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Chapter 1 · 3 hours

Principle of Power System Protection

IOE past exam questions

Past questions and answers

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

  • Asked 3 times
  • 2070 Magh · 4 marks
  • 2072 Magh · 4 marks
  • 2073 Bhadra · 4 marks

What are the basic requirements of any protection scheme in a power system? Explain them with the help of suitable example.

Answer

A protection scheme must detect every fault quickly and remove only the faulty part, while staying stable during normal operation. The basic requirements (qualities) are:

  1. Selectivity (discrimination): only the circuit breakers nearest to the fault should trip, so that the healthy part keeps supplying load. Example: for a fault on one outgoing 11 kV feeder of a substation, only that feeder breaker should open, not the incoming transformer breaker.
  2. Speed: the fault must be cleared as fast as possible to limit damage to equipment, reduce burning at the fault point and keep generators in synchronism (transient stability). Example: a 132 kV line fault is cleared in about 3–5 cycles (relay ≈ 1 cycle + breaker 2–3 cycles).
  3. Sensitivity: the relay must operate for the smallest fault current expected (e.g. a high-resistance earth fault at minimum generation). Example: a restricted earth-fault relay on a transformer is set at 10–20 % of rated current so it can see winding faults near the neutral.
  4. Reliability: the scheme must operate correctly every time it is needed (dependability) and must not operate when not needed (security). Good design, quality CTs/relays, DC supply and regular testing give reliability.
  5. Stability: the scheme must remain inoperative for faults outside its zone (through faults), for load swings and for magnetising inrush. Example: a transformer differential relay must not trip for an external fault on the LV feeder or for inrush current at switching-on.
  6. Simplicity and economy: the scheme should be simple to maintain and its cost should be justified by the value of the protected equipment. A small distribution transformer is protected by fuses, while a large generator gets differential, earth fault, loss-of-excitation and other relays.
  7. Adequateness: it should cover all the likely faults and abnormal conditions (overload, unbalance, over-voltage) of the equipment.

These requirements partly conflict (more speed and sensitivity can reduce stability), so the designer reaches a balance suitable for each element.

  • Asked 3 times
  • 2071 Magh · 2 marks
  • 2077 Chaitra · 6 marks
  • 2080 Chaitra · 5 marks

What are primary and back up protection? Explain the necessity of back up protection with an example.

Answer

Primary protection is the first line of defence: the protection installed for a particular zone (line, transformer, bus, generator) that is expected to clear any fault in that zone quickly and with the smallest possible outage.

Back-up protection is the second line of defence: a protection that operates, after a deliberate time delay, only when the primary protection or its breaker fails to clear the fault.

Why primary protection can fail

  • Failure of the relay itself or its setting.
  • Open or shorted CT/PT secondary leads.
  • Failure of the DC tripping supply or trip coil.
  • Circuit breaker mechanism stuck (breaker failure).
  • Failure of the communication channel in pilot schemes.

Necessity of back-up protection

  • No protective equipment is 100 % reliable, and an uncleared fault can burn equipment, cause fire, collapse system voltage and make generators lose synchronism.
  • It protects the system while primary protection is out for maintenance.
  • It covers parts of the system not fully covered by primary protection (e.g. the remote end of a line).
  • It must be time-graded so that primary protection always has the first chance.

Example

 Gen   A          B          C
 ──●───┼──[CB1]───┼──[CB2]───┼──[CB3]──► load
       R1         R2         R3
  • A fault F occurs on section B–C (beyond CB2).
  • The primary protection is relay R2 with breaker CB2; it should trip instantly.
  • If R2 or CB2 fails, relay R1 at the upstream station (set with a higher time, e.g. 0.4 s more) trips CB1 as remote back-up.
  • The cost is that section A–B and its loads are also disconnected, but the fault is removed and the rest of the system survives.

In modern substations, breaker-failure (local back-up) protection is also used: if CB2 has not opened within about 150–200 ms after its trip signal, all other breakers on the same bus are tripped.

