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Chapter 6 · 12 hours

Circuit Breakers

IOE past exam questions

Past questions and answers

35 questions set from this chapter, 3 of them more than once. Most asked first.

  • Asked 2 times
  • 2071 Bhadra · 8 marks
  • 2076 Bhadra · 6 marks

A 50 Hz 3-phase alternator with grounded neutral has an inductance of 1.6 mH per phase and is connected to the bus-bars through a circuit breaker. The capacitance to earth of the circuit between the alternator and the circuit breaker is 0.0032 μF per phase. Due to a short on the bus-bars the breaker opens when the rms value of the current is 8000 A. Determine the following (i) frequency of oscillation (ii) Active recovery voltage (iii) time for maximum RRRV and (iv) maximum RRRV.

Answer

When the breaker interrupts a bus fault fed by an alternator, the voltage across its contacts rises as a transient oscillation of the system L and C. The fault current lags the voltage by nearly 90°, so at current zero the system voltage is at its peak V_m (the active recovery voltage). The restriking voltage is v = V_m(1 − cos ωₙt), which peaks at 2V_m.

The system voltage is not given, so it is found from the fault current. With only the alternator reactance limiting the current:

V_ph(rms) = I × X_L,   V_m = √2 × I × X_L

Data

L = 1.6 mH, C = 0.0032 μF, I = 8000 A rms, f = 50 Hz.

(i) Frequency of oscillation

√(LC) = √(1.6×10⁻³ × 0.0032×10⁻⁶)
      = 2.2627 × 10⁻⁶ s
f_n   = 1 / (2π √(LC))
      = 1 / (2π × 2.2627×10⁻⁶)
      = 70 337 Hz ≈ 70.34 kHz
ω_n   = 2π f_n = 4.4194 × 10⁵ rad/s

(ii) Active recovery voltage

X_L = 2π × 50 × 1.6×10⁻³ = 0.5027 Ω
V_m = √2 × 8000 × 0.5027
    = 5687 V ≈ 5.69 kV (peak, phase value)

(The maximum restriking voltage would be 2V_m = 11.37 kV.)

(iii) Time for maximum RRRV

v      = V_m (1 − cos ω_n t)
dv/dt  = V_m ω_n sin ω_n t
Maximum when ω_n t = π/2:
t = π / (2 ω_n) = 1 / (4 f_n)
  = 1 / (4 × 70 337)
  = 3.554 × 10⁻⁶ s = 3.55 μs

(iv) Maximum RRRV

RRRV_max = V_m ω_n
         = 5687 × 4.4194×10⁵
         = 2.513 × 10⁹ V/s
         = 2513 V/μs ≈ 2.51 kV/μs

Answer: (i) f_n = 70.34 kHz; (ii) active recovery voltage = 5.69 kV (peak); (iii) t = 3.55 μs; (iv) maximum RRRV = 2.51 kV/μs.

  • Asked 2 times
  • 2075 Baisakh · 8 marks
  • 2077 Chaitra · 6 marks

Describe the construction, principle of operation and application of vacuum circuit breaker.

Answer

A vacuum circuit breaker (VCB) is a breaker whose contacts open inside a sealed chamber evacuated to about 10⁻⁵ to 10⁻⁷ torr. Since vacuum has very high dielectric strength and no gas to ionise, the arc (formed only of metal vapour) is extinguished at the first current zero.

Construction

        Fixed terminal
            |
     +------+------+   ceramic/glass
     |   [fixed]   |   envelope
     |  ( shield ) |   metal vapour
     |   [moving]  |   condensing shield
     +------+------+
          bellows      metal bellows
            |
      Moving stem --> operating mechanism
  1. Vacuum interrupter (bottle): sealed envelope of glass or ceramic, with metal end plates.
  2. Fixed and moving contacts: made of copper–chromium (Cu-Cr) alloy, with spiral or axial-magnetic-field (AMF) shaped faces to keep the arc moving and diffuse.
  3. Metal bellows: stainless-steel bellows let the moving contact travel (about 8–20 mm) without breaking the vacuum.
  4. Vapour condensing shield: metal shield around the contacts on which the metal vapour condenses, protecting the envelope insulation.
  5. Operating mechanism: spring-charged (motor-wound) mechanism with closing and trip coils.
  6. Frame, insulators and terminals.

Principle of operation

  1. When the trip coil is energised, the mechanism separates the contacts in vacuum.
  2. At separation, the last point of contact is heated and metal vapour is released; the arc burns in this vapour (there is no gas).
  3. Because of contact shape, the arc is driven around or spread over the contact face, keeping it diffuse and limiting erosion.
  4. As the current approaches zero, vapour production falls; the vapour condenses rapidly on the contacts and shield within microseconds.
  5. At current zero the gap recovers its very high dielectric strength within a few microseconds (much faster than in oil or air), so the arc does not restrike. Interruption occurs at the first current zero.

Applications

  • Medium voltage systems: 3.3 kV to 33 kV (mainly 11 kV and 33 kV indoor/outdoor switchgear in Nepal's distribution substations).
  • Capacitor bank and reactor switching (restrike-free).
  • Motor switching, arc furnace switching (frequent operation).
  • Railway traction, mining and ring main units.

Advantages (brief)

  • Compact, silent, no fire or explosion risk, no gas/oil maintenance.
  • Very long contact life; suitable for frequent operation.
  • Fast recovery; handles high RRRV.
  • Limitation: current chopping can produce overvoltage on inductive loads (reduced by Cu-Cr contacts or surge suppressors); economic mainly up to about 36 kV.
  • Asked 2 times
  • 2071 Bhadra · 8 marks
  • 2073 Bhadra · 6 marks

Describe with a neat sketch the construction and principle of operation of a SF6 circuit breaker.

Answer

An SF6 circuit breaker is a breaker in which the contacts separate in sulphur hexafluoride (SF6) gas under pressure. SF6 is electronegative: it captures free electrons to form heavy negative ions, which rapidly de-ionises the arc path and restores dielectric strength after current zero.

Construction

        Fixed contact (hollow)
     +---------[  ]---------+
     |   SF6 gas at about    |
     |   2.8 to 5 bar        |
     |   arc  ->  nozzle     |
     |       [moving]        |
     |   insulating nozzle   |
     +----------+------------+
                |
  HP SF6 tank --valve-- LP tank --compressor
                |
       operating mechanism

Main parts:

  1. Interrupter unit: fixed (hollow) contact and moving contact with arcing probes/contacts and current-carrying main contacts; an insulating (PTFE) nozzle directs gas flow.
  2. Gas system: in the two-pressure design, a high-pressure reservoir (about 14 bar) and low-pressure chamber (about 2.8 bar), with compressor, filters, valves and gas monitoring (density switches). Modern puffer breakers use a single pressure (about 5–6 bar).
  3. Arc-quenching chamber of porcelain or metal enclosure (live tank, dead tank or GIS).
  4. Operating mechanism: spring, hydraulic or pneumatic.
  5. Alarms and lockouts for low gas pressure, and absorbers (activated alumina) to take up decomposition products.

Principle of operation

  1. In the closed position the contacts are surrounded by SF6 at low pressure.
  2. On a trip signal the moving contact is pulled away; an arc forms between the arcing contacts.
  3. At the same time a valve opens (two-pressure type) or the puffer piston compresses gas, and SF6 is blown axially along the arc through the nozzle.
  4. SF6 molecules capture the free electrons and form heavy, slow negative ions (SF6⁻, SF5⁻), so the conductivity of the arc column falls sharply.
  5. The gas also cools the arc (high thermal conductivity near arc temperatures).
  6. At current zero the arc path de-ionises quickly and the dielectric strength rises faster than the restriking voltage, so the arc is extinguished.
  7. The gas returns to the low-pressure side and is recompressed and reused; it recombines after the arc, so little is lost.

Applications

  • 33 kV to 765 kV outdoor switchyards and gas-insulated substations (GIS).
  • 132 kV and 220 kV substations in Nepal's grid widely use SF6 breakers.

Advantages

  • Excellent arc quenching and dielectric strength (about 2.5–3 times air); silent operation.
  • Compact, closed system, no moisture or fire risk; low maintenance.
  • No current chopping problem; suitable for high RRRV.
  • Disadvantage: SF6 is a strong greenhouse gas and needs careful handling; costly, needs leak monitoring.
  • 2070 Bhadra · 2+2+2+2 marks

A circuit breaker is rated as 2500 A, 1500 MVA, 33 KV, 3 sec, 3 phase OCB. Determine (i) the rated symmetrical breaking current (ii) rated making current (iii) short time rating and (iv) rated service voltage.

Answer

The ratings of a CB follow from its breaking MVA and rated voltage.

Definitions and formulas used

  • Rated symmetrical breaking current I_b = rated breaking MVA / (√3 × rated line kV), rms.
  • Rated making current = 1.8 × √2 × I_b = 2.55 × I_b (peak), since the first major peak of an asymmetrical current can be 1.8 times the peak of the symmetrical current.
  • Short time rating = the current the CB can carry while closed for the stated time without damage; it equals I_b for the given duration.

Given

2500 A, 1500 MVA, 33 kV, 3 s, 3-phase oil CB.

(i) Rated symmetrical breaking current

I_b = MVA / (√3 × kV)
    = 1500 / (√3 × 33)
    = 26.24 kA (rms)

(ii) Rated making current

I_m = 2.55 × I_b
    = 2.55 × 26.24
    = 66.92 kA (peak)

(iii) Short time rating

= 26.24 kA (rms) for 3 seconds

(iv) Rated service voltage

= 33 kV (rms, line to line)

Answer: (i) 26.24 kA rms; (ii) 66.92 kA peak; (iii) 26.24 kA for 3 s; (iv) 33 kV rms (line). Rated normal current is 2500 A.

