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Collection Technique .......................................................................... Cahier technique no. 158 Calculation of short-circuit currents B. De Metz-Noblat F. Dumas G. Thomasset "Cahiers Techniques" is a collection of documents intended for engineers and technicians, people in the industry who are looking for more in-depth information in order to complement that given in product catalogues. Furthermore, these "Cahiers Techniques" are often considered as helpful "tools" for training courses. They provide knowledge on new technical and technological developments in the electrotechnical field and electronics. They also provide better understanding of various phenomena observed in electrical installations, systems and equipments. Each "Cahier Technique" provides an in-depth study of a precise subject in the fields of electrical networks, protection devices, monitoring and control and industrial automation systems. 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(please specify) is compulsory. no. 158 Calculation of short-circuit currents Benoît de METZ-NOBLAT Graduate engineer from ESE (Ecole Supérieure d’Electricité), he worked for Saint-Gobain first as a research engineer, then in maintenance and new projects on a production site. He joined Schneider Electric in 1986, and is currently responsible for the Electrical System Analysis Group at the Research and Development Department. Frédéric DUMAS After graduating as doctor engineer from UTC (Université de Technologie de Compiègne) in 1993, he joined Schneider Electric. Works for the the Electrical System Analysis Group at the Research and Development Department, and is in charge of research projects in the field of industrial and distribution networks, including the development of software for electrotechnical calculations. Georges THOMASSET A graduate engineer from IEG (Institut d’Electrotechnique de Grenoble) in 1971, has since conducted numerous research, design and implementation of complex industrial network projects in Merlin Gerin’s Technical Management Department. He has long and valuable experience as responsible of the technical section for the industrial unit in the Engineering and Contracting Department from 1984 to 1996. He is now responsible of the general technical development of offers at the Schneider Electric Applications and Services Activity. ECT 158 updated June 2000 Lexicon Abbreviations λ Factor depending on the saturation BC Breaking capacity. inductance of a generator. MLVS Main low voltage switchboard. k et K Constants (tables and graphs). Ra Equivalent resistance of the upstream Symbols network. A Cross sectional area of conductors. RL Line resistance per unit length. α Angle between the initiation of the fault S Cross-sectional area of conductors. and zero voltage. Sn Transformer kVA rating. c Voltage factor. Ssc Short-circuit power. cos ϕ Power factor (in the absence of harmonics). tmin Minimum dead time for short-circuit development time, often equal to the e Electromotive force. time delay of a circuit breaker. E Electromotive force (maximum value). u Instantaneous voltage. ϕ Phase angle (current with respect to usc Transformer short-circuit voltage in %. voltage). U Network phase-to-phase voltage with i Instantaneous current. no load. ia Alternating sinusoidal component of Un Network rated voltage with load. the instantaneous current. x Reactance, in %, of the rotating idc Aperiodic component of the machines instantaneous current. Xa Equivalent reactance of the upstream ip Maximum current value (first peak of network. the fault current). XL Line reactance per unit length. I Maximum r.m.s. current. Xsubt Reactance in % of rotating machines. Ib Short-circuit breaking current ZL Link impedance. (IEC 60909). Zsc Network upstream impedance for a Ik Steady-state short-circuit current three-phase fault. (IEC 60909). Zup Equivalent impedance of the Ik ” Initial short-circuit current upstream network (IEC 60909). Z(1), Z(2), Z(0) Ir Rated current of a generator. Positive-sequence, negative- Is Service current. øsequence and zero-sequence Isc Steady-state short-circuit current (Isc3 impedances of a network or an = three-phase, Isc2 = two-phase, …). element. Cahier Technique Schneider Electric no. 158 / p.2 Calculation of short-circuit currents In view of sizing electrical installation and the required equipment, as well as determining the means required for the protectiuon of life and property, short-circuit currents must be calculated for every point in the network. This "Cahier Technique" reviews the calculation methods for short-circuit currents as laid down by standards such as IEC 60909. It is intended for radial low-volktage (LV) and high-voltage (HV) circuits. The aim is to provide a further understanding of the calculation methods, essential when determining short-circuit currents, even when computerised methods are employed. Summary 1 Introduction p. 4 1.1 The main types of short-circuits p. 5 1.2 Etablishing the short-circuit current p. 7 1.3 Standardised Isc calculations p. 10 1.4 Methods presented in this document p. 11 1.5 Basic assumptions p. 11 2 Calculation of Isc by the impedance method 2.1 Isc depending on the different types of short-circuit p. 12 2.2 Determining the various short-circuit impedances p. 13 2.3 Relationships between impedances at the different voltage levels in an installation p. 18 2.4 Calculation example p. 19 3 Calculation of Isc values in a radial 3.1 Advantages of this method p. 23 network using symmetrical components 3.2 Symmetrical components p. 23 3.3 Calculation as defined by IEC 60909 p. 24 3.4 Equations for the various currents p. 26 3.5 Calculation example p. 27 4 Computerised calculations and conclusion p. 31 Bibliography p. 32 Cahier Technique Schneider Electric no. 158 / p.3 1 Introduction Electrical installations almost always require are used. Two values of the short-circuit current protection against short-circuits wherever there must be evaluated: is an electrical discontinuity. This most often c the maximum short-circuit current, used to corresponds to points where there is a change in conductor cross-section. The short-circuit determine: current must be calculated at each level in the v the breaking capacity of the circuit breakers; installation in view of determining the v the making capacity of the circuit breakers; characteristics of the equipment required to v the electrodynamic withstand capacity of the withstand or break the fault current. wiring system and switchgear. The flow chart in figure 1 indicates the procedure for determining the various short- The maximum short-circuit current corresponds circuit currents and the resulting parameters for to a short-circuit in the immediate vicinity of the the different protection devices. downstream terminals of the protection device. It In order to correctly select and adjust the must be calculated accurately and used with a protection devices, graphs of figures 2, 3 & 4 safety margin. Upstream Ssc HV / LV transformer rating usc (%) Isc at transformer terminals Power factor, Breaking capacity Main coincidence factor, duty factor, Conductor characteristics Inst. trip setting circuit breaker foreseeable expansion c Busbars: factor - length, Isc - width, of main LV switchboard - thickness. outgoers c Cables: Main LV Breaking capacity switchboard - type of insulation, - single-core or multicore, Inst. trip setting distribution - length, circuit breakers - cross-section, Isc c Environment : at head of secondary - ambient temperature, switchboards - installation method, Secondary Breaking capacity - number of contiguous circuits. distribution Inst. trip setting circuit breakers Feeder current Isc ratings at head of final voltage drops switchboards Breaking capacity Final distribution Inst. trip setting circuit breakers Load Isc at end of final rating outgoers Fig. 1: short-circuit (Isc) caculation procedure when designing an electrical installation. Cahier Technique Schneider Electric no. 158 / p.4 t 1 2 t Cable or I2t characteristic θ1 > θ2 Fuse time-current a5 s curve Transient I2t = k2A2 overload Iz1 < Iz2 I IB Ir Iz I Fig. 2: the I2t characteristics of a conductor depending Fig. 4: circuit protection using an aM fuse. on the ambient temperature (temperature of case 1: t1 being higher that temperature of case 2 : t2, the admissible current limit Iz1 is lower than the admissible v protection of life depends on circuit breaker or current limit Iz2). fuse operation, essentially the case for TN and IT electrical systems. t Note that the minimum short-circuit current corresponds to a short-circuit at the end of the Cable or I2t protected line, generally phase-to-earth for LV and Design current characteristic phase-to-phase for HV (neutral not distributed), under the least severe operating conditions (fault at the end of a feeder and not just downstream from a protection device, one transformer in Circuit breaker service when two can be connected, etc.). Transient time-current Note also that whatever the case, for whatever overload curve type of short-circuit current (minimum or maximum), the protection device must clear the short-circuit within a time tc that is compatible with the thermal stresses that can be withstood by the protected cable: IB Ir Iz Isc BC I ∫i dt ≤ k 2 A 2 (see fig. 2, 3 and 4 ) 2 Fig. 3: circuit protection using a circuit breaker. where A is the cross-sectional area of the conductors and k is a constant calculated on the c the minimum short-circuit current, essential basis of different correction factors for the cable when selecting the time-current curve for circuit installation method, contiguous circuits, etc. breakers and fuses, in particular when: Further practical information may be found in the v cables are long and/or the source impedance "Electrical Installation Guide" published by is relatively high (generators, UPSs); Schneider Electric (see the bibliography). 