See4423 chapter1 introduction[1]

3,126 views
3,004 views

Published on

introduction to power system protection

Published in: Education, Technology, Business
0 Comments
4 Likes
Statistics
Notes
  • Be the first to comment

No Downloads
Views
Total views
3,126
On SlideShare
0
From Embeds
0
Number of Embeds
13
Actions
Shares
0
Downloads
203
Comments
0
Likes
4
Embeds 0
No embeds

No notes for slide

See4423 chapter1 introduction[1]

  1. 1. OFFICE: CENTER OF ELECTRICAL ENERGY SYSTEM (CEES) FACULTY ELECTRICAL ENGINEERING UNIVERSITI TEKNOLOGI MALAYSIA 81310 UTM JOHOR BAHRU JOHOR DARUL TAKZIMROOM NO: P07-412Tel: 07-5536262
  2. 2. WELCOME • POWER SYSTEM ENGINEERING • SEE 4423• LECTURER: ASSOC PROF MD SHAH MAJID• ROOM NO: P08-409/P07-417(CEES)• TEL: 07 -5535295/5536262 (CEES)• E-mail: mdshah@fke.utm.my, mdshah@ieee.org
  3. 3. SEE 4423 POWER SYSTEM ENGINEERINGSEMESTER II SESSION 2009/2010 Assoc Prof MD SHAH MAJID
  4. 4. SEE 4423 POWER SYSTEM ENGINEERINGSEMESTER 1 SESSION 2009/2010 4
  5. 5. CHAPTER 1INTRODUCTION TO PROTECTION SYSTEM
  6. 6. Introduction:Need for Protective Systems• Power system consists of; generators, transformers, transmission and distribution lines, etc…..etc. – Sistem kuasa mengandungi: penjana, pengubah, talian penghantaran dan pengagihan dsbnya• Short circuits and other abnormal conditions often occur on a power system – Litar pintas dan keadaan tidak normal kerap berlaku pada sistem kuasa• Heavy current associated with short circuits is likely to cause damage to equipments – Arus yg tinggi ketika litar pintas mungkin menyebabkan kerosakan pada peralatan jika geganti perlindungan dan pemutus litar tidak di bekalkan untuk perlindungan bagi setiap seksyen sistem kuasa
  7. 7. Need for Protective Systems• Short circuit is a faults (power engineer)• Failure of conducting path due to a break in a conductor is a type of fault• Fault occurs – automatic protective device is needed to isolate the faulty element as quickly as possible to keep the healthy section of the system in normal operation• Fault must be cleared within fraction of a second• Uncleared short circuits may cause total failure of the system
  8. 8. • A protective scheme includes transducers, protective relays and circuit breakers to isolate the faulty section of the system from the healthy section• Protection is needed not only against short circuit but also against any abnormal condition – Overspeed of generators and motors – Overvoltage – Underfrequency – Loss of excitation – Overheating of stator or rotor of an alternator – Etc…etc…etc• Protection is a pre-requisite for an effective and reliable system• A protective relay does not anticipate or prevent the occurrence of fault, rather it takes action only after a fault has occurred – except Buchholz relay; a gas actuated relay
  9. 9. Nature and causes of Faults• Faults are caused either by; – Insulation failure – – Conducting path failure • results in short circuit• Faults on transmission and distribution lines are caused by overvoltage – Lightning and switching surges – External conducting objects falling on overhead line.• Birds also may cause faults on overhead line if their bodies touch one of the phase and earth wire
  10. 10. • If conductors are broken – failure of conducting path and the conductor becomes open-circuited – If broken conductor fall to the ground, results in short circuit• Joint failure on cables and overhead lines also cause a failure of the conducting path• Opening of one or two of the three phases makes the system unbalanced – Set-up harmonics
  11. 11. • Other causes of faults in o/h line – Direct lightning strokes – Aircraft – Snakes – Ice and snow loading – Abnormal loading – Storm – Earthquakes – Creepers – Etc…..etc……etc……etc
  12. 12. • Cables, transformer generators and other equipment: – Failure of solid insulation due to aging – Heat – Moisture – Overvoltage – Mechanical damage – Accidental with earth – Flashover due to overvoltages – Etc…etc…..etc