  • 2073 Bhadra · 3+5 marks

Explain what do you understand by: i) Rated symmetrical breaking current ii) Making capacity iii) Short time current rating. Obtain their value for a three phase CB rated as 500 A, 500 MVA, 11 kV, 3 sec.

Answer

i) Rated symmetrical breaking current

The rms value of the AC (symmetrical) component of the short-circuit current that the CB can interrupt at its rated voltage, at the instant of contact separation. It is the basis of the breaking capacity:

Breaking capacity (MVA) = √3 × V(kV) × I_b(kA)

ii) Making capacity

The peak value of current (including the DC offset) during the first cycle that the CB can safely make when it closes on a short circuit. The contacts must withstand the large electromagnetic forces at the first major peak, which can be 1.8 times the symmetrical peak:

I_making = 1.8 × √2 × I_b = 2.55 × I_b (peak)

iii) Short time current rating

The rms current the CB can carry in the closed position for a stated short time (1 s or 3 s) without thermal or mechanical damage, while a downstream breaker clears the fault. It is usually equal to the rated symmetrical breaking current for the given time.

Calculation for 500 A, 500 MVA, 11 kV, 3 s CB

I_b = 500 / (√3 × 11)
    = 26.24 kA (rms)

I_making = 2.55 × 26.24
         = 66.92 kA (peak)

Short time rating = 26.24 kA (rms) for 3 s

Answer: Rated symmetrical breaking current = 26.24 kA rms; making capacity = 66.92 kA peak; short time rating = 26.24 kA for 3 s. (Rated normal current = 500 A, rated voltage = 11 kV.)

Note that the 33 kV, 1500 MVA breaker and this 11 kV, 500 MVA breaker have the same breaking current, because MVA and voltage are in the same ratio.

  • 2077 Chaitra · 4 marks

Determine the rated symmetrical breaking current, making current and short time rating of a circuit breaker with following parameters: 33 kV, 1500 A, 3000 MVA, 1 sec.

Answer

The breaking, making and short-time currents of a CB are fixed by its rated MVA and voltage.

Definitions and formulas used

  • Rated symmetrical breaking current I_b = rated breaking MVA / (√3 × rated line kV), rms.
  • Rated making current = 1.8 × √2 × I_b = 2.55 × I_b (peak), since the first major peak of an asymmetrical current can be 1.8 times the peak of the symmetrical current.
  • Short time rating = the current the CB can carry while closed for the stated time without damage; it equals I_b for the given duration.

Given

33 kV, 1500 A, 3000 MVA, 1 s.

I_b = 3000 / (√3 × 33)
    = 52.49 kA (rms)

I_making = 2.55 × 52.49
         = 133.84 kA (peak)

Short time rating = 52.49 kA (rms) for 1 s

Answer: Rated symmetrical breaking current = 52.49 kA rms; making current = 133.8 kA peak; short time rating = 52.49 kA for 1 s.

  • 2078 Chaitra · 4 marks

Calculate breaking capacity in MVA and making capacity in kA of a 3 phase circuit breaker with following parameters: Rated current = 800 A; Rated voltage = 33 kV; Ibsym = 25 kA.

Answer

Breaking capacity is the MVA corresponding to the symmetrical breaking current at rated voltage, and making capacity is the peak current the CB can close onto, taken as 2.55 times the symmetrical breaking current.

Given

Rated current = 800 A (normal current, not used for these values), V = 33 kV, I_b(sym) = 25 kA.

Breaking capacity

Breaking MVA = √3 × V × I_b
             = √3 × 33 kV × 25 kA
             = 1428.9 MVA

Making capacity

I_making = 1.8 × √2 × I_b = 2.55 × I_b
         = 2.55 × 25
         = 63.75 kA (peak)

Answer: Breaking capacity ≈ 1429 MVA; making capacity = 63.75 kA (peak).

  • 2075 Baisakh · 4 marks

Explain the terms related to Circuit Breaker: i) Making Capacity ii) Breaking Capacity iii) Operation Duty iv) Rated Current.

Answer

These terms define the ratings of a circuit breaker as per IEC 62271-100 / IS 13118.

i) Making capacity

The peak value of current that the CB can make (close on to) at rated voltage when closing on a short circuit, including the DC offset of the first cycle. The contacts must withstand the large electromagnetic force at this first peak.

Making current = 1.8 × √2 × I_b = 2.55 × I_b (peak)

ii) Breaking capacity

The largest current the CB can break at rated voltage under specified conditions. Expressed as the rms symmetrical breaking current I_b (kA) or as MVA:

Breaking capacity (MVA) = √3 × V(kV) × I_b(kA)

Asymmetrical breaking current includes the DC component at contact separation.

iii) Operating duty

The sequence of operations the CB must perform at rated breaking current, with given time intervals. Standard duties:

  • O – t – CO – t′ – CO, e.g. O – 3 min – CO – 3 min – CO (non-auto-reclose duty)
  • O – 0.3 s – CO – 3 min – CO (rapid auto-reclosing duty)

Here O = open, CO = close followed immediately by open, t = dead time. The CB must keep its rated breaking capacity throughout this sequence.

iv) Rated current

The rms value of normal current the CB can carry continuously at rated frequency without the temperature of any part exceeding its permitted limit. Standard values: 400, 630, 800, 1250, 1600, 2000, 2500, 3150 A, etc.

Example: a 33 kV CB rated 1250 A, 25 kA, duty O–0.3 s–CO–3 min–CO has breaking capacity √3 × 33 × 25 = 1429 MVA and making capacity 2.55 × 25 = 63.75 kA peak.

  • 2074 Bhadra · 8 marks

A 50 Hz, three-phase synchronous generator has an inductance per phase of 1.7 mH and its neutral is grounded. It feeds a line through a circuit breaker. The total stray capacitance to ground of the generator and circuit breaker is 0.0025 μF. A fault occurs just beyond the circuit breaker, which opens when the symmetrical short circuit current is 7500 A (rms). Ignoring the first pole to clear factor, determine the following: (i) Natural frequency of oscillations (ii) Peak value of TRV (iii) Time at which peak value of TRV occurs (iv) Maximum rate of rise of TRV (v) Time at which the maximum in part (iv) occurs.

Answer

The transient recovery voltage (TRV) is the oscillatory voltage that appears across the CB contacts just after the arc is extinguished at current zero.

Method

At fault current zero the system voltage is at its peak V_m (current lags by about 90°). The voltage across the opening contacts then follows the L–C transient (resistance neglected):

v(t) = V_m (1 − cos ω_n t),   ω_n = 1/√(LC)
peak  = 2 V_m      at ω_n t = π     (t = 1/(2 f_n))
dv/dt = V_m ω_n sin ω_n t
max   = V_m ω_n    at ω_n t = π/2   (t = 1/(4 f_n))

The source voltage is not given, so it is obtained from the fault current limited by the generator reactance: V_m = √2 × I × ωL (phase peak).

Data

L = 1.7 mH, C = 0.0025 μF, I = 7500 A rms, f = 50 Hz, neutral grounded (first-pole-to-clear factor ignored).

X_L = 2π × 50 × 1.7×10⁻³ = 0.5341 Ω
V_m = √2 × 7500 × 0.5341 = 5665 V (peak)
√(LC) = √(1.7×10⁻³ × 0.0025×10⁻⁶) = 2.0616×10⁻⁶ s

(i) Natural frequency

f_n = 1 / (2π × 2.0616×10⁻⁶)
    = 77 201 Hz ≈ 77.2 kHz
ω_n = 4.8507 × 10⁵ rad/s

(ii) Peak value of TRV

V_peak = 2 V_m = 2 × 5665 = 11 329 V ≈ 11.33 kV

(iii) Time of peak TRV

t = π / ω_n = 1 / (2 f_n)
  = 1 / (2 × 77 201) = 6.477 × 10⁻⁶ s = 6.48 μs

(iv) Maximum rate of rise of TRV

RRRV_max = V_m ω_n
         = 5665 × 4.8507×10⁵
         = 2.748 × 10⁹ V/s ≈ 2.75 kV/μs

(v) Time of maximum RRRV

t = π / (2 ω_n) = 1 / (4 f_n)
  = 3.238 × 10⁻⁶ s = 3.24 μs

Answer: (i) 77.2 kHz; (ii) 11.33 kV; (iii) 6.48 μs; (iv) 2.75 kV/μs; (v) 3.24 μs.

  • 2080 Chaitra · 8 marks

A 3 phase 50 Hz alternator with grounded neutral is connected to a bus bar through a circuit breaker. Due to short circuit on the busbar the circuit breaker trips when the current is 7000 A (rms) as shown in figure. Determine (i) Active recovery voltage (ii) Frequency of oscillation (iii) Time to reach the peak restriking voltage (iv) Maximum RRRV. [Figure: Alternator A with grounded neutral, series inductance L = 1.5 mH/phase, shunt capacitance C = 0.0025 μF/phase to ground, then the circuit breaker and a fault on the busbar.]

Answer

When the CB interrupts the busbar fault, the voltage across its contacts rises from zero towards the system voltage as an L–C oscillation (the restriking voltage). Its steady value after the transient is the recovery voltage.

  A (alternator)    L=1.5 mH      CB
 (~)---------------^^^^^^----+---o/ o---+ bus
  |                          |          |  fault
 earth                    C=0.0025 μF   X
                             |
                            earth

Method

At fault current zero the system voltage is at its peak V_m (current lags by about 90°). The voltage across the opening contacts then follows the L–C transient (resistance neglected):

v(t) = V_m (1 − cos ω_n t),   ω_n = 1/√(LC)
peak  = 2 V_m      at ω_n t = π     (t = 1/(2 f_n))
dv/dt = V_m ω_n sin ω_n t
max   = V_m ω_n    at ω_n t = π/2   (t = 1/(4 f_n))

The source voltage is not given, so it is obtained from the fault current limited by the generator reactance: V_m = √2 × I × ωL (phase peak).