1.1 The main types of short-circuits Various types of short-circuits can occur in foreign conducting body such as a tool or an electrical installations. animal); v internal or atmospheric overvoltages; Characteristics of short-circuits v insulation breakdown due to heat, humidity or The primary characteristics are: a corrosive environment; c duration (self-extinguishing, transient and steady-state); c location (inside or outside a machine or an electrical switchboard). c origin: v mechanical (break in a conductor, accidental Short-circuits can be: electrical contact between two conductors via a c phase-to-earth (80 % of faults); Cahier Technique Schneider Electric no. 158 / p.5 c phase-to-phase (15 % of faults). This type of v excessive temperature rise due to an increase fault often degenerates into a three-phase fault; in Joule losses, with the risk of damage to c three-phase (only 5 % of initial faults). insulation; These different short-circuit currents are c on other circuits in the network or in near-by presented in figure 5 . networks: Consequences of short-circuits v voltage dips during the time required to clear the fault, ranging from a few milliseconds to a The consequences are variable depending on few hundred milliseconds; the type and the duration of the fault, the point in the installation where the fault occurs and the v shutdown of a part of the network, the extent of short-circuit power. Consequences include: that part depending on the design of the network c at the fault location, the presence of electrical and the discrimination levels offered by the arcs, resulting in: protection devices; v damage to insulation; v dynamic instability and/or the loss of machine v welding of conductors; synchronisation; v fire and danger to life; v disturbances in control / monitoring circuits, c on the faulty circuit: v etc... v electrodynamic forces, resulting in: - deformation of the busbars; - disconnection of cables; a) Symmetrical three-phase short- b) Phase-to-phase short-circuit clear of circuit. earth. L3 L3 L2 L2 L1 L1 Ik" Ik" c) Phase-to-phase-to-earth short- d) Phase-earth short-circuit. circuit. L3 L3 L2 L2 L1 L1 Ik" Ik" Ik" " Ik Short-circuit current, Partial short-circuit currents in conductors and earth. In calculations, the various currents (Ik") are identified by an index. Fig. 5: different types of short-circuits and their currents. The direction of current is chosen arbitrarily. (See IEC 60909). Cahier Technique Schneider Electric no. 158 / p.6 1.2 Establishing the short-circuit current A simplified network comprises a source of R and the R / X ratio is between 0.1 and 0.3. The constant AC power, a switch, an impedance Zsc ratio is virtually equals cos ϕsc for low values: that represents all the impedances upstream of R the switch, and a load impedance Zs (see fig. 6 ). cos ϕ sc = R + X2 2 In a real network, the source impedance is made However, the transient conditions prevailing while up of everything upstream of the short-circuit the short-circuit current develops differ depending including the various networks with different on the distance between the fault location and the voltages (HV, LV) and the series-connected generator. This distance is not necessarily wiring systems with different cross-sectional physical, but means that the generator areas (A) and lengths. impedances are less than the link impedance In figure 6 , when the switch is closed, the between the generator and the fault location. design current Is flows through the network. When a fault occurs between A and B, the Fault away from the generator negligible impedance between these points This is the most frequent situation. The transient results in a very high short-circuit current Isc that conditions are those resulting from the application is limited only be impedance Zsc. of a voltage to a reactor-resistance circuit. This The current Isc develops under transient voltage is: e = E sin (ω t + α ) conditions depending on the reactances X and the resistances R that make up impedance Zsc: Current i is then the sum of the two components: Zsc = R2 + X 2 i = ia + idc . In power distribution networks, reactance c The first (ia) is alternating and sinusoidal: X = L ω is normally much greater than resistance ia = Ι sin (ω t + α ) where I = maximum current = E , Zsc R X α = angle characterising the difference between the initiation of the fault and zero voltage. c The second (idc) is an aperiodic component, R - t A idc = - Ι sin α e L . Its initial value depends on Zsc a and its decay rate is proportional to R / L. Zs At the initiation of the short-circuit, i is equal to e zero by definition (the design current Is is negligible), hence: B i = ia + idc = 0 Figure 7 shows the graphical composition of i as Fig. 6: simplified network diagram. the algebraic sum of its two components ia et idc. ia = I sin (ω t + α) - R t idc = - I sin α e L I t θ ω i = ia + idc Fault initiation Fig. 7: graphical presentation and decomposition of a short-circuit current occuring away from the generator. Cahier Technique Schneider Electric no. 158 / p.7 a) Symmetrical The moment the fault occurs or the moment of closing, with respect to the network voltage, is characterised by i its closing angle a (occurrence of the fault). The voltage I = 2 Ia can therefore be expressed as: u = E sin (ω t + α ) . The current therefore develops as follows: E R - t i = sin (ω t + α - ϕ ) - sin (α - ϕ ) e L Z u with its two components, one being alternating with a shift equal to ϕ with respect to the voltage and the second aperiodic and decaying to zero as t tends to infinity. Hence the two extreme cases defined by: c α = ϕ ≈ π/2 , said to be symmetrical (or balanced), b) Asymmetrical (see figure a ). E The fault current can be defined by: i = sin ω t Z i idc which, from the initiation, has the same shape as for steady state conditions with a peak value E / Z. ip c α = 0, said to be asymmetrical (or unbalanced), (see figure b ). The fault current can be defined by: E u R - t i = sin (ω t - ϕ) - sin ϕ e L Z Its initial peak value ip therefore depends on ϕ, i.e. on the R / X = cos ϕ ratio of the circuit. Fig. 8: graphical presentation of the two extreme cases (symmetrical and asymmetrical) for a short-circuit current. Figure 8 illustrates the two extreme cases for assumed to be constant and the internal the development of a short-circuit current, reactance of the machine variable. The presented, for the sake of simplicity, with a reactance develops in three stages: single-phase, alternating voltage. c subtransient (the first 10 to 20 milliseconds of R the fault); - t The factor e Lis inversely proportional to the c transient (up to 500 milliseconds); aperiodic component damping, determined by the R / L or R / X ratios. c steady-state (or synchronous reactance). The value of ip must therefore be calculated to Note that in the indicated order, the reactance determine the making capacity of the required acquires a higher value at each stage, i.e. the circuit breakers and to define the electrodynamic subtransient reactance is less than the transient forces that the installation as a whole must be reactance, itself less than the steady-state capable of withstanding. reactance. The successive effect of the three Its value may be deduced from the rms value of reactances leads to a gradual reduction in the the symmetrical short-circuit current Ia using the equation: ip = K 2 Ιa where the coefficient K is K indicated by the curve in figure 9 , as a function 2.0 of the ratio R / X or R / L. 1.8 Fault near the generator 1.6 When the fault occurs in the immediate vicinity of the generator supplying the circuit, the variation 1.4 in the impedance of the generator, in this case 1.2 the dominant impedance, damps the short-circuit current. 1.0 0 0.2 0.4 0.6 0.8 1.0 1.2 R / X The transient current-development conditions are complicated by the variation in the Fig. 9: variation of coefficient K depending on R / X or electromotive force resulting from the short- R / L (see IEC 60909). circuit. For simplicity, the electromotive force is Cahier Technique Schneider Electric no. 158 / p.8 short-circuit current which is the sum of four Practically speaking, information on the components (see fig. 10 ): development of the short-circuit current is not c the three alternating components essential: (subtransient, transient and steady-state); c in a LV installation, due to the speed of the c the aperiodic component resulting from the breaking devices, the value of the subtransient development of the current in the circuit short-circuit current, denoted Ik", and of the (inductive). maximum asymmetrical peak amplitude ip is a) 0 t (s) b) 0 t (s) c) 0 t (s) d) 0 t (s) 0.1 0.3 0.5 e) 0 t (s) Subtransient Transient Steady-state Fig. 10: total short-circuit current Isc (e), and contribution of its components: a) subtransient of each reactance; b) transient reactance; c) steady-state reactance; d) aperiodic component. Note that the decrease in the generator reactance is faster than that of the aperiodic component. This is a rare situation that can cause saturation of the magnetic circuits and interruption problems because several periods occur before the current passes through zero. Cahier Technique Schneider Electric no. 158 / p.9 sufficient when determining the breaking interruption is effective, i.e. following a time t after capacities of the protection devices and the the development of the short-circuit, where electrodynamic forces; t = tmin Time tmin (minimum time delay) is the sum c in LV power distribution and in HV applications, of the minimum operating time of a protection however, the transient short-circuit current is relay and the shortest opening time of the often used if breaking occurs before the steady- associated circuit breaker, i.e. the shortest time state stage, in which case it becomes useful to between the appearance of the short-circuit use the short-circuit breaking current, denoted Ib, current and the initial separation of the pole which determines the breaking capacity of the contacts on the switching device. time-delayed circuit breakers. Ib is the value of Figure 11 presents the various currents of the the short-circuit current at the moment short-circuits defined above. i Symmetrical Subtrans. Transient Steady-state Asymmetrical Fig. 11: short-circuit currents near a generator (schematic diagram). 