  13. 13. Types of faults• Symmetrical faults• Unsymmetrical faults• Symmetrical faults – Kerosakan tiga fasa atau tiga fasa ke bumi• Unsymmetrical faults – Single line to earth, line-to-line, double line to earth, open-circuited phases
  14. 14. • Faults can interrupt power system in several ways: – Heavy current to flow • Effect : overheating of power system component – Fault is a short circuit and exist as electrical arc and liquid e.g air • Effect: equipment faulty and fire
  15. 15. • Fault – can increase/decrease voltage system outside its acceptable range – Can cause unstable three phase system, improper operation of three phase equipments – Prevent power flow – Can cause system to be unstable and collapse
  16. 16. • Faults incur MONEY.• If fault is isolated as quickly as possible and accurate –less money required• Cost of protective equipment is 5% of the total cost of the system
  17. 17. • PROTECTION DOES NOT MEAN PREVENTION – A protective relay does not anticipate or prevent the occurrence of fault, rather it takes action only after a fault has occurred – except Buchholz relay; a gas actuated relay
  18. 18. Functions of Protective System• Fast and automatically opens the faulty section in the power system• Increase system reliability and security – only affected area will be isolated and maintain the healthy line – Ensure consumers receive continuity of supply• Delay in isolating the fault – System unstable, loss of synchronism, total failure of the system – Fire
  19. 19. Requirements of Protective System• The basic requirements of a protective system are as follows: – Discrimination/selectivity – Sensitivity – Reliable – Stability – Speed
  20. 20. (Discrimination/Selectivity)• Ability to select either to operate or not – Keupayaan untuk memilih sama ada bekerja atau tidak• Select to isolate the faulty section only (the rest normal condition) – Memilih untuk mengasingkan bahagian rosak sahaja (yang lain berkeadaan normal)• The nearest circuit breaker will trip – Pemutus litar yang hampir sahaja terbelantik (trip)
  21. 21. Radial protection system SD2 SD3SD1SD1 SD2 SD3
  22. 22. Sensitivity• Relay should operate when the magnitude of the current exceeds the preset value• This value is called pick up current• Should not operate when current is below its pick-up value• Should be sufficiently sensitive to operate when the operatinfg current just exceeds the pick-up value – relate with minimum operating current
  23. 23. Reliability• A protective system must operate reliably when a fault occurs in its zone of protection• Failure may due to its protective element system; CT, PT, CB, relay etc….• Reliability of protective system 95%
  24. 24. Stability• A protective system should remain stable even when a large current is flowing through its protective zone due to an external fault• Concerned circuit breaker is supposed to clear the fault
  25. 25. No Yes SD1 SD2 SD3 Radial system protection
  26. 26. Speed-fast operation• Isolate fault as quickly as possible (at shortest time possible )• Isolate disturbances before loss of synchronism and plant stop operation – time should not exceed critical clearing time• Speed balanced with economy – Cost of protective equipment should be relevant to the cost of the protected zone• Avoid fire to equipments, interruption of supply to consumer, voltage drop operating time; 1 cycle, ½ cycle are also available; distribution system > 1 cycle
  27. 27. Operating time• Total time to accomplish the isolation• Calculated at the instance of fault until the given trip signal• Must be low for the sake of plant and equipment safety• Delay time – for discrimination
  28. 28. Protection economy• Protection system can be designed as: – Simple and cheap – Complex and expensive• Baesd on: – The cost of fault – Safety level requirement
  29. 29. Protection economics…• Cost of fault – Cost of damage toward the plant – Cost loss of revenue due to cut-out supply – Cost of customers confidence• Higher the cost of fault, more expensive the protection system• Higher the plant kVA, more complex protection system required
