV) where insulation is cheap.
EarthingCeC_eArresterInsulation level
Isolated≈1.0100%Full
Resistance/reactance0.8–1.085–100%Full
Effective (solid)≤0.880%Reduced

Practice

  • 132 kV and above: neutrals are effectively (solidly) earthed to use reduced insulation; this is standard in Nepal and India.
  • 11 kV and 33 kV distribution: often resistance or solid earthing; insulation is cheap, so full insulation is common.
  • Effective earthing also gives fast, selective earth-fault relaying and avoids arcing grounds, but causes high earth-fault current and possible interference with communication lines.

Thus neutral earthing directly lowers or raises the voltage stress on insulation and is a key input to insulation coordination and to the number of discs chosen for an overhead line.

  • Asked 3 times
  • 2074 Magh · 1+3 marks
  • 2071 Bhadra · 1+3 marks
  • 2071 Magh · 1+3 marks

State whether the following statement is TRUE or FALSE. Justify your answer with a brief explanation: The insulators are designed so that it will punctured before it gets flashover.

Answer

FALSE. Insulators are designed so that they flash over before they puncture.

  • Flashover: an arc through the air over the insulator surface. The insulator is not damaged (perhaps slight glazing burn) and service resumes after the breaker recloses.
  • Puncture: breakdown through the porcelain/glass body. The disc is permanently damaged and must be replaced.
  • Therefore the puncture strength is kept well above the flashover voltage. The ratio
Safety factor=Puncture strengthFlashover voltage\text{Safety factor} = \frac{\text{Puncture strength}}{\text{Flashover voltage}}

is about 10 for pin insulators and 1.3–1.5 or more for suspension discs (impulse).

  • Asked 2 times
  • 2078 Kartik · 6 marks
  • 2071 Magh · 6 marks

Compare the properties of porcelain and annealed tough glass.

Answer

Both porcelain and toughened (annealed and then heat-tempered) glass are used for overhead line insulator discs.

PropertyPorcelainToughened glass
MaterialClay, quartz, feldspar, glazedSoda-lime glass, heat tempered
Dielectric strengthAbout 60 kV/cmHigher, about 140 kV/cm
Compressive strengthHigher, about 70,000 kg/cm²About 10,000 kg/cm²
Tensile strengthLow, about 500 kg/cm²Much higher, about 35,000 kg/cm²
Thermal coefficient of expansionLow; withstands temperature changesHigher; stress due to sudden temperature change (reduced by toughening)
Defect detectionInternal cracks/punctures invisible; need testingA damaged disc shatters visibly ("self-indicating"); easy inspection
Moisture/condensationGlazed surface sheds waterMoisture condenses easily, dust collects
Shape/sizeLarge and complex shapes easy to makeLimited to simpler shapes and lower voltages per unit
CostHigherCheaper
WeightHeavierLighter for same strength
Life/ageingLong life, stableLong life; no ageing of glass body
Inner defectsPossible air voidsTransparent, so bubbles/defects seen in manufacture

Summary

  • Toughened glass gives higher dielectric and tensile strength, lower cost and easy detection of damaged discs (broken shell is visible from the ground).
  • Porcelain has better resistance to thermal shock, is less affected by moisture/condensation and pollution, and can be made in complex shapes; it is preferred in polluted or humid areas and for station post insulators.
  • 2080 Chaitra · 4 marks

State whether the following statement is TRUE or FALSE and justify your answer with brief explanation: Lightning over voltage is more dominating in distribution system than in transmission system.

Answer

TRUE.

  • Distribution lines (11 kV, 33 kV) have low insulation levels (BIL about 75–200 kV). Even an induced surge from a nearby lightning stroke (100–300 kV) is enough to flash them over; direct strokes always do.
  • They usually have no overhead shield wire, short and low poles, and run through open areas where many strokes occur.
  • Transmission lines have high BIL (550–1550 kV), shield (earth) wires and low tower footing resistance, so most strokes are intercepted and induced surges are harmless.
  • In EHV lines (400 kV and above) the insulation is governed mainly by switching surges, not lightning.

Hence lightning is the dominant over-voltage for distribution systems, while switching surges become dominant as transmission voltage increases.