Data

L = 1.5 mH/phase, C = 0.0025 μF/phase, I = 7000 A rms, f = 50 Hz.

(i) Active recovery voltage

X_L = 2π × 50 × 1.5×10⁻³ = 0.4712 Ω
V_m = √2 × 7000 × 0.4712
    = 4665 V ≈ 4.67 kV (peak, per phase)

(ii) Frequency of oscillation

√(LC) = √(1.5×10⁻³ × 0.0025×10⁻⁶) = 1.9365×10⁻⁶ s
f_n   = 1 / (2π × 1.9365×10⁻⁶)
      = 82 187 Hz ≈ 82.19 kHz
ω_n   = 5.164 × 10⁵ rad/s

(iii) Time to reach peak restriking voltage

t = 1 / (2 f_n) = 1 / (2 × 82 187)
  = 6.084 × 10⁻⁶ s = 6.08 μs
Peak restriking voltage = 2 V_m = 9.33 kV

(iv) Maximum RRRV

RRRV_max = V_m ω_n
         = 4665 × 5.164×10⁵
         = 2.409 × 10⁹ V/s ≈ 2.41 kV/μs
(occurs at t = 1/(4 f_n) = 3.04 μs)

Answer: (i) active recovery voltage = 4.67 kV (peak); (ii) f_n = 82.19 kHz; (iii) t = 6.08 μs; (iv) maximum RRRV = 2.41 kV/μs.

  • 2072 Asoj · 8 marks

In a short circuit test on a 132 kV three phase system, the breaker gave the following results: pf of the fault is 0.4, recovery voltage is 0.9 of full line value and the natural frequency is 16 kHz. Determine the rate of rise of restriking voltage in following cases: i) Assume that symmetrical fault is grounded. ii) Assume that symmetrical fault is not grounded.

Answer

The rate of rise of restriking voltage (RRRV) is the slope of the transient voltage across the CB contacts after current zero; it decides whether the arc re-ignites.

Method

  • Peak phase voltage: V_m = √2 × V_L/√3.
  • At current zero the instantaneous system voltage is V_m sin φ (fault current lags by φ, cos φ = fault p.f.).
  • Active recovery voltage: V_r = k × sin φ × V_m, where k is the recovery voltage factor (0.9 of full line value).
  • For an ungrounded symmetrical fault the first pole to clear sees 1.5 times this (first-pole-to-clear factor 1.5).
  • Restriking voltage peak = 2 V_r, reached at t = 1/(2f_n).
  • RRRV (average up to the peak, the usual textbook value) = 2V_r / t = 4 f_n V_r. The maximum instantaneous rate is ωₙV_r.

Data

V_L = 132 kV, cos φ = 0.4, k = 0.9, f_n = 16 kHz.

sin φ = √(1 − 0.4²) = 0.9165
V_m   = √2 × 132/√3 = 107.78 kV
t     = 1/(2 × 16000) = 31.25 μs

i) Fault grounded

V_r   = 0.9 × 0.9165 × 107.78 = 88.90 kV
Peak  = 2 × 88.90             = 177.80 kV
RRRV  = 177.80 kV / 31.25 μs  = 5.69 kV/μs
(max instantaneous = ω_n V_r
 = 2π×16000×88.90 kV = 8.94 kV/μs)

ii) Fault not grounded (first-pole factor 1.5)

V_r   = 1.5 × 88.90           = 133.35 kV
Peak  = 2 × 133.35            = 266.71 kV
RRRV  = 266.71 kV / 31.25 μs  = 8.53 kV/μs
(max instantaneous = 13.41 kV/μs)

Answer: RRRV ≈ 5.69 kV/μs with grounded fault and ≈ 8.53 kV/μs with ungrounded fault (average up to first peak).

An ungrounded fault therefore puts 50% more stress on the breaker.

  • 2078 Chaitra · 6 marks

In a short circuit test on a 220 kV 3-phase system, the breaker gave the following results: p.f. of the fault 0.45, recovery voltage 0.9 of full line value; the breaking current is symmetrical and the restriking transient had a natural frequency of 18 kHz. Determine the rate of rise of restriking voltage. Assume that the fault is grounded.

Answer

The RRRV is the rate at which the voltage across the CB contacts rises after the arc is extinguished at current zero.

Method

  • Peak phase voltage: V_m = √2 × V_L/√3.
  • At current zero the instantaneous system voltage is V_m sin φ (fault current lags by φ, cos φ = fault p.f.).
  • Active recovery voltage: V_r = k × sin φ × V_m, where k is the recovery voltage factor (0.9 of full line value).
  • For an ungrounded symmetrical fault the first pole to clear sees 1.5 times this (first-pole-to-clear factor 1.5).
  • Restriking voltage peak = 2 V_r, reached at t = 1/(2f_n).
  • RRRV (average up to the peak, the usual textbook value) = 2V_r / t = 4 f_n V_r. The maximum instantaneous rate is ωₙV_r.

Data

V_L = 220 kV, cos φ = 0.45, k = 0.9, f_n = 18 kHz, fault grounded (factor 1).

sin φ = √(1 − 0.45²)          = 0.8930
V_m   = √2 × 220/√3           = 179.63 kV
V_r   = 0.9 × 0.8930 × 179.63 = 144.37 kV
Peak restriking = 2 V_r       = 288.75 kV
t = 1/(2 f_n) = 1/(2×18000)   = 27.78 μs
RRRV = 288.75 kV / 27.78 μs   = 10.39 kV/μs

Maximum instantaneous rate (at t = 1/(4f_n) = 13.9 μs):

ω_n V_r = 2π × 18000 × 144.37 kV
        = 16.33 kV/μs

Answer: RRRV ≈ 10.39 kV/μs (average up to first peak of 288.75 kV); maximum instantaneous RRRV ≈ 16.33 kV/μs.

  • 2072 Magh · 8 marks

In a 220 kV system the reactance and capacitance up to the location of circuit breaker is 8 Ω and 0.025 μF, respectively. A resistance of 600 Ω is connected across the contacts of the circuit breaker. Determine the following: (i) Natural frequency of oscillation (ii) Damped frequency of oscillation (iii) Critical value of resistance which will give no transient oscillation and (iv) The value of resistance which will give damped frequency of oscillation one-fourth of the natural frequency of oscillation.

Answer

Resistance switching means connecting a resistor across the CB contacts (or arc) so that the restriking voltage transient is damped, its frequency is lowered and RRRV is reduced.

Formulas (resistance R across the contacts, parallel R–L–C circuit)

Natural frequency  f_n = 1 / (2π √(LC))
Damped frequency   f_d = (1/2π) √(1/LC − 1/(4R²C²))
Critical resistance R_c = ½ √(L/C)
   (R ≤ R_c gives no oscillation)
For f_d = f_n / k:
   1/(4R²C²) = (1/LC)(1 − 1/k²)
   R = R_c / √(1 − 1/k²)

Inductance from reactance: L = X/(2π × 50).

Data

X = 8 Ω, C = 0.025 μF, R = 600 Ω, f = 50 Hz.

L = 8 / 314.16 = 25.46 mH

(i) Natural frequency

1/LC = 1/(25.46×10⁻³ × 0.025×10⁻⁶) = 1.5708×10⁹
f_n  = √(1.5708×10⁹) / 2π
     = 6308 Hz ≈ 6.31 kHz

(ii) Damped frequency

1/(4R²C²) = 1/(4 × 600² × (0.025×10⁻⁶)²) = 1.1111×10⁹
f_d = √(1.5708×10⁹ − 1.1111×10⁹) / 2π
    = 3412 Hz ≈ 3.41 kHz

(iii) Critical resistance

R_c = ½ √(25.46×10⁻³ / 0.025×10⁻⁶)
    = ½ × 1009.25 = 504.6 Ω

(iv) Resistance for f_d = f_n/4

R = R_c / √(1 − 1/16) = 504.6 / 0.9682
  = 521.2 Ω
Check: f_d = 1577 Hz = 6308/4 ✓

Answer: (i) f_n = 6.31 kHz; (ii) f_d = 3.41 kHz; (iii) R_c = 504.6 Ω; (iv) R = 521.2 Ω.

  • 2073 Magh · 8 marks

For a 132 kV system, the reactance and capacitance up to the location of circuit breaker is 5 Ω and 0.035 μF, respectively. A resistance of 350 Ω is connected across the contacts of the circuit breaker. Determine the (a) Natural frequency of oscillation (b) Damped frequency of oscillation (c) Critical value of resistance which will give no transient oscillation and (d) The value of resistance which will give damped frequency of oscillation one-third of the natural frequency of oscillation.

Answer

A resistance across the CB contacts damps the restriking voltage oscillation; if it is at or below the critical value, the voltage rises without oscillation.

Formulas (resistance R across the contacts, parallel R–L–C circuit)

Natural frequency  f_n = 1 / (2π √(LC))
Damped frequency   f_d = (1/2π) √(1/LC − 1/(4R²C²))
Critical resistance R_c = ½ √(L/C)
   (R ≤ R_c gives no oscillation)
For f_d = f_n / k:
   1/(4R²C²) = (1/LC)(1 − 1/k²)
   R = R_c / √(1 − 1/k²)

Inductance from reactance: L = X/(2π × 50).

Data

X = 5 Ω, C = 0.035 μF, R = 350 Ω, f = 50 Hz.

