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Chapter 5 · 6 hours

System Earthing

IOE past exam questions

Past questions and answers

12 questions set from this chapter, 4 of them more than once. Most asked first.

  • Asked 5 times
  • 2073 Magh · 4 marks
  • 2074 Bhadra · 4 marks
  • 2075 Baisakh · 5 marks
  • 2075 Bhadra · 5 marks
  • 2080 Chaitra · 5 marks

A 33 kV, 3 phase, 50 Hz overhead line 60 km long has a capacitance to ground of each line equal to 0.015 μF per km. Determine the inductance and kVA rating of the Peterson coil (arc suppression coil).

Answer

A Peterson coil (arc suppression coil) is an iron-cored reactor connected between the system neutral and earth, tuned to the line-to-earth capacitance so that the earth-fault current is neutralised.

Principle: during an earth fault on one phase, the healthy phases send a capacitive current I_C = 3 V_ph ωC into the fault. The Peterson coil in the neutral sends an equal inductive current I_L = V_ph / (ωL) through the fault, which cancels I_C, so the arc goes out. For resonance:

I_L = I_C
V_ph/(ωL) = 3 V_ph ωC
        L = 1 / (3 ω² C)

Rating of the coil = V_ph × I_L (the coil sees phase voltage during the fault).

Data

  • Line voltage V_L = 33 kV, f = 50 Hz, ω = 2π × 50 = 314.16 rad/s
  • Capacitance per phase C = 0.015 μF/km × 60 km = 0.9 μF = 0.9 × 10⁻⁶ F

Inductance of the coil

L = 1 / (3 ω² C)
  = 1 / (3 × 314.16² × 0.9 × 10⁻⁶)
  = 1 / 0.26648
  = 3.753 H

kVA rating

V_ph = 33000/√3          = 19052.6 V
I_L  = I_C = 3 V_ph ω C
     = 3 × 19052.6 × 314.16 × 0.9×10⁻⁶
     = 16.16 A
Rating = V_ph × I_L
       = 19052.6 × 16.16
       = 307.9 kVA

Answer: Inductance of Peterson coil = 3.753 H; kVA rating ≈ 307.9 kVA (coil current 16.16 A at 19.05 kV).

  • Asked 4 times
  • 2070 Bhadra · 6 marks
  • 2073 Magh · 4 marks
  • 2074 Bhadra · 6 marks
  • 2075 Baisakh · 5 marks

Why is the neutral in a power system grounded? What are the methods of grounding a neutral? Explain them (at least one in detail).

Answer

Neutral grounding (earthing) means connecting the neutral point of a star-connected generator, transformer or system to earth, either directly or through an impedance. An ungrounded (isolated neutral) system has serious problems, so modern systems above a few kV are grounded.

Why the neutral is grounded

  1. Stable neutral voltage: the neutral is held at earth potential, so phase voltages to earth stay balanced and near V_ph under normal and fault conditions.
  2. No arcing grounds: in an ungrounded system, an earth fault draws capacitive current that produces intermittent arcs; repeated charging/discharging raises voltage to 5–6 times normal. Grounding removes this.
  3. Reduced overvoltage: healthy-phase voltages do not rise to √3 V_ph during an earth fault; insulation can be graded for lower levels, which saves cost.
  4. Reliable protection: a definite earth-fault current flows, which operates earth-fault relays quickly and selectively.
  5. Safety: equipment frames and persons are protected; static and induced charges are discharged.
  6. Lightning: surges are discharged to earth through arresters connected to a grounded neutral.

Methods of neutral grounding

  1. Solid (effective) grounding: neutral connected directly to earth with no impedance. Used at LV (400 V) and at high voltages (132 kV and above).
  2. Resistance grounding: a resistor in the neutral limits earth-fault current (e.g. to rated full-load current). Used for 3.3–33 kV systems and generators.
  3. Reactance grounding: a reactor in the neutral; used where solid grounding gives too high a fault current but X₀/X₁ must stay low.
  4. Resonant grounding (Peterson coil / arc suppression coil): a tuned reactor whose inductive current cancels capacitive fault current, L = 1/(3ω²C). Used on overhead lines with many transient faults (e.g. 33 kV).
  5. Grounding transformer grounding: where no neutral is available (delta system), a zig-zag or star-delta earthing transformer provides an artificial neutral.