  • 2080 Chaitra · 8 marks

Compute the number of insulator discs required for a single circuit, 220 kV, 50 Hz overhead transmission line considering the various external and internal over voltages. Use the related data given in the appendix.
[Given Table A-2: Withstand voltage capability for different system voltages:]
Maximum system voltage (kV)1 minute dry withstand (kV)1 minute wet withstand (kV)Impulse withstand (kV)
123215185450
145265230550
255435395900
4207606801550
[Given Table A-3: Flashover voltages for 254 × 154 mm disc insulators:]
No. of discs1 minute dry FOV (kV)1 minute wet FOV (kV)Impulse FOV (kV)
18050150
215590255
3215130355
4270170440
5325210525
6380250610
7435290695
8485330780
9535370860
10585410945
116354501025
126854851105
137305201185
147755551265
158205901345
168656201425
179106501505
189556801585
1910007101665
2010457401745
[Minimum air clearance: 6.5 inch per 10 kV (rms) and factor of safety = 8 inch.]

Answer

Data: 220 kV, single circuit, 50 Hz. From Table A-2 (220 kV class row): 1-min dry withstand 435 kV, 1-min wet withstand 395 kV, impulse withstand 900 kV. Highest system voltage taken as 245 kV (standard for 220 kV). Usual design factors assumed: flashover withstand ratio (FWR) = 1.15, non-atmospheric condition factor (NACF) = 1.1, factor of safety (FOS) = 1.1, so

k=1.15×1.1×1.1=1.3915k = 1.15 \times 1.1 \times 1.1 = 1.3915

1. External over-voltage: lightning

Vreq=900×1.3915=1252.35 kVV_{req} = 900 \times 1.3915 = 1252.35\ \text{kV}

Table A-3 (impulse FOV): 13 discs = 1185 kV < 1252.35; 14 discs = 1265 kV. → 14 discs

2. Internal over-voltage: switching surge

Assume switching surge ratio SSR = 2.8 (usual for 220 kV).

Vss=SSR×2 Vmax3=2.8×2×2453=560.12 kV (peak)Vreq=560.12×1.3915=779.40 kV (peak)=779.40/2=551.12 kV (rms)\begin{aligned} V_{ss} &= SSR \times \frac{\sqrt{2}\,V_{max}}{\sqrt{3}} = 2.8 \times \frac{\sqrt{2} \times 245}{\sqrt{3}} = 560.12\ \text{kV (peak)} \\ V_{req} &= 560.12 \times 1.3915 = 779.40\ \text{kV (peak)} \\ &= 779.40/\sqrt{2} = 551.12\ \text{kV (rms)} \end{aligned}

Compared with 1-min wet FOV: 13 discs = 520 kV, 14 discs = 555 kV. → 14 discs

3. Power-frequency over-voltage

Wet: 395×1.3915=549.64395 \times 1.3915 = 549.64 kV → wet FOV 14 discs = 555 kV → 14 discs Dry: 435×1.3915=605.30435 \times 1.3915 = 605.30 kV → dry FOV 11 discs = 635 kV → 11 discs

Result

ConditionRequired (kV)Discs
Lightning impulse1252.3514
Switching surge551.12 (rms eq.)14
Power freq., wet549.6414
Power freq., dry605.3011

The largest requirement governs: 14 discs (254 × 154 mm) per suspension string. String length ≈14×0.154=2.16\approx 14 \times 0.154 = 2.16 m. (Tension strings commonly use one extra disc, i.e. 15.)

Check of air clearance: a=6.5×245/310+8=99.94a = 6.5 \times \frac{245/\sqrt{3}}{10} + 8 = 99.94 in ≈2.54\approx 2.54 m.

  • 2079 Chaitra · 1+3 marks

State and justify whether the following statement is TRUE or FALSE: The value of Protection ratio (PR) increases on increasing system voltage for designing purpose.

Answer

FALSE. The protection ratio used for design is reduced (or kept about the same), not increased, at higher system voltages.

  • Protection ratio PR=insulation withstand level (BIL/SIL)protective level of surge arresterPR = \dfrac{\text{insulation withstand level (BIL/SIL)}}{\text{protective level of surge arrester}}.
  • At low and medium voltages insulation is cheap, so a generous margin (PR about 1.4 or more) is used.
  • At EHV the cost of insulation (string length, clearances, tower size, transformer insulation) rises steeply, and modern metal-oxide arresters have accurate, low protective levels. So designers use a smaller margin, about 1.15–1.25, to obtain reduced insulation levels.

Hence PR tends to decrease with increasing system voltage.

  • 2078 Chaitra · 1+3 marks

State and justify whether the following statement is TRUE or FALSE: Temporary overvoltage is more dominating than transient overvoltage in transmission line.

Answer

FALSE.

  • Temporary over-voltages (power-frequency, e.g. load rejection, Ferranti effect, earth-fault rise of healthy phases) are only about 1.2–1.5 pu and last for seconds. They decide the arrester rating and power-frequency withstand, but rarely the insulation size.
  • Transient over-voltages: lightning (several MV on direct strokes) and switching surges (2–3.5 pu). They are far larger in magnitude and govern the number of discs, air clearances and BIL: lightning dominates up to about 220 kV, switching surges at 400 kV and above.