L = 5 / 314.16 = 15.915 mH

(a) Natural frequency

1/LC = 1/(15.915×10⁻³ × 0.035×10⁻⁶) = 1.7952×10⁹
f_n  = √(1.7952×10⁹) / 2π
     = 6743 Hz ≈ 6.74 kHz

(b) Damped frequency

1/(4R²C²) = 1/(4 × 350² × (0.035×10⁻⁶)²) = 1.6660×10⁹
f_d = √(1.7952×10⁹ − 1.6660×10⁹) / 2π
    = 1809 Hz ≈ 1.81 kHz

(c) Critical resistance

R_c = ½ √(15.915×10⁻³ / 0.035×10⁻⁶)
    = ½ × 674.34 = 337.2 Ω

Since 350 Ω > 337.2 Ω, the given resistor still allows a (heavily damped) oscillation, as found in (b).

(d) Resistance for f_d = f_n/3

R = R_c / √(1 − 1/9) = 337.2 / 0.9428
  = 357.6 Ω
Check: f_d = 2248 Hz = 6743/3 ✓

Answer: (a) f_n = 6.74 kHz; (b) f_d = 1.81 kHz; (c) R_c = 337.2 Ω; (d) R = 357.6 Ω.

  • 2075 Bhadra · 4 marks

The system inductance per phase of a system is 8 mH and the frequency of transient oscillation is 10.273 kHz. If a 50 Hz, 6.6 kV generator is connected to that system, calculate: i) Max voltage across the contacts of the CB at an instant when it passes through zero. ii) System capacitance per phase. iii) Average rate of rise of voltage up to the first peak of oscillation, neglect resistance.

Answer

When the CB current passes through zero, the voltage across its contacts oscillates at the natural frequency of the system and can reach twice the peak system voltage.

Method (resistance neglected)

  • Peak phase voltage V_m = √2 × V_L/√3.
  • Restriking voltage v = V_m(1 − cos ωₙt); maximum across contacts = 2V_m.
  • Time to first peak t = π√(LC) = 1/(2f_n).
  • Average rate of rise up to first peak = 2V_m / t.

Data

L = 8 mH, f_n = 10.273 kHz, V_L = 6.6 kV, 50 Hz.

i) Maximum voltage across contacts

V_m = √2 × 6600/√3 = 5389 V
v_max = 2 V_m = 10 778 V ≈ 10.78 kV

ii) System capacitance per phase

f_n = 1/(2π√(LC))  →  C = 1/(4π² f_n² L)
C = 1/(4π² × 10273² × 8×10⁻³)
  = 3.0 × 10⁻⁸ F = 0.03 μF

iii) Average rate of rise up to first peak

t = 1/(2 f_n) = 1/(2 × 10273) = 48.67 μs
Average RRRV = 10 778 V / 48.67 μs
             = 221.4 V/μs ≈ 0.221 kV/μs

Answer: i) 10.78 kV; ii) C = 0.03 μF per phase; iii) 0.221 kV/μs (2.214 × 10⁸ V/s).

  • 2079 Chaitra · 6 marks

A 50 Hz, 11 kV generator is connected to a power system. The system inductance and capacitance per phase are 10 mH and 0.02 μF respectively. Calculate: i) Maximum voltage across the contacts of the CB at an instant when it passes through zero. ii) Frequency of transient oscillation. iii) Average rate of rise of voltage up to the first peak of oscillation, neglect the resistance.

Answer

The restriking voltage across the CB contacts after current zero is an L–C oscillation that rises to twice the peak phase voltage when resistance is neglected.

Method (resistance neglected)

  • Peak phase voltage V_m = √2 × V_L/√3.
  • Restriking voltage v = V_m(1 − cos ωₙt); maximum across contacts = 2V_m.
  • Time to first peak t = π√(LC) = 1/(2f_n).
  • Average rate of rise up to first peak = 2V_m / t.

Data

V_L = 11 kV, 50 Hz, L = 10 mH, C = 0.02 μF per phase.

i) Maximum voltage across the contacts

V_m = √2 × 11000/√3 = 8981 V
v_max = 2 V_m = 17 963 V ≈ 17.96 kV

ii) Frequency of transient oscillation

√(LC) = √(10×10⁻³ × 0.02×10⁻⁶) = 1.4142×10⁻⁵ s
f_n = 1/(2π × 1.4142×10⁻⁵)
    = 11 254 Hz ≈ 11.25 kHz

iii) Average rate of rise up to first peak

t = 1/(2 f_n) = 44.43 μs
Average RRRV = 17 963 V / 44.43 μs
             = 404.3 V/μs ≈ 0.404 kV/μs

Answer: i) 17.96 kV; ii) 11.25 kHz; iii) 0.404 kV/μs (4.04 × 10⁸ V/s).

  • 2079 Chaitra · 4 marks

What is the difference between restriking voltage and recovery voltage?

Answer

Restriking voltage is the high-frequency transient voltage that appears across the CB contacts immediately after the arc is extinguished at current zero. Recovery voltage is the normal-frequency (50 Hz) steady voltage that appears across the contacts after the transient has died out.

 v |      restriking (transient)
   |     /\  /\
   |    /  \/  \/~~~~~~ recovery voltage
   |   /                 (50 Hz, steady)
   |__/__________________ t
     current zero
PointRestriking voltageRecovery voltage
NatureTransient, oscillatorySteady, power frequency
FrequencyNatural frequency 1/(2π√LC), kHz rangeSystem frequency (50 Hz)
TimeFirst few μs to ms after current zeroAfter the transient decays
Peak valueUp to 2 V_m (resistance neglected)About V_m sin φ
Depends onSystem L, C, R and p.f.System voltage and p.f. of fault
EffectIts RRRV decides whether the arc restrikesDecides final dielectric stress on gap

In short, the restriking voltage is the oscillation superimposed on the recovery voltage; the breaker succeeds only if the dielectric strength of the gap rises faster than the restriking voltage.

  • 2077 Chaitra · 6 marks

Define following terms: (i) Recovery voltage (ii) Re-striking voltage (iii) RRRV. Also, derive an expression for the restriking voltage in terms of system inductance and capacitance.

Answer

(i) Recovery voltage

The normal-frequency (50 Hz) rms voltage that appears across the CB contacts after the final arc extinction and after the high-frequency transients have died out. Its peak is about V_m sin φ.

(ii) Restriking voltage

The transient, high-frequency voltage that appears across the contacts at or near current zero during arc interruption. It oscillates about the recovery voltage and can reach 2V_m.

(iii) RRRV

The rate of rise of restriking voltage, dv/dt, in kV/μs. If the voltage across the gap rises faster than the gap regains dielectric strength, the arc restrikes. It is the main factor in arc extinction.

Expression for restriking voltage

Consider a short circuit at the CB terminals of a generator with inductance L and stray capacitance C across the contacts (resistance neglected).

  e = V_m cos ωt
 (~)----^^^L^^^----+----o/ o----X fault
  |                |
 earth            === C
                   |
                  earth
  1. Before interruption, C is shorted by the fault. The fault current lags e by 90°, so at current zero e = V_m (taken constant during the short transient).
  2. After the arc goes out at current zero, L and C form a series circuit driven by V_m. Let v be the voltage across C (= across the contacts) and i the current:
V_m = L di/dt + v,    i = C dv/dt
V_m = LC d²v/dt² + v
  1. Taking the Laplace transform with v(0) = 0, i(0) = 0:
V_m/s = LC s² V(s) + V(s)
V(s)  = V_m / [ s (1 + s² LC) ]
      = V_m ω_n² / [ s (s² + ω_n²) ],  ω_n = 1/√(LC)
      = V_m [ 1/s − s/(s² + ω_n²) ]
  1. Inverse transform:
v(t) = V_m (1 − cos ω_n t),   ω_n = 1/√(LC)

Results

f_n          = 1 / (2π √(LC))
Peak value   = 2 V_m   at t = π √(LC)
RRRV         = dv/dt = V_m ω_n sin ω_n t
RRRV (max)   = V_m / √(LC)  at t = (π/2) √(LC)

So a smaller capacitance or inductance raises the natural frequency and the RRRV, making interruption harder.

  • 2072 Asoj · 8 marks

Describe arc extinction process in vacuum circuit breakers. Also explain the preferred voltage levels of its application and its advantages over OCB and air blast circuit breakers.

Answer

A vacuum circuit breaker interrupts current between contacts in a sealed vacuum of about 10⁻⁵ to 10⁻⁷ torr. With no gas present, the arc is a metal-vapour arc that is extinguished at the first current zero because the vapour condenses almost instantly.

Arc extinction process

   [fixed contact]
        | | |  metal vapour + ions
        |~|~|  (arc in vapour)
   [moving contact] -> separates
   condensing shield around gap
  1. Contact separation: the moving Cu-Cr contact is pulled away by the mechanism. Contact pressure falls; the last bridging point melts and evaporates.
  2. Metal-vapour arc: the current continues through the ionised metal vapour emitted from cathode spots. At low current the arc is diffuse (many cathode spots spread over the contact face).
  3. Arc control: at high currents the arc tends to constrict. Specially shaped contacts (spiral/radial-magnetic-field or axial-magnetic-field) make the arc rotate or stay diffuse, preventing local overheating and erosion.
  4. Current approaching zero: vapour production falls with current; the cathode spots die out.
  5. Condensation: the vapour and ions spread out and condense on the contacts and on the metal shield within microseconds, so the gap again becomes a near-perfect vacuum.
  6. Dielectric recovery: a vacuum gap has very high breakdown strength; it recovers within a few microseconds, faster than the restriking voltage rises. The arc is cleared at the first current zero with a short contact gap (about 10–20 mm).
  7. Current chopping: if the arc becomes unstable before natural zero at very low currents, the current may be chopped, causing overvoltage in inductive circuits; Cu-Cr contact material keeps the chopping current very low (a few amperes).