Resistance grounding in detail

   Generator / transformer star
        a   b   c
         \  |  /
          \ | /
           (N)
            |
          [ R ]  neutral resistor
            |
           ===  earth
  • Working: during a fault on phase a, fault current flows from the phase to earth, through R, back to neutral. R limits it to a safe value, usually 1 to 2 times full-load current, while still enough to operate earth-fault relays.
  • The resistor also damps the oscillations that cause arcing grounds, and reduces burning of machine cores and the electromagnetic stress at the fault.
  • Advantages: controlled fault current, no arcing ground, lower damage, less interference with communication lines, reliable relaying.
  • Disadvantages: healthy-phase voltages rise above normal (between V_ph and √3 V_ph), so insulation must be slightly higher; resistor losses (I²R) during fault; costlier than solid grounding.
  • Use: 3.3 kV to 33 kV systems, and generator neutrals (high-resistance grounding through a distribution transformer).
  • Asked 2 times
  • 2076 Bhadra · 5 marks
  • 2080 Chaitra · 5 marks

What are the causes of over voltages on power system? List the merits and demerits of solid grounding.

Answer

An overvoltage is any voltage above the maximum normal operating voltage that stresses the insulation of power system equipment.

Causes of overvoltages

A. External causes

  1. Direct lightning strokes on lines, towers or substation equipment, giving surges of several MV.
  2. Indirect (induced) lightning: charged clouds near a line induce bound charges; when the cloud discharges elsewhere, the released charge travels as a surge.
  3. Electrostatic charging of lines by wind, dust, fog and changing atmospheric conditions.

B. Internal causes

  1. Switching surges: opening or closing of CBs; energising long lines, switching off unloaded lines and capacitor banks (restriking), current chopping when breaking transformer magnetising current.
  2. Arcing grounds in ungrounded systems, giving 5–6 times normal voltage.
  3. Insulation failure: sudden phase-to-ground fault produces travelling waves.
  4. Resonance: series LC resonance or ferro-resonance at certain frequencies or harmonics.
  5. Load rejection and Ferranti effect on long lightly loaded lines.

Merits of solid grounding

  1. Neutral is held at earth potential; healthy phases stay near phase voltage during an earth fault, so insulation can be graded to lower levels (cheaper equipment, e.g. 80% arresters).
  2. No arcing grounds, because the fault current is large and the arc is cleared by the CB.
  3. Large, definite fault current makes earth-fault relaying simple, fast and selective.
  4. Simple and cheapest method; no extra equipment in the neutral.
  5. Effective for single-phase loads connected phase-to-neutral (LV distribution).

Demerits of solid grounding

  1. Very high earth-fault current (can exceed three-phase fault current), which stresses equipment thermally and mechanically and needs CBs of high breaking capacity.
  2. Large fault current causes greater interference with nearby communication circuits.
  3. Every earth fault, even transient, causes a CB trip and loss of supply (poor continuity).
  4. Higher step and touch voltages near the fault; danger to persons.
  5. Heavy fault current can damage generator windings, so generators are usually not solidly grounded.

Solid grounding is used at LV (400/230 V) and for EHV systems (132 kV and above) where insulation saving outweighs these demerits.

  • Asked 2 times
  • 2071 Bhadra · 8 marks
  • 2077 Chaitra · 5 marks

Define causes of overvoltage and explain methods of overvoltage protection of electrical equipments.

Answer

An overvoltage is a voltage, normally a short-duration surge, above the highest normal operating voltage of the system. Insulation must be protected from it by shielding, by diverting the surge to earth and by controlling switching.

Causes of overvoltage

External:

  • Direct lightning stroke on conductors, towers or substation.
  • Induced lightning (bound charge released when a nearby cloud discharges).
  • Electrostatic build-up on lines due to wind, dust and atmospheric changes.

