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Cahier technique no. 158

Calculation of short-circuit currents

B. de Metz-Noblat F. Dumas C. Poulain

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Benoît de METZ-NOBLAT Graduate Engineer from ESE (Ecole Supérieure d’Electricité), he worked first for Saint-Gobain, then joined Schneider Electric in 1986. He is now a member of the Electrical Networks competency group that studies electrical phenomena affecting power system operation and their interaction with equipment.

Frédéric DUMAS After completing a PhD in engineering at UTC (Université de Technologie de Compiègne), he joined Schneider Electric in 1993, initially developing software for electrical network calculations in the Research and Development Department. Starting in 1998, he led a research team in the field of industrial and distribution networks. Since 2003, as a project manager, he has been in charge of the technical development of electrical distribution services.

Christophe POULAIN Graduate of the ENI engineering school in Brest, he subsequented followed the special engineering programme at the ENSEEIHT institute in Toulouse and completed a PhD at the Université Pierre et Marie Curie in Paris. He joined Schneider Electric in 1992 as a research engineer and has worked since 2003 in the Electrical Networks competency group of the Projects and Engineering Center.

ECT 158 updated September 2005 Lexicon

Abbreviations Ra Equivalent resistance of the upstream BC Breaking capacity. network. MLVS Main low voltage switchboard. RL Line resistance per unit length. Sn Transformer kVA rating. Symbols ACross-sectional area of conductors. Scc Short-circuit power α Angle between the initiation of the tmin Minimum dead time for short-circuit fault and zero voltage. development, often equal to the time delay of a circuit breaker. c Voltage factor. uInstantaneous voltage. cos ϕ Power factor (in the absence of harmonics). usc Transformer short-circuit voltage in %. eInstantaneous electromotive force. U Network phase-to-phase voltage with no load. EElectromotive force (rms value). Un Network nominal voltage with load. ϕ Phase angle (current with respect to voltage). x Reactance, in %, of rotating machines. iInstantaneous current. Xa Equivalent reactance of the upstream network. i Alternating sinusoidal component of ac X Line reactance per unit length. the instantaneous current. L Xsubt Subtransient reactance of a generator. idc Aperiodic component of the instantaneous current. Z(1) Posititve-sequence impedance of a network i Maximum current (first peak of p or an the fault current). Z(2) Negative-sequence impedance element. I Current (rms value). Z(0) Zero-sequence impedance Ib Short-circuit breaking current (IEC 60909). ZL Line impedance. Zsc Network upstream impedance for a Ik Steady-state short-circuit current (IEC 60909). three-phase fault. Zup Equivalent impedance of the upstream Ik"Initial symmetrical short-circuit current network. (IEC 60909). Subscripts Ir Rated current of a generator. G Generator. Is Design current. k or k3 3-phase short circuit. Isc Steady-state short-circuit current k1 Phase-to-earth or phase-to-neutral (Isc3 = three-phase, short circuit. Isc2 = phase-to-phase, etc.). λ Factor depending on the saturation k2 Phase-to-phase short circuit. inductance of a generator. k2E / kE2E Phase-to-phase-to-earth short circuit. k Correction factor (NF C 15-105) S Generator set with on-load tap K Correction factor for impedance changer. (IEC 60909). SO Generator set without on-load tap changer. κ Factor for calculation of the peak short- circuit current. T Transformer.

Cahier Technique Schneider Electric n° 158 / p.2 Calculation of short-circuit currents

In view of sizing an electrical installation and the required equipment, as well as determining the means required for the protection 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 and meshed low-voltage (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 Development of 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 2.1 Isc depending on the different types of short-circuit p. 12 the impedance method 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. 27 3.5 Examples of short-circuit current calculations p. 28 4 Conclusion p. 32 Bibliography p. 32

Cahier Technique Schneider Electric n° 158 / p.3 1 Introduction

Electrical installations almost always require and 4 are used. Two values of the short-circuit protection against short-circuits wherever there current 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 determine in conductor cross-section. The short-circuit v The breaking capacity of the circuit breakers current must be calculated at each level in the v The making capacity of the circuit breakers installation in view of determining the v The electrodynamic withstand capacity of the characteristics of the equipment required to wiring system and switchgear withstand or break the fault current. The maximum short-circuit current corresponds The flow chart in Figure 1 indicates the to a short-circuit in the immediate vicinity of the procedure for determining the various short- downstream terminals of the protection device. circuit currents and the resulting parameters for It must be calculated accurately and used with a the different protection devices of a low-voltage safety margin. installation. c The minimum short-circuit current, essential In order to correctly select and adjust the when selecting the time-current curve for circuit protection devices, the graphs in Figures 2, 3 breakers and fuses, in particular when