So in transmission line insulation design, transient over-voltages are more dominating than temporary ones.

  • 2077 Chaitra · 6 marks

Discuss the criterion for single and double earth wire selection for a high voltage overhead transmission line.

Answer

The earth (shield/ground) wire is strung above the phase conductors to intercept lightning strokes and drain them to earth through the towers. The choice between one and two earth wires depends on how well the phase conductors are shielded.

Criteria

  1. Shielding angle: the angle between the vertical through the earth wire and the line joining the earth wire to the outermost phase conductor. It should be about 30° (20–30° at EHV and high-lightning areas). If one earth wire on the tower peak cannot give this angle to all phases, two are used.
  2. Conductor configuration:
    • Vertical or triangular formation (narrow tower; 66–132 kV S/C and D/C) → single earth wire at the peak is enough.
    • Horizontal formation (220 kV and 400 kV S/C, wide spacing) → two earth wires, one over each outer phase. The middle phase is shielded if the angle between the two earth wires is within about 60° (earth wires spaced not more than about twice their height above the conductors).
  3. System voltage and importance: EHV lines (≥ 220 kV) and important evacuation/interconnection lines use two earth wires for better shielding and lower outage rate.
  4. Lightning intensity (isokeraunic level): in areas of high thunderstorm days (e.g. hilly Nepal regions), double earth wires reduce shielding failures.
  5. Tower height and width: taller/wider towers attract more strokes and need better coverage.
  6. Tower footing resistance: two earth wires share stroke current and lower the effective surge impedance, reducing back-flashover.
  7. Cost: a second earth wire adds conductor, peak steel and tower loading; used only when needed.
  Single earth wire       Two earth wires
        o                  o           o
       /|\                 |\         /|
   30°/ | \              30| \       / |
     o  |  o               o    o    o
     o  |  o            (horizontal phases)
     o  |  o
  (vertical D/C)
  • 2075 Bhadra · 1+3 marks

State whether the following statement is TRUE or FALSE. Justify your answer with a brief explanation: Impulse withstand voltage of an overhead transmission line insulator depends on system (neutral) earthing condition.

Answer

TRUE.

  • The insulation level (impulse withstand voltage, BIL) is chosen as protection ratio × protective level of the surge arrester.
  • The arrester rating depends on the earthing coefficient: with effectively (solidly) earthed neutral the healthy-phase voltage rises to only about 0.8 of line voltage, so an "80% arrester" with a lower protective level is used. With isolated neutral, a "100% arrester" is needed.
  • So effective earthing allows a reduced BIL (e.g. 220 kV: 900 kV instead of 1050 kV), and fewer insulator discs; an ungrounded system needs full insulation.

The impulse strength of a given disc does not change, but the required impulse withstand level of the line does depend on neutral earthing.

  • 2075 Bhadra · 16 marks

In design of a transmission line for delivering power of 180 MW for a distance of 200 km. Considering that the system is to be operated at power factor of 0.85, find the required numbers of insulator disc per string to be selected.
[Given: Most economical voltage empirical formula: V = [Lt/1.6 + (P × 1000)/(cosφ × Nc × 150)]^0.5 kV, with Lt in km and P in MW. Standard voltages: 66 kV, 132 kV, 220 kV, 400 kV.] [In this paper's appendix the formula is printed without the 5.5 multiplier: V = [L/1.6 + (P × 1000)/(cosφ × Nc × 150)]^0.5.]
[Given Table A-1: Transmission line capability curve with assumption of single circuit transmission line surge impedance of 400 Ω (m.f. = power transmission capability/SIL):]
Length (km)Multiplying factor
802.75
1602.25
2401.75
3201.35
4801.0
6400.75
[Given Table A-2: Withstand voltage capability for different system voltages:]
Maximum system voltage (kV)1 minute dry withstand (kV)1 minute wet withstand (kV)Impulse withstand (kV)
123215185450
145265230550
255435395900
4207606801550
[Given Table A-3: Flashover voltages for 254 × 154 mm disc insulators:]
No. of discs1 minute dry FOV (kV)1 minute wet FOV (kV)Impulse FOV (kV)
18050150
215590255
3215130355
4270170440
5325210525
6380250610
7435290695
8485330780
9535370860
10585410945
116354501025
126854851105
137305201185
147755551265
158205901345
168656201425
179106501505
189556801585
191000 (printed as 100)7101665
2010457401745
[Other given data: Minimum air clearance: 6.5 inch per 10 kV (rms) and factor of safety = 8 inch. Maximum insulator string swing: 45°. Minimum ground clearance Hg = (V − 33)/33 + 17 feet, V is L-L voltage in kV.]