Internal:

  • Switching surges: closing/opening long lines, switching capacitor banks (restrike), current chopping of transformer magnetising current.
  • Arcing grounds in isolated-neutral systems.
  • Insulation failure (sudden earth fault).
  • Resonance and ferro-resonance.
  • Load rejection and Ferranti effect.

Methods of overvoltage protection

  1. Ground (earth/shield) wires: bare conductors run above the phase conductors and earthed at every tower. They intercept direct strokes and shield the line within a protective angle (about 30°). Low tower-footing resistance is needed to avoid back flashover.
  2. Lightning conductors / masts / shield wires over substations: tall rods or overhead wires earthed through a good earth mat protect outdoor equipment from direct strokes.
  3. Lightning (surge) arresters: connected between line and earth near equipment. They act as an open circuit at normal voltage, conduct at a surge to divert it to earth, then reseal. Types: rod gap, horn gap, expulsion type, valve type (SiC) and metal-oxide (ZnO) gapless arresters, now standard.
  4. Rod gaps and horn gaps: simple spark gaps across insulators; cheap back-up, but they cause a short circuit until the CB trips.
  5. Surge absorbers: series inductor with parallel resistor or a capacitor (ferranti absorber) that flatten the steep front of travelling waves before they reach transformer or machine windings.
  6. Neutral grounding: solid, resistance or Peterson coil grounding removes arcing ground overvoltages.
  7. Control of switching surges: CBs with pre-insertion/opening resistors, synchronised (point-on-wave) switching, restrike-free breakers (vacuum, SF6), shunt reactors to limit Ferranti rise.
  8. Insulation coordination: insulation levels (BIL) of equipment are chosen higher than the protective level of arresters by a safety margin (about 20–25%).
  9. Good earthing: low earth-mat and tower-footing resistance so that surge current is discharged safely.
 Line ---+-------------+---- Transformer
         |             |
      [Arrester]   [Surge absorber]
         |             |
        ===           ===
 Shield wire above line, earthed at each tower

A combination of shield wires against direct strokes and metal-oxide arresters near every important equipment gives the most reliable protection.

  • 2076 Bhadra · 5 marks

Calculate the inductance and kVA rating of the arc suppression coil of a 50 km long 33 kV, 50 Hz, 3 phase overhead transmission system having line to earth capacitance of each phase equal to 0.0125 μF.

Answer

An arc suppression coil (Peterson coil) is a reactor between neutral and earth whose inductive current cancels the capacitive earth-fault current, so the fault arc is self-extinguished.

Principle: during an earth fault on one phase, the healthy phases send a capacitive current I_C = 3 V_ph ωC into the fault. The Peterson coil in the neutral sends an equal inductive current I_L = V_ph / (ωL) through the fault, which cancels I_C, so the arc goes out. For resonance:

I_L = I_C
V_ph/(ωL) = 3 V_ph ωC
        L = 1 / (3 ω² C)

Rating of the coil = V_ph × I_L (the coil sees phase voltage during the fault).

Data and assumption

The length (50 km) is given, so the capacitance 0.0125 μF is taken as per km of each phase:

  • C = 0.0125 μF/km × 50 km = 0.625 μF per phase
  • V_L = 33 kV, f = 50 Hz, ω = 314.16 rad/s

Inductance

L = 1 / (3 ω² C)
  = 1 / (3 × 314.16² × 0.625 × 10⁻⁶)
  = 5.404 H

kVA rating

V_ph = 33000/√3              = 19052.6 V
I_L  = 3 V_ph ω C
     = 3 × 19052.6 × 314.16 × 0.625×10⁻⁶
     = 11.22 A
Rating = V_ph × I_L = 19052.6 × 11.22
       = 213.8 kVA

Answer: L = 5.404 H; kVA rating ≈ 213.8 kVA.

(If 0.0125 μF were the total capacitance of the line, the same formula gives L ≈ 270 H and about 4.3 kVA, which is unrealistically small; the per-km reading is the intended one.)

  • 2077 Chaitra · 6 marks

A 90 km long 66 kV, 50 Hz, 3 phase overhead line has capacitance to earth equal to 0.02 μF per km. What should be the inductance and kVA rating of the arc suppression coil?