  • Asked 2 times
  • 2071 Bhadra · 4 marks
  • 2072 Asoj · 4 marks

What do you understand by a zone of protection? Draw the protection zone diagram for a sample (modern) power system network and explain its rules.

Answer

A zone of protection is the part of the power system (generator, transformer, bus, line or motor) protected by a particular protection scheme. A fault anywhere inside the zone makes all the circuit breakers on its boundary trip, so that the zone is fully isolated. The boundary of a zone is defined by the location of the CTs and the circuit breakers.

Protection zones of a sample power system

 ┌──────────────┐
 │  G    Gen    │ 1 Generator zone
 └──┬───────────┘
 ┌──┴─[CB]──────┐
 │  Gen-transf. │ 2 Transformer zone
 └──┬───────────┘
 ┌──┴─[CB]──────┐
 │  HV bus      │ 3 Bus zone
 └──┬───────────┘
 ┌──┴─[CB]──────┐
 │  Line        │ 4 Transmission line zone
 └──┬───────────┘
 ┌──┴─[CB]──────┐
 │  Rx bus      │ 3 Bus zone
 └──┬───────────┘
 ┌──┴─[CB]──────┐
 │  Step-down Tr│ 2 Transformer zone
 └──┬───────────┘
 ┌──┴─[CB]──────┐
 │  Feeders/Mtr │ 5 Feeder / motor zone
 └──────────────┘
 Each box is a zone; adjacent boxes
 overlap around the circuit breaker.

Rules for drawing zones

  1. Each zone is drawn around one main element: generator, transformer, bus, line or motor.
  2. Every element of the system must lie in at least one zone; there should be no blind (unprotected) spot.
  3. Adjacent zones overlap, and the overlap region contains a circuit breaker. This is done by placing the CTs of the two zones on opposite sides of the breaker.
  4. A fault inside the overlap region trips the breakers of both zones; the overlap is kept as small as possible so that only a small extra part is disconnected.
  5. Zone boundaries are set by CT location; a zone includes the breakers that must trip to isolate it.

Using zones makes the scheme selective: a fault on the line trips only the two line breakers, while generator, transformers and buses stay in service.

  • 2075 Bhadra · 6 marks

What are the basic requirements of protection scheme? Differentiate between a protective system and protective scheme.

Answer

Basic requirements of a protection scheme

  1. Selectivity: only the faulty section is disconnected by the nearest breakers.
  2. Speed: fault is cleared quickly to limit damage and preserve stability.
  3. Sensitivity: operates for the minimum fault current (high-resistance faults, minimum generation).
  4. Reliability: operates whenever required (dependability) and never wrongly (security).
  5. Stability: stays inactive for through-faults, load swings and transformer inrush.
  6. Simplicity and economy: cost proportional to the importance of the equipment; easy to maintain.
  7. Adequateness: covers all likely faults and abnormal conditions of the equipment.

Protective system vs protective scheme

A protective system is the complete set of equipment that together performs the protective function for an element: CTs and VTs, relays, DC battery supply, trip circuit, wiring, communication channel and the circuit breaker. If any component fails, protection fails.

A protective scheme is the method or principle used to protect a particular element, i.e. how the relays are applied and connected, e.g. differential protection of a transformer, distance protection of a line, or overcurrent protection of a feeder. A single protective system may contain several protective schemes.

PointProtective systemProtective scheme
MeaningWhole hardware chainPrinciple/method of protection
ContentsCT, VT, relay, battery, trip coil, CB, wiringRelay type, connection, settings logic
NaturePhysical equipmentConcept / arrangement
ScopeAll schemes of a zone plus CBOne type of protection
Failure causeComponent failure (CB stuck, battery down)Wrong choice or setting of principle
ExampleCT + IDMT relay + DC supply + CB on a feederMerz-Price differential, distance, overcurrent
 Fault ─► CT/VT ─► Relay ─► Trip coil ─► CB opens
                     ▲          ▲
                Scheme logic   DC battery
     └────────── protective system ──────────┘
  • 2076 Bhadra · 6 marks

List down the basic requirements of good protective scheme. What do you understand by coordination of protective device and protection zone?

