Industrial Power System Protection POWER SYSTEM PROTECTION

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1 Industrial Power System Protection POWER SYSTEM PROTECTION

2 Power System Components Generation Transmission Distribution

3 WHY IT IS REQUIRED The risk if there is no protection system Electric Shock Damage etc

4 Electric Shock 1 ma 2 ma Barely perceptible, no harmful effects 5 ma 10 ma Throw off, painful sensation 10 ma- 15 ma Muscular contraction (kemungkinan bahaya) 20 ma 30 ma Impaired breathing (Melemahkan pernafasan) 50 ma and above Ventricular fibrillation and death Body resistance : 500 Ω 50 kω Indirect accident (Kemalangan secara tak langsung)

5 RISK There are two ways we can be at risk : 1. Touching live parts of equipment or system that are intended to be live, this is called direct contact. 2. Touching conductive parts which are not meant to be live, but have become live due to a fault, this is called indirect contact.

Indirect contact And corresponding protective

6 BRIEF THEORY Standards and regulations distinguish two kinds of dangerous contact : Direct contact Indirect contact And corresponding protective measures Fig 2 : Direct contact Fig 3 : Indirect contact

7 Akibat

8 Berita Harian Jumaat 13 Julai 2007

9 AKIBAT (CONSEQUNECES) Protection system can avoid this damage. Right setting. Maintenance etc

10 MEDIUM INDUSTRY - TNB ELECTRIC ARC EFFECT Accident happened because a lack of management and the personnel does not follows the working procedures

11 MEDIUM INDUSTRY - TNB ELECTRIC EXPLOSIVE EFFECT

12 The Role of Protection in Power System Industrial Power System Protection The investment involved in a Power System is so great that proper precaution must be taken to ensure that the equipment not only operates as nearly as possible to peak efficiency, but also that it is protected from accidents.

13 Industrial Power System Protection Faults Faults can damage or disrupt power system in several ways : Large fault currents will cause overheating of power system components. The extremely high temperatures in arcs due to short circuits will vapourize any known substance, causing equipment destruction and/or fire. The system voltages can be lowered or raised outside their acceptable ranges. Three phase systems can become unbalances thus causing three phase equipment to operate improperly. Faults block the flow of power. Faults can cause the system to become unstable and 'break-out'.

14 Industrial Power System Protection Faults Faults cannot be avoided due to: Human error Ageing Natural disaster Animal

15 Industrial Power System Protection Protective Relaying It should be realised that protective relaying can minimise:- The cost of repairing the damage. The spread of fault to other equipment causing damage from overheating and abnormal mechanics forces. The propagation of the effects of fault to other parts of the system resulting in widespread outages or consequential damages to other relatively unimportant equipment. The time equipment is out of service. The loss in revenue.

16 Industrial Power System Protection Selection And Application Of Power System Equipment The current that occurs in power systems is usually large and dangerous. The selection and application of power system equipment is, therefore, made with a view to promoting : Safety of personnel Uninterrupted power supply production Long life of equipment Ease and low cost of maintenance Amongst all the points mentioned above, safety is the foremost requirement and no compromise can be made on this aspect.

17 Industrial Power System Protection The Role of A Transmission Protection System The basic role of a transmission protection system is to sense faults or lines or at substation and to rapidly isolate those faults by opening all incoming current paths. Designed not to prevent fault, but to respond and minimise their effect.

18 The Protection System must fulfill the requirements of :- 1. Rapidly and automatically disconnecting the faulty section of the power network, and 2.Minimising disconnection of healthy plant, thus ensuring maximum security of supply to consumers

19 Industrial Power System Protection Fault detection are:- Fault Detection Magnitude of current flowing. Current in abnormal path (e.g. Earth). Current balance (e.g. current in = current out?). Voltage balance (usually derived from current). Direction of Power. Change of parameter (e.g. Z1 in distance protection). Damage (e.g. Buchholz protection). Non electrical parameters.

20 Causes Of Failure In Power System Components Industrial Power System Protection Power systems are subject to many kinds of fault. The principal types are (short circuits): Three Phase With And Without Earth Connection; Phase To Phase (Two-Phase) ; Phase To Earth (Single-Phase); Double Phase To Earth (Phase-Phase-Earth) The occurrence of fault is inevitable because of ageing, human error, external interference etc.