Upstream Ssc

HV / LV transformer rating usc (%)

Isc at transformer terminals b Breaking capacity Power factor Conductor characteristics Main b Coincidence factor b Busbars ST and inst. trip setting circuit breaker b Duty factor v Length b Foreseeable expansion v Width Isc factor v Thickness of main LV switchboard b outgoers Cables Main LV v Type of insulation Breaking capacity switchboard v Single-core or multicore distribution v ST and inst. trip setting Length circuit breakers v Cross-section Isc b Environment at head of secondary v Ambient temperature switchboards v Installation method Secondary v Breaking capacity Number of contiguous circuits distribution ST and inst. trip setting circuit breakers b Feeder current Isc ratings at head of final b Voltage drops switchboards Breaking capacity Final distribution Inst. trip setting circuit breakers Load Isc rating at end of final outgoers

Fig. 1 : Short-circuit (Isc) calculation procedure when designing a low-voltage electrical installation (ST = short time; Inst. = instantaneous)

Cahier Technique Schneider Electric n° 158 / p.4 v Cables are long and/or the source impedance where A is the cross-sectional area of the is relatively high (generators, UPSs) conductors and k is a constant calculated on the v Protection of life depends on circuit breaker or basis of different correction factors for the cable fuse operation, essentially the case for TN and installation method, contiguous circuits, etc. IT electrical systems Further practical information may be found in the “Electrical Installation Guide” published by Note that the minimum short-circuit current Schneider Electric (see the bibliography). corresponds to a short-circuit at the end of the protected line, generally phase-to-earth for LV and phase-to-phase for HV (neutral not t distributed), under the least severe operating conditions (fault at the end of a feeder and not just downstream from a protection device, one Design Cable or I2t transformer in service when two can be current characteristic connected, etc.). Note also that whatever the case, for whatever type of short-circuit current (minimum or Circuit breaker maximum), the protection device must clear the time-current short-circuit within a time tc that is compatible Transient overload curve with the thermal stresses that can be withstood by the protected cable: 2 i 22 ∫ i dt k A (see Fig. 2, 3, and 4)

IB Ir Iz Isc BC I t 12 (tri) Fig. 3 : Circuit protection using a circuit breaker.

t

Cable or I2t characteristic a5 s

I2t = k2S2 Furse time-current curve Transient overload

Iz1 < Iz2 I Fig. 2 : The I2t characteristics of a conductor depending on the ambient temperature (1 and 2 represent the rms value of the current in the conductor at different temperatures θ1 and θ2, with θ1 > θ2; Iz being the limit of IB Ir Iz I the permissible current under steady-state conditions). Fig. 4 : Circuit protection using an aM fuse.

1.1 The main types of short-circuits

Various types of short-circuits can occur in c Origin electrical installations. v Mechanical (break in a conductor, accidental Characteristics of short-circuits electrical contact between two conductors via a The primary characteristics are: foreign conducting body such as a tool or an c Duration (self-extinguishing, transient and animal) steady-state) v Internal or atmospheric overvoltages

Cahier Technique Schneider Electric n° 158 / p.5 v Insulation breakdown due to heat, humidity or v Fire and danger to life a corrosive environment c On the faulty circuit c Location (inside or outside a machine or an v Electrodynamic forces, resulting in electrical switchboard) - Deformation of the busbars Short-circuits can be: - Disconnection of cables c Phase-to-earth (80% of faults) v Excessive temperature rise due to an increase c Phase-to-phase (15% of faults). This type of in Joule losses, with the risk of damage to fault often degenerates into a three phase fault insulation c Three-phase (only 5% of initial faults) c On other circuits in the network or in near-by These different short-circuit currents are networks presented in Figure 5 . v Voltage dips during the time required to clear the fault, ranging from a few milliseconds to a Consequences of short-circuits few hundred milliseconds The consequences are variable depending on v Shutdown of a part of the network, the extent the type and the duration of the fault, the point in of that part depending on the design of the the installation where the fault occurs and the network and the discrimination levels offered by short-circuit power. Consequences include: the protection devices c At the fault location, the presence of electrical v Dynamic instability and/or the loss of machine arcs, resulting in synchronisation v Damage to insulation v Disturbances in control / monitoring circuits v Welding of conductors v etc.

a) Three-phase short-circuit b) Phase-to-phase short-circuit clear of earth

L3 L3

L2 L2

L1 L1

" " Ik3 Ik2

c) Phase-to-phase-to-earth short-circuit d) Phase-to-earth short-circuit

L3 L3

L2 L2

L1 L1

" " Ik2EL3 Ik2EL2 " Ik1

" IkE2E

Short-circuit current, Partial short-circuit currents in conductors and earth.