  30. 30. Protection system at customer level• Simple fuse – protect equipment and certain circuits• Fius ringkas melindungi alat-alat dan litar tertentu• Miniature circuit breaker- MCB
  31. 31. Protection system at distribution level• Fuse and switch fuse• Automatic reclosing circuit breaker – For rural area – Transient fault is corrected by self clearing
  32. 32. Protection system at transmission level• Technical consideration outweighs the economy• 275kV dan 400kV System require a protection system which is: – very reliable – Full discrimation – High speed• Expensive and complex• Provide back-up protection
  33. 33. Zone of Protection GENERATOR PROTECTION CIRCUIT BREAKER• Power system is divided into HV SWITCHGEAR several zones PROTECTION• Each zone of protection is TRANSFORMER provided with 2 types of PROTECTION protection: EHV SWITCHGEAR – Primary protection PROTECTION – Backup protection TRANSMISSION LINE PROTECTION EHV SWITCHGEAR PROTECTION ZONE OF PROTECTION
  34. 34. Protection zone• Region or area encompass a protection system
  35. 35. Protection zone CB3 CB3CB1 CB2 SD1 SD2 SD3 Radial protection system
  36. 36. Main/primary protection• In general primary protection is provided for each transmission line segment, major piece of equipment and switchgear• If fault occurs, it is the duty of the primary protective scheme to clear the fault• First line of defence - Responsible to isolate fault as quickly as possible• Fails, back-up protection clears the fault
  37. 37. Backup Protection• Act when primary protection fails – second line of defence• Usually several back up protection scheme will act to control the power system• Longer time delay
  38. 38. Radial protection systemSD1 SD2 SD3 Main protection Back-up protection
  39. 39. Zone of protection• Zone of protection usually overlap
  40. 40. Fault detection• Current magnitude• Current in abnormal path (earth)• Current balance (current out#current in)• Voltage balance• Changes in impedance• Protective system damage(Buchholz), power flow direction, temeprature & pressure
  41. 41. Protective system components1. Circuit breaker2. Transducer – current transformer (CT), voltage transformer (VT), potential transformer (PT)3. Communication links – Pilot wire – Radio link – Overlapping signal, power line carrier (PLC)
  42. 42. Protective system components4. Relay – Electromagnetic – Static – improve reliability,versatile, fast response (1/4 cycle) – Microprocessor- VLSI teknologi – current interest to power engineers; adv:attractive flexibility due to programmable approach, can provide protection at low cost and compete with conventional relays5. Fuse
  43. 43. Transducer Serves as a sensor to detect abnormal system conditions and to transform the high values of shortcircuit current and voltage to a lower values 43
  44. 44. Protective relays• Process the signals provided by the transducers which may be in the form of current, voltage or a combination of current and voltage 44
  45. 45. Circuit breaker• Mechanical device used to energize and interrupt an electric circuit 45
  46. 46. Ex. Protective system: power flow balance 2. CT, PT1. PL 1. P.L 4. Relay R1 R2 3
  47. 47. Classifications of relay based on its function• Overcurrent relay• Undervoltage relay• Impedance relay• Under frequency relay• Directional relay etc….etc….etc
  48. 48. Classification of protective scheme• Overcurrent protection• Distance protection• Current carrier protection• Differential protection
  49. 49. Transformer in a protective system • CT assume ideal Feeder Ip • Normal rating: conductor – 1:1, 2:1. 2.5:1, 4:1, 5:1 – 20:1, 40:1, 100:1, 200:1, 300:1, 600:1 Is CT – 1000:1, 2000:1, 4500:1 • Secondary coil is connected to a Vp “sensitive device” eg voltmeterVs • Low Stray impedance so that voltage drop small VT
  50. 50. Current transformer (CT)• Normal rating – 1A (Europe) – 5A (USA)• Example : – 50:5, 100:5, 150:5, 200:5, 250:5 – 450:5, 500:5,……1000:5, ……, 6000:• Secondary coil is connected to a “current sensing device” of zero impedance• Shunt impedance high so that I0 low.