Answer

The arc suppression (Peterson) coil in the neutral must supply an inductive fault current equal to the capacitive current fed by the two healthy phases during a line-to-ground fault.

Principle: during an earth fault on one phase, the healthy phases send a capacitive current I_C = 3 V_ph ωC into the fault. The Peterson coil in the neutral sends an equal inductive current I_L = V_ph / (ωL) through the fault, which cancels I_C, so the arc goes out. For resonance:

I_L = I_C
V_ph/(ωL) = 3 V_ph ωC
        L = 1 / (3 ω² C)

Rating of the coil = V_ph × I_L (the coil sees phase voltage during the fault).

Data

  • V_L = 66 kV, f = 50 Hz, ω = 314.16 rad/s
  • C = 0.02 μF/km × 90 km = 1.8 μF per phase

Inductance

L = 1 / (3 ω² C)
  = 1 / (3 × 314.16² × 1.8 × 10⁻⁶)
  = 1.876 H

kVA rating

V_ph = 66000/√3            = 38105.1 V
I_L  = 3 V_ph ω C
     = 3 × 38105.1 × 314.16 × 1.8×10⁻⁶
     = 64.64 A
Rating = V_ph × I_L = 38105.1 × 64.64
       = 2463.3 kVA ≈ 2.46 MVA

Answer: Inductance = 1.876 H; kVA rating ≈ 2463 kVA.

  • 2072 Magh · 4 marks

A transmission line has a capacitance of 0.1 microfarad per phase. Determine the inductance of Peterson coil to neutralize the effect of capacitance of (i) complete length of (ii) 90% length of the line. f = 50 Hz.

Answer

For a Peterson coil to neutralise the capacitive earth-fault current of a line of capacitance C per phase, ωL = 1/(3ωC), i.e. L = 1/(3ω²C). If only a fraction k of the line length is to be neutralised, the capacitance to be tuned is kC, so L = 1/(3ω²kC).

Data

  • C = 0.1 μF per phase (whole line), f = 50 Hz
  • ω = 2π × 50 = 314.16 rad/s, ω² = 98696

(i) Complete length (k = 1)

L = 1 / (3 × 98696 × 0.1×10⁻⁶)
  = 1 / 0.029609
  = 33.77 H

(ii) 90% of the length (k = 0.9)

C' = 0.9 × 0.1 μF = 0.09 μF
L  = 1 / (3 × 98696 × 0.09×10⁻⁶)
   = 1 / 0.026648
   = 37.53 H

Answer: (i) L = 33.77 H for the complete line; (ii) L = 37.53 H for 90% of the line.

A smaller part of the line gives less capacitive current, so a larger inductance (smaller inductive current) is needed. In practice the coil has tappings so it can be retuned when the network length changes.

  • 2079 Chaitra · 5 marks

In a 50 Hz overhead line, the capacitance of one line to earth was 1.5 μF. It was decided to use an earth fault neutralizer. Calculate the reactance to neutralize the capacitance of (i) 100% of length of the line (ii) 90% of length of line.

Answer

An earth fault neutraliser (Peterson coil) is a reactor connected between the neutral and earth. During a line-to-earth fault its inductive current equals the capacitive current 3V_phωC of the healthy phases, so:

V_ph / X_L = 3 V_ph ω C   →   X_L = 1 / (3 ω C)

For a fraction k of the line, C is replaced by kC.

Data

  • C = 1.5 μF (line to earth, per phase), f = 50 Hz, ω = 314.16 rad/s

(i) 100% of the line

X_L = 1 / (3 × 314.16 × 1.5×10⁻⁶)
    = 1 / 1.4137×10⁻³
    = 707.4 Ω
(L = X_L/ω = 2.252 H)

(ii) 90% of the line

C' = 0.9 × 1.5 = 1.35 μF
X_L = 1 / (3 × 314.16 × 1.35×10⁻⁶)
    = 1 / 1.2723×10⁻³
    = 786.0 Ω
(L = 2.502 H)

Answer: Reactance of the neutraliser = 707.4 Ω for 100% of the line and 786.0 Ω for 90% of the line.