Fig. 5 : Different types of short-circuits and their currents. The direction of current is chosen arbitrarily (See IEC 60909).

Cahier Technique Schneider Electric n° 158 / p.6 1.2 Development of the short-circuit current

A simplified network comprises a source of the R / X ratio is between 0.1 and 0.3. The ratio constant AC power, a switch, an impedance Zsc is virtually equals cos ϕ for low values: that represents all the impedances upstream of R the switch, and a load impedance Zs cos ϕ = (see Fig. 6 ). R22 + X However, the transient conditions prevailing In a real network, the source impedance is made while the short-circuit current develops differ up of everything upstream of the short-circuit depending on the distance between the fault including the various networks with different location and the generator. This distance is not voltages (HV, LV) and the series-connected necessarily physical, but means that the wiring systems with different cross-sectional generator impedances are less than the areas (A) and lengths. impedance of the elements between the In Figure 6, when the switch is closed and no generator and the fault location. fault is present, the design current Is flows through the network. Fault far from the generator This is the most frequent situation. The transient When a fault occurs between A and B, the conditions are those resulting from the negligible impedance between these points application of a voltage to a reactor-resistance results in a very high short-circuit current Isc that circuit. This voltage is: is limited only be impedance Zsc. e = E 2 sin (ωα t + ) The current sc develops under transient I Current i is then the sum of the two components: conditions depending on the reactances X and i = i+ i the resistances R that make up impedance Zsc: ac dc c The first (iac) is alternating and sinusoidal Zsc = R22 + X Ι ωαϕ− i=ac 2 sin ( t + ) ϕ In power distribution networks, reactance X = L E is normally much greater than resistance R and where I = , Zsc α = angle characterising the difference between the initiation of the fault and zero voltage. c The second (i ) is an aperiodic component R X dc R - t =−ΙαϕL idc -2 sin ( ) e . Its initial value depends on a and its decay rate is proportional A to R / L. Zsc At the initiation of the short-circuit, i is equal to zero by definition (the design current Is is e Zs negligible), hence:

i = iac + idc = 0 Figure 7 shows the graphical composition of i as B the algebraic sum of its two components iac and idc Fig. 6 : Simplified network diagram.

(ω α − ϕ) R iac = I sin t + - t (α − ϕ) L idc = - I sin e

I t

α-ϕ ω i = iac + idc

Fault initiation

Fig. 7 : Graphical presentation and decomposition of a short-circuit current occuring far from the generator.

Cahier Technique Schneider Electric n° 158 / p.7 a) Symmetrical The moment the fault occurs or the moment of closing, with respect to the network voltage, is characterised by its i closing angle a (occurrence of the fault). The voltage can Ir therefore be expressed as: u = E 2 . sin (ωα t + ). The current therefore develops as follows:  R  E2 - t i = sin (ωαϕ t + - ) - sin ( αϕ - ) e L    u Z   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) (see Fig. a ) b) Asymmetrical E2 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 Fig. b ) The fault current can be defined by: u  R  E2 - t i = sin ()ωϕt - + sin ϕ e L    Z  

Its initial peak value ip therefore depends on ϕ 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 The transient current-development conditions the development of a short-circuit current, are in this case modified by the variation in the presented, for the sake of simplicity, with a electromotive force resulting from the single-phase, alternating voltage. shortcircuit. R For simplicity, the electromotive force is − t assumed to be constant and the internal The factor e L is inversely proportional to the reactance of the machine variable. The aperiodic component damping, determined by reactance develops in three stages: the R / L or R / X ratios. c Subtransient (the first 10 to 20 milliseconds of The value of ip must therefore be calculated to the fault) determine the making capacity of the required c Transient (up to 500 milliseconds) circuit breakers and to define the electrodynamic c Steady-state (or synchronous reactance) forces that the installation as a whole must be capable of withstanding. Its value may be deduced from the rms value of Ι the symmetrical short-circuit current a using the κ equation: 2.0 ip = κ . r . Ia, where the coefficient κ is indicated by the curve in Figure 9 , as a function 1.8 of the ratio R / X or R / L, corresponding to the expression: 1.6 R −3 1.4 κ =+102.. 098 e X 1.2 Fault near the generator 1.0 When the fault occurs in the immediate vicinity of 0 0.2 0.4 0.6 0.8 1.0 1.2 R/X the generator supplying the circuit, the variation in the impedance of the generator, in this case Fig. 9 : Variation of coefficient κ depending on the dominant impedance, damps the short-circuit R / X or R / L (see IEC 60909). current.