Answer

The number of discs depends on the line voltage. So the voltage level and number of circuits are fixed first, and then the string is sized against the power-frequency (wet and dry), impulse and switching-surge withstand levels from Tables A-2 and A-3.

Step 1: Most economical voltage

The appendix prints Still's formula without its multiplier. That form gives about 39 kV, which is clearly not a transmission voltage. So the standard form with the factor 5.5 is used:

V=5.5L1.6+P×1000cos⁡ϕ×Nc×150 kVV = 5.5\sqrt{\frac{L}{1.6} + \frac{P\times 1000}{\cos\phi \times N_c \times 150}}\ \text{kV}

With L=200L = 200 km, P=180P = 180 MW and cos⁡ϕ=0.85\cos\phi = 0.85:

Circuits NcN_cValue under rootVV (kV)Nearest standard
1125+1411.76=1536.76125 + 1411.76 = 1536.76215.6220 kV
2125+705.88=830.88125 + 705.88 = 830.88158.5132 kV or 220 kV

Step 2: Technical check with SIL (Table A-1)

SIL=V2Zc=V2400SIL = \frac{V^2}{Z_c} = \frac{V^2}{400}

For 200 km, interpolate between 160 km (2.25) and 240 km (1.75):

m.f.=2.25−200−160240−160(2.25−1.75)=2.0m.f. = 2.25 - \frac{200-160}{240-160}(2.25-1.75) = 2.0
OptionSIL (MW)Capability = m.f. × SIL (MW)≥ 180 MW?
132 kV, 1 circuit43.5687.1No
132 kV, 2 circuits2 × 43.56174.2No
220 kV, 1 circuit121.0242.0Yes
220 kV, 2 circuits2 × 121.0484.0Yes (over-designed)

Selected: 220 kV, single circuit. It is the economic voltage for Nc=1N_c = 1 and it carries 180 MW with a margin.

Step 3: Withstand levels required (Table A-2)

For 220 kV nominal, the table row is the 255 kV maximum system voltage:

TestRequired withstand (kV)
1 min dry, power frequency435
1 min wet, power frequency395
Impulse (BIL)900

The string's flashover voltage must be at least the required withstand.

Step 4: Number of discs for each criterion (Table A-3, 254 × 154 mm discs)

  1. Dry power frequency: 435 kV needs FOV ≥ 435 kV. 7 discs give 435 kV, so 7 discs.
  2. Wet power frequency: 395 kV needs FOV ≥ 395 kV. 9 discs give only 370 kV and 10 discs give 410 kV, so 10 discs.
  3. Impulse (lightning): 900 kV needs FOV ≥ 900 kV. 9 discs give 860 kV and 10 discs give 945 kV, so 10 discs.
  4. Switching surge: take a switching surge ratio of 2.8 (usual for a 220 kV line without closing resistors).
Vph,peak=2×2453=200.0 kVVss=2.8×200.0=560.1 kV (peak)Equivalent rms=560.12=396.1 kV\begin{aligned} V_{ph,peak} &= \frac{\sqrt{2}\times 245}{\sqrt{3}} = 200.0\ \text{kV} \\ V_{ss} &= 2.8\times 200.0 = 560.1\ \text{kV (peak)} \\ \text{Equivalent rms} &= \frac{560.1}{\sqrt{2}} = 396.1\ \text{kV} \end{aligned}

Comparing 396.1 kV with the wet FOV column gives 10 discs (410 kV).

CriterionDiscs needed
Dry 1-min7
Wet 1-min10
Impulse10
Switching surge10

Step 5: Final selection

  • Governing requirement: 10 discs, set by the wet, impulse and switching-surge criteria.
  • Add one disc to allow for a cracked or punctured disc in service, which keeps the withstand level intact.

Answer: 220 kV single-circuit line with 11 discs (254 × 154 mm) per suspension string, and 12 discs per tension string. The extra disc in the tension string covers its heavier mechanical duty and the greater exposure of dead-end and angle positions. The suspension string length is about 11×0.154=1.6911 \times 0.154 = 1.69 m plus hardware.

Assumptions: standard atmospheric conditions (no altitude correction) and light pollution. For heavy pollution or high-altitude routes in Nepal, add discs to meet creepage and air-density corrections.

  • 2074 Bhadra · 4 marks

Explain the different types of insulator failure.