21 Power System Protection Requirements Industrial Power System Protection Power system protection equipment must have the ability to trip circuit breakers when faults occur. Reliability When a fault occurs the relay is required to operate with a fast response. In order to get a reliable system, power system equipment must be designed, installed and maintained correctly. Selectivity When a fault occurs, the protection system is required to select and trip only the nearest circuit breakers. This property of selective tripping is also called 'discrimination' and is achieved by two general methods : time graded systems and unit systems. The relay must also not respond to abnormal but harmless system conditions such as switching transients or sudden changes in load.

22 Industrial Power System Protection Sensitivity Power System Protection Requirements The relay must not fail to operate even in border line situations when operation is required. A protective system is said to be sensitive when it will operate for very small internal fault currents. Speed The relay should make the decision to act as fast as possible. This is done in order to maintain a healthy system by removing faulty sections as quickly as possible. By so doing, damage to the equipment as well as the probability that synchronism will be lost be minimized.

23 Industrial Power System Protection Back-Up Protection Operate when the main protection fails to operate. Often to act as main protection when the main protection is taken out of service for maintenance. Some delay must be introduced.

24 Industrial Power System Protection Components Of Protection Circuit breakers To switch on and switch off Transducers (e.g. CTs, VTs) To provide currents/voltages in the secondary which are proportional to those in the primary. Communication links Compare electrical quantities at the two ends of the protected section. Pilot wires Radio links Superimposed signals on power conductors (power line carrier, PLC)

25 Industrial Power System Protection Components Of Protection Relays To isolate a faulty section with the least interruption to service. Electromechanical - solenoid, induction Electronic or static relays Fuses A fuse is a combination of a protective device and a circuit breaker.

26 Industrial Power System Protection Protection Economics Can be relatively simple and cheap, or they can be extremely complex and expensive The choice of the complexity depends mainly on two factors. The cost of faults. The desired level of supply security.

27 Industrial Power System Protection Protection Economics Faults costs are a combination of the:- Cost of damage to plant The cost of lost revenue through supply disconnection The cost of lost consumer goodwill The higher the fault costs the more expensive may be the protection. The higher the MVA the more complex will be the protection system. The protection scheme applied at various levels are:- At Consumer Level Simple fuse Also extensive use of miniature circuit breaker (MCB)

28 Industrial Power System Protection Protection Economics At Distribution Level Fuses Overcurrent protection Security of supply is high priority (ring & duplication feeders). At Transmission Level Complex protection schedules are used. Must be fast to maintain system stability. Must possess good discrimination to minimise supply disruption.

29 Protection Zone

30 Power System Network Radial feeders Parallel Feeders

31 Power System Network (cont.) Ring System Interconnected System

32 2.9 CURRENT TRANSFORMER CT s behave in principle like any other transformer, but they are supplied by a current source The primary winding is in series with the power system network and carries the full system network.

33 CURRENT TRANSFORMER (CONT.) The secondary winding feeds the protection system with a current (and a voltage), this current being as near as possible a faithful replica of the power system current but reduced by some factor N, the CT ratio. Besides reducing the current level, the C.T. also isolates the relay circuit from the primary circuit which is a high voltage power circuit, and allows the use of standardized current rating for relays. CT s are usually designed to have rated secondary current of 5 amp or 1 amp.

34 CURRENT TRANSFORMER (CONT.) CT ratio is required to be maintained constant over a wide range of primary current, up to say 10 to 20 times rating under conditions, to ensure accurate reproduction of the primary current in magnitude and phase. Different from instrument CTs because it is required to be accurate over the normal working range of currents. For protection, CT has to perform reliably at normal currents as well as at fault currents.

35 CURRENT TRANSFORMER (CONT.) This demands that the load on the secondary winding - the burden - must be of sufficiently low impedance so that the voltage, and hence the flux, does not rise excessively under maximum primary current conditions. Typically CT ratings are VA.

36 CURRENT TRANSFORMER (CONT.) Precaution must be taken to never open the secondary circuit of a current transformer while current is flowing in the primary circuit. Inducing voltage peaks of several hundred volts across the open-circuited secondary. This is dangerous situation, could easily receive a bad shock First short-circuit the secondary winding and then remove the component

37 Core Material Magnetization.jpg The magnetization characteristic of (a) cold-rolled grain-oriented silicon steel (b) hot-rolled silicon silicon steel ( c) nickel-iron.