1.3 Standardised Isc calculations The standards propose a number of method. The results obtained may be different from those presented in the next chapter, because these c Application guide C 15-105, which factors are taken into account; supplements NF C 15-100 (Normes Françaises) v the "composition" method, which may be used (low-voltage AC installations), details four when the characteristics of the power supply are methods: not known. The upstream impedance of the v the "impedance" method, used to calculate given circuit is calculated on the basis of an fault currents at any point in an installation with a estimate of the short-circuit current at its origin. high degree of accuracy. Power factor cos ϕsc = R / X is assumed to be This method involves adding the various identical at the origin of the circuit and the fault resistances and reactances of the fault loop location. In other words, it is assumed that the separately, from (and including) the source to the elementary impedances of two successive given point, and then calculating the sections in the installation are sufficiently similar corresponding impedance. The Isc value is in their characteristics to justify the replacement finally obtained by applying Ohm's law: of vectorial addition of the impedances by Isc = Un / ∑(Z) algebraic addition. This approximation may be All the characteristics of the various elements in used to calculate the value of the short-circuit the fault loop must be known (sources and wiring current modulus with sufficient accuracy for the systems). addition of a circuit. This very approximate Note that in the application guide, a number of method should be used only for installations factors are not taken into account, notably: rated up to 800 kVA; - the reactances of the circuit breakers and the v the "conventional" method, which can be used, busbars; when the impedances or the Isc in the - the resistances of rotating machines. installation upstream of the given circuit are not Cahier Technique Schneider Electric no. 158 / p.10 known, to calculate the minimum short-circuit composition and conventional methods. This currents and the fault currents at the end of a method may be used to determine the line. It is based on the assumption that the characteristics of a circuit to be added to an voltage at the circuit origin is equal to 80 % of existing installation for which sufficient the rated voltage of the installation during the information is not available. It is directly short-circuit or the fault. applicable to LV installations, and can be used This method considers only the resistance of the with correction coefficients if the voltage is not conductors and applies a coefficient greater than 230 / 400 V. 1 to conductors with large cross-sectional areas c Standard IEC 909 (VDE 0102) applies to all to take into account their inductance (1.15 for networks, radial or meshed, up to 230 kV. This 150 mm2, 1.20 for 185 mm2, etc.). It is mainly method, based on the Thevenin theorem, used for final circuits with their origin at a calculates an equivalent voltage source at the distance that is sufficiently far from the power short-circuit location and then determines the source (network or power-station unit); corresponding short-circuit current. All network v the "simplified" method (presented in detail in feeders as well as the synchronous and this application guide), which, via tables based asynchronous machines are replaced in the on numerous simplifying assumptions, indicates calculation by their impedances (positive for each conductor cross-sectional area: sequence, negative-sequence and zero- - the current rating of the overload protection sequence). All line capacitances and the parallel device; admittances of non-rotating loads, except those - maximum lengths of wiring systems to maintain of the zero-sequence system, are neglected. protection against indirect contact; c Other methods use the superposition principle - permissible lengths in view of line voltage and require that the load current first be drops. calculated. Note also the method proposed by The data in the tables is in fact the result of standard IEC 865 (VDE 0103) which calculates calculations run using essentially the the thermally equivalent short-circuit current. 1.4 Methods presented in this document In this "Cahier Technique" publication, two c the IEC 60909 method, used primarily for HV methods are presented for the calculation of networks, was selected for its accuracy and its short-circuit currents in radial networks: analytical character. More technical in nature, c the impedance method, reserved primarily for it implements the symmetrical-component LV networks, was selected for its high degree of principle. accuracy and its instructive value, given that virtually all characteristics of the circuit are taken into account. 1.5 Basic assumptions To simplify the short-circuit calculations, a fault remains three-phase and a phase-to-earth number of assumptions are required. These fault remains phase-to-earth; impose limits for which the calculations are valid c for the entire duration of the short-circuit, the but usually provide good approximations, facilitating comprehension of the physical voltages responsible for the flow of the current phenomena and consequently the short-circuit and the short-circuit impedance do not change current calculations. They nevertheless maintain significantly; a fully acceptable level of accuracy, "erring" c transformer regulators or tap-changers are systematically on the conservative side. assumed to be set to a medium position (if the The assumptions used in this document are as short-circuit occurs away from the generator, the follows: actual position of the transformer regulator or c the given network is radial with rated voltages tap-changers does not need to be taken into ranging from LV to HV, but not exceeding account; 230 kV, the limit set by standard IEC 60909; c arc resistances are not taken into account; c the short-circuit current, during a three-phase c all line capacitances are neglected; short-circuit, is assumed to occur simultaneously on all three phases; c load currents are neglected; c during the short-circuit, the number of phases c all zero-sequence impedances are taken into involved does not change, i.e. a three-phase account. Cahier Technique Schneider Electric no. 158 / p.11 2 Calculation of Isc by the impedance method 2.1 Isc depending on the different types of short-circuit Three-phase short-circuit flows from the generator to the location of the This fault involves all three phases. Short-circuit fault, i.e. the impedances of the power sources current Isc3 is equal to: and the lines (see fig. 12 ). This is, in fact, the "positive-sequence" impedance per phase: U/ 3 Ιsc 3 = ∑ R + ∑ X 2 2 Zsc Zsc = where where U (phase-to-phase voltage) corresponds to the transformer no-load voltage which is 3 to ∑R = the sum of series resistances, 5 % greater than the on-load voltage across the terminals. For example, in 390 V networks, the ∑X = the sum of series reactances. phase-to-phase voltage adopted is U = 410, and It is generally considered that three-phase faults the phase-to-neutral voltage is U / 3 = 237 V . provoke the highest fault currents. The fault Calculation of the short-circuit current therefore current in an equivalent diagram of a polyphase requires only calculation of Zsc, the impedance system is limited by only the impedance of one equal to all the impedances through which Isc phase at the phase-to-neutral voltage of the Three-phase fault ZL Zsc U/ 3 V Ιsc 3 = ZL Zsc ZL Phase-to-phase fault ZL Zsc U U Ιsc 2 = ZL 2 Zsc Zsc Phase-to-neutral fault ZL Zsc U/ 3 ZLn V Ιsc1 = Zsc + Z Ln ZLn Phase-to-earth fault ZL Zsc U/ 3 V Ιsc(0) = Zsc + Z(0) Z(0) Z(0) Fig. 12: the various short-circuit currents. Cahier Technique Schneider Electric no. 158 / p.12 network. Calculation of Isc3 is therefore essential source is less than Zsc (for example, at the for selection of equipment (maximum current and terminals of a star-zigzag connected transformer electrodynamic withstand capability). or of a generator under subtransient conditions). In this case, the phase-to-neutral fault current Phase-to-phase short-circuit clear of earth may be greater than that of a three-phase fault. This is a fault between two phases, supplied with a phase-to-phase voltage U. Phase-to-earth