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n° 164 current transformers for HV protection Michel Orlhac Graduated from the Ecole Centrale de Paris in 1977. After one year's specialisation at the university of Stuttgart (Germany), he entered the overseas projects department of Stein Heurtey (iron and steel engineering). In 1980 he joined Merlin Gerin, becoming part of the technical section of the High Voltage Prefabricated Switchgear Department (P.S.H.T.) where he completed a study on current transformers. This Cahier Technique publishes the results of this study. At present he is the marketing manager for France-Transfo, a subsidiary of the Merlin Gerin Group. E/CT 164 first issued, march 1995 Cahier Technique Merlin Gerin n° 164 / p.2 current transformers Current transformers or CTs take up a lot of space in HV cubicles. Thorough for HV protection knowledge of how they work makes it possible to: c reduce their dimensions and thus their cost, c use standard CTs in a larger number of configurations. content The purpose of this study is to learn more about CT operation in association with protection relays and to lay down a few rules for sizing them properly. 1. Theoretical review p. 4 After a brief theoretical review of CT operation and current protection Hysteresis - Saturation p. 5 devices, the behaviour of the CT- Characterisation of CTs p. 6 protection relay combination is studied 2. General current protection Current transformers p. 7 in two particularly important cases information Functional CTs p. 8 in HV: Protection relays p. 8 c overcurrent relay supplied by a heavily saturated CT, Technological evolution p. 8 c protection relay connected to two 3. Response of a CT in saturated state Experiments - Wiring p. 9 CTs in parallel in duplex cubicles. Testing with symmetrical p. 10 This study is completed by constant currents experimental results. Testing with asymmetrical p. 11 currents Conclusions on CTs delivering p. 12 on an overcurrent relay 4. Parallel cubicle operation p. 14 5. General conclusions p. 16 Appendix: CT standards NF C 42-502 (Norme Française) p. 17 IEC 185 p. 19 Cahier Technique Merlin Gerin n° 164 / p.3 1. theoretical review Current transformers consist of a di 2 magnetic circuit in toroid form. The e 2 = v 2 + R 2 i2 + l 2 dt i1 primary is made up of n1 turns or simply dϕ dϕ a single conductor crossing the toroid e 1 = n1 and e 2 = − n 2 dt dt (n1 = 1). The secondary is wound in n2 If all the functions described are → regular turns around this toroid (see dI n sinusoidal of pulsation ω, the following fig. 1 and 2). can be written vectorially: Ampere's theorem states that the sum → → → of the ampere-turns is equal to the V1 = E1 + (R1 + jl1 ω ) I 1 → → → circulation of the magnetic field vector. E2 = V2 + R2 I 2 → → → → → → n1 i1 + n2 i2 = ∫ H n dI E1 = jn1 ω Φ E2 = − jn2 ω Φ → Toroid → H = magnetic field I1 → → → + I2 = Ie i2 n = tangent unit vector n A transformer is said to be perfect The wiring diagram in figure 3 and when equations (1) result in the vectorial fig. 1. → → representation in figure 4. → ∫ Toroid n dI = 0 H The exciting current I e is broken down → → on the axes Φ and E into: In the real transformer, this term refers → → → to the error introduced by the magnetic I e = Ia + Im i1 → circuit and defines the exciting current c where I a represents the part of this ie formed at the secondary by: current lost in the magnetic circuit (iron n1 i1 + n 2 i 2 = n 2 i e losses due to hysteresis and eddy n2 currents). If n = is the winding ratio, the → n1 c and I m is the magnetising current relationship is written as: which transfers power from one winding i1 to the other by creation of a i2 + i2 = ie magnetomotive force which induces the n → flux Φ . fig. 2. The transformer can then be represented (see fig. 3) as having two parallel elements: c a perfect transformer of ratio n delivering a current i1/n at the secondary, l1 l2 i1 R1 i1/ n R2 i2 c an impedance which consumes a current ie. Moreover, each winding, both primary ie and secondary, creates a slight voltage n = n 2/ n 1 drop due to the resistance of the winding (R1 and R2) and to the leakage e1 n2 n1 e2 inductances ( l 1 and l 2 ). Since, in the V1 V2 Z case of the CT, the secondary winding im ia is tight and regular, l 2 need not be perfect transformer considered. If ϕ is the flux common to both windings, the following can be written between the emf e1, e2 and the real transformer difference in potential v1, v2: di1 fig. 3: CT schematic diagram. v 1 = e 1 + R 1 i1 + l 1 dt Cahier Technique Merlin Gerin n° 164 / p.4 hysteresis - saturation Magnetic circuit quality is defined by l 1 ω I1 I1 «order of creation» the relationship it imposes between the of values: induction vector B and the magnetic R1 I1 I1, V1 I1, Ie field vector H. At a given moment and in a fixed point, V1 E1 these two vectors are linked by the