38 Core Material (CT) Nickel-iron core : highest permeability, low exciting current, low errors,and saturation at relatively low flux density. Suitable for CTs used for meters and instruments. Cold-rolled grain-oriented silicon steel : high permeability, high saturation level, reasonably small exciting current and low errors is used for protective relays. Hot-rolled silicon steel : lowest permeability. Not suitable for CTs.

39 CT (cont.) I p ' x r I s I m ' I o ' I c ' E s V s Beban R Circuit connection Circuit model referred to secondary

40 CURRENT TRANSFORMER CONSTRUCTION Either bar primary or wound primary

41 bar primary CURRENT TRANSFORMER CONSTRUCTION (Cont.) Primary side one turn only Toroidal core..

43 CURRENT TRANSFORMER CONSTRUCTION (Cont.) Wound Primary A CT has been treated as being of conventional construction. The primary winding conductor is of large section, being designed to carry the short-circuit current of the system. A CT of 100/5 amperes rating might have 4 primary turns and 80 secondary turns (neglecting turns compensation)

45 Current Transformer Errors The secondary referred equivalent circuit is shown below (neglecting primary leakage reactance and resistance) I p ' x r I s I m ' I o ' I c ' E s V s Beban R CT equivalent circuit

46 Current Transformer Errors (cont.) If CT has a nominal ratio Kn, then I p ' = I p /K n... (2.1) and I o ' = I o /K n... (2.2) where I p = primary current I o = no load primary excitation current

47 Current Transformer Errors (cont.) The phasor diagram for the CT, taking flux φ as reference is shown below E s I p ' I s θ I c ' I o ' I m ' φ

48 Current Transformer Errors (cont.) The relationship between E s and φ is obtained from e = N (dφ / dt)... (2.3) If φ = φ m sinωt... (2.4) Then e = N s ωφ m Cos ωt... (2.5) where N s is the number of secondary turns Error occur due to burden is shunted with excitation impedance. A small amount of input current is used to excite core, which reduce the current flow to the burden. so, I s = I p ' - I o '... (2.6) where I s is the secondary current.

49 CT ERROR DEFINITIONS (cont.) nominal ratio actual ratio Ratio error = x 100 (2.7) actual ratio If nominal ratio = K n and turns ratio = K t, then I p ' K n -K t I s ratio error = x 100% I p ' K t I s

50 CT ERROR DEFINITIONS (cont.) K n I s -K t I p ' Ratio error = x 100% (2.8) K t I p ' If the CT has no turns compensation, then K n = K t, thus I s - I p ' Ratio error = x 100% (2.9) I p ' This is the difference of magnitude between I p ' and I s.

51 CT ERROR DEFINITIONS(cont) Phase error is the angular difference between I s and I p ' Phase error = θ sin θ, for small θ (in radians)

52 Composite Error This is defined in BS3938 : 1973 as the r.m.s value of the difference between the ideal secondary current and the actual current; it includes current and phase errors and the effects of harmonics in the exciting current. In a CT with negligible leakage flux and no turns correction, composite error corresponds to the rms value of the exciting current, usually expressed as a percentage of the primary current. If the exciting impedance were in fact linear, the vectorial error I o would be the composite error. In practice the the exciting current contains some harmonics which increase its rms value and thus increase the composite error

53 Composite error (cont.) The accuracy class of measuring current transformers used in British practice, in accordance with BS3938 : 1973 is shown in Table 2.1. Table 2.1 : Limits of error for accuracy class 3 and 5. Class + percentage current (ratio) error at percentage of rated current shown below For class 3 and class 5 the current vector at rated frequency shall not exceed the values given in Table 2.1 when the secondary burden is any value from 50% to 100% of the rated burden.

54 2.9.3 Accuracy Limit Current of Protective Current Transformers Protective equipment is intended to respond to fault conditions (above the normal rating) CT must retain a reasonable accuracy up to the largest relevant value. This value is known as accuracy limit current (arus had kejituan), and may be expressed in primary term or equivalent secondary terms. The ratio of the accuracy limit current to the rated current is known as ` accuracy limit factor' (faktor had kejituan).