fault (one or two phases) In this case, the short-circuit current Isc2 is less This type of fault brings the zero-sequence than that of a three-phase fault: impedance Z(0) into play. U 3 Ιsc2 = = Ιsc 3 ≈ 0.86 Ιsc3 Except when rotating machines are involved 2 Zsc 2 (reduced zero-sequence impedance), the short- Phase-to-neutral short-circuit clear of earth circuit current Isc(0) is less than that of a three- This is a fault between one phase and the phase fault. neutral, supplied with a phase-to-neutral voltage Calculation of Isc(0) may be necessary, V = U/ 3. depending on the neutral system (system The short-circuit current Isc1 is: earthing arrangement), in view of defining the U/ 3 setting thresholds for the zero-sequence (HV) or Ιsc1 = earth-fault (LV) protection devices. Zsc + ZLn In certain special cases of phase-to-neutral Figure 12 shows the various short-circuit faults, the zero-sequence impedance of the currents 2.2 Determining the various short-circuit impedances This method involves determining the short- circuit currents on the basis of the impedance As, Xup = Zup2 - Rup2 , represented by the "circuit" through which the 2 Xup Rup short-circuit current flows. This impedance may = 1 - be calculated after separately summing the Zup Zup various resistances and reactances in the fault 2 Therefore, for 20 kV, loop, from (and including) the power source to Xup the fault location. = 1 - (0,2) = 0, 980 2 (The circled numbers X may be used to come Zup Xup = 0.980 Zup at 20 kV, back to important information while reading the example at the end of this section.) hence the approximation Xup ≈ Zup . c Internal transformer impedance Network impedances The impedance may be calculated on the basis c Upstream network impedance of the short-circuit voltage usc expressed as a Generally speaking, points upstream of the power percentage: source are not taken into account. Available data U2 on the upstream network is therefore limited to 3 Z T = usc where that supplied by the power distributor, i.e. only the Sn short-circuit power Ssc in MVA. U = no-load phase-to-phase voltage of the The equivalent impedance of the upstream transformer; network is: Sn = transformer kVA rating; U2 U usc = voltage that must be applied to the 1 Zup = primary winding of the transformer for the rated Ssc where U is the no-load phase-to-phase voltage current to flow through the secondary winding, of the network. when the LV secondary terminals are short- circuited. The upstream resistance and reactance may be deduced from Rup / Zup (for HV) by: For public distribution MV / LV transformers, the Rup / Zup ≈ 0.3 at 6 kV; values of usc have been set by the European Rup / Zup ≈ 0.2 at 20 kV; Harmonisation document HD 428-1S1 issued in Rup / Zup ≈ 0.1 at 150 kV. October 1992 (see fig. 13 ). Rating (kVA) of the HV / LV transformer ≤ 630 800 1,000 1,250 1,600 2,000 Short-circuit voltage usc (%) 4 4.5 5 5.5 6 7 Fig. 13: standardised short-circuit voltage for public distribution transformers. Cahier Technique Schneider Electric no. 158 / p.13 Note that the accuracy of values has a direct The relative error is: influence on the calculation of Isc in that an error ∆Ιsc Ι' sc - Ιsc Zup U2 / Ssc of x % for usc produces an equivalent error (x %) = = = for ZT. Ιsc Ιsc ZT usc U2 / Sn 4 In general, RT << XT , in the order of 0.2 XT, i.e.: and the internal transformer impedance may be ∆Ιsc 100 Sn = considered comparable to reactance XT. For low Ιsc usc Ssc power levels, however, calculation of ZT is required because the ratio RT / XT is higher. Figure 14 indicates the level of conservative The resistance is calculated using the joule error in the calculation of Isc, due to the fact that losses (W) in the windings: the upstream impedance is neglected. The figure demonstrates clearly that it is possible to neglect W W = 3 RT Ιn2 ⇒ RT = the upstream impedance for networks where the 3 Ιn2 short-circuit power Ssc is much higher than the Notes: transformer kVA rating Sn. For example, when 5 Ssc / Sn = 300, the error is approximately 5 %. v when n identically-rated transformers are c Link impedance connected in parallel, their internal impedance The link impedance ZL depends on the values, as well as the resistance and reactance resistance per unit length, the reactance per unit values, must be divided by n. length and the length of the links. v the resistance per unit length of overhead v particular attention must be paid to special lines, cables and busbars is calculated as: transformers, for example, the transformers for ρ rectifier units have usc values of up to 10 to RL = 12 % in order to limit short-circuit currents. A where When the impedance upstream of the A = cross-sectional area of the conductor; transformer and the transformer internal ρ = conductor resistivity, however the value used impedance are taken into account, the short- varies, depending on the calculated short-circuit circuit current may be expressed as: current (minimum or maximum). U Ιsc = 6 The table in figure 15 provides values for 3 (Zup + Z T ) each of the above-mentioned cases. Initially, Zup and ZT may be considered Practically speaking, for LV and conductors with comparable to their respective reactances. The cross-sectional areas less than 150 mm2, only short-circuit impedance Zsc is therefore equal to the resistance is taken into account the algebraic sum of the two. (RL < 0.15 mΩ / m when A > 150 mm2). The upstream network impedance may be v the reactance per unit length of overhead lines, neglected, in which case the new current value is: cables and busbars may be calculated as: U d Ι' sc = XL = L ω = 15.7 + 144.44 Log 3 ZT r ∆Isc/Isc (%) 12 Psc = 250 MVA 10 Psc = 500 MVA 5 0 500 1,000 1,500 2,000 Pn (kVA) Fig. 14 : resultant error in the calculation of the short-circuit current when the upstream network impedance Zup is neglected. Cahier Technique Schneider Electric no. 158 / p.14 expressed as mΩ / km for a single-phase or 8 - 0.09 mΩ / m for touching, single-conductor three-phase delta cable system, where (in mm): r = radius of the conducting cores; cables (flat or triangular ); d = average distance between conductors. 9 - 0.15 mΩ / m as a typical value for busbars N.B. Above, Log = decimal logarithm. For overhead lines, the reactance increases ( ) and spaced, single-conductor cables slightly in proportion to the distance between d ( ) ; For "sandwiched-phase" busbars conductors (Log ), and therefore in t (e.g. Canalis - Telemecanique), the reactance is proportion to the operating voltage. considerably lower. 7 the following average values are to be used: Notes: X = 0.3 Ω / km (LV lines); v the impedance of the short links between the X = 0.4 Ω / km (MV or HV lines). distribution point and the HV / LV transformer may be neglected. This assumption gives a The table in figure 16 shows the various conservative error concerning the short-circuit reactance values for conductors in LV current. The error increases in proportion to the applications, depending on the wiring system. transformer rating. The following average values are to be used: v the cable capacitance with respect to the earth - 0.08 mΩ / m for a three-phase cable ( ), (common mode), which is 10 to 20 times greater than that between the lines, must be taken into and, for HV applications, between 0.1 and account for earth faults. Generally speaking, the 0.15 mΩ / m. Current Resistivity Resistivity value Concerned (*) (Ω mm2 / m) conductors Copper Aluminium Maximum short-circuit current ρ1 = 1.25 ρ20 0.0225 0.036 PH-N Minimum short-circuit current ρ1 = 1.5 ρ20 0.027 0.043 PH-N Fault current in TN and IT ρ1 = 1.25 ρ20 0.0225 0.036 PH-N (**) systems PE-PEN Voltage drop ρ1 = 1.25 ρ20 0.0225 0.036 PH-N (**) Overcurrent for conductor ρ1 = 1.5 ρ20 0,027 0.043 Phase-Neutral thermal-stress checks PEN-PE if incorporated in same multiconductor cable ρ1 = 1.25 ρ20 0.0225 0.036 Separate PE (*) ρ20 is the resistivity of the conductors at 20 °C. 0.018 Ωmm2 / m for copper and 0.029 Ωmm2 / m for aluminium. (**) N, the cross-sectional area of the neutral conductor, is less than that of the phase conductor Fig. 15: conductor resistivity ρ values to be taken into account depending on the calculated short-circuit current (minimum or maximum). See UTE C 15-105. Wiring system Busbars Three-phase Spaced single-core Touching single- 3 touching 3 "d" spaced cables (flat) cable espacés core cables (triangle) cables (flat) d = 2r d = 4r d d r Diagram Average reactance 0.15 0.08 0.15 0.085 0.095 0.145 0.19 per unit lengt values (mΩ / m) Extreme reactance 0.12-0.18 0.06-0.1 0.1-0.2 0.08-0.09 0.09-0.1 0.14-0.15 0.18-0.20 per unit length values (mΩ / m) Fig. 16: cables reactance values depending on the wiring system. Cahier Technique Schneider Electric no. 158 / p.15 capacitance of a HV three-phase cable with a - a resistance for cable cross-sectional areas cross-sectional area of 120 mm2 is in the order less than 74 mm2; of 1 µF / km, however the capacitive current - a reactance for cable cross-sectional areas remains low, in the order of 5 A / km at 20 kV. greater than 660 mm2. c The reactance or resistance of the links may v second case. Consider a three-phase cable, at be neglected. 