Answer

Basic requirements of a good protective scheme

  1. Selectivity – only the faulty section is isolated.
  2. Speed – minimum fault-clearing time to limit damage and keep stability.
  3. Sensitivity – operates on the smallest expected fault current.
  4. Reliability – operates correctly whenever needed and never falsely.
  5. Stability – no operation for external faults, load swings or inrush.
  6. Simplicity – fewer components, easy testing and maintenance.
  7. Economy – cost matched to the value of the protected equipment.

Coordination of protective devices

Coordination is the process of choosing the settings (pick-up current and time) of relays, fuses and breakers in series so that the device nearest to the fault operates first, and each upstream device acts only as time-delayed back-up.

  • Achieved by time grading, current grading or a combination (IDMT relays).
  • A grading margin (coordination time interval) of about 0.3–0.5 s is kept between successive relays, allowing for breaker opening time, relay overshoot and errors.

Example: On a radial feeder A→B→C with relays R1, R2, R3, relay R3 at the far end is set at 0.1 s, R2 at 0.5 s and R1 at 0.9 s for a fault at C. R3 clears the fault; R2 and R1 only reset.

Protection zone

A zone of protection is the region of the system (generator, transformer, bus, line, motor) guarded by a particular protection; its boundary is fixed by the CTs, and all breakers on that boundary trip for a fault inside it.

  • Every part of the system is covered by at least one zone.
  • Neighbouring zones overlap around a circuit breaker so there is no blind spot.
 ┌─Gen─┐┌─Transf─┐┌─Bus─┐┌──Line──┐┌─Bus─┐
 G ──[CB]─────[CB]────[CB]──────[CB]──── load
      overlaps at each CB
  • 2079 Chaitra · 6 marks

Why the protective zones are arranged in overlap fashion? Discuss in brief the various essential qualities of Protective Relaying.

Answer

Why protective zones are overlapped

A zone boundary is fixed by the CT location. If the CTs of two neighbouring zones were placed on the same side of a breaker, the small region between them (including the breaker itself) would belong to no zone. A fault there would be a blind spot and could be cleared only by slow back-up protection.

Therefore the CTs of each zone are placed on the far side of the common breaker, so the zones overlap across the breaker:

       Zone 1 (bus)
   ◄──────────────────►
 ──●──CT2──[ CB ]──CT1──●── line
         ◄──────────────────►
             Zone 2 (line)
 CT1 feeds bus relay, CT2 feeds line relay
  • A fault inside the overlap trips the breakers of both zones; this is acceptable because the overlap is very small.
  • Overlap ensures that every point of the system, including the breakers, is protected.
  • With dead-tank breakers and bushing CTs, overlap is easily obtained.

Essential qualities of protective relaying

  1. Selectivity: trips only the breakers needed to isolate the faulty element.
  2. Speed: clears the fault quickly (a few cycles) to reduce damage and maintain system stability.
  3. Sensitivity: responds to the minimum fault current, e.g. high-resistance earth faults.
  4. Reliability: dependable (always trips when required) and secure (never trips when not required).
  5. Stability: remains inoperative for through-faults, power swings and transformer inrush.
  6. Simplicity: least number of relays and contacts; easy maintenance.
  7. Economy: cost of protection proportional to the importance of the equipment.

These qualities often conflict, e.g. higher speed and sensitivity may reduce stability; good relaying balances them for each application.

  • 2070 Bhadra · 4 marks

Why protection system is needed in electrical power system? State and explain various causes of electrical faults.

Answer

Need for a protection system

A power system cannot be designed to be fault-free. When a fault occurs, very high current flows and voltage collapses. Protection is required to:

  • Disconnect the faulty part quickly, so that equipment (generators, transformers, cables) is not damaged by heating and mechanical forces.
  • Maintain continuity of supply to healthy parts of the system.
  • Preserve system stability (keep generators in synchronism).
  • Prevent fire, explosion and danger to human life.
  • Detect abnormal conditions like overload, under/over voltage and unbalance before they become faults.