  51. 51. (current transformer CT)• Working principle similar to voltage transformer• Supply by current source• Primary winding connected in series with power circuit – Carries full load current• Transformer impedance (referred to secondary and can be neglected)• Secondary coil connected to load (burden)
  52. 52. (current transformer CT)• Secondary winding feeds the protective system• Current is reduced but must be almost the same with the power system current• Secondary rated current 5 A or 1 A• Fault current 10 – 20 X rated current. Transformer must be capable to operate reliably/accurately at this value• Burden usually usually small – flux do not saturate
  53. 53. Current transformer construction Bar Type Core Power system current Primary winding (1 turn)- from power systemSecondary winding woundon the core
  54. 54. Current transformer constructionWound type
  55. 55. design• CT – similar to a normal transformer emf equation• Average induced voltage = product of No. of turn and rate of change of flux magnet; eaverage = N(dΦ/dt)• Knee point voltage(rms) = 4.44 BAfN
  56. 56. Burden• Defined as the load connected across its secondary CT – express in VA (VA taken as nominal secondary current of CT) or – Impedance (at the rated secondary current at a given power factor usually 07. lagging• Increase in impedance – increase burden• CT unloaded if secondary winding is short circuited
  57. 57. Example• 5VA burden at 1A transformer, gives 5 Ohms impedance.• 5VA/1A = 5V• impedance = 5V/1A = 5Ω• or• At 5A CT• 5VA/5A• impedance = 1V/5A = 0.2 Ω
  58. 58. Example• CT ratio 300/1, core area 40x30 mm• Voltage at knee point?• V = 4.44 x 0.0018 x 300 x 50 = 120 V• 1.5 x 40 x 30 x10-6 = 0.0018 Wb ( B for sheet steel, 1.5 tesla at knee point)
  59. 59. CT open circuit• If burden high, Es high– exceed Vkp (knee point voltage)• Io high ; I2 less• Limited value when the secondary CT open circuit; I2 = 0;• Then N1I1 = N2 (I2+Io) = N2Io• This will drive the CT to saturation level• dΦ/dt = 100 x Vkp (induced = 100 Vkp )• Cause insulation failure and overheated
  60. 60. Current transformer equivalent circuit Ip’ x r Is I0’ Ic ’ Load (burden) Es Im’ Vs R Ip = primary current(nominal ratio) Kn I0 = primary excitation current (no load) I p = I p / Kn I0 = I0 / Kn
  61. 61. Current transformer equivalent circuit Ip’ x r Is I0’ Im’ Ic ’ Load (burden) Es Vs REs Ip’ I p = Is + I0 Is θ Is = I p − I 0 I0’Ic ’ φ Im’
  62. 62. relationship Es and φ (flux)N=no of turn in secondary dφ es = N dt π es = Nωφm sin(ωt + ) 2φ = φ m sin ωt Nωφ mes = Nωφm cos ωt Es = ∠90° 2
  63. 63. (Nominal Turn Ratio)• Determine from the given current ratio – Eg.1000/5 (bar type) Ip 200 K n = nominal ratio = = Is 1
  64. 64. Actual Ratio• Actual ratio value Ip and Is ′ Ip actual ratio = winding ratio × Is ′ Ip = Kt × Is
  65. 65. • Winding ratio= secondary coil turn if transformer is of bar type l – Kt = N2/1• Ideally ′ Ip =1 Is• Practically ′ Ip 〉1 Is Actual turn≥ nominal ratio
  66. 66. CT error• Current error or ratio error• Phase error• Composite error
  67. 67. Ratio error or current error nominal ratio -actual ration ratio error = × 100% actual ratioNominal ratio= Kn ′ Ip actual ratio = K t ×Turn ratio= Kt IsNo compensation Kn=Kt
  68. 68. Cont’d Ip K n − Kt Is ratio error = Ip Kn=Kt Kt Is (no compensation) K n I s − Kt I = p Kt I p difference in I s − I p magnitude betweenratio error = × 100% Ip’ and Is I p
  69. 69. Phase ErrorDifference in angle between Ip’ and Is Phase error= θ Es Ip’ Is α θ I0’ φ Im’
  70. 70. Compensating winding• Secondary coil winding is reduced (1 or 2 turn) to compensate current error due by I0’.• Secondary coil current will be high, but the value will be minimised by the excitation current component• This can minimised the current error