20 °C, with aluminium conductors. As above, the If one of the values, RL or XL, is low with respect impedance ZL curve may be considered identical to the other, it may be neglected because the to the asymptotes, but for cable cross-sectional resulting error for impedance ZL is consequently areas less than 120 mm2 and greater than very low. For example, if the ratio between RL 1,000 mm2 (curves not shown). and XL is 3, the error in ZL is 5.1 %. The curves for RL and XL (see fig. 17 ) may be Impedance of rotating machines used to deduce the cable cross-sectional areas c Synchronous generators for which the impedance may be considered The impedances of machines are generally comparable to the resistance or to the reactance. expressed as a percentage, for example: Examples: Isc / In = 100 / x where x is the equivalent of the transformer usc. v first case. Consider a three-phase cable, at 20 °C, with copper conductors. Their reactance Consider: is 0.08 mΩ / m. The RL and XL curves x U2 (see fig. 17 ) indicate that impedance ZL 10 Z = where approaches two asymptotes, RL for low cable 100 Sn cross-sectional areas and XL = 0.08 mΩ / m for U = no-load phase-to-phase voltage of the high cable cross-sectional areas. For the low and generator, high cable cross-sectional areas, the impedance Sn = generator VA rating. ZL curve may be considered identical to the asymptotes. 11 What is more, given that the value of R / X is The given cable impedance is therefore low, in the order of 0.05 to 0.1 for MV and 0.1 considered, with a margin of error less than to 0.2 for LV, impedance Z may be considered 5.1 %, comparable to: comparable to reactance X. Values for x are given in the table in figure 18 for turbo- generators with smooth rotors and for "hydraulic" generators with salient poles (low speeds). mΩ / m On reading the table, one may be surprised to note that the steady-state reactance for a short- 1 0.8 circuit exceeds 100 % (at that point in time, Isc < In) . However, the short-circuit current is essentially inductive and calls on all the reactive power that the field system, even over-excited, can supply, whereas the rated current essentially 0.2 ZL carries the active power supplied by the turbine 0.1 (cos ϕ from 0.8 to 1). 0.08 c Synchronous compensators and motors 0.05 XL The reaction of these machines during a short- circuit is similar to that of generators. RL 12 They produce a current in the 0.02 network that depends on their reactance in % 0.01 (see fig. 19 ). 10 20 50 100 200 500 1,000 c Asynchronous motors Cross-sectional area A (in mm2) When an asynchronous motor is cut from the Fig. 17: impedance ZL of a three-phase cable, network, it maintains a voltage across its at 20 °C, with copper conductors. terminals that disappears within a few Subtransient Transient Steady-state reactance reactance reactance Turbo-generator 10-20 15-25 150-230 Salient-pole generators 15-25 25-35 70-120 Fig. 18: generator reactance values, in x %. Cahier Technique Schneider Electric no. 158 / p.16 Subtransient Transient Steady-state reactance reactance reactance High-speed motors 15 25 80 Low-speed motors 35 50 100 Compensators 25 40 160 Fig. 19: synchronous compensator and motor reactance values, in x %. hundredths of a second. When a short-circuit It is therefore unlikely, except for very powerful occurs across the terminals, the motor supplies a capacitor banks, that superposition will result in current that disappears even more rapidly, an initial peak higher than the peak current of an according to time constants in the order of: asymmetrical fault. v 0.02 seconds for single-cage motors up to It follows that when calculating the maximum 100 kW; short-circuit current, capacitor banks do not need v 0.03 seconds for double-cage motors and to be taken into account. motors above 100 kW; However, they must nonetheless be considered v 0.03 to 0.1 seconds for very large HV slipring motors (1,000 kW). when selecting the type of circuit breaker. During opening, capacitor banks significantly reduce the In the event of a short-circuit, an asynchronous circuit frequency and thus produce an effect on motor is therefore a generator to which an current interruption. impedance (subtransient only) of 20 to 25 % is attributed. c Switchgear Consequently, the large number of LV motors, 14 Certain devices (circuit breakers, contactors with low individual outputs, present on industrial with blow-out coils, direct thermal relays, etc.) sites may be a source of difficulties in that it is not have an impedance that must be taken into easy to foresee the average number of in-service account, for the calculation of Isc, when such a motors that will contribute to the fault when a device is located upstream of the device short-circuit occurs. Individual calculation of the intended to break the given short-circuit and reverse current for each motor, taking into remain closed (selective circuit breakers). account the impedance of its link, is therefore a tedious and futile task. Common practice, notably 15 For LV circuit breakers, for example, a in the United States, is to take into account the reactance value of 0.15 mΩ is typical, with the combined contribution to the fault current of all the resistance negligible. asynchronous LV motors in an installation. For breaking devices, a distinction must be made 13 They are therefore thought of as a unique depending on the speed of opening: source, capable of supplying to the busbars a v certain devices open very quickly and thus current equal to (Istart / In) times the sum of the significantly reduce short-circuit currents. This is rated currents of all installed motors. the case for fast-acting, limiting circuit breakers and the resultant level of electrodynamic forces Other impedances and thermal stresses, for the part of the c Capacitors installation concerned, remains far below the A shunt capacitor bank located near the fault theoretical maximum; location discharges, thus increasing the short- v other devices, such as time-delayed circuit circuit current. This damped oscillatory discharge breakers, do not offer this advantage. is characterised by a high initial peak value that c Fault arc is superposed on the initial peak of the short- The short-circuit current often flows through an circuit current, even though its frequency is far arc at the fault location. The resistance of the arc greater than that of the network. is considerable and highly variable. The voltage Depending on the coincidence in time between drop over a fault arc can range from 100 to 300 V. the initiation of the fault and the voltage wave, For HV applications, this drop is negligible with two extreme cases must be considered: respect to the network voltage and the arc has v if the initiation of the fault coincides with zero no effect on reducing the short-circuit current. voltage, the discharge current is equal to zero, For LV applications, however, the actual fault whereas the short-circuit current is asymmetrical, current when an arc occurs is limited to a much with a maximum initial amplitude peak; lower level than that calculated (bolted, solid v conversely, if the initiation of the fault coincides fault), because the voltage is much lower. with maximum voltage, the discharge current superposes itself on the initial peak of the fault 16 For example, the arc resulting from a short- current, which, because it is symmetrical, has a circuit between conductors or busbars may low value. reduce the prospective short-circuit current by Cahier Technique Schneider Electric no. 158 / p.17 ; 20 to 50 % and sometimes by even more than c Various impedances 50 % for rated voltages under 440 V. Other elements may add non-negligible However, this phenomenon, highly favourable impedances. This is the case for harmonics filters in the LV field and which occurs for 90 % of faults, and inductors used to limit the short-circuit current. may not be taken into account when determining They must, of course, be included in calculations, the breaking capacity because 10 % of faults take as well as wound-primary type current place during closing of a device, producing a solid transformers for which the impedance values vary fault without an arc. This phenomenon should, depending on the rating and the type of however, be taken into account for the calculation construction. of the minimum short-circuit current. 2.3 Relationships between impedances at the different voltage levels in an installation Impedances as a function of the voltage c For the system as a whole, after having The short-circuit power Ssc at a given point in calculated all the relative impedances, the short- the network is defined by: circuit power may be expressed as: U2 1 Ssc = U Ι 3 = Ssc = from which it is possible to deduce Zsc ΣZR This means of expressing the short-circuit power the fault current Isc at a point with a voltage U: implies that Ssc is invariable at a given point in Ssc 1 Ιsc = = 3 U ΣZR the network, whatever the voltage. And the 3 U equation Ιsc 3 = U 3 Zsc implies that all ΣZR is the composed vector sum of all the impedances must be calculated with respect to relative upstream imedances. It is therefore the the voltage at the fault location, which leads to relative impedance of the upstream network as certain complications that often produce errors in seen from a point at U voltage. calculations for networks with two or more voltage Hence, Ssc is the short-circuit power, in VA, at a levels. For example, the impedance of a HV line point where voltage is U. must be multiplied by the square of the reciprocal For example, if we consider the simplified of the transformation ratio, when calculating a diagram of figure 20 : fault on the LV side of the transformer: 2 ULV 2 At point A, Ssc = 2 U ULV 17 Z LV = Z HV LV ZT + ZL UHV UHV A simple means of avoiding these difficulties is the 1 relative impedance method proposed by H. Rich. Hence, Ssc = ZT ZL 2 + 2 Calculation of the relative impedances UHV ULV This is a