Causes of electrical faults

  1. Insulation failure: ageing, overheating, moisture, contamination and over-stress break down insulation of cables, transformers and machines.
  2. Lightning: direct strokes or induced surges cause flashover of line insulators (the most common cause of overhead line faults).
  3. Switching surges: over-voltages during switching of lines and capacitor banks can flash over weak insulation.
  4. Mechanical causes: falling trees, birds, animals and kites bridging conductors; conductor breakage by wind or ice; excavation damage to cables.
  5. Environmental pollution: salt or industrial dust deposits on insulators cause flashover in wet conditions.
  6. Human error: wrong switching operations, closing on earthed equipment, poor maintenance.
  7. Equipment failure: defects in breakers, CTs, joints and terminations.
  8. Overloading: sustained overload overheats conductors and insulation, leading to breakdown.

Most faults (about 70–80 % on overhead lines) are single line-to-ground faults; three-phase symmetrical faults are rare (about 5 %) but are the most severe.

  • 2073 Magh · 4 marks

State the various methods used to provide backup protection.

Answer

Back-up protection operates, after a time delay, only when the primary protection or its breaker fails to clear a fault. The common methods are:

  1. Remote back-up: relays at the adjacent (upstream) station act as back-up. They see the fault but are time-graded so that the primary relay has the first chance.
    • Example: zone-2 and zone-3 of a distance relay, or a time-graded IDMT relay at the source end.
    • Independent of the primary station's CTs, DC supply and breaker, so very reliable; but it is slow and disconnects a larger part of the system.
  2. Relay back-up (local duplication): a second, independent relay at the same station (preferably a different principle, fed from separate CT cores and DC supply) trips the same breaker. Example: main-1 and main-2 distance relays on EHV lines.
  3. Breaker back-up (breaker failure protection): if the breaker does not open within a set time (about 150–200 ms) after the trip signal while current still flows, the breaker-failure relay trips all other breakers on the same bus that feed the fault.
  4. Centrally coordinated back-up: in modern substations with a central computer, the system state is monitored and, on failure of primary protection, the most suitable breakers are tripped.
 A          B          C
 ─[CB1]────┼──[CB2]────┼── F
  R1 (remote back-up)  R2 primary + local
                       BF relay trips bus B

Usually remote back-up is used on distribution systems and local relay + breaker back-up on EHV systems.

  • 2071 Magh · 2 marks

Define co-ordination with reference to protection system.

Answer

Coordination (discrimination) in a protection system is the selection of settings (pick-up current and operating time) of all protective devices connected in series, so that for any fault only the device nearest to the fault operates first, and the upstream devices operate only as time-delayed back-up if it fails.

  • It is achieved by time grading, current grading or both (IDMT relays).
  • A coordination time interval of about 0.3–0.5 s is kept between successive relays.
  • Example: for a fault on a distribution feeder, the feeder relay (0.2 s) trips before the transformer relay (0.6 s).
  • 2075 Baisakh · 6 marks

Explain the fault clearing process with the help of trip circuit.

Answer

Fault clearing is the sequence of events from the instant the fault occurs to the final interruption of fault current by the circuit breaker. It uses the trip circuit, which links the relay to the breaker.

Trip circuit

  Line ═══[ CB main contacts ]═══════
        │                       ▲
       CT                       │ opens
        │                       │
  ┌─────▼─────┐      ┌──────────┴──┐
  │ Relay coil│      │  Trip coil  │
  └─────┬─────┘      └──┬──────────┘
        │ closes        │
  (+)───┤ Relay contact ├──[52a]──(−)
       DC battery         aux. switch
  • CT/VT: reduce line current and voltage to relay level.
  • Relay operating coil: energised by the CT secondary.
  • Relay contact: in the DC trip circuit.
  • Trip coil: solenoid that releases the breaker mechanism.
  • Auxiliary switch (52a): closed when breaker is closed; opens with the breaker and breaks trip-coil current.
  • Battery (station DC supply): independent of the AC system, which may collapse during the fault.