  • 2079 Chaitra · 5 marks

Explain in detail about the need and different types of lightning protection scheme.

Answer

Lightning protection is the set of devices that intercept a lightning stroke or divert the resulting surge safely to earth so that system insulation, equipment and people are not damaged.

Need for lightning protection

  • A lightning stroke carries 10–200 kA with very steep front (1–10 μs), producing surges of several MV on lines, far above insulation levels (BIL).
  • Without protection it causes insulator flashover, puncture of transformer and machine windings, damage to substation equipment, prolonged outages, fires and danger to life.
  • Nepal's hilly terrain has high thunderstorm activity, so lines and substations are frequently hit.

Types of lightning protection schemes

  1. Earthing screen (for substations): a network of earthed copper/steel wires stretched over the substation, connected to the earth mat. Direct strokes hit the screen and go to earth. It does not protect against surges coming in along the lines.
  2. Overhead ground (shield) wires (for lines): earthed wires above phase conductors, bonded at each tower. They shield conductors against direct strokes within a 30°–45° protective angle; low tower-footing resistance (below about 10 Ω) avoids back flashover.
  3. Lightning conductors/masts: vertical rods or masts with good earthing protect buildings and outdoor yards within a protective cone.
  4. Lightning arresters (surge diverters): protect equipment from travelling waves that reach terminals. Types:
    • Rod gap: two rods set at a gap; cheap but no follow-current control.
    • Horn gap: arc rises on horn-shaped electrodes and extinguishes itself; with series resistor.
    • Multi-gap: series of small gaps with resistors.
    • Expulsion type: fibre tube; gases blow out the arc; used on distribution lines.
    • Valve type (SiC): series gaps with non-linear silicon carbide resistor blocks.
    • Metal oxide (ZnO) arrester: gapless, highly non-linear; now the standard at all voltages.
  5. Surge absorbers: reduce steepness of the incoming wave before it reaches transformer windings.
   Shield wire
  ===========  (earthed at tower)
  |  a  b  c |
  |  |  |  | |
  Tower    Line --+--> Substation
                  |
               [Arrester] --- earth mat

A complete scheme uses shield wires and earthing screens against direct strokes, and arresters at line entrances and at transformer terminals against travelling waves, all connected to a low-resistance earth.

  • 2078 Chaitra · 6 marks

Describe the protection of stations and sub-stations against direct lightning strokes.

Answer

Stations and substations contain costly equipment (transformers, breakers, busbars), so a direct lightning stroke must be intercepted before it reaches live parts and conducted to earth through a path of low impedance. The common methods are earthing screens, overhead shield wires and lightning masts, all bonded to the station earth mat.

1. Earthing screen

   ==+=====+=====+=====+==   earthed wire mesh
     |     |     |     |     above equipment
  [T/F]  [CB]  [Bus] [CT]
     |     |     |     |
  ======== earth mat ========
  • A network (mesh) of copper or galvanised steel wires is erected over the whole substation on the steel structures.
  • It is connected at many points to the low-resistance earth mat of the substation.
  • A direct stroke hits the screen instead of the equipment and the current flows to earth.
  • Advantages: simple and cheap; covers the whole yard.
  • Limitations: it does not protect against travelling waves entering through incoming lines; if the earth resistance is high, the potential of the screen rises and can cause back flashover to equipment.

2. Overhead ground wires extended over the station

  • The shield wires of incoming lines are carried over the substation and earthed on the gantries.
  • Equipment lies within the protective zone (shielding angle of about 30°–45° under the wire).

3. Lightning masts / spikes

  • Tall earthed steel masts or spikes on gantries (Franklin rods) placed so that all equipment is within the cone or rolling-sphere protection zone.
  • Each mast has its own earth connection bonded to the earth mat.

Design points

  1. Earth mat resistance should be low (typically below 1 Ω for large substations).
  2. Adequate clearance between shield wires/masts and live parts to avoid back flashover.
  3. Down conductors short and straight to reduce inductance.
  4. Protective zone checked by the protective-angle or rolling-sphere method.

Protection against direct strokes is always used with lightning arresters at line entrances and transformer terminals, which protect against surges travelling in along the lines.