  71. 71. Example T1• A current transformer 50 Hz has a secondary current of 3A,Sebuah• Secondary impedance circuit (0.6+j0.45),• Maximum flux 0.253 weber• Magnetisation current and core loss current (primary circuit) at this load are 0.2 and 0.15 A respectively• Turn ratio1:10, assume no compensation• Determine no of turn, ratio error and phase error of transformer
  72. 72. CT equivalent circuit r Ip’ Is=3 ∠ αº I0’ Im’ Ic ’ Load (burden) Es Vs = 0.6+j0.45 Es Ip’ I p = I p / Kn I0 = I0 / Kn Is θ I p = Is + I0 I0’ Ic ’ φ Is = I p − I 0 Im’α
  73. 73. N1:N2 = 1:10 Kt = 10 = KnIs=3 ∠ αº Im = 0.2 A Es = I s Z Ic = 0.15 A = 3∠α ° × 0.75∠36.87° N s × φm × ω = 2.25∠(α + 36.87)° Es = ∠90° 2 Es (α + 36.87)° = 90° Ip’ α = 90 − 36.87α Is = 53.13° θ I0’ Ic’ Im’ φ
  74. 74. N s × φm × ω = 2.25 2 2.25 × 2 Ns = 0.253 × 2π × 50 = 40 ∴ N1:N2 = 4:40
  75. 75. I 0 = 0.20 + j 0.15 Is=3 ∠ αº ′ 1 I0 = (0.20 + j 0.15) × Im = 0.2 A 10 Ic = 0.15 A = 0.02 + j 0.015 ′ ′ I p = I0 + Is Es = 0.020 + j 0.015 + 3∠53.13° Ip’ = 3.024∠52.9975° Is θ I0’Ic’ φ Im’
  76. 76. ′ Is + I p 3 − 3.024ratio error = = × 100% ′ 3.024 Ip = −0.794 % phase error = 53.13 − 52.9975 = 0.1325
  77. 77. Ex 1• A current transformer has 15 turn on the primary and 75 turn on the secondary. The relay burden is (1.2+j2.6)Ω and secondary winding impedance of the transformer is (3.6+j0.4)Ω. If the magnitude of the secondary voltage is 26 volt and magnetization current referred to secondary is 0.24∠-45o amp. Determine the ratio error and phase error.
  78. 78. Ex 2• A new non compensated CT with a cross sectional area of 40 cm2 has a nominal ratio of 2000/5. When 5A is supplied to a relay, the combination of load relay and secondary winding impedance is (10+j5.77) Ω. If the magnetisation amp-turn is 100 and core loss amp-turn is 50, determine the ratio error and phase error. Using the same data, calculate also the compensating turn to reduce the ratio error to a minimum.
  79. 79. Ex. 3• A CT, 50 Hz has a primary winding of 10 turn and its cross sectional area is 8 cm2. Secondary current is 5A. The total load and secondary impedance is 0.8+j0.5 Ω. The excitation current required at the primary winding to establish a working flux with the secondary open is 0.6∠45o amp. If the flux density is 0.1878 wb/m2,determine the nominal ratio of the transformer. Determine also the ratio error and phase error of the transformer.
  80. 80. Composite error• BS 3938:1973 – (r.m.s.) value; difference between ideal secondary current (i.e CT is ideal ; no excitation component) with actual secondary current.• Encompass (current error, phase error and harmonic effect)
  81. 81. Accuracy limit current of a protection system CT• Accuracy limit current (saturation current) – The maximum current a CT can sustain before exceed its accuracy – Specify either in term of primary current or secondary current• Accuracy limit factor (saturation factor) – Ratio of accuracy limit primary current and rated primary current – Standard values are 5, 10, 15, 20 and 30
  82. 82. CT specification• Expressed in – VA at rated current /class accuracy/accuracy limit factor – Burden standard rated values • 2.5,5,7.5,10,15 and 30 VA – 2 accuracy class: 5P and 10P – gives composite error at rated accuracy limit 5% and 10% respectively – Standard accuracy limit factor: • 5, 10, 15, 20 and 30
  83. 83. Rating of CT • Method to explain CT – 10VA/5P/25 Accuracy limit current is 25 x rated current Transformer works in good condition/reliably until 25 x itsRated burden Class – error rated curent do not exceed 5%
  84. 84. Ex 1• CT 10VA/5P/15• Calculate the accuracy limit factor when the burden is half of its initial value.• Secondary resistance of CT 0.15 ohm