I 1/ n relative permeability of the magnetic ϕ1 material µr such that: E1 → → B = µ o µr H → B A magnetic circuit is thus characterised → Im Φ by the curve b = f (h) known as the magnetising curve. α Ia Φ Ie According to the different material → types, the curves in figure 5 are H obtained, the results of sinusoidal ϕ2 I2 excitation (primary current). E2 V2 In sinusoidal state, b represents voltage since: E2 → Φ→ B= n S V2 I2 → → E2 = n2 jω Φ R2 I2 l 2 ω I2 → → N.B.: the real proportions, between the representative vectors of primary and secondary V ≈ E2 values, are not repected. h represents the exciting current since fig. 4: vectorial representation of a CT. → → n2 Ie = ∫ H n dI Toroid assuming that hypotheses: magnetising curves ie and B as a function of time b B, ie →→ H n = H = constant n 2 Ie = L H perfect transformer h t Perfect transformer Permeability of the medium is assumed infinite → → → → I1 linear H = 0 hence I e = 0 and I 2 = transformer n h t This hypothesis approaches the real situation with CTs since they normally «work» far below saturation. I2 is then saturable trans- the mirror image of I1. former without Linear transformer hysteresis h t Permeability of the medium is constant B = Cste x H hence ie and i2 are sinusoidal functions. saturable Saturable transformer without transformer with hysteresis h t hysteresis Saturation is the sudden variation of µr from a high value to a low value at the exciting current: ie induction B point known as the «saturation bend». Induction b then increases only slowly fig. 5: magnetising curves and their incidence on ie. and ie deforms to form a peak. Cahier Technique Merlin Gerin n° 164 / p.5 Saturable transformer with c for measurement CTs Bear in mind that the less the CT is hysteresis The module error loaded (the more it is below its accuracy The magnetising curve is undoubled, I / n − I2 level power Y), the greater its accuracy. thus indicating the resistance of the εM = 1 Its real accuracy level is therefore I1 / n magnetic circuit to the induction greater than its rated accuracy level Fp. The phase error variations. Curve ie then exhibits a This point is developed in chapter 3. −2 characteristic «swing». ε ϕ = (I1, I2 ) 10 rd Admissible short term current The magnetising curve of a CT can An accuracy class X is given (generally Expressed in kA it is the maximum easily be observed using an 0.5 or 1) which expresses limit values current admissible Ith for one second oscilloscope. A sinusoidal voltage V2 (t) of the module error εM and of the phase (the secondary being short-circuited). It is applied to the secondary (the primary shift error εϕ as a function of the load represents CT thermal overcurrent is not charged). The current ie (t) ratio N: withstand. absorbed then represents the exciting (standard values are given in the I current and is proportional to the N = 1 (N var ies from 0.1 to 1. 2) standards in the appendix). I1n For times other than 1 second, the heat magnetic field vector H. for N = 1 εM = X (in class 0.5 for I1 = I1n, conservation law I2 t = cste can be Integration of voltage V2 represents the εM = 0.5 %) applied: flux ϕ2 which is proportional to the (for value details refer to the standards → for t < 1 sec. the calculation gives I > Ith, magnetic induction vector B (see in the appendix). thus increasing electrodynamic forces. fig. 6a). c for protection CTs However, the limit guaranteed value is Integration of a sinusoidal value causes The composite error εc Idyn = 2.5 Ith. a rotation of π/2 (90°). It is thus 2 1 1 T i1 ∫o sufficient on an oscilloscope: εc = i2 − n dt c to sweep with ie, I1 / n T c to apply voltage V2 to the vertical Protection CTs are characterised by amplifier. V2(t) 3 symbols: Y, P, Fp: The magnetising curve of the material Y = error rate (5 or 10), is thus obtained (see fig. 6b). P = protection, Fp = accuracy limit factor which gives ie(t) characterisation of CTs the limit values of errors εM, εϕ and εc as a function of the load ratio N. CTs are characterised in practice by the following values (according to For N = Fp standards NF C 42-502 and IEC 185). εc = Y (in class 10P5 for I1 = 5 l1n: εc = 10 %) a - scales: ie = 0.25 A per square CT voltage (for value details, refer to the standards V2 = 50 V per square. This is the operating voltage applied to in the appendix). the CT primary. Note that the primary is at the HV potential level and that one of For a CT working at a rated induction ϕ 2 = ∫ V2 dt (or B) Bn, a saturation coefficient Ks such that: the terminals of the secondary (which must never be opened) is normally Bs earthed. Ks = Bn Just as for all equipment, a maximum where Bs is the saturation induction 1 min withstand voltage at standard characterising the core material. frequency and a maximum impulse In practice K s ≈ Fp and they are often ie (or H) withstand voltage are defined (refer to treated as the same in calculations. the standards in the appendix). e.g. for a rated voltage of 24 kV, the CT Accuracy level power must withstand 50 kV for 1 mn at 50 Hz Expressed in VA, it indicates the power and 125 kV impulse voltage. that the secondary can supply while respecting the rated accuracy class Y, Rated winding ratio P, Fp. Normally takes the form: l1/l2. 