Preferred voltage levels

  • Most economical and widely used for medium voltage, 3.3 kV to 36 kV (11 kV and 33 kV in Nepal's distribution system).
  • For higher voltages, several interrupters must be connected in series, so SF6 breakers are preferred above about 36 kV (though VCBs up to 72.5 kV and some 145 kV designs exist).

Advantages over OCB and ABCB

PointVCBOCBAir blast CB
Arc mediumVacuum, sealedOil, needs replacementCompressed air system
Fire/explosion riskNoneOil fire riskNone but high-pressure air
NoiseSilentQuietVery noisy blast
MaintenanceAlmost noneFrequent oil filtrationCompressor upkeep
Size/weightCompact, lightBulky tankLarge with air plant
Contact lifeVery long (10,000+ ops)ShortMedium
Arc timeShort, first current zeroLongerShort
Frequent switchingExcellentPoorFair

Other advantages: no exhaust of gas or flame, suitable for any position, unaffected by atmosphere and altitude, handles high RRRV and capacitor switching without restrike.

  • 2074 Bhadra · 8 marks

Describe the construction and working of a puffer type SF6 circuit breaker. Also mention the properties of SF6 gas which make it a good arc quenching medium.

Answer

A puffer type SF6 circuit breaker is a single-pressure SF6 breaker in which the gas flow needed to quench the arc is produced by the motion of the moving contact itself: a cylinder attached to the moving contact compresses gas against a fixed piston and blows it through a nozzle onto the arc.

Construction

   Fixed contact (hollow, arcing tip)
           |    |
      =====|nozzle|=====   PTFE nozzle
           |  arc |
   +-------[moving]------+
   |  puffer cylinder    |  moves down with contact
   |  (gas compressed)   |
   +====[fixed piston]===+
           |
     operating rod -> spring/hydraulic mechanism
   Whole unit in SF6 at about 5-6 bar
  1. Interrupter enclosure (porcelain or earthed metal tank) filled with SF6 at one pressure (about 5–6 bar).
  2. Fixed contact with arcing tip, and main contacts that carry normal current.
  3. Moving contact fixed to a puffer cylinder; a stationary piston sits inside the cylinder.
  4. Insulating PTFE nozzle that guides the gas jet along the arc.
  5. Operating mechanism, gas density monitor, absorbent (alumina) for decomposition products.

Working

  1. On tripping, the moving contact and cylinder move away from the fixed contact.
  2. The main contacts part first; current transfers to arcing contacts, which then part and draw an arc.
  3. As the cylinder moves over the fixed piston, the gas inside is compressed (pressure rises 2–3 times).
  4. When the arcing contact leaves the nozzle throat, the compressed SF6 is blown axially through the nozzle across the arc.
  5. The gas cools the arc and its electronegative molecules capture electrons; at current zero the gap de-ionises rapidly and dielectric strength builds up faster than the restriking voltage.
  6. The used gas recombines and remains in the enclosure; no compressor is needed.
  7. Modern self-blast (auto-puffer) designs also use arc heat to raise pressure, reducing the operating energy.

Properties of SF6 that make it a good arc quencher

  1. Electronegative: captures free electrons to form heavy, slow negative ions, so arc conductivity falls very quickly after current zero.
  2. High dielectric strength: about 2.5–3 times air at the same pressure; at a few bar it is comparable to oil.
  3. Fast dielectric recovery: strength builds up about 100 times faster than air after current zero.
  4. Good heat transfer: high thermal conductivity in the 2000–3000 K range cools the arc boundary.
  5. Chemically stable and inert: non-flammable, non-toxic, non-corrosive in pure form; dissociated products recombine after the arc.
  6. Low arc voltage and energy, so low contact erosion.
  7. Heavy molecular weight (about 5 times air), so it stays in the enclosure and can be stored as a liquid.

These properties allow short arcing time, compact interrupters and use from 33 kV up to 765 kV and in GIS.

  • 2076 Bhadra · 6 marks

Explain the construction and working of SF6 circuit breaker. Also enumerate the dielectric properties and arc quenching characteristic of SF6 circuit breaker with their limitations.

Answer

An SF6 circuit breaker is a breaker in which the contacts separate inside a sealed chamber filled with sulphur hexafluoride (SF6) gas under pressure, and the arc is quenched by blowing SF6 over it. It is the standard breaker for 33 kV to 765 kV.

Construction

  • Interrupter unit (arc chamber): a sealed metal/porcelain chamber filled with SF6 at about 3–5 bar (gauge); it holds a fixed contact and a hollow moving contact with a nozzle.
  • Contacts: main current-carrying contacts and arcing contacts (copper-tungsten tips) so that the arc does not burn the main contacts.
  • Gas system: high-pressure reservoir (about 14–16 bar in the older double-pressure type), low-pressure chamber, compressor, filters (activated alumina to absorb decomposed products), pressure and density monitors.
  • Operating mechanism: spring, pneumatic or hydraulic, linked to the moving contact through an insulated rod.
  Fixed contact      Nozzle      Moving contact
   ┌──────┐        ┌──\/──┐      ┌──────────┐
   │      │  arc   │      │◄═════│ (hollow) │◄── drive rod
   └──────┘ ~~~~~~ └──────┘      └──────────┘
        ▲ SF6 blast from piston/reservoir ▲
  ─────── sealed chamber filled with SF6 ───────

Working

  1. In the closed position the contacts touch and the chamber is filled with SF6 at rated pressure.
  2. On a trip signal the moving contact is pulled away; an arc forms between the arcing contacts.
  3. In the puffer type the same motion compresses gas in a cylinder; in the double-pressure type a valve opens from the high-pressure reservoir. Either way SF6 flows axially through the nozzle along the arc.
  4. SF6 is electronegative: its molecules capture free electrons to form heavy negative ions (SF6⁻, F⁻) that move slowly. The arc column loses conductivity very fast.
  5. At the next current zero the arc is extinguished and the gap regains dielectric strength quickly, so restriking does not occur.
  6. The gas recombines almost fully into SF6 after cooling; small by-products are absorbed by filters.

Dielectric properties of SF6

  • Dielectric strength about 2.5–3 times air at the same pressure; at 3 bar it is close to transformer oil.
  • High electronegativity, so fast recovery of dielectric strength after current zero.
  • Colourless, odourless, non-toxic, non-inflammable and chemically inert at normal temperature.
  • High thermal conductivity and heat transfer, good cooling of the arc.

Arc quenching characteristics

  • Arc time constant very small (a few μs), so even high rates of rise of restriking voltage are handled.
  • Low arc voltage and low arc energy, hence little contact erosion.
  • No current chopping, so low switching overvoltages.
  • Silent operation; no exhaust to atmosphere.

Limitations

  • SF6 is a very strong greenhouse gas (GWP about 23,500 times CO₂); leaks must be prevented and gas recovered.
  • Needs leak-proof sealing and gas handling plant; costly.
  • Arc by-products (SF4, S2F2, metal fluorides) are toxic and corrosive; moisture must be kept out.
  • Gas liquefies at low temperature and high pressure (about 10 °C at 15 bar), so heaters may be needed.
  • Dielectric strength falls sharply if pressure drops, so continuous pressure monitoring and lock-out are required.
  • 2079 Chaitra · 6 marks

Describe with neat sketch the construction and principle of operation of an air blast circuit breaker.

Answer

An air blast circuit breaker (ABCB) is a breaker in which a blast of high-pressure compressed air (about 20–30 kg/cm²) is directed on the arc to cool it, sweep away ionised gas and extinguish it at current zero. It was widely used from 66 kV to 400 kV before SF6 breakers became common.

Construction

  • Compressed air reservoir charged by an air compressor plant.
  • Blast valve that admits air from the reservoir to the arc chamber on a trip signal.
  • Arc extinction chamber containing a fixed contact (often a nozzle) and a moving contact held closed by a spring.
  • Exhaust to atmosphere through a silencer.
  • Isolating switch / resistor connected in parallel or series in some designs to control restriking voltage.
 Air reservoir ──► Blast valve ──► Arc chamber ──► Exhaust
  (20-30 bar)                         │
                     ┌─────────┐    ┌─┴────────┐
          Fixed ─────│ nozzle  │ ~~ │ moving   │◄─ spring
          contact    └─────────┘arc └──────────┘
                     air flows ───────────────►

Types (by direction of air flow)

  1. Axial blast: air flows along the arc through a nozzle; moving contact is pushed back by the air pressure itself. Most common.
  2. Cross blast: air flows across the arc and drives it into arc splitters, lengthening it; suited to medium voltages and high current.
  3. Radial blast: air flows radially into the gap.

Principle of operation

  1. When the breaker trips, the blast valve opens and high-pressure air enters the arc chamber.
  2. The air pressure pushes the moving contact away against its spring; an arc is drawn.
  3. The fast air stream cools the arc, removes hot ionised particles and lengthens the arc.
  4. The arc resistance rises and at the next current zero the arc is extinguished.
  5. Fresh dry air fills the gap, so the dielectric strength builds up faster than the restriking voltage.
  6. After extinction a series isolator opens (in some designs) and the blast valve closes; the contacts are held open.

Advantages

  • No fire risk (no oil); suitable for indoor use.
  • Very fast arc extinction and short arcing time; little contact burning.
  • Suitable for frequent operation and repetitive auto-reclosing.
  • Arcing products are blown away; low maintenance of the medium.

Disadvantages

  • Needs a compressor plant and air storage; costly upkeep.
  • Very noisy operation (needs silencers).
  • Prone to current chopping at low currents and high RRRV, so resistors across contacts are used.
  • Air leakage at fittings must be watched.
  • 2072 Magh · 8 marks

Describe with a neat sketch the principle of operation of a minimum oil circuit breaker. Why is it called so?