55 2.9.3 Accuracy Limit Current of Protective Current Transformers (cont.) The accuracy class of protective current transformers used in British practice, in accordance with BS 3938:1973 is shown in Table 2.2. Table 2.2 : Limits of errors for accuracy class 5P and class 10P Class Current error at rated primary current (%) Phase displacement at rated current (minutes) Composite error at rated accuracy limit primary current (%) 5P ±1 ± P ±13 10 For class 5P and class 10P the current error, the phase displacement and the composite error (at rated frequency) shall not exceed the values given in Table 2.2 when the secondary burden is 100% of rated burden

56 2.9.3 Accuracy Limit Current of Protective Current Transformers (cont.) During overcurrrent grading process, the current transformer accuracy must be maintained to a certain level. This is possible with the burden, express in VA at the rated current, and the wiring should be reasonable low Protective ratings are expressed in terms of rated burden, class, and accuracy limit factor, for example 10VA/5P/25 or 10VA Class 5P25

57 Example 2.1 The rating of a current transformer is given as 10VA/5P/15. Calculate new accuracy limit factor when the burden becomes half. Assume that the secondary resistance of the CT is equal to 0.15 ohm Burden impedance (10VA at rated current 5 amp) = 10/5 2 = 0.4 ohm the secondary resistance of the CT is given as 0.15 ohm. So, the total secondary impedance is = ohm = 0.55 ohm (arithmetic summation is adequate) Secondary emf at accuracy limit current = 0.55 x 5 x 15 = volt

58 Example 2.1 (cont.) If the burden becomes half, it means that the impedance is equal to 0.2 ohm. So, the secondary impedance = ohm = 0.35 ohm New accuracy limit current = 41.25/0.35 = amp The new accuracy limit factor = 117.8/5 = 23.56

59 2.9.4 Class X Current Transformer The classification of Table 2.2 is only really useful for overcurrent protection For the application of current transformers to earth fault protection, and for the majority of other protection applications it is better to refer directly to the maximum useful e.m.f. which can be obtained from CT. In this context, the knee-point of the excitation curve is defined as that point at which a further increase of 10% of secondary e.m.f. would require an increment of exciting current of 50%, as in Figure Design requirements for CTs for general protective purposes are frequently laid out in terms of knee-point e.m.f., exciting current at the knee-point (or some other specific point) and secondary winding resistance. Such current transformers are designated Class X.

60 Figure 2.11 : Definition of knee-point of excitation curve Rajah 2.11 : Definisi titik-lutut lengkung ujaan

61 CT UNDER TRANSIENT CONDITION CT must be able to withstand asymmetrical condition during the fault. The main asymmetrical current is shown below Asymmetircal AC = Symmetrical + Exponential DC current current current

62 CT UNDER TRANSIENT CONDITION(cont.) The current waveform can be expressed as:- I p ' = I ac + I dc = -I m Cos ωt + I m e -αt (2.11) where α = R p /L p, the ratio of the power system resistance R p to inductance L p Consider a simplified circuit model of the CT and its (resistive) burden R as shown, neglecting CT winding leakage reactance and core losses

63 CT UNDER TRANSIENT CONDITION(cont.) I p I o ' I s L R V s Under unsaturated conditions the magnetising inductance L is such that ωl>>r, and so I o ' may be neglected initially.

64 CT UNDER TRANSIENT CONDITION(cont.) Hence, I s R= V s = I p 'R = N (dφ/dt) (2.12) Where N = Number of secondary turns φ = CT flux Then RI p ' = R [I ac + I dc ] = N (dφ/dt) N (dφ/dt) = R [I ac + I dc ] (dφ/dt) = (R/N) [I ac + I dc ]

65 CT UNDER TRANSIENT CONDITION(cont.) Φ = t R N Φ = Φ ac t 0 I ac dt + Φ dc R N φ = R R I ( ) ac N ω = N ω ω 0 t R m [ ] -αt R I -αt Φdc = Ime dt = 1 e N N α 0 + t 0 m Im cos t dt sin t I d c dt (2.13) (2.14) (2.15)

66 CT UNDER TRANSIENT CONDITION(cont.) The ratio (φ dc )m/(φ ac )m is given by R = N R N ω = α Im α Im ω ωl p = α= = R X R p p p dimana R L p p (2.16)

67 CT UNDER TRANSIENT CONDITION(cont.) This value is the power system X/R ratio. This ratio is usually between about 3 and 16, depending on the type of power network (o/h line, cable, voltage level,etc) The implication of this result is that to avoid saturation of the CT core, the knee point flux level must be 3 to 16 times greater than ac working flux level. For this approximation the CT secondary current waveform would faithfully reproduce the primary current until φ reached φ sat, where Is would fall to zero.