Cahier Technique Schneider Electric n° 158 / p.8 Note that in the indicated order, the reactance c The three alternating components (subtransient, acquires a higher value at each stage, i.e. the transient and steady-state) subtransient reactance is less than the transient c The aperiodic component resulting from the reactance, itself less than the synchronous development of the current in the circuit (inductive) reactance. The successive effect of the three reactances leads to a gradual reduction in the This short-circuit current i(t) is maximum for a short-circuit current which is the sum of four closing angle corresponding to the zero-crossing components (see Fig. 10 ): of the voltage at the instant the fault occurs.

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 reactance = X”d b) transient reactance = X’d c) synchronous reactance = Xd 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 n° 158 / p.9 It is therefore given by the following expression:    11 −−'' 11 ' 1E 2− =− tT/ dd+− tT/ +  ω − tT/ a it() E 2  '' ' e  ' e cos t '' e  XXdd  Xd Xd  Xd  Xd Where: c In LV power distribution and in HV applications, E: Phase-to-neutral rms voltage across the however, the transient short-circuit current is often generator terminals used if breaking occurs before the steady-state stage, in which case it becomes useful to use the X"d: Subtransient reactance short-circuit breaking current, denoted Ib, which X'd: Transient reactance determines the breaking capacity of the time- X : Synchronous (steady-state) reactance d delayed circuit breakers. Ib is the value of the T": Subtransient time constant d short-circuit current at the moment interruption is T'd: Transient time constant effective, i.e. following a time t after the beginning T : Aperiodic time constant a of the short-circuit, where t = tmin. Time tmin Practically speaking, information on the (minimum time delay) is the sum of the minimum development of the short-circuit current is not operating time of a protection relay and the shortest essential: opening time of the associated circuit breaker, i.e. c In a LV installation, due to the speed of the the shortest time between the appearance of the breaking devices, the value of the subtransient short-circuit current and the initial separation of the pole contacts on the switching device. short-circuit current, denoted I"k , and of the Figure 11 presents the various currents of the maximum asymmetrical peak amplitude ip is sufficient when determining the breaking capacities short-circuits defined above. of the protection devices and the electrodynamic forces

i Symmetrical

Subtrans. Transient Steady-state Asymmetrical

r 2 I"k ip r 2 Ik

Fig. 11 : short-circuit currents near a generator (schematic diagram).

1.3 Standardised Isc calculations

The standards propose a number of methods. corresponding impedance. The Isc value is c Application guide C 15-105, which finally obtained by applying Ohm’s law: supplements NF C 15-100 (Normes Françaises) Un (low-voltage AC installations), details three Isc = ( ) . methods 3∑ Z v The “impedance” method, used to calculate All the characteristics of the various elements in fault currents at any point in an installation with a the fault loop must be known (sources and wiring high degree of accuracy. systems). This method involves adding the various v The “composition” method, which may be used resistances and reactances of the fault loop when the characteristics of the power supply are separately, from (and including) the source to not known. The upstream impedance of the the given point, and then calculating the given circuit is calculated on the basis of an