Answer

A minimum oil circuit breaker (MOCB) is an oil circuit breaker in which oil is used only for arc extinction, while insulation between live parts and earth is provided by solid insulation (porcelain or organic material). It is used from 3.3 kV to about 132 kV.

Why it is called "minimum oil"

In a bulk oil CB a large earthed steel tank is filled with oil, which serves both for arc quenching and for insulating live parts from the earthed tank, so a lot of oil is needed. In the MOCB the interrupter is housed in a small insulated chamber, which is itself kept live, so only the small quantity of oil needed for arc quenching is used (about 10% of a bulk oil CB of the same rating). Hence the name "minimum" or "small" oil circuit breaker.

Construction

  • Supporting chamber: porcelain, filled with oil, physically separated from the interrupting chamber; it carries the breaker and gives insulation to earth.
  • Circuit breaking chamber: a porcelain/insulating tube holding the fixed contact, a moving contact (rod), and a turbulator (side-vent or axial-vent arc control device).
  • Top chamber: a metal chamber with an oil separator to cool and separate gases from oil; a vent releases gas.
  • Operating mechanism: spring or solenoid, connected to the moving contact through an insulating rod.
          ┌──────────────┐  Top chamber
          │ gas separator│──► vent
          ├──────────────┤
          │ fixed contact│
          │   ┌──┴──┐    │  Breaking chamber
          │   │turbu│    │  (oil, porcelain)
          │   │lator│    │
          │   └──┬──┘    │
          │ moving contact│
          ├──────┼───────┤
          │ support      │  Supporting chamber
          │ chamber (oil)│  (porcelain)
          └──────┼───────┘
            operating rod

Principle of operation

  1. In the closed position the moving contact rod is held engaged in the fixed contact.
  2. On a trip signal the mechanism pulls the moving contact down, and an arc is struck between the contacts in oil.
  3. The arc heat decomposes oil into gas (mainly hydrogen, about 70%) forming a gas bubble at high pressure.
  4. The turbulator directs this gas and oil flow through side vents (cross blast) or axially across the arc, which cools it, lengthens it and deionises the gap.
  5. Hydrogen has high thermal conductivity, so the arc loses heat fast; at current zero the arc is extinguished and fresh oil fills the gap.
  6. Gases rise to the top chamber where oil is separated and the gas is vented.

Advantages

  • Very small oil requirement, less fire risk, lighter and smaller.
  • Lower cost and less space than bulk oil CB.
  • Suitable for indoor and outdoor substations.

Disadvantages

  • Small oil volume gets carbonised quickly, so frequent oil replacement is needed.
  • Not suitable for frequent operation or high-speed reclosing.
  • Danger of explosion if oil level falls.
  • Difficult to remove gases from the small space.
  • 2070 Magh · 8 marks

Compare the merits and demerits of bulk oil circuit breakers, minimum oil circuit breakers and air blast circuit breakers.

Answer

Bulk oil (BOCB), minimum oil (MOCB) and air blast (ABCB) circuit breakers differ in the quenching medium and in what gives insulation to earth. BOCB uses a large earthed oil tank, MOCB uses a little oil in insulated live chambers, and ABCB uses compressed air.

Merits and demerits

BreakerMeritsDemerits
Bulk oilSimple and reliable; oil absorbs arc energy and gives good insulation; arc produces hydrogen which cools well; CTs can be mounted inside bushingsLarge oil volume, fire and explosion risk; heavy and bulky; slow; oil carbonises and needs maintenance; not good for frequent operation
Minimum oilNeeds only about 10% of oil; small, light, cheaper; less fire risk; good for indoor and outdoor use up to 132 kVOil deteriorates quickly (carbonisation); frequent oil change; not suited for frequent switching or high-speed reclosing; explosion risk if oil level is low
Air blastNo fire risk; very fast arc extinction and short arc time; low contact erosion; suitable for repeated operation and auto-reclosing; clean mediumNeeds compressor and air storage plant; very noisy; current chopping and high overvoltages need resistors; sensitive to RRRV; air leakage

Comparison of characteristics

PointBulk oilMinimum oilAir blast
Quenching mediumOilOilCompressed air (20–30 bar)
Insulation to earthOil in earthed tankPorcelain/solidPorcelain/air
Medium quantityVery largeSmallAir from reservoir
Speed of operationSlowMediumFast
Fire hazardHighLowNone
MaintenanceOil filtering, highOil change oftenCompressor upkeep
NoiseLowLowVery high
Voltage rangeUp to 33 kV (old up to 132 kV)3.3–132 kV66–400 kV
Auto-reclosingPoorPoorVery good

Summary

  • BOCB is cheap but bulky and hazardous; now mostly obsolete.
  • MOCB saves oil and space; popular for medium voltage before vacuum CBs.
  • ABCB is best for EHV and frequent switching where speed matters, at the cost of the air plant and noise.
  • 2071 Magh · 8 marks

Compare the performance and characteristics of i) minimum oil circuit breaker and air blast CB ii) Air-blast circuit breaker and bulk oil circuit breaker.

Answer

The comparison is based on how each breaker quenches the arc and the practical consequences in speed, safety, maintenance and application.

i) Minimum oil CB vs Air blast CB

PointMinimum oil CBAir blast CB
Arc quenching mediumSmall quantity of oil; arc produces hydrogen gasCompressed air at 20–30 bar
Insulation to earthPorcelain/solid materialPorcelain and air
Arc extinctionBy gas bubble and turbulator (cross/axial vent)By air blast cooling and sweeping the arc
Arcing timeLongerVery short
SpeedMediumHigh
Fire riskSmall (some oil present)None
Repeated operation / auto-recloseNot suitableVery suitable
Auxiliary plantNoneCompressor and air receiver
NoiseQuietVery noisy
Current choppingLowSignificant; resistors needed
MaintenanceOil change after a few fault clearancesCompressor and valves
Range3.3–132 kV66–400 kV

ii) Air blast CB vs Bulk oil CB

PointAir blast CBBulk oil CB
MediumCompressed airLarge volume of oil in earthed steel tank
InsulationAir/porcelainOil insulates contacts from tank
Size and weightSmaller, lighter for the same ratingVery heavy and bulky
Fire and explosion hazardNilHigh
Speed of operationFast (2–3 cycles)Slow (5–8 cycles)
Arc energy and contact erosionLowHigh
Medium after operationFresh air every timeOil carbonises, needs filtering
Frequent operationSuitableUnsuitable
NoiseVery highLow
Auxiliary plantCompressor neededNone
CostHigh (with air plant)Low
ApplicationEHV transmission, frequent switchingOld distribution systems, up to 33 kV

Remarks

  • Oil breakers are simple and need no auxiliary plant, but carry fire risk and are slow.
  • Air blast breakers are fast and clean, but costly and noisy; they have largely been replaced by SF6 breakers at EHV.
  • 2075 Bhadra · 8 marks

Compare the performance and characteristics and working principle of: i) Minimum oil circuit breaker vs Bulk oil circuit breaker ii) Air blast circuit breaker vs puffer type SF6 circuit breaker.

Answer

The comparison covers working principle, performance and characteristics of each pair.

i) Minimum oil CB vs Bulk oil CB

Working principle (common): the arc between separating contacts decomposes oil into a gas bubble (mainly hydrogen); high pressure and cooling by hydrogen deionise the gap so that the arc goes out at current zero.

PointMinimum oil CBBulk oil CB
Role of oilArc quenching onlyArc quenching and insulation to earth
TankSmall porcelain chamber, kept liveLarge earthed steel tank
Oil quantityAbout 10% of BOCBVery large
Arc controlTurbulator (axial/side vent) always usedPlain break or explosion pot
Size, weightSmall, lightHeavy, bulky, large foundations
Fire riskLowHigh
Oil deteriorationFast (small volume)Slow
MaintenanceFrequent oil changeOil filtering, tank cleaning
Range3.3–132 kVUp to 33 kV

ii) Air blast CB vs Puffer type SF6 CB

Working principle:

  • ABCB: a blast of compressed air from a separate reservoir is blown on the arc; it cools the arc and sweeps away ions, and air is exhausted to atmosphere.
  • Puffer SF6: the moving contact drives a piston that compresses SF6 in a cylinder and blows it through a nozzle over the arc. SF6 is electronegative and absorbs free electrons, so the gap recovers very fast. Gas stays in a sealed enclosure.
PointAir blast CBPuffer SF6 CB
MediumAir at 20–30 barSF6 at about 5–7 bar
Gas supplyExternal compressor and receiverSelf-generated by piston; no compressor
Dielectric strengthLow (needs high pressure)2.5–3 times air
Arc time constantLargerVery small
Current choppingHigh; resistors neededNegligible
NoiseVery loud exhaustSilent (closed system)
Breaks per pole at EHVMoreFewer
MaintenanceCompressor, valves, leaksVery low; gas leak monitoring
EnvironmentClean (air)SF6 is a strong greenhouse gas
CostHigh (with air plant)Moderate
Present useMostly replacedStandard for 33–765 kV

Remarks

SF6 puffer breakers combine the speed of air blast breakers with silent, compact and low-maintenance operation, which is why they have replaced both oil and air blast types at transmission voltages.

  • 2070 Bhadra · 8 marks

What is the most important basis for the classification of circuit breaker? How they are classified? Describe briefly. Mention their field of application.

Answer

The most important basis for classifying circuit breakers is the medium used for arc extinction, because the medium decides the breaking capacity, voltage range, size, speed and maintenance of the breaker.