2 I a = constant It represents the total consumption of I2 is very generally 5 A or 1 A (for rated the secondary circuit (except for CT), b - scales: ie = 0.25 A per square values of I1, refer to the standards in i.e. the power consumed by all the ϕ2 = 0.077 V.s per square. the appendix). connected devices as well as the fig. 6 : oscillographic reading of curves i(t) Rated accuracy class connecting wires. V2(t) and h(b) of a CT, 50/5, 15 VA, 10P20 This depends on whether the CT is used (for rated values, refer to the standards where: V2 = 83 V and le = 0.26 A. for measurement or protection: in the appendix). Cahier Technique Merlin Gerin n° 164 / p.6 2. general current protection information Protection devices have many functions since they have to: c protect equipment from destruction or damage as a result of faults (short- circuit, overload...), c ensure normal operation of the installation and its equipment (control, load shedding...), c guarantee safety of personnel. current transformers CT with cross primary Wound type CT with Wound type CT with winding (cable) wound primary wound primary Since relays cannot be connected 1 secondary - 600/1 winding winding directly onto the MV network, the 1 secondary - 200/5 2 secondaries - 200/5 and 100/5 information they receive comes from fig.7: different types of CTs. current transformers or CTs (see fig. 7) and from voltage transformers or VTs. When primary current is high, the CTs are of the cross bar type, and when it is from part of the network to be low they are of the wound primary type. monitored (a motor, a transformer, a CTs have a number of roles to play in busbar...) to quickly detect and isolate relay electrical networks: any faults inside that part. I2 = I1/m c supplying at their secondary a current Zero sequence protection exactly mirroring the one flowing in the This monitors the zero sequence I1 HV conductor concerned, component Io of the three-phase c providing galvanic insulation between current which appears during phase- fig. 8. the HV and the measuring and earth faults. There are two possible protection circuits, configurations: c protecting the measuring and c a toroid transformer encircling the I1 I'1 protection circuits from damage when a three phase conductors (if possible). fault occurs on the HV network. This configuration (see fig. 10a) I'1 - I1 Using this current image in the HV enables detection of small zero relay conductor, the relay generates in turn a sequence currents (1 to 100 A). tripping order according to the type of c three CTs achieving in the neutral protection it provides and the values at connection of their secondary the sum fig. 9. which it has been preset [threshold(s), of the three phase currents. This time delay(s)....]. configuration (see fig. 10b) is the only This order is transmitted to one or more one possible for large and numerous cables or busbar ducts. It is not a) breaking devices (circuit-breaker, contactor, switch). recommended when the zero sequence CT configurations vary according to the current to be detected is 5% less than type of protection to be provided. ln (or even 12% for consumer substations according to standard Overcurrent protection (see fig. 8) NF C 13-100 (French Standard)). Io relay This directly uses the «current» information supplied at the CT secondary to detect short-circuit or overload functional CTs b) currents or calculate the thermal status of In HV cubicles, the «current a machine. Note that this configuration transformer» function takes on a new type must also contain the protection dimension as a result of its content and devices using in addition to VTs: shape. c directional overcurrent protection, Thus: c power protection (active or reactive). c a number of CTs can be moulded in Io Earth leakage protection (see fig. 9) the same enclosure: one core for the relay This measures the current difference measurement function, one core for the between two CTs, one connected protection function and sometimes even fig. 10. downstream and the other upstream a third core for earth leakage protection, Cahier Technique Merlin Gerin n° 164 / p.7 c the enclosure is used to ensure c replace relays (automation) in the insulation between two compartments cubicle, and plugging-in of the breaking device: c provide operators with measurement the CT is then said to be «functional». of electrical parameters. An application example is given in the These units, with their increased metalclad cubicles for withdrawable vocation, are: switchgear (see fig. 11 and 12). c flexible (protections are chosen Overall dimensions are thus reduced by simply by programming), using one insulating enclosure (the most c parameterisable (large choice of appropriate), thus also reducing costs. settings), c reliable (they are fitted with self- monitoring or with watchdog and the protection relays self-test), The equipment currently available is c economic (reduced wiring and based on the three technologies: implementation time). electromechanical, analog and digital. Their digital communication and The oldest of these is powerful algorithms also enable electromechanical technology: relays additional functions such as logic are simple and specialised (current, discrimination to be performed. fig. 11: functional CT for HV metalclad voltage, frequency, ... monitoring) but This communication capacity means cubicles (Merlin Gerin). their accuracy is poor as their settings that genuine network operation (similar may be altered over time. to technical management of industrial The last two technologies benefit from installations) is now possible. the advantages provided by electronics Finally, their ability to acquire and (see fig. 13): process the information provided by c compact dimensions of the device, sensors allows them to make full use of c low power required for acquisition of the performances of the new non- «current» information (a few fractions of magnetic sensors. VA), c response time not dependent on the current received by the relay, technological evolution c reliability increased by lack of In this current sensor field, sensors with mechanical parts (no dirt accumulation wide measuring bands are being or corrosion, not affected by impacts), increasingly used instead of current c low cost since they use mass produced transformers (1 or 5 A). These sensors non-specific electronic components. based on Rogowski's principle (non- Finally, in the nineteen eighties, digital magnetic sensors) are currently on the technology made it possible, thanks to market and provide distributors with microprocessor processing power, to optimised solutions (fewer alternative produce information processing units versions and simplified choice) which fig. 12: installation example of functional able to: are far more efficient (improved CTs in a Fluair 200 12 kV HV metalclad c globally provide the various response curve linearity) than cubicle (Merlin Gerin). protections, traditional transformers. fig. 13: Vigirack static relays (Merlin Gerin). Cahier Technique Merlin Gerin n° 164 / p.8 3. response of a CT in saturated state The emergence of static relays leads to revision of protection behaviour as a i2 whole in the case of strong currents: as i1 the CT saturates beyond a certain threshold, the first reaction is often to i2 avoid this by raising the threshold. However, this results in both additional resistance relay costs (more efficient, larger, more tested CT R space consuming CT) and in the risk of excessive temperature rise of the relays. measuring oscilloscope On the contrary, saturation plays a useful role for the «measurement» function since primary current image accuracy is only useful up to the value u2 of the rated current I1n. Beyond this standard CT point, the measurement ceases to be of i1 any use and saturation must occur for a u1 low current (2 to 3 I1n) in order to limit the secondary current and protect the fig. 14: diagram for checking proper relay operation. measuring instruments. It is thus necessary to know the I2 response of the CT in saturated state to (A) ensure the protection device works properly when the primary current 500 1 exceeds rated current strength, particularly for the high values which appear if a short-circuit occurs. In theory, induction in the core reaches a plateau at the saturation bend, thus 2 100 limiting current strength at the 3 secondary. In actual fact the experiment performed will show that current strength at the secondary slightly increases and that protection relay operation is quite satisfactory. 10 experiments - wiring A current i1 is injected in the CT primary, and the current supplied by the secondary in a load Z containing a relay R and a resistance is analysed (see fig. 14). 