Answer

An insulator fails when it can no longer keep the live conductor isolated from the earthed tower, or can no longer hold the conductor mechanically. The common types are:

  1. Flashover: an arc forms through the air around the insulator surface between the conductor and the pin or cap. The porcelain is usually not damaged unless the heat cracks or glazes it. Causes include lightning or switching surges and wet or polluted surfaces.
  2. Puncture: the discharge passes through the body of the porcelain. The insulator is permanently destroyed. Good design makes the puncture strength greater than the flashover voltage. The ratio of the two is the safety factor, about 10 for pin and 5 for suspension insulators.
  3. Cracking of porcelain: caused by unequal expansion of the porcelain, cement and steel parts under temperature changes, by cement growth, or by mechanical impact (for example stone throwing).
  4. Defective material: porous or badly vitrified porcelain absorbs moisture, which lowers its insulation resistance and leads to leakage and puncture.
  5. Improper glazing: cracks or uneven glaze let dirt and moisture collect, which increases leakage current.
  6. Surface leakage and pollution flashover: salt, dust and industrial deposits form a conducting film in fog or drizzle. This causes tracking and flashover at normal voltage.
  7. Mechanical failure: the string breaks or the pin or cap fails under excess tension, wind or ice, or from corrosion of the metal fittings.
  8. Short-circuited or punctured discs in a string: these put more voltage on the remaining discs and can cause a cascade failure of the string.
  • 2074 Magh · 6 marks

Explain the basic configuration and advantage of string insulator used in HV line in brief.

Answer

A string (suspension) insulator is a chain of porcelain or glass discs joined by metal caps and pins. It hangs from the tower cross-arm, and the conductor is clamped at the bottom. It is used for all lines above about 33 kV.

Basic configuration

     cross-arm ===========
                  |  <- hanger / clevis
                [===]  disc 1
                  |
                [===]  disc 2
                  :
                [===]  disc n
                  |
           ------(o)------ conductor clamp
  • Each disc is designed for about 11 kV. The number of discs is chosen to match the line voltage.
  • The common arrangements are I-string, V-string, tension (strain) string, and double or multiple strings for heavy conductors.

Advantages

  1. Each disc works at a low voltage (about 11 kV), so a string can be built for any voltage by adding discs. One large pin insulator would not be practical.
  2. It is cheaper than a pin insulator above 33 kV, because the cost of a pin insulator rises steeply with voltage.
  3. A damaged disc can be replaced alone. The whole string need not be changed.
  4. The string is flexible. It swings with the wind, so the mechanical stress on it is lower.
  5. The conductor hangs below the cross-arm, so the cross-arm and tower partly shield it from lightning.
  6. Line voltage can later be raised just by adding discs.
  7. Strings can be used in parallel to get higher mechanical strength.
  • 2073 Bhadra · 6 marks

Explain the factors affecting the choice of BIL of a transmission line for insulator discs selections.

Answer

The basic insulation level (BIL) is the crest value of the standard 1.2/50 µs lightning impulse voltage that the insulation must withstand without flashover. It sets the minimum impulse flashover voltage of the insulator string, and so it largely sets the number of discs. The main factors are:

  1. System voltage: the BIL rises with the nominal and maximum system voltage. Standard values (IEC/IS) are, for example, 550 kV for 132 kV, 900–1050 kV for 220 kV and 1425–1550 kV for 400 kV.
  2. Lightning performance required (isokeraunic level): in areas with many thunderstorm days, more lightning surges reach the line. A higher BIL, or better shielding, is chosen to keep tripouts within the target.
  3. Shielding and tower footing resistance: the earth wire, its shielding angle and the footing resistance decide the back-flashover voltage across the string. With high footing resistance (rocky hills), a higher BIL is needed.
  4. Switching surges at EHV: above about 300 kV, switching surges (SSR × peak phase voltage) may need more insulation than lightning does. The insulation is then coordinated for switching (BSL), and the BIL follows from it.
  5. Surge arresters and protective level: arresters at line ends limit the surge. The BIL must keep a margin of about 15–25% above the arrester's protective level (insulation coordination).
  6. Earthing of the neutral: effectively earthed systems have lower temporary overvoltage, so a reduced BIL can be used. Insulated or resonant-earthed systems need full BIL.
  7. Altitude and atmosphere: low air density at high altitude (as in Nepal's hills) lowers flashover voltage. The required BIL is corrected by the air-density factor δ\delta.
  8. Pollution and rain: contaminated or wet conditions lower the flashover voltage, so more discs or a higher margin are needed.
  9. Economics and reliability: a higher BIL means more discs, longer strings, larger towers and more cost. It is balanced against the cost of outages.
  • 2073 Magh · 6 marks

Discuss the various factors affecting the air clearance for a transmission line of a given voltage.