Classification by arc quenching medium

  1. Air break circuit breaker

    • Arc is drawn in air at atmospheric pressure and lengthened by arc runners, blow-out coils and arc splitters (high resistance method).
    • Application: low and medium voltage up to about 11 kV; LV switchboards, motor control, DC circuits.
  2. Oil circuit breaker

    • Arc in oil decomposes it into hydrogen gas which cools and deionises the arc.
    • Bulk oil: oil for quenching and insulation; up to 33 kV (old installations).
    • Minimum oil: oil for quenching only, porcelain for insulation; 3.3–132 kV.
    • Application: outdoor distribution substations (now being replaced).
  3. Air blast circuit breaker

    • Compressed air (20–30 bar) blown axially or across the arc.
    • Application: 66–400 kV transmission; where frequent operation and fast auto-reclosing are needed; arc furnaces.
  4. SF6 circuit breaker

    • Electronegative SF6 gas blown over the arc (puffer or double-pressure type).
    • Application: 33–765 kV substations, GIS (gas-insulated switchgear).
  5. Vacuum circuit breaker

    • Contacts open in a vacuum of 10⁻⁶ to 10⁻⁷ torr; the arc is a metal-vapour arc that vanishes at current zero.
    • Application: 3.3–33 kV indoor switchgear, capacitor and motor switching, railways.

Other bases of classification

  • By voltage: low voltage (below 1 kV), medium/high voltage (1–66 kV), extra high voltage (above 220 kV).
  • By location: indoor and outdoor breakers.
  • By operating mechanism: spring, pneumatic, hydraulic, solenoid.
  • By external design: dead tank (enclosure at earth potential) and live tank (interrupter at line potential).
  • By interrupting principle: high resistance interruption (DC and LV) and low resistance or current-zero interruption (AC HV).

Summary table

TypeVoltage rangeTypical use
Air breakup to 11 kVLV/MV panels, DC
Bulk oilup to 33 kVOld distribution
Minimum oil3.3–132 kVDistribution substations
Air blast66–400 kVTransmission, arc furnaces
SF633–765 kVTransmission, GIS
Vacuum3.3–33 kVIndoor MV switchgear
  • 2070 Magh · 8 marks

Figure shows a 132/66 kV power transformer is fed from an infinite bus. If the 66 kV bus is a load bus, specify for the breaker 'X': i) Rated continuous current ii) Breaking capacity in MVA. [Figure: Infinite bus – breaker Y – 3-phase 132/66 kV, 50 MVA transformer with Z = 14% – breaker X – load bus.]

Answer

Breaker X is on the 66 kV side between the transformer and the load bus. It must carry the full-load current of the transformer continuously and must interrupt the fault current that flows through it for a fault just beyond it (on the load bus).

Assumptions: the source is an infinite bus (zero impedance), the load bus has no other source, and fault current is limited only by the transformer impedance (14% on 50 MVA base).

i) Rated continuous current

I_rated = S / (√3 × V)
        = 50 × 10⁶ / (√3 × 66 × 10³)
        = 437.39 A

So breaker X must have a continuous current rating of at least 437.4 A; the next standard rating, 630 A, is chosen (IEC R10 series).

ii) Breaking capacity

For a three-phase fault on the load bus (just after X), the only impedance is the transformer reactance:

Fault MVA = Base MVA / Z(pu)
          = 50 / 0.14
          = 357.14 MVA

Fault current at 66 kV:
I_f = 357.14 × 10⁶ / (√3 × 66 × 10³)
    = 3124.2 A  ≈ 3.12 kA

A fault on the transformer side of X is fed directly from the infinite bus and is cleared by breaker Y, so X only has to break the current fed through the transformer.

Making current (peak, first cycle) = 2.55 × 3.124 = 7.97 kA.

Answer: rated continuous current = 437.4 A (choose 630 A standard); breaking capacity = 357.14 MVA (about 3.12 kA at 66 kV). In practice a standard breaker of 72.5 kV, 630 A, with a breaking current of at least 3.15 kA (or the next higher standard, e.g. 12.5 kA) would be specified.

  • 2071 Magh · 8 marks

Select a circuit breaker for the following application: 6.6 kV, 650 MVA (breaking capacity) breaker. It is supposed to be connected to primary of 6600/433 V, 2000 kVA transformer. Justify your choice and prepare the specifications for the breaker.

Answer

The breaker controls the 6.6 kV primary of a 2000 kVA, 6600/433 V distribution transformer and must interrupt 650 MVA at 6.6 kV.

Calculations

Rated (normal) current:
I_n = 2000 × 10³ / (√3 × 6600) = 174.95 A

Symmetrical breaking current:
I_b = 650 × 10⁶ / (√3 × 6600) = 56 860 A ≈ 56.86 kA

Making current (peak):
I_m = 2.55 × I_b = 2.55 × 56.86 = 144.99 kA ≈ 145 kA

Short-time current (1 s or 3 s) = I_b ≈ 56.86 kA

Choice of breaker and justification

A vacuum circuit breaker (indoor, withdrawable type) is chosen:

  • 6.6 kV is in the medium voltage range where vacuum breakers are standard and cheapest over life.
  • Vacuum gives fast recovery of dielectric strength, very low contact erosion and long life (10,000+ operations) with almost no maintenance.
  • No fire hazard (no oil) and no gas handling (unlike SF6), so it suits indoor switchgear near the transformer.
  • Compact size; suitable for transformer switching. An SF6 breaker of the same rating is an acceptable alternative; old practice would have used a minimum oil breaker.
  • Transformer magnetising current is small; modern vacuum interrupters have low chopping current, and surge arresters at the transformer terminals take care of switching surges.

Specification of the breaker (as per IEC 62271-100 standard values)

ParameterValue
TypeIndoor, 3-pole vacuum CB, withdrawable
Rated voltage7.2 kV (system 6.6 kV)
Rated frequency50 Hz
Rated normal current630 A (required 175 A)
Rated short-circuit breaking current63 kA symmetrical (required 56.86 kA)
Rated breaking capacity≥ 650 MVA at 6.6 kV
Rated making current160 kA peak (2.5 × 63; required 145 kA)
Rated short-time withstand current63 kA for 3 s
Rated operating sequenceO – 0.3 s – CO – 3 min – CO
Power frequency withstand20 kV rms for 1 min
Lightning impulse withstand60 kV peak
Opening / break timeabout 3 cycles (60 ms)
Operating mechanismMotor-wound spring, 110 V DC trip and close coils
ProtectionIDMT overcurrent + earth fault relay through CTs 200/5 A

Answer: rated current 174.95 A (630 A standard), breaking current 56.86 kA, making current 145 kA; a 7.2 kV, 630 A, 63 kA vacuum circuit breaker is specified.

  • 2073 Magh · 8 marks

Explain the high resistance and current zero interruption method of arc extinction in the circuit breakers. What is the criteria for successful arc interruption?

Answer

An arc between separating contacts is extinguished either by raising its resistance until the current cannot be maintained (high resistance method) or by letting it go out at a natural current zero and preventing it from restriking (low resistance or current zero method).

High resistance method

  • The arc resistance is increased with time so that the current falls to a value too small to maintain the arc.
  • Arc resistance is raised by:
    1. Lengthening the arc (arc runners, separating contacts further).
    2. Cooling the arc (contact with cold surfaces, air).
    3. Reducing cross-section (forcing it through narrow slots).
    4. Splitting it into many short arcs in series (arc splitter plates).
  • Disadvantage: a lot of energy is dissipated in the arc, so it is used only in DC breakers and low/medium voltage air break AC breakers.

Low resistance (current zero) method

  • Used in all AC high-voltage breakers (oil, air blast, SF6, vacuum).
  • The arc resistance is kept low until current zero, so arc energy is small. AC current passes through zero 100 times per second (50 Hz); at each zero the arc goes out by itself.
  • The task is to prevent restriking after current zero. Two quantities race against each other:
    • the dielectric strength of the gap, which must rise quickly, and
    • the restriking voltage across the contacts, which also rises quickly.
  • Gap strength is built up by:
    1. Lengthening the gap quickly (fast contact separation).
    2. High pressure in the gap, which increases recombination of ions.
    3. Cooling (e.g. hydrogen from oil, SF6).
    4. Blast effect: sweeping out ionised particles with fresh oil, air or SF6.

Two theories explain this: the energy balance theory (arc restrikes if heat produced exceeds heat removed) and the voltage race / recovery rate theory (Slepian's theory: arc does not restrike if the rate of build-up of dielectric strength is faster than the rate of rise of restriking voltage).

 Voltage
   │      dielectric strength
   │        ___________----
   │     _-'      ___
   │   _-'     _-'  restriking voltage
   │  /     _-'
   │ /   _-'
   │/_-'
   └────────────────────── time
   current zero

Criteria for successful arc interruption

  1. At current zero, the rate of rise of dielectric strength of the gap must always exceed the rate of rise of restriking voltage (RRRV); the dielectric strength curve must stay above the restriking voltage curve.
  2. The energy removed from the arc space by cooling, deionisation and blast must be greater than the energy supplied by the restriking voltage (energy balance).
  3. The gap must withstand the full recovery voltage after transients die out.
  4. In practice this needs fast contact separation, effective deionisation, and control of RRRV (e.g. resistance switching).
  • 2075 Bhadra · 4 marks

Explain the operation of auto reclosure.

Answer

Auto reclosure is the automatic re-closing of a circuit breaker a short time after it has tripped on a fault, so that the line is restored without an operator if the fault was temporary.

Why it is used

About 80–90% of faults on overhead lines are transient (lightning flashover, birds, tree branches touching in wind). Once the arc is cleared and the air deionises, the line can be safely re-energised.