Fault clearing process

  1. Fault occurs: current in the line rises sharply.
  2. Sensing: the CT secondary current rises proportionally and flows through the relay coil.
  3. Relay operation: when current exceeds the pick-up setting, the relay (after its set time) closes its contacts. This time is the relay time.
  4. Trip coil energised: DC current flows from the battery through relay contacts, trip coil and the auxiliary switch.
  5. Breaker opens: the trip coil plunger releases the latch; stored energy (spring/pneumatic) separates the main contacts. This is the opening time.
  6. Arc extinction: an arc forms between the contacts and is extinguished at a current zero by the arc-quenching medium (SF₆, vacuum, oil, air). This is the arcing time.
  7. Trip circuit reset: the auxiliary switch opens, de-energising the trip coil to prevent it burning.

Total fault clearing time = relay time + breaker opening time + arcing time, e.g. 1 cycle + 1.5 cycles + 0.5–1 cycle ≈ 3–4 cycles (60–80 ms at 50 Hz) for a modern EHV breaker.

  • 2074 Bhadra · 6 marks

Consider the system portion shown in figure below. (i) Sketch the zone of protection. (ii) Describe a possible backup scheme for failure of breaker 7 for a fault on line BC; is load 1 interrupted? (iii) Describe a means of clearing a fault on line BC without a momentary interruption of load 1. [Figure: Bus A is fed through breaker 1. Two parallel lines run from bus A to bus B: the upper line has breaker 2 at A and breaker 4 at B, the lower line has breaker 3 at A and breaker 5 at B. Load 1 is supplied from bus B through breaker 6. Line BC leaves bus B through breaker 7 and reaches bus C, from which Load 2 is supplied through breaker 8.]

Answer

Assumption: breakers 1–8 are as described; each line has a breaker at both ends and bus B connects breakers 4, 5, 6 and 7.

(i) Zones of protection

       [1]
  ┌─────┼──────── Bus A zone ──────┐
  │  ───┴──┬──────────────┬───     │
  └────────┼──────────────┼────────┘
          [2]            [3]
  ┌ Line AB1 ┐      ┌ Line AB2 ┐
  │    │     │      │    │     │
  └───[4]────┘      └───[5]────┘
  ┌────────┼──────────────┼────────┐
  │  ──────┴───┬──────┬───┴──      │ Bus B zone
  └───────────[6]────[7]───────────┘
           Load 1 ┌──┼────────┐ Line BC
                  │  │        │ zone
                  └─[8]───────┘
                     │ Bus C / Load 2
 Each zone overlaps its neighbours at
 every breaker (1,2,3,4,5,6,7,8).

Zones: bus A (1, 2, 3); line AB-1 (2, 4); line AB-2 (3, 5); bus B (4, 5, 6, 7); line BC (7, 8); load 1 feeder (6); load 2 / bus C (8).

(ii) Back-up for failure of breaker 7 (fault on line BC)

  • Normally relays of line BC trip breakers 7 and 8.
  • If breaker 7 fails to open, the fault is still fed from bus B.
  • Local back-up (breaker failure protection): after about 150–200 ms, the breaker-failure relay of 7 trips all other breakers connected to bus B: 4, 5 and 6.
  • Remote back-up (alternative): zone-2/zone-3 distance relays or time-graded overcurrent relays at bus A trip 2 and 3.
  • In both cases bus B is de-energised, so load 1 is interrupted (in the local scheme breaker 6 is tripped; in the remote scheme bus B loses supply).

(iii) Clearing BC fault without momentary interruption of load 1

The fault current from bus B must be interrupted by a breaker other than the one that isolates load 1. Possible means:

  • Provide two breakers in series on the BC line at B (or a breaker-and-a-half / ring bus arrangement), so that if breaker 7 fails the second breaker clears the fault and bus B remains healthy.
  • Use a double-bus with bus coupler arrangement: line BC and load 1 on different buses; the breaker-failure relay of 7 trips only the bus coupler and the breakers of BC's bus.