calculation method used to establish a relationship between the impedances at the different voltage levels in an electrical installation. This method proposes dividing the impedances (in ohms) by the square of the network line-to- line voltage (in volts) at the point where the impedances exist. The impedances therefore become relative. UHV c For lines and cables, the relative resistances and reactances are defined as: ZT R X RR = 2 and XR = 2 where R is in ohms U U and U in volts. ULV c For transformers, the impedance is expressed on the basis of their short-circuit voltages usc and ZL their kVA rating Sn: A U2 usc Z = Sn 100 c For rotating machines, the equation is identical, with x representing the impedance Fig. 20: calculating Ssc at point A. expressed in %. Cahier Technique Schneider Electric no. 158 / p.18 2.4 Calculation example (with the impedances of the power sources, are identical and all motors are running when the the upstream network and the power supply fault occurs. transformers as well as those of the electrical The Isc value must be calculated at the various links) fault locations indicated in the network diagram (see fig. 21 ), that is: Problem c point A on the HV busbars, with a negligible Consider a 20 kV network that supplies a impedance; HV / LV substation via a 2 km overhead line, and c point B on the LV busbars, at a distance of a 1 MVA generator that supplies in parallel the 10 meters from the transformers; busbars of the same substation. Two 1,000 kVA c point C on the busbars of an LV sub- parallel-connected transformers supply the LV distribution board; busbars which in turn supply 20 outgoers to c point D at the terminals of motor M. 20 motors, including the one supplying motor M. Then the reverse current of the motors must be All motors are rated 50 kW, all connection cables calculated at C and B, then at D and A. Upstream network U1 = 20 kV 3L Psc = 500 MVA Overhead line 3 cables, 50 mm2, copper, G length = 2 km A Generator 1 MVA Xsubt = 15 % 2 transformers 1,000 kVA secondary winding 237 / 410 V usc = 5 % 10 m B Main LV switchboard busbars 3 bars, 400 mm2 / ph, copper, length = 10 m 3L Link 1 3 single-core cables, 400 mm2, aluminium spaced, flat, C length = 80 m LV sub-distribution board Link 2 3L 3 three-phase cables, 35 mm2, copper, length = 30 m D Motor M 50 kW (efficiency: 0.9, cos ϕ: 0.8) usc = 25 % Fig. 21: diagram for calculation of Isc values at points A, B, C and D. Cahier Technique Schneider Electric no. 158 / p.19 In this example, reactances X and resistances R the installation (see figure 22 ). The relative are calculated with their respective voltages in impedance method is not used. Solution Section Calculations Results (the circled numbers X indicate where explanations may be found in the preceding text) 20 kV↓ X (Ω) R (Ω) ( ) 2 1. upstream network Zup = 20 x 103 / 500 x 106 1 Xup = 0.98 Zup 2 0.78 Rup = 0.2 Zup ≈ 0.2 Xup 0.15 2. overhead line Xc o = 0.4 x 2 7 0.8 (50 mm2) 2, 000 Rc o = 0.018 x 6 0.72 50 ( ) 2 15 20 x 103 3. generator XG = x 10 60 100 106 R G = 0.1 X G 11 6 20 kV↑ X (mΩ) R (mΩ) Fault A 1 5 4102 4. transformers ZT = x x 3 5 2 100 106 XT ≈ ZT 4.2 R T = 0.2 X T 4 0.84 410 V↓ 5. circuit-breaker X cb = 0.15 15 0.15 6. busbars XB = 0.15 x 10-3 x 10 9 1.5 (3 x 400 mm2) 10 RB = 0.0225 x 6 ≈0 3 x 400 Fault B 7. circuit-breaker X cb = 0.15 0.15 −3 8. cable link 1 Xc1 = 0.15 x 10 x 80 12 (3 x 400 mm2) 80 Rc1 = 0.036 x 6 2.4 3 x 400 Fault C 9.circuit-breaker X cb = 0.15 0.15 10. cable link 2 Xc 2 = 0.09 x 10 −3 x 30 8 2.7 (35 mm2) 30 Rc 2 = 0.0225 x 19.2 35 Fault D 25 4102 12 Xm = x 11. motor 50 kW 100 (50 / 0.9 x 0.8) 103 605 Rm = 0.2 Xm 121 Fig. 22: impedance calculation. Cahier Technique Schneider Electric no. 158 / p.20 I- Fault at A (HV busbars) These values make clear the importance of Isc Elements concerned: 1, 2, 3. limitation due to the cables. The "network + line" impedance is parallel to that of the generator, however the latter is much ZC = RC + XC ≈ 19 mΩ 2 2 greater and may be neglected: 410 X A = 0.78 + 0.8 ≈ 1.58 Ω ΙC = ≈ 12,459 A 3 x 19 x 10−3 RA = 0.15 + 0.72 ≈ 0.87 Ω RC = 0.19 hence k = 1.55 on the curve in XC ZA = R2 A + X2 A ≈ 1.80 Ω hence figure 9 and therefore the peak Isc is equal to: 3 20 x 10 ΙA = ≈ 6,415 A 1.55 x 2 x 12,459 ≈ 27,310 A 3 x 1.80 IA is the "steady-state Isc" and for the purposes IV - Fault at D (LV motor) of calculating the peak asymmetrical Isc: Elements concerned: RA (1, 2, 3) + (4, 5, 6) + (7, 8) + (9, 10). = 0.55 The reactances and the resistances of the circuit XA breaker and the cables must be added to XC hence k = 1.2 on the curve in figure 9 and therefore Isc is equal to: and RC. 1.2 x 2 x 6,415 = 10,887 A . XD = (XC + 0,15 + 2, 7) 10-3 = 21, 52 mΩ and II - Fault at B (main LV switchboard busbars) Elements concerned: (1, 2, 3) + (4, 5, 6). RD = (RC + 19.2) 10-3 = 22.9 mΩ The reactances X and resistances R calculated for the HV section must be recalculated for the ZD = RD + XD ≈ 31.42 mΩ 2 2 LV network via multiplication by the square of 410 the voltage ratio 17 , i.e.: ΙD = ≈ 7, 534 A 3 x 31.42 x 10-3 (410 / 20,000)2 = 0.42 x 10−3 hence [(XA 0.42) + ] RD XB = 4.2 + 0.15 + 1.5 10-3 = 1.06 hence k ≈ 1.05 on the curve in XD XB = 6.51 mΩ and figure 9 and therefore the peak Isc is equal to: RB = [(RA 0.42) + ] 0.84 10-3 1.05 x 2 x 7,534 ≈ 11,187 A RB = 1.2 mΩ As each level in the calculations makes clear, These calculations make clear, firstly, the low the impact of the circuit breakers is negligible importance of the HV upstream reactance, with compared to that of the other elements in the respect to the reactances of the two parallel network. transformers, and secondly, the non-negligible impedance of the 10 meter long, LV busbars. V - Reverse currents of the motors ZB = 2 RB + 2 XB = 6.62 mΩ hence It is often faster to simply consider the motors as independent generators, injecting into the fault a 410 ΙB = ≈ 35,758 A "reverse current" that is superimposed on the 3 x 6.62 x 10-3 network fault current. RB c Fault at C = 0.18 hence k = 1.58 on the curve in XB The current produced by the motor may be figure 9 and therefore the peak Isc is equal to: calculated on the basis of the "motor + cable" impedance: 1.58 x 2 x 35,758 = 79,900 A What is more, if the fault arc is taken into XM = (605 + 2.7) 10-3 ≈ 608 mΩ account (see § c fault arc section 16 ), IB is RM = (121 + 19.2) 10-3 ≈ 140 mΩ reduced to a maximum value of 28,606 A and a ZM = 624 mΩ , hence minimum value of 17,880 A. 410 ΙM = ≈ 379 A III - Fault at C (busbars of LV sub-distribution 3 x 624 x 10-3 board) For the 20 motors Elements concerned: (1, 2, 3) + (4, 5, 6) + (7, 8). Ι MC = 7,580 A . The reactances and the resistances of the circuit Instead of making the above calculations, it is breaker and the cables must be added to XB and RB. possible (see 13 ) to estimate the current XC = (XB + 0.15 + 12) 10-3 = 18.67 mΩ injected by all the motors as being equal to and (Istart / In) times their rated current (95 A), i.e. RC = (RB + 2.4) 10-3 = 3.6 mΩ (4.8 x 95) x 20 = 9,120 A. Cahier Technique Schneider Electric no. 158 / p.21 This estimate therefore allows protection by increases from 35,758 A to 43,198 A and the excess value with respect to IMC (7,580 A). peak Isc from 79,900 A to 94,628 A. On the basis of However, as mentioned above, if the fault arc is R / X = 0.3 => k = 1.4 and the peak taken into account, Isc is reduced between 45.6 to 75 kA. Ιsc = 1.4 x 2 x 7, 580 ≈ 15, 005 A . c Fault at A (HV side) Consequently, the short-circuit current (subtransient) on the LV busbars increases from Rather than calculating the equivalent 12,459 A to 20,039 A and Isc from 27,310 A to impedances, it is easier to estimate 42,315 A. (conservatively) the reverse current of the motors at A by multiplying the value at B by the c Fault at D The impedance to be taken into account is LV / HV transformation value 17 , i.e.: 1 / 19th of ZM, plus that of the cable. 410 7, 440 x = 152.5 A 605 20 x 10-3 X MD = + 2.7 10-3 ≈ 34, 5 mΩ 19 This figure, compared to the 6,415 A calculated previously, is negligible. 121 RMD = + 19.2 10-3 ≈ 25.5 mΩ 19 Rough calculation of the fault at D ZMD = 43 mΩ hence This calculation makes use of all the 410 approximations mentioned above (notably 15 Ι MD = = 5,505 A and 16). 3 x 43 x 10-3 giving a total at D of: ΣX = 4.2 + 1.5 + 12 + 0.15 7,534 + 5,505 = 13,039 A rms, and Isc ≈ 20,650 A. ΣX = 17.85 mΩ = X'D c Fault en B ΣR = 2.4 + 19.2 = 21.6 mΩ = R'D As for the fault at C, the current produced by the motor may be calculated on the basis of the Z'D = R'D + X'D ≈ 28.02 mΩ 2 2 "motor + cable" impedance: 410 XM = (605 + 2.7 + 12) 10-3 ≈ 620 mΩ Ι' D = ≈ 8,448 A 3 x 28.02 x 10-3 RM = (121 + 19.2 + 2.4) 10-3 ≈ 142.6 mΩ hence the peak Isc: ZM = 636 mΩ hence 2 x 8,448 ≈ 11,945 A 410 ΙM = ≈ 372 A To find the peak asymmetrical Isc, the above 3 x 636 x 10-3 value must be increased by the contribution of For the 20 motors IMB = 7,440 A. the energised motors at the time of the fault 13 Again, it is possible to estimate the current injected by all the motors as being equal to 4.8 i.e. 4.8 times their rated current of 95 A: times their rated current (95 A), i.e. 9,120 A. The approximation again overestimates the real ( Ιsc = 11,945 + 4.8 x 95 x 2 x 20 ) value of IMB. Using the fact that = 24, 842 A . R / X = 0.3 => k = 1.4 and the peak Compared to the figure obtained by the full Ιsc = 1.4 2 x 7.440 = 14, 728 A calculation (20,039), the approximate method Consequently, the short-circuit current allows a quick evaluation with an error remaining (subtransient) on the main LV switchboard on the side of safety. Cahier Technique Schneider Electric no. 158 / p.22 3 Calculation of Isc values in a radial network using symmetrical components 3.1 Advantages of this method Calculation using symmetrical components is moduli and imbalances exceeding 120°).This is particularly useful when a three-phase network is the case for phase-to-earth or phase-to-phase unbalanced, because, due to magnetic short-circuits with or without earth connection; phenomena, for example, the traditional c the network includes rotating machines and/or "cyclical" impedances R and X are, normally special transformers (Yyn connection, for speaking, no longer useable. This calculation method is also required when: example). c a voltage and current system is not This method may be used for all types of radial symmetrical (Fresnel vectors with different distribution networks at all voltage levels. 