Answer

Air clearance is the minimum distance that must be kept in air between a live conductor (or its fittings) and earthed parts such as the tower body, cross-arm or earth wire, and between phases. Its purpose is to prevent flashover. For a given line voltage it depends on:

  1. Maximum system voltage and overvoltages: clearance is set by the highest stress, not by the nominal voltage. This means the power-frequency maximum voltage, the lightning impulse and, at EHV, the switching surge level.
  2. Switching surge ratio: lines with higher switching overvoltages (no closing resistors, long lines, lossless lines) need larger gaps. Above 300 kV this usually governs the clearance.
  3. Insulator string swing: wind blows the suspension string through an angle θ\theta (typically 15°–45°). The clearance must still hold in the swung position. So both the swing angle and the string length affect the cross-arm length. V-strings restrict swing.
  4. Wind pressure and conductor size: a larger diameter and a higher design wind give a larger swing angle.
  5. Altitude and atmospheric conditions: air density, humidity and temperature change the breakdown strength of air. Clearances are increased at high altitude.
  6. Gap geometry: rod-plane gaps (conductor to tower leg) and conductor-to-cross-arm gaps have different breakdown strengths (gap factor). Sharp fittings reduce the breakdown voltage.
  7. Rain, fog, pollution and birds: these effectively reduce the gap. A safety margin is added (for example 8 inch or 30 cm).
  8. Factor of safety and standards used: for example, 6.5 inch per 10 kV (phase rms) plus 8 inch, or 1 cm per kV (phase peak) plus 30 cm. The IS 5613 or IEC clearances are another source.
  9. Live-line maintenance: if hot-line work is planned, a larger clearance is kept for workers and tools.
  10. Conductor bundling and corona rings: these change the field distribution and the effective gap.
  • 2072 Asoj · 8 marks

Describe the steps to be followed while selecting the number of insulator discs for a transmission line design.

Answer

The number of discs is chosen so that the string withstands every voltage stress it will meet: power frequency in dry and wet conditions, lightning impulse and switching surges. Pollution and spare-disc allowances are then added. The steps are:

  1. Fix the system voltage. Take the nominal voltage VV and the maximum system voltage VmV_m (about 1.1V1.1V, for example 245 kV for a 220 kV line).
  2. Find the required withstand levels from standards or the given table for VmV_m:
    • 1-minute power-frequency dry withstand
    • 1-minute power-frequency wet withstand
    • lightning impulse withstand (BIL)
    • switching impulse withstand (BSL), for Vm>245V_m > 245 kV
  3. Correct for site conditions. Divide the required withstand values by the air-density factor δ\delta (altitude, temperature) and any humidity factor. This gives the equivalent values at standard conditions.
  4. Power-frequency dry test: from the disc flashover table, pick the smallest number of discs whose 1-minute dry FOV is at least the required dry withstand.
  5. Power-frequency wet test: do the same with the wet FOV column. This usually governs over the dry test.
  6. Lightning impulse: choose the discs whose impulse FOV is at least the BIL. Allow for the uneven voltage distribution (string efficiency) along the string.
  7. Switching surge: compute the surge Vss=SSR×2Vm/3V_{ss} = SSR \times \sqrt{2}V_m/\sqrt{3}. Use a switching surge ratio of 2–3.5 depending on breaker type and line length. Choose discs whose switching flashover, or equivalent wet FOV, is at least VssV_{ss}.
  8. Pollution (creepage) check: the number of discs ≥ (specific creepage in mm/kV × VmV_m) / (creepage distance per disc). Use 16–31 mm/kV from light to very heavy pollution.
  9. Take the largest number from steps 4–8.
  10. Add allowances: one extra disc for a defective or punctured disc. Use one or two more discs in tension (dead-end) strings than in suspension strings.
  11. Check the string length and air clearance against the tower window, the swing angle and the cross-arm length. If the string becomes too long, consider a V-string, anti-fog discs or composite insulators.

Example: for a 220 kV line (Vm=255V_m = 255 kV row of the table: wet 395 kV, BIL 900 kV), 10 discs meet the wet and impulse levels. Adding one spare disc gives 11 discs per suspension string.

  • 2072 Magh · 6 marks

Classify the overhead line insulators. Write advantages of string insulators for HV lines and state the various configuration of string insulator.

Answer

Overhead line insulators support the conductors and isolate them electrically from the towers.

Classification

BasisTypes
Construction / usePin type, suspension (disc/string), strain (tension), shackle, post type, stay (egg) insulator
MaterialPorcelain, toughened glass, composite (polymer, silicone rubber)
VoltageLV/MV (pin, shackle, post) and HV/EHV (string, long-rod, composite)
Position on lineSuspension (straight towers) and tension (angle and dead-end towers)
  • Pin type: fixed on a pin on the cross-arm. Used up to 33 kV.
  • Suspension / disc: discs hung in a string, used above 33 kV.
  • Strain: a disc string in the horizontal plane at dead-ends, sharp angles and river crossings.
  • Shackle: used on LV distribution lines and service drops.
  • Post: used in substations and compact lines.
  • Stay insulator: fitted in stay wires to keep the lower part dead.