10 100 500 N = I1/I1n The currents at the secondary l2 are given, according to the current supplied Fig. 15: I2 = f(N) for 1 CT only (15 VA 10P5 100/5). at the primary (represented by the Load Z at the secondary: I 1. relay only, parameter N = 1 ) for various loads Z 2. Z = rated Z of CT, i.e. 0.6 Ω and cos ϕ = 1, I1n 3. Z = rated Z of CT, i.e. 0.6 Ω and cos ϕ = 0.8. and various CTs (see fig. 15). Cahier Technique Merlin Gerin n° 164 / p.9 testing with symmetrical Nevertheless, the rms current I2 v both relays trip from their threshold ß continues to increase as is shown in right up to Nmax. constant currents line 2 in figure 15. Testing at reduced load Testing at resistive rated load As I2 increases, the power supplied at The secondary load only comprises the The test was carried out using a CT with the secondary P2 = Z l2 and the power 2 relay and the connecting wires. low performance: 10P5, 50/5 with a delivered at each relay Pr = R l2 also 2 Compared with the rated load of 15 VA, rated load Z of 15 VA (at 5 A) made up increase. This accounts for the this represents a load of roughly 9%. of an overcurrent relay and a resistance. tripping of both relay types as from Two relays were used: c results threshold ß to which they were set right v a Vigirack static relay, Curves i2 (t) (see fig. 16c) and i2 (N) up to Nmax. v an electromechanical relay. (see line 1 in figure 15) show that the Testing at rated partly inductive load saturation bend is far higher than at As both these relays have a low This test resembles the previous one. rated load. internal resistance, a resistance was However, a choke is placed in the This bend follows the law: added to reach roughly 0.6 Ω, i.e. K s (P2 + R 2 I2 ) = constant secondary circuit to represent the case 2 15 VA at 5 A (connecting wiring of an electromechanical relay included). Because the inductance of with P2 = Z l 2 is the total power connected by itself to the secondary 2 the electromechanical relay was low supplied at the secondary (consumed which would consume the rated power (15 µH, i.e. cos ϕ = 0.95 for the relay by the relay and the connecting wires). of the CT. In practice, these relays only), the load can be considered to be R2 = internal resistance of the CT never fall below cos = 0.8. purely resistive in both cases. secondary winding, In this test, the current I1 explored the The test consisted in making current l1 Ks = saturation coefficient (real or rated). range I1n = 50 A to I1 max = 16,400 A, vary in the range I1n = 50 A Thus, in practice, when a CT delivers into at I1max = 54 kA N max i.e. N max = 328 and η = = 65.6 a load less than its rated accuracy level Fp power (in VA), saturation occurs at a far 54,000 I.e. N max = = 1,080 and c results higher overcurrent level than the rated 50 Current i2 (t) assumes the curve given saturation coefficient Ks. N max 1,080 η= = = 216 in figure 16b. The presence of a choke This phenomenon must be taken into Fp 5 spreads out the peak, hence the lower . consideration and calculated for each (the latter value indicates the level of value I2 (see line 3 in figure 15). application since it may generate saturation to which the CT was With respect to testing at pure resistive overcurrents in the secondary which subjected). load: are incompatible with the thermal and c results v I2 is multiplied by a factor of 0.65, dynamic withstands of the relays The current i2 (t) collected at the v the total power supplied at the connected to the CT secondary (for secondary takes the form of a peak secondary is multipled by a factor calculation, refer to the conclusions above: N = 10 (see fig. 16a). of 0.4, given below). i2 i2 i2 v2 v2 v2 fig. 16 a - CT 15 VA 10P5 50/5 fig. 16 b - CT 15 VA 10P5 50/5 fig. 16 c - CT 15 VA 10P5 50/5 testing at purely resistive rated load testing at rated load with cos ϕ = 0.8 testing at reduced rated load I1 = 16,400 A I1 = 16,400 A relays + connecting wires - I1 = 14,200 A scale: i2 = 100 A/square; v2 = 100 V/square. scale: i2 = 25 A/square; v2 = 50 V/square. scale: i2 = 100 A/square; v2 = 5 V/square. Cahier Technique Merlin Gerin n° 164 / p.10 testing with asymmetrical currents a) The test was performed using an asymmetrical current, i.e. the sum of a scales: symmetrical sinusoidal current and a v1 DC component with the following characteristics: i1 500 A/mm Î ≈ 2.3 Irms i2 10 A/mm These values are slightly less than those in standard NF C 64-100 for which no saturation Î = 205 = 1.8 2 b) Irms scales: i.e. 20% of asymmetry at 70 ms. The secondary load is identical to that of the main test at resistive rated load v1 comprising an electromechanical or static relay. i1 1,000 A/mm c results Both relays correctly respond in a few ms and in the same manner as in symmetrical testing throughout the i2 10 A/mm range explored (up to Î1 = 140 kA peak with Irms = 54 kA). Remarks: c the first peak seen at the secondary by the relays is enough to make them trip, if its energy is sufficient: this is the case for Irms greater than 2 kA but c) below this value (see fig. 17a) the third peak is required; scales: c the CT does not saturate during the first negative peak of the primary current for Î1 = 4 kÂ ; c the response delivered by the CT on the first negative peak of the primary v1 (or even secondary) is normally shorter than the responses in steady state (which is reached as from the sixth peak); c the above points show that for higher peak factors (case of off-load i1 2,000 A/mm energising of transformers with an i2 20 A/mm Î = 3.7 ), there is a risk of the