  • 2072 Magh · 4 marks

What is chemical earthing? Write the factors affecting the earth resistance.

Answer

Chemical earthing is an earthing method in which the electrode (GI or copper-bonded rod/pipe) is surrounded by a conductive chemical compound (bentonite, graphite-based or other earth-enhancing compound) instead of the traditional salt and charcoal, to obtain a low and stable earth resistance with little maintenance.

Chemical earthing

  • A pit or bore is made and the electrode is placed in the centre.
  • The space around it is filled with earth-enhancing compound (e.g. bentonite clay mixed with graphite/carbon), which absorbs and holds moisture and has very low resistivity.
  • Advantages: low resistance even in rocky or dry soil, stable value through seasons, no need for regular watering, longer life with less corrosion, compact.
  • Used in telecom towers, substations, data centres and buildings where soil is poor.

Factors affecting earth resistance

  1. Soil resistivity: the main factor; clay and loam have low resistivity, sand, gravel and rock very high.
  2. Moisture content: resistivity falls sharply as moisture rises (dry soil can be 10 to 100 times worse).
  3. Salt (chemical) content: dissolved salts increase conductivity.
  4. Temperature: resistivity rises as temperature falls, and very sharply when soil freezes.
  5. Electrode size and depth: longer and deeper electrodes reach moist, stable soil layers and have lower resistance.
  6. Number and spacing of electrodes: electrodes in parallel, spaced at least twice their length apart, lower the combined resistance.
  7. Contact between electrode and soil: loose backfill, voids or corrosion raise resistance.
  8. Soil compaction and grain size: compact, fine-grained soil gives better contact.

Example: a 3 m copper-bonded rod in dry sandy soil may give 40–50 Ω; with chemical backfill the same rod can give below 5 Ω.

  • 2075 Bhadra · 5 marks

What are the factors affecting soil resistivity? How can we reduce earth resistance?

Answer

Soil resistivity (ρ, in Ω·m) is the resistance of a 1 m cube of soil between opposite faces. It is the main quantity deciding the resistance of an earth electrode, since for a rod electrode R ≈ (ρ / 2πl) · ln(4l/d), where l is the rod length and d its diameter.

Factors affecting soil resistivity

  1. Type of soil: clay and marshy soil 10–100 Ω·m; sand and gravel 500–1000 Ω·m or more; rock above 1000 Ω·m.
  2. Moisture content: water dissolves salts and carries current; resistivity falls very quickly up to about 15–20% moisture.
  3. Dissolved salts (chemical content): more salt means more ions and lower resistivity.
  4. Temperature: lower temperature increases resistivity; frozen soil has very high resistivity.
  5. Grain size and compactness: fine, closely packed grains give lower resistivity than loose coarse soil.
  6. Seasonal variation: resistivity is highest in the dry season, so earth resistance must be designed for that time.
  7. Depth (layering): different layers have different values; deeper layers are usually moister and more stable.

Methods to reduce earth resistance

  1. Increase the length (depth) of the electrode: reaches moist layers; resistance falls roughly inversely with length.
  2. Use multiple electrodes in parallel, spaced at least equal to (preferably twice) their length, or use an earth mat/grid in substations.
  3. Increase electrode diameter or use plate/strip electrodes, giving larger contact area (gain is smaller than increasing length).
  4. Chemical treatment: surround the electrode with salt and charcoal in layers, or with bentonite or other earth-enhancing compounds.
  5. Keep the soil moist: provide a watering pipe in the pit, especially in the dry season.
  6. Replace high-resistivity soil near the electrode with low-resistivity soil (clay, loam).
  7. Use counterpoise wires (buried horizontal conductors) for tower footings in rocky areas.
  8. Ensure good contact and corrosion-free joints: compact backfill and use copper or copper-bonded electrodes.

The aim is usually below 1 Ω for large substations, below 5 Ω for distribution substations and below 10 Ω for tower footings (typical values).

Questions from Old Question Collection (EE 651) (IOE EE 651 exam papers from 2070 Bhadra to 2080 Chaitra (16 papers)). Answers are written for this site; check them against your class notes.

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