Advantages of string insulators for HV lines

  1. Each disc is rated about 11 kV, so any voltage is handled by adding discs.
  2. Cheaper than pin insulators above 33 kV.
  3. A failed disc is replaced alone.
  4. The flexible string reduces mechanical stress.
  5. The conductor below the cross-arm is partly shielded from lightning.
  6. The voltage can be upgraded later by adding discs.

Configurations of string insulators

  1. I-string (single suspension): one vertical string. It is simple and cheap but swings freely, so it needs a longer cross-arm.
  2. Double I-string: two parallel strings for heavy conductors or important crossings (higher mechanical strength and reliability).
  3. V-string: two strings in a V shape. The conductor cannot swing, so the tower window and right-of-way are smaller. Used at 400 kV and above.
  4. Tension (strain) string: horizontal, single or double, at angle and dead-end towers.
  5. Y-string or inverted V, and long-rod / composite strings: used for compact EHV lines.
  6. Jumper (pilot) string: holds the jumper loop at tension towers to keep clearance.
  • 2071 Magh · 1+3 marks

State whether the following statement is TRUE or FALSE. Justify your answer with a brief explanation: On increasing the voltage level, the cost of pin type insulator varies linearly.

Answer

FALSE.

The cost of a pin insulator does not rise linearly with voltage. It rises much faster than linearly, roughly with a power of the voltage between 2 and 3.

  • A higher voltage needs more creepage and flashover distance. So the insulator needs more porcelain sheds and a greater height and diameter. Its volume, and therefore its weight and cost, grows roughly with the cube of its linear size.
  • Large porcelain pieces are hard to fire uniformly. Above about 25–33 kV the insulator must be made in two or three parts cemented together, which adds cost.
  • A taller insulator also needs a stronger and longer pin to resist the bending moment.

For this reason pin insulators become uneconomical above about 33 kV. Suspension (string) insulators are used instead, because their cost rises nearly linearly with voltage: one disc is added per about 11 kV.

  • 2071 Magh · 1+3 marks

State whether the following statement is TRUE or FALSE. Justify your answer with a brief explanation: For the same operating voltage level, lossless line and lossy line have same switching over voltage.

Answer

FALSE.

The switching overvoltage depends on how the travelling waves set up by switching are reflected and damped.

  • On a lossless line there is no resistance or conductance to absorb energy. Waves reflect fully at the open end and add up, so the overvoltage can approach 2 p.u. from one reflection, and 3 p.u. or more with trapped charge or reclosing.
  • On a lossy line, series resistance (including skin effect) and corona losses attenuate and distort the wavefront as it travels. Each reflection is smaller, so the peak switching overvoltage is lower.

Therefore, for the same operating voltage, the lossy line has a lower switching surge ratio than the lossless line. The lossless case is the worst case and is used for conservative insulation design.

  • 2070 Magh · 6 marks

Explain the basic configurations of insulating string in brief.

Answer

An insulator string can be arranged in different ways, depending on the voltage, the conductor load, the tower type and the allowable swing.

  1. Single I-string (suspension): one vertical string at each phase of a straight-line (suspension) tower.
    • Simple, light and cheap.
    • It swings freely with the wind (up to about 45°), so a longer cross-arm and a larger clearance are needed.
  2. Double I-string: two parallel I-strings joined by a yoke plate.
    • Twice the mechanical strength, and it stays secure if one string fails.
    • Used for bundled or heavy conductors and for crossings of roads, rivers and railways.
  3. V-string: two strings fixed at separate points on the cross-arm and meeting at the conductor.
    • The conductor cannot swing transversely, so the tower window, phase spacing and right-of-way are smaller.
    • Each leg carries part of the load, which gives higher strength.
    • Common at 400 kV and above, and for compact lines.
  4. Tension (strain) string: horizontal, single or double, at angle, section and dead-end towers. It takes the full conductor tension and usually has one or two more discs than the suspension string.
  5. Jumper (pilot) string: a short I-string at tension towers. It holds the jumper loop so that the loop keeps its clearance from the tower.
  6. Y-string or inverted-V, and long-rod / composite: used in EHV and compact designs to control swing and to save weight.
  I-string        V-string        Tension string
  ===+===       ===+---+===      tower|=[]=[]=[]=o== cond.
     |              \ /
    [ ]             [ ]
    [ ]              o  cond.
     o  cond.
  • 2069 Bhadra (old course) · 4 marks

For a given amount of power to be transmitted over a given distance, state whether the following statement is TRUE or FALSE. Justify your answer with a brief explanation: The major reason that V-type insulator string configurations are used for a transmission line of 400 kV and above is to improve the mechanical strength of the string.