Irms response at the secondary disappearing during the first peaks. If, in addition, the time constant of the primary current DC component is high (t = 80 ms in the case quoted), this disappearance fig. 17: CT secondary responses on an asymmetrical primary current for: continues until the primary current a) lrms. ≈ 1.4 kA, crosses the 0 axis. This phenomenon is b) lrms. ≈ for 14 kA and Î1 ≈ 32 kA, shown on the curves in figure 17 c) lrms. ≈ 54 kA and Î1 ≈ 140 kA. (tripping time moves to 68 ms). Cahier Technique Merlin Gerin n° 164 / p.11 conclusions on CTs current value, even if the CT is strongly saturated. delivering on an Thus, the CT saturation coefficient Ks overcurrent relay must be calculated not according to the The above tests show that for both short-circuit current lcc but according to electromagnetic and static relays, the maximum setting threshold of the tripping is obtained whatever the associated relay (see fig. 18 and 19). choice of Ks Icc I1n ITIn I1r = β I1n Ith = Icc θ ∞ ∞ I2n I2max θ I2 max Ir min Ir max Ithr of the I1n = rated current network I1r = β I1n setting current Icc = short-circuit current θ = maximum short-circuit time of the relay Ir min to Ir max = setting range Ithr = admissible short term current (1s) of the CT ITIn = rated primary rating I2n = rated secondary current I2 max = CT response to Icc fig. 18: characteristics to be considered for defining a CT. Cahier Technique Merlin Gerin n° 164 / p.12 1. the saturation threshold Ks must correspond to the maximum setting value of the relay. 2. the CT must thermally withstand 3. this CT must the current Icc for a time θ at least electrodynamically equal to the breaking time of the withstand the peak short-circuit by the circuit-breaker. value 2.5 Icc. 4. the secondary circuit must thermally withstand the maximum secondary rms current I2 max created by Icc at a primary for the time θ. 5. the relay setting range (Ir mini, Ir maxi) must be large enough to cover the CT response at the setting current of network B I1n. fig. 19: general rules for sizing a CT. Cahier Technique Merlin Gerin n° 164 / p.13 4. parallel cubicle operation Power supplies with double busbars encountered in this system is highly provide insulation between are frequently used in HV network complex lockings: compartments and to plug in the configurations. c cubicles connected in duplex (see breaking device. This arrangement fig. 20). Using standard elements, this makes it necessary to connect the There are currently two solutions for relays (which are not backed up) on most cubicles: solution can advantageously replace each CT secondary. This has resulted c the double busbar cubicle: the circuit- the double busbar, as it is more in the study below concerning breaker may be connected to either reliable. operation of two identical CTs busbar without discontinuity of service. As on the new cubicle generations, the connected in parallel on the same One of the drawbacks often CTs are standard elements used to load. 1455 1300 1455 relay 1800 1060 fig. 20: connection of 2 cubicles in duplex relay Wiring diagram I1 Connection of two cubicles in duplex, standard CT as shown in figure 18, results in the diagram in figure 21 for protection. resistance relay One of the CTs (said to be «live») is tested «live» CT R supplied at the primary by the shunt i2 ir shunt Ir HV network; its secondary supplies a current i2 broken down into a current im im on the secondary of the other CT (said recorder to be «dead») and a current ir on the Im rated load of 15 VA made up of an electromagnetic or static relay and a pure resistance. I2 tested «dead» CT The tests were performed on two identical CTs of the same series (15 VA fig. 21: wiring diagram for study of a parallel-connected CT. 50/5 10P5 as in the above paragraphs). Cahier Technique Merlin Gerin n° 164 / p.14 Results These are given in the curves of scales: figures 22 (currents as a function of time) and figure 23 (root mean square im 9.4 A/mm currents and tripping times). The following observations are made: c both relays quickly respond from their ir 9.55 A/mm tripping threshold ß to η = 72, c the static relay trips in a constant time T ≈ 20 ms, whereas the i2 19 A/mm electromagnetic relay reacts as a function of I2 (T ≈ 80 ms at the tripping i1 955 A/mm threshold to T ≈ a few ms at η = 72); c the secondary current I2 continues to increase but two separate zones appear: Vigirack relay v before η = 10, Ir ≈ I2 and lm << lr the secondary current flows entirely into the relays since the «dead» CT acts as fig. 22: currents at the secondary of 2 parallel-connected CTs. I1 = 12,500 A. an infinite impedance, v after η = 10 Im ¡ I2, which means that the secondary current mostly I(A) flows into the «dead» CT but, T(ms) I2 (total secondary current) however, Ir continues to grow, thus Im (current flowing in the «dead» CT) causing the relays