  85. 85. SolutionBurden impedance 10VA at ratedcurrent 5A r Es VA V ×I = Burden (R) = I × R× I V = I 2R 10 R = 2 = 0.4 ohm 5
  86. 86. SolutionTotal impedance referred to rsecondary winding= 0.15 + 0.4 = 0.55 ohm Es V Burden (R)Accuracy limit factor= ratio accuracylimit current and rated currentAccuracy limit current= Accuracy limit factor × Rated current accuracy limit current = 15 × 5 = 75 A
  87. 87. secondary emf = 0.55 × 75 V = 41.25 V r Es Burden (R=0.4) When burden is half R = 0.2 ohmTotal secondary impedance= 0.15 + 0.2 = 0.35 ohm
  88. 88. new accuracy limit current = 41.25 / 0.35 = 117.8 Anew accuracy limit factor = 117.8 / 5 = 23.56 Smaller the burden, the higher the accuracy Therefore : burden has to be low possible
  89. 89. CT Class X• BS 3938 – CT for special application is known as Class “X’• Expressed in turn ratio, knee point voltage, excitation current at the specified voltage and secondary winding resistance – Application – scheme where the phase fault stability and/or accurate time grading required• Transformer rating specify in term of e.m.f maximum that can be use by the transformer• knee point – Point where an increase of 10% in secondary e.m.f require an addition of 50% excitation current
  90. 90. Definition of knee pointExcitationvoltage +10% Knee point +50% Excitation current
  91. 91. Voltage transformer/potential transformer(VT/PT)• Supply voltage lower than the sistem voltage• Nominal sekunder voltage110V• 2 types – Wound ( electromagnetic) type(> 132 kV non economical) – Capacitor type- CVT• Element X and C consist of tuning circuit – to reduce the ratio and phase angle error secondary voltage
  92. 92. 1/ωC2 = ωL pd 50 Hz; V2 = V’ Talian C1 LVp Rb V’ Vs C2 V2 Xb V’ = 12 kV, C1 = 2000pF L C2 V2 110 V V’
  93. 93. Protection techniques• Protection scheme: – Arrangements of CT, VT, pilot wire and relay so that the required operating characteristics can be obtained• Each section of the power system network is monitored by several protection scheme.
  94. 94. Main/primary scheme usually consists of: Unit SchemeBackup scheme consists of: Non Unit Scheme
  95. 95. Unit scheme• ZONE OF PROTECTION IS WELL DEFINED
  96. 96. Unit protection scheme CT CT Protected zone P Pilot wire R relay QDifference current scheme: circulating currentFAULT:Non similar current flow in the relayCB will trip R
  97. 97. CT Healthy and fault e1 Protected zon e2 external zone (through fault): No current in relay, e1=e2 R R FAULT: Current flow in circuit, result e1≠e2 Differencec current scheme: voltage balance
  98. 98. Balance power direction CT, PT PL P.L Relay R1 R2 Trip TripRestrain Restrain
  99. 99. CB1 CB2A B R1 R2 Trip Trip Restrain Restrain Power flow entering the feeder (from point A) will cause relay (R1) close trip contact CB1 Power flow exit fron the feeder (at point B) will cause relay R2 rotate and close restrain contact of the further circuit breaker (remote) (CB1) (relay R2 send restrain signal CB1)
  100. 100. CB1 CB2A B R1 R2 Trip Trip Restrain Restrain Power enter at A and exit at B, block trip will prevent the CB from opening.
  101. 101. CB1 CB2A B R1 R2 Trip Trip Restrain Restrain Power enter at A and at B (fault occurs), both CB open
  102. 102. CB1 CB2A B R1 R2 Trip TripRestrain Restrain Power enter at A and no power at B, circuit breaker A, CB1 will open
  103. 103. NON UNIT SCHEME supervise/monitor one or several points in the power system signal will be detected at the points this signal is the input to the relay section monitored/supervised is not clearly definedZONE OF PROTECTION IS NOT WELL DEFINED
  104. 104. 2 examples of non unit scheme (1.) Over Current Protection (2.) Distance Protection
  105. 105. ‘click here to add title•Q & A

×