68 CT UNDER TRANSIENT CONDITION(cont.)

69 PROTECTION TECHNIQUES A protection scheme is a particular arrangement of CTs, VTs, pilots and relays which results in a desired operating characteristics They fall in two main groups : UNIT schemes NON-UNIT schemes The general philosophy of protection engineering is that every item of plant or section of network is under the supervision of several network schemes, is often achieved by having non-unit scheme acting as back-up to main unit schemes.

70 Industrial Power System Protection Protection Schemes Unit Scheme Non-Unit Scheme

71 Industrial Power System Protection Non-Unit Scheme Not clearly defined. Requires some form of protection coordination to achieve coordination. Current Time Time & Current

72 Industrial Power System Protection Example Of Non-Unit Protection e.g.1 e.g.2 First Example of non-unit protection A B C D X Z 1A (20 ms) X X X Z 2A (150 ms) Z 1C (20 ms) Second example of non-unit protection

73 Industrial Power System Protection Unit Scheme Defined precisely by the location of the instrument transformers. Will operate if a fault occurs within a particular zone Costly

74 Industrial Power System Protection Example Of Unit Protection Unit Protection - protects on exactly defined part of system. e.g.1: Difference in currents causes trip 87 Examples of unit protection

75 Industrial Power System Protection Example Of Unit Protection e.g.2: Sum of currents in should be zero 64 High Impedance Relay Another examples of unit protection

76 Power Direction balance Another unit scheme technique is power direction balance, utilizing the inherent characteristic of an induction disc or cup relay to rotate in one direction for power flow from A to B, and the other direction for power flow from B to A. If such relays are fitted with two sets of contacts and are situated one at end A and one at end B of the (say) feeder to be protected, the operating principle will be as follows: CB CT FEEDER CT CB VT A B VT TRIP TRIP BLOCK TRIP BLOCK TRIP a) any flow of power into the feeder will cause the relay at that end to rotate and close the TRIP contacts for the local circuit breaker; b) any flow of power out of the feeder will cause the relay at that end to rotate and close the BLOCK TRIP contacts for the remote circuit breaker; c) thus if power flows in at A and out at B, the block trip signal from B prevents circuit breaker A from opening; d) but if power flows in at A and in at B, both circuit breakers are opened; e) or if power flows in at A and no power flows at B, then circuit breaker A opens.

77 Latihan 2 1. Suatu Pengubah arus 50 hz mempunyai lilitan utama 20 dan lilitan sekunder 60, pada satu teras dengan keratan rentas 8 cm 2. Arus sekunder ialah 5 amp. Jumlah galangan beban dan galangan sekunder ialah j 0.4 ohm. Arus ujaan yang diperlukan pada belitan utama untuk menghasilkan fluks kerja dengan litar sekunder dibuka ialah 0.6/45 amp. Kira ketumpatan fluks dalam teras, ralat nisbah dan ralat fasa.

78 2. Suatu pengubah arus bar yang tidak dipampas mempunyai nisbah namaan 1000/5. Bila membekalkan 5 amp ke suatu geganti, beban geganti ialah j 2.45 ohm dan belitan sekunder ialah j 0.35 ohm, amp-lilit pemagnetan ialah 95 dan amp-lilit kehilangan teras ialah Tentukan ralat nisbah, ralat fasa, lilitan pampasan untuk mengurangkan ralat nisbah menjadi minimum, dengan menganggap data-data tidak berubah.

Electrical Protection System Design and Operation

ELEC9713 Industrial and Commercial Power Systems Electrical Protection System Design and Operation 1. Function of Electrical Protection Systems The three primary aims of overcurrent electrical protection