to trip (do not forget 100 π that ir is dephased by with respect Ir (current flowing in the relay) 2 to im); c the current lr flowing in the relays T: static relay tripping during testing with two CTs is lower than for testing with one CT (at 10 N = 300, roughly - 40%). T: electromagnetic relay tripping Conclusion The connection of two CTs in parallel presents no problem: c for low currents: hardly any current 1 1 10 100 N = I1/I1n flows into the «dead» CT, c for high currents: sufficient current fig. 23: rms currents and tripping times of the relays connected to 2 CTs in parallel (10P5 50/5 15 VA). flows into the relay to trip it. Cahier Technique Merlin Gerin n° 164 / p.15 5. general conclusions The conclusions in chapters 4 and 5 Nevertheless, CT saturation, as shown show that: in this experiment, should not be c the relays operate correctly in both considered a handicap: cases studied: c when a CT «supplies» one or more v high CT saturation, measuring instruments, saturation, by v parallel-connection of two CTs; limiting rms current at the secondary, c static relays give the most reliable protects the devices which, moreover, response (constant operating time for do not generally need to be very all currents greater than the setting threshold). accurate above l1n. Moreover, static relays generally have c when a CT «supplies» a protection a very small acquisition time, thus device, operation is ensured even if meaning operation is more reliable saturation occurs. The idea of sizing when the CT is strongly saturated and a CT according to the highest current it supplies a very short current impulse. may have to withstand at the primary Do not forget, however, that the must therefore be rejected. Moreover, transient phenomena considered were this oversizing is risky for the relay and limited to the asymmetrical current less cabling which could be seriously than: damaged. Î = 2.5 Irms Cahier Technique Merlin Gerin n° 164 / p.16 appendix: CT standards NF C 42-502 (French Standard) Normal rated current values Rated insulation levels c at the primary (in A): 10 - 12.5 - 15 - The insulation levels recommended by the standard are given in table II A presented 20 - 25 - 30 - 40 - 50 - 60 - 75 in figure 24. and their decimal multiples or submultiples. Preferential values are given in bold. highest voltage for withstand voltage c at the secondary (in A): 1 - 5 equipment (kV) 1 minute at standard frequency to impulse voltage Accuracy class (rms value) (kV) (peak value) (kV) c measurement CTs 0.6 3 The normal accuracy classes are: 1.2 6 0.1 - 0.2 - 0.5 - 1 - 3 - 5. 2.4 11 The rated frequency operating range is 3.6 16 45 96% to 102% of rated frequency. 7.2 22 60 For transformers of accuracy classes 12 28 75 0.1 - 0.2 - 0.5 and 1, the current error 17.5 38 95 and phase shift in the rated frequency 23 45 95 range must not exceed the values in 24 50 125 table III (see fig. 25) when the 36 70 170 secondary load is between 25% and 52 95 250 100% of accuracy load. 72.5 140 325 For transformers of accuracy classes 3 and 5, the current error in the rated fig. 24: insulation levels (table II A). frequency range must not exceed the values in table IV (see fig. 26) when the secondary load is between 50% and accuracy current error phase shift, ± class (ratio error) for current values 100% of accuracy load. as a percentage, ±, given as a percentage In all cases, the load used must be for current values of rated current inductive with a power factor of 0.8, given as a percentage unless the corrresponding power is less of rated current minutes centiradians than 5 VA, in which case its power factor % I1n 10 20 100 120 10 20 100 120 10 20 100 120 is the unit. On no account must the load 0.1 0.25 0.20 0.1 0.1 10 8 5 5 0.30 0.24 0.15 0.15 be less than 1 VA. 0.2 0.5 0.35 0.2 0.2 20 15 10 10 0.60 0.45 0.3 0.3 0.5 1.0 0.75 0.5 0.5 60 45 30 30 1.8 1.35 0.9 0.9 1 2.0 1.5 1.0 1.0 120 90 60 60 3.6 2.7 1.8 1.8 Note: after agreement between manufacturer and user, guarantees can be provided for accuracy and phase shift, between 120% and 200% of In n. fig. 25: error limits (table III). accuracy current error class (ratio error) as a percentage, ±, for current values given as a percentage of rated current % I1n 50 120 3 3 3 5 5 5 There is no phase shift limit for classes 3 and 5. fig. 26: error limits (table IV). Cahier Technique Merlin Gerin n° 164 / p.17 c Protection CTs The normal accuracy limit factor values are: 5 - 10 - 15 - 20 - 30 - 40. The rated frequency operating range is accuracy ratio error phase shift composite error 90% to 110% of rated frequency. class for currents for rated current for accuracy between In limit current The normal accuracy classes are 5P and 2 In (as a %) and 10P. (as a %) minutes centiradians For accuracy level power and in the 5P ±1 ± 60 ± 1.8 5 rated frequency range, the current 10P ±3 10 error, phase shift