With these arrangements a fault on BC, even with a stuck breaker, is removed without de-energising the bus that supplies load 1.

  • 2078 Chaitra · 6 marks

The system of fig. 1(a) is rearranged as shown in fig. 1(b) where the bus arrangement at B has been changed to a ring bus arrangement. (i) Compare the operation of the two systems when a fault occurs on line BC with a. a failure of breaker 7, b. a failure of breaker 6. (ii) Sketch the zone of protection. [Fig. 1(a): Bus A is fed through breaker 1. Two parallel lines run from A to bus B: upper line with breaker 2 at A and breaker 4 at B, lower line with breaker 3 at A and breaker 5 at B. Load 1 is supplied from bus B through breaker 6. Line BC leaves bus B through breaker 7 and ends at breaker 8 at bus C; Load 2 is supplied from C through breaker 9.] [Fig. 1(b): Same system, but B is a ring bus with four breakers 4, 5, 6 and 7 in the ring. The line from breaker 2 joins the ring between breakers 4 and 5, the line from breaker 3 joins between breakers 5 and 6, Load 1 is taken from the ring between breakers 4 and 7, and line BC leaves the ring between breakers 7 and 6, ending at breaker 8 at bus C; Load 2 is fed from C through breaker 9.]

Answer

Ring at B (Fig. 1(b)) in order 4 → 5 → 6 → 7 → 4: line from breaker 2 between 4 and 5, line from breaker 3 between 5 and 6, line BC between 6 and 7, and load 1 between 7 and 4.

(i) Comparison of operation for a fault on line BC

Normal clearing

  • Fig. (a): breakers 7 and 8 trip. Load 1 continues.
  • Fig. (b): line BC is fed from the ring through 6 and 7, so breakers 6, 7 and 8 trip. The ring is opened, but load 1 is still fed through 4 from the line of breaker 2; no load is lost.

(a) Failure of breaker 7

ArrangementBack-up actionLoad 1
Fig. (a) single busBreaker-failure relay trips all bus-B breakers 4, 5, 6 (or remote 2, 3 trip)Interrupted
Fig. (b) ring busBreaker adjacent to 7 in ring, i.e. 4, trips (with 6 and 8)Interrupted, because load 1 lies between 7 and 4

In the ring bus fewer breakers trip and the lines from A remain connected to the ring, but load 1 is still lost because it is placed next to breaker 7.

(b) Failure of breaker 6

ArrangementBack-up actionLoad 1
Fig. (a)Breaker 6 is the load-1 breaker and is not involved in clearing a BC fault; 7 and 8 clear itNot interrupted
Fig. (b)Breaker adjacent to 6, i.e. 5, trips (with 7 and 8); line from 3 is isolatedNot interrupted; load 1 fed through 4 from line 2

Conclusion: the ring bus isolates a stuck breaker by opening only its neighbour in the ring, so a breaker failure removes at most one extra circuit instead of the whole bus. To protect load 1 fully, it should be placed in the ring between two line-in-feeds (e.g. between 4 and 5), not next to line BC.

(ii) Zones of protection

 Bus A zone (1,2,3)
      ─┬──[2]─────────[4]─┬─
       │  line AB1 zone   │ ring B:
      ─┴──[3]────┐        │
                 │  ┌─[4]─┴─[5]─┐
     line AB2 ──►└──┤ (2)   (3) ├
                    │           │
          Load 1 ◄──┤[7]─────[6]│
                    └─────┬─────┘
                          │ line BC zone
                         [8]── bus C ─[9]─ Load 2
  • Line zones: AB1 (2 and ring breakers 4, 5), AB2 (3 and 5, 6), BC (6, 7, 8), load 1 (7, 4), load 2 (9).
  • In a ring bus there is no separate bus zone: each ring section between two breakers belongs to the circuit connected to it.
  • Each zone overlaps its neighbour at the breaker between them.

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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