Operation

  1. A fault occurs; the protective relay trips the breaker.
  2. The breaker stays open for a dead time (typically 0.2–0.5 s for high-speed reclosing on EHV lines; a few seconds on distribution lines) so the fault arc path deionises.
  3. The auto-reclose relay sends a close signal to the breaker.
  4. If the fault has cleared (transient fault), the breaker stays closed and the relay resets after a reclaim time.
  5. If the fault is still present (permanent fault), the protection trips the breaker again.
  6. After a preset number of attempts (usually 1–3 shots), the breaker locks out and stays open until it is reset manually.
 Fault  Trip   dead time   Reclose
   │     │◄────────────►│
 ──┴─────┘               └─────── (stays closed: transient)
                         └─ trip again → lockout (permanent)

Types

  • Single-shot / multi-shot (number of reclosures).
  • Three-pole (all phases) or single-pole (only the faulted phase, for single line-to-ground faults; keeps power transfer and stability).
  • High-speed (below 0.5 s) and delayed auto reclosing.

Advantages

  • Improves continuity of supply and reduces outage time.
  • Improves transient stability of the system.
  • Reduces operating staff and patrolling.
  • 2075 Baisakh · 4 marks

Explain operation of HVDC circuit breaker.

Answer

An HVDC circuit breaker interrupts direct current in an HVDC line. DC is hard to break because, unlike AC, it has no natural current zero, and the large inductive energy (½ L I²) of the circuit must be absorbed during interruption.

Requirements

  1. Create an artificial current zero in the main interrupter.
  2. Absorb the stored energy of the system inductance.
  3. Withstand the transient recovery voltage that appears after interruption.

Operation (resonance / commutation type)

The common mechanical HVDC breaker has three parallel branches:

        ┌──── Main interrupter (SF6/vacuum) ────┐
 I_dc ──┤                                       ├──►
        ├──── L ──── C ──── (switch S) ─────────┤
        │      commutation / resonant branch    │
        └──── Surge arrester (ZnO) ─────────────┘
  1. In normal operation the DC current flows through the main interrupter.
  2. On a fault the main contacts open, and an arc forms.
  3. A pre-charged capacitor C in the commutation branch is discharged through L (switch S closes), or the arc's negative resistance makes the LC branch oscillate. This injects a current opposite to the line current.
  4. The current in the main interrupter is driven to zero, and the arc is extinguished there.
  5. The line current is then commutated into the LC branch and charges the capacitor; the voltage rises rapidly.
  6. When the voltage reaches the protective level, the surge arrester (energy absorber) conducts, absorbs the stored energy ½ L I² and forces the current to zero.

Hybrid HVDC breaker

Modern breakers (e.g. for multi-terminal VSC-HVDC) use a fast mechanical disconnector with a low-loss IGBT load commutation switch in the main path and a large IGBT main breaker in parallel, plus arresters. They interrupt in a few milliseconds.

Uses

Multi-terminal HVDC systems, metallic return transfer breakers and DC grids. In point-to-point HVDC links, faults are usually cleared by converter control instead.

  • 2076 Bhadra · 4 marks

What are the main parts of equipment which are used for the testing of a circuit breaker in a laboratory type testing station?

Answer

A laboratory (direct) type testing station tests circuit breakers for short-circuit making and breaking capacity using its own short-circuit generator, so tests can be done under controlled conditions.

Main parts of the equipment

  1. Short-circuit generator: a special alternator designed to withstand repeated short circuits; driven by an induction motor. Its excitation is controlled to set the test current and voltage.
  2. Flywheel: coupled to the generator shaft; it stores kinetic energy so that speed stays almost constant during the short circuit.
  3. Driving motor: brings the generator and flywheel up to speed; it is disconnected before the test.
  4. Exciter: supplies controlled DC field current to the generator.
  5. Master (back-up) breaker: connected between generator and test circuit; it protects the generator and clears the circuit if the test breaker fails.
  6. Make switch: closes the circuit at a chosen instant on the voltage wave, so a symmetrical or asymmetrical current can be produced.
  7. Short-circuit transformer: steps the generator voltage up to the test voltage; it is built to withstand short-circuit forces.
  8. Reactors and resistors: adjust the magnitude and power factor of the test current.
  9. Capacitors: adjust the rate of rise of restriking voltage (RRRV).
  10. Test cell: a strong enclosure where the breaker under test is installed, for safety.
  11. Measuring and recording equipment: potential dividers, shunts/CTs and oscillographs (now digital recorders) to record current, voltage and timing.
  12. Control room: for sequence control of all switches and observation.
 Motor ─ Flywheel ─ SC Generator ─ Master CB ─ Make
                         │                   switch
                      Exciter                  │
          Reactor ─ SC Transformer ─ Test CB ─ Load
          (control of I and pf)     (test cell)
  • 2078 Chaitra · 6 marks

Describe the working of cross jet explosion pot in bulk oil circuit breaker with a neat sketch.

Answer

A cross jet explosion pot is an arc control device used in bulk oil circuit breakers to quench the arc with jets of oil blown across it, using the pressure created by the arc itself. It is suited to high currents.

Construction

  • A strong insulating pot (fibreglass or similar) mounted around the fixed contact, inside the oil tank.
  • One side of the pot has a set of arc splitter / side vents (exit ports) arranged one above the other.
  • The moving contact is a rod that slides down through the bottom throat of the pot.
  • The upper part of the pot forms a closed space where gas pressure builds up.
        ┌────────────────────┐
        │   Fixed contact    │
        │  ┌──────────────┐  │
 gas    │  │ pressure     │  │
 pocket │  │ space        │  │
        │  └──────┬───────┘  │
        │  ~arc~  │   ═══►   │ side vents
        │  ~arc~  │   ═══►   │ (oil + gas
        │  ~arc~  │   ═══►   │  jets out)
        │         │          │
        └─────────┼──────────┘
              moving contact
                  ▼ moves down

Working

  1. In the closed position the moving contact sits in the fixed contact inside the pot.
  2. When the breaker trips, the moving contact moves downward and an arc is drawn inside the pot.
  3. The arc decomposes oil into gas (mostly hydrogen). Because the pot is closed, a high-pressure gas bubble forms in its upper part.
  4. As the moving contact travels down, it uncovers the side vents one after another.
  5. The high-pressure gas forces oil out through these vents, producing jets of oil and gas across the arc (cross jet).
  6. The cross jets push the arc sideways into the vents, lengthen it, cool it and remove ionised products.
  7. The arc is extinguished at a current zero, usually before the moving contact leaves the pot. Fresh oil then fills the gap and builds up dielectric strength.

Features

  • Self-generated pressure, so the higher the fault current, the stronger the blast; good for high fault currents.
  • At low currents pressure is small and arcing time is long; for this the self-compensated explosion pot (cross jet plus plain pot) is used.
  • Short arc length and arcing time compared with a plain break oil CB.
  • 2080 Chaitra · 8 marks

What is the purpose of incorporating arcing contacts alongside main contacts in circuit breakers? Also explain the function of arc runners and arc splitters within an air-break circuit breaker, supported by a clear diagram.

Answer

Purpose of arcing contacts

In medium and high current breakers each pole has main contacts (to carry normal current with very low resistance) and arcing contacts in parallel with them.

  • On opening, the main contacts separate first; the current shifts to the arcing contacts, which separate a little later. So the arc forms only on the arcing contacts.
  • On closing, the arcing contacts touch first and the main contacts close later, so any pre-arc also strikes on the arcing contacts.
  • Arcing contacts are made of hard, heat-resistant material (copper-tungsten, silver-tungsten); main contacts are of high-conductivity silver-plated copper.
  • Benefits: main contacts are protected from burning and pitting, keep low contact resistance and low heating; only the cheap arcing contacts need replacement.

Air break circuit breaker

An air break CB opens its contacts in air at atmospheric pressure and extinguishes the arc by the high resistance method: the arc is lengthened, cooled and split until its voltage exceeds the system voltage.

Arc runners (arc horns)

  • Two diverging metal horns extended upward from the arcing contacts.
  • Once the arc forms, it moves upward along the runners because of (i) the thermal convection of hot air and (ii) the electromagnetic force of its own current loop (aided by a blow-out coil in some designs).
  • As the runners diverge, the arc becomes longer and thinner; its resistance and voltage increase and it cools.
  • The arc root is moved off the contacts, protecting them.

Arc splitters (arc chutes)

  • A stack of insulated or metal plates placed above the runners inside an arc chute.
  • The rising arc is driven into the plates, where it is split into many short arcs in series.
  • Each short arc needs a minimum cathode-anode voltage drop (about 20–40 V), so the total arc voltage rises sharply above the supply voltage and the current is forced to zero.
  • The plates also cool the arc and absorb its heat; insulating plates lengthen it in a zig-zag path.
         ┌─────────────────┐
         │ ║ ║ ║ ║ ║ ║ ║   │  arc splitter plates
         │  \  arc chute /  │  (arc split into
         │   \  ~~~~~~  /   │   many short arcs)
         │    \ ~~~~~~ /    │
         │     \ ~~~~ /     │  arc runners
         │      \ ~~ /      │  (diverging horns)
         │   ┌──┐    ┌──┐   │
         │   │A │    │A │   │  A = arcing contacts
         │   │M │    │M │   │  M = main contacts
         └───┴──┴────┴──┴───┘
         fixed      moving

Sequence of arc extinction

  1. Main contacts open; current transfers to arcing contacts.
  2. Arcing contacts open; arc is struck.
  3. Arc rises along the runners, getting longer.
  4. Arc enters the chute and is split by the splitter plates.
  5. Arc voltage exceeds the system voltage; current falls to zero and the arc is extinguished.

Air break breakers with these devices are used up to about 11 kV and in DC circuits.

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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