3.2 Symmetrical components Similar to the Leblanc theorem which states that and by using the following operator a rectilinear alternating field with a sinusoidal 2π j 1 3 amplitude is equivalent to two rotating fields a = e 3 = - + j between Ι1, Ι2 , turning in the opposite direction, the definition of 2 2 symmetrical components is based on the and Ι3 . equivalence between an unbalanced three- This principle, applied to a current system, is phase system and the sum of three balanced confirmed by a graphical representation three-phase systems, namely the positive- (see fig. 23 ). For example, the graphical sequence, negative-sequence and zero- addition of the vectors produces, for, the sequence (see fig. 23 ). following result: The superposition principle may then be used to calculate the fault currents. Ι2 = a 2 Ι1(1) + a Ι1(2) + Ι1(3) In the description below, the system is defined Currents Ι1 and Ι3 may be expressed in the using current Ι1 as the rotation reference, where: same manner, hence the system: c Ι1(1) is the positive-sequence component; Ι1 = Ι1(1) + Ι1(2) + Ι1(0) c Ι1(2) is the negative-sequence component; Ι2 = a 2 Ι11 + aΙ1(2) + Ι1(0) c Ι1(0) is the zero-sequence component; Ι3 = a Ι1(1) + a 2 Ι1(2) + Ι1(0) Positive-sequence Negative-sequence zero-sequence I3(1) I1(0) I2(2) I3 I1 I1(1) + I1(2) + I2(0) I3(0) ωt = I2 ωt I3(2) I2(1) ωt ωt Geometric construction of I1 Geometric construction of I2 I1(1) I1 I1(0) I2 I1(1) I1(2) I1(0) a2 I1(1) a I1(2) I1(2) Fig. 23: graphical construction of the sum of three balanced three-phase systems (positive-sequence, negative-sequence and zero-sequence). Cahier Technique Schneider Electric no. 158 / p.23 These symmetrical current components are related to the symmetrical voltage components Elements Z(0) by the corresponding impedances: Transformer V(1) V(2) V(0) Z(1) = , Z( 2 ) = and Z(0) = (seen from secondary winding) Ι(1) Ι(2) Ι(0) No neutral ∞ These impedances may be defined from the Yyn or Zyn free flux ∞ characteristics (supplied by the manufacturers) forced flux 10 to 15 X(1) of the various elements in the given electrical Dyn or YNyn X(1) network. Among these characteristics, we can primary D or Y + zn 0.1 to 0.2 X(1) note that Z(2) ≈ Z(1), except for rotating machines, Machine whereas Z(0) varies depending on each element (see fig. 24 ). Synchronous ≈ 0.5 Z(1) Asynchronous ≈0 For further information on this subject, a detailed presentation of this method for calculating solid Line ≈ 3 Z(1) and impedance fault currents is contained in the Fig. 24: zero-sequence characteristic of the various "Cahier Technique" n° 18 (see the appended elements in an electrical network. bibliography). 3.3 Calculation as defined by IEC 60909 Standard IEC 60909 defines and presents a Depending on the required calculations and the method implementing symmetrical components, given voltage levels, the standardised voltage that may be used by engineers not specialised in levels are indicated in figure 25 . the field. 2- Determine and add up the equivalent positive- sequence, negative-sequence and zero- The method is applicable to electrical networks sequence impedances upstream of the fault with a rated voltage of less than 230 kV and the location. standard explains the calculation of minimum 3- Calculate the initial short-circuit current using and maximum short-circuit currents. The former the symmetrical components. Practically is required in view of calibrating overcurrent speaking and depending on the type of fault, the protection devices and the latter is used to equations required for the calculation of the Isc determine the rated characteristics for the are indicated in the table in figure 26 . electrical equipment. 4- Once the Isc (Ik") value is known, calculate the In view of its application to LV networks, the other values such as the peak Isc value, the standard is accompanied by application guide steady-state Isc value and the maximum, steady- IEC 60781. state Isc value. Procedure Effect of the distance separating the fault from the generator 1- Calculate the equivalent voltage at the fault When using this method, two different location, equal to c Un / 3 where c is a voltage possibilities must always be considered: factor required in the calculation to account for: c the short-circuit is away from the generator, c voltage variations in space and in time; the situation in networks where the short-circuit c possible changes in transformer tappings; currents do not have a damped, alternating component. c subtransient behaviour of generators and motors. This is generally the case in LV networks, except when high-power loads are supplied by special HV substations; c the short-circuit is near the generator Rated Voltage factor c (see fig. 11), the situation in networks where the voltage for calculation of short-circuit currents do have a damped, Un Isc max. Isc min. alternating component. This generally occurs in LV HV systems, but may occur in LV systems when, 230 - 400 V 1 0.95 for example, an emergency generator supplies priority outgoers. Others 1.05 1 The main differences between these two cases HV are: 1 to 230 kV 1.1 1 c for short-circuits away from the generator: Fig. 25: values for voltage factor c (see IEC 60909). v the initial (Ik" ), steady-state (Ik) and breaking (Ib) short-circuit currents are equal (Ik" = Ik = Ib); Cahier Technique Schneider Electric no. 158 / p.24 Type Ik” Fault occurring of short-circuit General situation far from the generators c Un c Un Three-phase (any Ze) = = 3 Z(1) 3 Z(1) In both cases, the short-circuit current depends only on Z(1), which is generally replaced by Zk, the short-circuit impedance at the fault location, defined by Zk = Rk 2 + Xk 2 where Rk is the sum of the resistances of one phase, connected in series; Xk is the sum of the reactances of one phase, connected in series. c Un c Un = = Phase-to-phase clear of earth (Ze = ∞) Z(1) + Z(2) 2 Z(1) c Un 3 c Un 3 Phase-to-earth = = Z(1) + Z(2) + Z(0) 2 Z(1) + Z(0) Phase-to-phase-to-earth c Un 3 Z(2) c Un 3 (Zsc between phases = 0) = = Z(1) Z(2) + Z(2) Z(0) + Z(1) Z(0) Z(1) + 2 Z(0) Symbols used in this table c phase-to-phase rms voltage of the three-phase network = U c short-circuit impedance = Zsc c modulus of the short-circuit current = Ik" c earth impedance = Ze. c symmetrical impedances = Z(1), Z(2), Z(0) Fig. 26: Short-circuit values depending on the positive-sequence, negative-sequence & zero-sequence impedances of the given network (see IEC 60909). v the positive-sequence (Z(1)) and negative- v the resistances per unit length RL of lines sequence (Z(2)) impedances are equal (overhead lines, cables, phase and neutral (Z(1) = Z(2)); conductors) should be calculated for a c for short-circuits near the generator: temperature of 20 °C; v the short-circuit currents are not equal, in fact c Calculation of the minimum short-circuit the relationship is Ik < Ib < Ik"; currents requires: v the positive-sequence impedance (Z(1)) is not v applying the voltage factor c corresponding to necessarily equal to the negative-sequence the minimum permissible voltage on the network; impedance (Z(2)). v selecting the network configuration, and in Note however that asynchronous motors may some cases the minimum contribution from also add to a short-circuit, accounting for up to 30 % of the network Isc for the first sources and network feeders, which result in the 30 milliseconds, in which case Ik" = Ik = Ib no lowest short-circuit current at the fault location: longer holds true. v taking into account the impedance of the busbars, the current transformers, etc.; Conditions to consider when calculating the v neglecting the motors; maximum and minimum short-circuit currents v considering resistances RL at the highest c Calculation of the maximum short-circuit foreseeable temperature: currents must take into account the following points: RL = 1 + 0.004 °C (θe - 20 °C) RL20 v application of the correct voltage factor c corresponding to calculation of the maximum where RL20 is the resistance at 20 °C; short-circuit currents; θe is the permissible temperature (°C) for the conductor at the end of the short-circuit. v among the assumptions and approximations mentioned in this document, only those leading The factor 0.004 / °C is valid for copper, to a conservative error should be used; aluminium and aluminium alloys. Cahier Technique Schneider Electric no. 158 / p.25 3.4 Equations for the various currents Initial short-circuit current Ik" Ik" / Ir ratio (see fig. 27 ) which expresses the The different initial short-circuit currents influence of the subtransient and transient Ik" are calculated using the equations in the table reactances with Ir as the rated current of the in figure 