and composite error must not exceed the values in table V (see fig. 27). fig. 27: error limits (table V) . To determine the current error and phase shift, the load must be inductive and equal to the accuracy load with a c the admissible times for the power factor of 0.8, unless the admissible short term current are set corresponding power is less than from the cold state. However, at the 10 VA; in this case the load could be user's request, the manufacturer is resistive (unit power factor). To obliged to indicate, for a given type of determine the composite error, the load device, the admissible short term power factor may be between 0.8 current based on a state corresponding (inductive circuit) and the unit, the value to operation, the heating current and being set by mutual agreement maximum ambient temperature. between manufacturer and user. However, in the latter case, verification Accuracy level power of admissible short term current cannot The normal accuracy level power be made mandatory as an acceptance values are: 2.5 - 5.0 - 10 - 15 - 30 - 75 - test. 100 VA. Admissible current peak value (Idyn). Admissible peak current and short The admissible current peak value is term current 2.5 Ith. However, another value can be Admissible peak current and short term accepted provided it is stated on the current (Ith). The short term current (Ith) identification plate. must be specified for each transformer. Their preferential values are given in paragraph 10.1 (see fig. 28). Notes highest Ith c for the highest network voltage less network (kA) than or equal to 36 kV, the admissible voltage short term current value is (kV) constructively linked to rated current 3.6 10 16 25 40 value. It is thus frequently expressed as 7.2 8 12.5 16 25 40 a multiple of rated current, for which the 12 8 12.5 16 25 40 preferential values are: 40 - 80 - 100 - 17.5 8 12.5 16 25 40 200 and 300. 23 8 12.5 16 25 40 c if no admissible values as a function 24 8 12.5 16 25 40 of time are given, it is accepted that the 36 8 12.5 16 25 40 transformer can withstand for a time t, 72.5 20 25 expressed in seconds, a current with a root mean square value given by the 100 20 formula: 245 20 31.5 420 40 I th I' th = t2 fig. 28: preferential values of Ith where t2 > t1 bearing in mind that Ith is (paragraph 10.1). given for t1 (= 1s). Cahier Technique Merlin Gerin n° 164 / p.18 IEC 185 This is the reference standard. The NF C 42-502 (Norme Française) differs only slightly from it. The differences are highest rated lightning impulse rated short term as follows: voltage withstand voltage withstand voltage for equipment Um (peak value) at standard frequency Rated insulation levels (rms value) network power (rms value) The IEC standard gives two tables: ≤ 500 kVA > 500 kVA c the same table as the NF C standard kV kV kV kV for European countries, 4.40 60 75 19 c another table as per USA practice with slightly more stringent values: refer 13.20 95 110 34 13.97 to table II B (see fig. 29). 14.52 Normal rated current values 26.4 150 50 Same preferential values at the primary. At the secondary possibility of a I2n = 2 A. 36.5 200 70 Accuracy class c measurement CT Current errors in module and phase are fig. 29: rated insulation voltages set for the U.S.A (table II B). the same in class 3 and 5. For classes 0.1 - 0.2 - 0.5 and 1, the errors are the same, except for the 10% of I1n column which is replaced by 5% of I1n with the errors listed in table IV A in figure 30. accuracy error εM error εϕ Moreover, the IEC standard defines two class for I1 = 5 % of I1n for I1 = 5 % of I1n additional classes, 0.2 S and 0.5 S for minutes centiradians CTs with special applications 0.1 0.4 15 0.45 (connection with special electrical energy 0.2 0.75 30 0.9 meters). In this table, the module and phase errors are given for I2n = 5 A only. 0.5 1.5 90 2.7 c protection CT 1 3 180 5.4 The IEC gives the same limit errors. The only difference is that the accuracy accuracy error εM for error εϕ for values as a % limit factor, Fp = 40, does not exist. class values as a % of of rated current I1n Accuracy level power rated current The IEC only gives the same normal I1n minutes centiradians values up to 30 VA. Beyond this point, % l1n 1 5 20 100 120 1 5 20 100 120 1 5 20 100 120 power can be chosen to meet needs. 0.2S 0.75 0.35 0.2 0.2 0.2 30 15 10 10 10 0.9 0.45 0.3 0.3 0.3 Peak current and short term current 0.5S 1.5 0.75 0.5 0.5 0.5 90 45 30 30 30 2.7 1.35 0.9 0.9 0.9 Unlike the NF C standard, the IEC standard does not define preferential values of Ith for each network voltage. However, application of the law i2 t = Cste fig. 30: accuracy class (table IV A). to define the Ith is limited to: 0.5 < t < 5 s. Cahier Technique Merlin Gerin n° 164 / p.19 Réal.: Sodipe - Valence - Photo.: IPV Edition: DTE - Grenoble Cahier Technique Merlin Gerin n° 164 / p.20 03-95 - 2500 - Printing.: Clerc