Answer

FALSE.

The main reason for using V-strings at 400 kV and above is electrical clearance and economy, not mechanical strength.

  • An I-string at 400 kV is long, about 3.5–4 m. Under wind it can swing up to about 45°, and the swung conductor would come too close to the tower. A very long cross-arm and a large tower window would be needed.
  • In a V-string the two legs hold the conductor fixed in the transverse direction, so there is practically no swing. The required clearance is therefore met with:
    • shorter cross-arms and a smaller tower window,
    • closer phase spacing,
    • a narrower right-of-way, and lighter and cheaper towers.

The two legs of a V-string do share the load, so its mechanical strength is higher. That is only a side benefit. When higher strength alone is needed, a double I-string is used instead.

  • 2069 Bhadra (old course) · 4 marks

For a given amount of power to be transmitted over a given distance, state whether the following statement is TRUE or FALSE. Justify your answer with a brief explanation: For a double circuit transmission line, it is always preferred to have two earth wires.

Answer

FALSE.

The number of earth wires is decided by the shielding angle needed to protect all phase conductors from direct lightning strokes, usually 20°–30°. It is not decided only by whether the line is single or double circuit.

  • A double-circuit line normally has a vertical configuration: three cross-arms on each side of a slender tower. The overall width is small, so one earth wire at the tower peak can often give a shielding angle of 30° or less to the outermost conductors. Many 66 kV, 132 kV and 220 kV double-circuit lines therefore use a single earth wire.
  • Two earth wires are used on a double-circuit line only when the cross-arms are wide (EHV, 400 kV and above). They are also used where lightning activity is high and a smaller shielding angle (or negative shielding) is wanted, or where footing resistance is poor.

So two earth wires are not always preferred. The choice depends on the geometry, the voltage and the lightning level. Two wires are used where they are needed.

  • 2068 Bhadra (old course) · 5 marks

State whether the following statement is TRUE or FALSE and give reasons briefly: In general, for a single circuit transmission line of 400 kV, double earth wire is selected.

Answer

TRUE.

A 400 kV single-circuit line normally uses a horizontal conductor configuration. The three phases are on one level, with large phase spacing of about 11–12 m between adjacent phases.

  • The outer phases are far from the tower centre. One earth wire at the centre would have to be placed very high to give the needed shielding angle of about 20°–30°, which is uneconomical.
  • Two earth wires placed above or slightly outside the outer phases give good shielding to all three conductors with low peaks. The middle phase is shielded from both sides (mutual shielding).
  • At EHV a lightning tripout causes a heavy loss of transmitted power. The lower lightning outage rate from two wires is worth their cost.
  • Two earth wires also lower the surge impedance of the earth-wire system. This improves back-flashover performance, and the wires can carry OPGW communication.

So, in general, two earth wires are selected for a 400 kV single-circuit line, as in IS practice and in Nepal's 400 kV designs.

  • 2067 Mangsir (old course) · 1+3 marks

For a given amount of power to be transmitted over a given distance, state whether the following statement is TRUE or FALSE. Justify your answer with a brief explanation: The choice of single or double earth wire depends not only whether the single circuit or double circuit transmission line but also the line voltage.

Answer

TRUE.

The number of earth wires is chosen to give a safe shielding angle (about 20°–30°) over all phase conductors at a reasonable cost. The geometry that sets this angle depends on both the number of circuits and the voltage.

  • Number of circuits: a double-circuit vertical tower is tall and narrow, so one earth wire can often shield all six conductors. A single-circuit horizontal tower is wide, so it may need two wires.
  • Line voltage:
    • Higher voltage means larger phase spacing and wider cross-arms, so wider shielding is needed.
    • EHV lines carry more power, so a lightning outage costs more and a lower outage rate is justified.
    • Typical practice: 66–132 kV lines use a single earth wire, and 400 kV and above (single or double circuit) use two earth wires. At 220 kV it depends on the configuration.
  • Other factors are the isokeraunic level, the tower footing resistance and the need for OPGW.

So the choice depends on the voltage as well as on the single or double circuit arrangement.

Questions from Old Question Collection (EE 754) (IOE exam papers from 2067 to 2080 (2067-2069 papers from the older Transmission and Distribution Design course)). Answers are written for this site; check them against your class notes.

Chapter titles and hours from the IOE syllabus ↗