26. generator. Peak value ip of the short-circuit current Steady-state short-circuit current Ik In no meshed systems, the peak value ip of the The amplitude of the steady-state short-circuit current Ik depends on generator saturation short-circuit current may be calculated for all influences and calculation is therefore less types of faults using the equation: accurate than for the initial symmetrical ip = K 2 Ιk " current Ik". The proposed calculation methods produce a sufficiently accurate estimate of the Ik” = is the initial short-circuit current; upper and lower limits, depending on whether K is a factor depending on the R / X ratio and the short-circuit is supplied by a generator or a defined in the graph in figure 9, or using the synchronous machine. following approximate calculation: c The maximum steady-state short-circuit R current, with the synchronous generator at its -3 highest excitation, may be calculated by: K = 1.02 + 0.98 e X Ikmax = λmax Ir Short-circuit breaking current Ib c The minimum steady-state short-circuit current Calculation of the short-circuit breaking current is calculated under no-load, constant (minimum) Ib is required only when the fault is near the excitation conditions for the synchronous generator and protection is ensured by time- generator and using the equation: delayed circuit breakers. Note that this current is Ikmin = λmin Ir where Ir is the rated current at the used to determine the breaking capacity of these generator terminals; circuit breakers. λ is a factor defined by the saturation inductance This current may be calculated with a fair degree Xd sat. of accuracy using the following equation: The λmax and λmin values are indicated in Ib = µ Ik” where where µ is a factor defined by figure 28 for turbo-generators and in figure 29 the minimum time delay tmin and the for machines with salient poles. µ 1.0 Minimum time delay tmin 0.02 s 0.9 0.05 s 0.8 0.1 s > 0.25 s 0.7 0.6 0.5 0 1 2 3 4 5 6 7 8 9 Three-phase short-circuit current Ik" / Ir Fig. 27: factor µ used to calculate the short-circuit breaking current Ib (see IEC 60909). Cahier Technique Schneider Electric no. 158 / p.26 λ λ 2.4 6.0 Xd sat λmax 2.2 1.2 5.5 1.4 2.0 5.0 1.6 1.8 1.8 2.0 4.5 Xd sat 2.2 1.6 λmax 0.6 4.0 1.4 3.5 0.8 1.2 3.0 1.0 1.2 1.0 2.5 1.7 2.0 0.8 2.0 0.6 λmin 1.5 0.4 λmin 1.0 0.2 0.5 0 0 0 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 " " Three-phase short-circuit current Ik / Ir Three-phase short-circuit current Ik / Ir Fig. 28: factors λmax and λmin for turbo-generators Fig. 29: factors λmax and λmin for generators with (see IEC 60909). salient poles (see IEC 60909). 3.5 Calculation example Problem Consider four networks, three 5 kV networks and 60 kV network one 15 kV network, supplied via a 30 kV network 290 MVA by transformers in substation E (see fig. 30 ). During construction of line GH, calculation of 10 MVA 10 MVA the breaking capacity of circuit breaker M is requested. E The following information is available: c only the secondary windings of the 30 kV 15 km 40 km transformers in substation E are earthed; c for a 30 kV line, the reactance value is 8 MVA 0.35 Ω / km (positive-sequence and negative- 5 kV 4 MVA 4 MVA 5 kV sequence conditions) and 3 0.35 Ω / km (zero- F 15 kV G sequence conditions); 2 MVA 2 MVA c the short-circuit reactance is 6 % for the cos ϕ: 0.8 6 MVA cos ϕ: 0.8 transformers in substation E and 8 % for the cos ϕ : 0.8 other transformers; 20 km 30 km c the factor c for U is set to 1; H c all loads connected to points F and G are M essentially passive; 4 MVA c all resistances are negligible with respect to the reactances. 5 kV 2 MVA cos ϕ: 0.8 Fig. 30 Cahier Technique Schneider Electric no. 158 / p.27 Solution v similarly, the transformers in the substations F, H and G, due to their delta windings, are not c On the basis of the positive-sequence and affected by the zero-sequence currents and, negative-sequence diagrams (see fig. 31 ), the therefore, have an infinite impedance for the following may be calculated: fault. b’ = b1 = j 5.4 Ω U2 30 2 a = = ⇒ j 3.1 Ω c’1 = 3 c1 = j 42 Ω Ssc 290 c’2 = 3 c2 = j 31.5 Ω U2 6 302 c’3 = 3 c3 = j 21 Ω b = usc = x ⇒ j 5.4 Ω c’4 = 3 c4 = j 15.75 Ω Sn 100 10 c1 = 0.35 x 40 ⇒ j 14 Ω d’ = ∞ c2 = 0.35 x 30 ⇒ j 10.5 Ω f’ = ∞ c3 = 0.35 x 20 ⇒ j 7 Ω c Calculations may therefore be made using two c4 = 0.35 x 15 ⇒ j 5.25 Ω simplified diagrams: v with line GH open (see fig. 33 ): U2 8 302 d = usc = x ⇒ j9Ω Z(1) = Z(2) = j 17.25 Ω Sn 100 8 This result is obtained by successive calculations U2 302 as shown in figure 34 . e = x 0.6 = x 0.6 ⇒ j 90 Ω S 6 A similar calculation for the zero-sequence U2 8 302 diagram gives the result: f = usc = x ⇒ j 18 Ω Sn 100 4 Z(0) = j 39.45 Ω c Un U2 302 Ιsc 3 = ≈ 1.104 kA g = x 0.6 = x 0.6 ⇒ j 270 Ω Z (1) 3 S 2 c Note on the zero-sequence diagram c Un 3 (see fig. 32 ): Ιsc1 = ≈ 0.773 kA Z (1) + Z (2) + Z (0) v the delta windings of the transformers in substation E block zero-sequence currents, i.e. Note the network is HV, hence coefficient the network is not affected by them; c = 1.1. a b b b' b' E E c4 d c1 c'4 d' c'1 g f e f g f' f' F G F G c3 c2 c'3 c'2 H H f f' g Fig. 31. Fig. 32. Cahier Technique Schneider Electric no. 158 / p.28 Positive-sequence and Zero-sequence diagram negative-sequence diagram j 3.1 j 5.4 j 5.4 j 17.25 j 5.4 j 5.4 j 39.45 E E j9 H H j 5.25 j 14 j 15.75 j 42 j 270 j 18 F j 90 Z(1), Z(2) F G Z(0) G j 18 j 270 j7 j 21 H H Z(1), Z(2) Z(0) Fig. 33. j 3.1 Z' Za j 5.4 j 5.4 j 5,4 j 5,4 Za = j 3,1 + j 5.4 + j 5.4 E = j 3.1 + j 2.7 = j 5.8 j9 Zc j 5.25 j 14 j 5.25 Z(1) = Zc = j 14 + j 18 j 270 j 90 j 18 j 9 + j 90 + j 270 F G = j 99 = j 302 j 18 j 270 j 288 Zb j7 H j7 Z(1), Z(0) H Za x Zb x Zc Z' = = j 5.381 Za Zb+Za Zc+Zb Zc j 5.25 j 10.631 j 288 Z= +j7 j 288 j 10.631 + j 288 = j 17.253 j7 H H Fig. 34. Cahier Technique Schneider Electric no. 158 / p.29 v with line GH closed (see fig. 35 ): Given the highest short-circuit current Z(1) = Z(2) = j 13.05 Ω (Isc3 = 1.460 kA), the line circuit breaker at Z(0) = j 27.2 Ω point M must be sized for: Isc3 = 1.460 kA P = U Ι 3 = 30 x 1.460 x 3 Isc1 = 1.072 kA P ≈ 76 MVA. Positive-sequence diagram Zero-sequence diagram j 3.1 j 5.4 j 5.4 j 13,05 j 5.4 j 5.4 j 27.2 Ω E E H H j 5.25 j9 j 14 j 15.75 j 42 j 270 j 18 Z(1), Z(2) Z(0) F G F G j 18 j 90 j 270 j7 j 10,5 j 21 j 31.5 H H Z(1), Z(2) Z(0) Z(1) = Z(2) = j 13.05 Ω Z(0) = j 27.2 Ω Fig. 35. Cahier Technique Schneider Electric no. 158 / p.30 4 Computerised calculations and conclusion Various methods for the calculation of short- Ecodial, a program designed and marketed by circuit currents have been developed. Some Schneider Electric. have been included in a number of standards All computer programs designed to calculate and are consequently included in this short-circuit currents are predominantly "Cahier Technique" publication as well. concerned with determining the required Several standardised methods were designed in breaking and making capacities of switchgear such a way that short-circuit currents could be and the electro-mechanical withstand capabilities calculated by hand or using a small calculator. of equipment. When computerised scientific calculations Other software is used by experts specialising in became a possibility in the 1970's, electrical- network design, for example, research on the installation designers devised software for their dynamic behaviour of electrical networks. Such particular needs. This software was initially run computer programs can be used for precise on mainframe computer systems, then on simulations of electrical phenomena over time minicomputers, but was difficult to use, and and their use is now spreading to include the therefore limited to a small number of experts. entire electro-mechanical behaviour of networks This software was finally transferred to the PC and installations. microcomputing environment, proving much Remember, however, that all software, whatever easier to use. Today, a wide range of software its degree of sophistication, is only a tool. To packages are available which comply with the ensure correct results, it should be used by applicable standards defining the calculation of qualified professionals who have acquired the Isc currents in LV applications, for example relevant knowledge and expertise. Cahier Technique Schneider Electric no. 158 / p.31 Bibliography Standards c IEC 60909: Short-circuit current calculation in three-phase AC systems. c IEC 60781: Application guide for calculation of short-circuit currents in low voltage radial systems. c NF C 15-100: Installations électriques à basse tension. c C 15-105: Guide pratique, Détermination des sections de conducteurs et choix des dispositifs de protection. Schneider Electric Cahiers Techniques c Analyse des réseaux triphasés en régime perturbé à l'aide des composantes symétriques, Cahier Technique n° 18 - B. DE METZ-NOBLAT c Neutral earthing in an industrial HV network. Cahier Technique no. 62 - F. SAUTRIAU. c LV circuit-breaker breaking capacity. Cahier Technique no. 154 - R. MOREL. Other publication by Institut Schneider Formation (ISF) c Electrical Installation Guide, Ref.: MD1ELG2E (information on this 400 page pbulication is obtainable on www.schneiderformation.com) Cahier Technique Schneider Electric no. 158 / p.32 © 2000 Schneider Electric Schneider Electric Direction Scientifique et Technique, DTP: Headlines Valence. Service Communication Technique Edition: Schneider Electric F-38050 Grenoble cedex 9 Printing: Imprimerie du Pont de Claix - Claix - France - 1000 Télécopie : (33) 04 76 57 98 60 - 100 FF 63594 07-00