Cahier Technique Schneider Electric n° 158 / p.10 estimate of the short-circuit current at its origin. sizes by increasing the resistance 15% for Power factor cos ϕ≈ R / X is assumed to be 150 mm2, 20% for 185 mm2, 25% for 240 mm2 identical at the origin of the circuit and the fault and 30% for 300 mm2. location. In other words, it is assumed that the This method is used essentially for final circuits elementary impedances of two successive with origins sufficiently far from the source. It is sections in the installation are sufficiently similar not applicable in installations supplied by a in their characteristics to justify the replacement generator. of vectorial addition of the impedances by c Standard IEC 60909 (VDE 0102) applies to all algebraic addition. This approximation may be networks, radial or meshed, up to 550 kV. used to calculate the value of the short-circuit This method, based on the Thevenin theorem, current modulus with sufficient accuracy for the calculates an equivalent voltage source at the addition of a circuit. short-circuit location and then determines the v The “conventional” method, which can be used corresponding short-circuit current. All network when the impedances or the Isc in the feeders as well as the synchronous and installation upstream of the given circuit are not asynchronous machines are replaced in the known, to calculate the minimum short-circuit calculation by their impedances (positive currents and the fault currents at the end of a sequence, negative-sequence and line. It is based on the assumption that the zerosequence). voltage at the circuit origin is equal to 80% of the All line capacitances and the parallel rated voltage of the installation during the short- admittances of non-rotating loads, except those circuit or the fault. of the zero-sequence system, are neglected. Conductor reactance is neglected for sizes under 150 mm2. It is taken into account for large

1.4 Methods presented in this document

In this “Cahier Technique” publication, two virtually all characteristics of the circuit are taken methods are presented for the calculation of into account short-circuit currents in radial networks: c The IEC 60909 method, used primarily for HV c The impedance method, reserved primarily for networks, was selected for its accuracy and its LV networks, was selected for its high degree of analytical character. More technical in nature, it accuracy and its instructive value, given that implements the symmetrical-component principle

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, voltages responsible for the flow of the current facilitating comprehension of the physical and the short-circuit impedance do not change phenomena and consequently the short-circuit significantly current calculations. They nevertheless maintain c Transformer regulators or tap-changers are a fully acceptable level of accuracy, “erring” assumed to be set to a main position (if the systematically on the conservative side. The short-circuit occurs away far from the generator, assumptions used in this document are as the actual position of the transformer regulator or follows: tap-changers does not need to be taken into c The given network is radial with nominal account voltages ranging from LV to HV, but not c Arc resistances are not taken into account exceeding 550 kV, the limit set by standard IEC 60909 c All line capacitances are neglected c The short-circuit current, during a three-phase c Load currents are neglected short-circuit, is assumed to occur simultaneously c All zero-sequence impedances are taken into on all three phases account c During the short-circuit, the number of phases involved does not change, i.e. a three-phase

Cahier Technique Schneider Electric n° 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 fault, i.e. the impedances of the power sources This fault involves all three phases. Short-circuit and the lines (see Fig. 12 ). This is, in fact, the “positive-sequence” impedance per phase: current Isc3 is equal to: U / 3   22  Ιsc = Zsc = ∑∑ R +  X where 3 Zcc where U (phase-to-phase voltage) corresponds ∑R = the sum of series resistances, to the transformer no-load voltage which is 3 to 5% greater than the on-load voltage across the ∑X= the sum of series reactances. terminals. For example, in 390 V networks, the It is generally considered that three-phase faults phase-to-phase voltage adopted is U = 410 V, provoke the highest fault currents. The fault and the phase-to-neutral voltage is current in an equivalent diagram of a polyphase U / 3 = 237 V. system is limited by only the impedance of one Calculation of the short-circuit current therefore phase at the phase-to-neutral voltage of requires only calculation of Zsc, the impedance thenetwork. Calculation of Isc3 is therefore equal to all the impedances through which Isc essential for selection of equipment (maximum flows from the generator to the location of the current and electrodynamic withstand capability).

Three-phase fault ZL Zsc

U / 3 Ιsc = V 3 Zsc ZL

ZL

Phase-to-phase fault ZL Zsc

Ι = U U sc 2 Z 2 . Zsc L Zsc

Phase-to-neutral fault ZL Zsc

U / 3 Ιsc = Z V 1 Ln Zsc + ZLn ZLn

Phase-to-earth fault ZL Zsc

Ι = U / 3 sco V Zsc + Zo

Zo Zo

Fig. 12 : The various short-circuit currents.

Cahier Technique Schneider Electric n° 158 / p.12 Phase-to-phase short-circuit clear of earth In certain special cases of phase-to-neutral This is a fault between two phases, supplied with faults, the zero-sequence impedance of the a phase-to-phase voltage U. In this case, the source is less than Zsc (for example, at the terminals of a star-zigzag connected transformer short-circuit current Isc2 is less than that of a three-phase fault: or of a generator under subtransient conditions). In this case, the phase-to-neutral fault current U 3 may be greater than that of a three-phase fault. ΙΙΙsc =≈ = sc 0.86 sc 2 2 Zsc 2 33 Phase-to-earth fault (one or two phases) For a fault occuring near rotating machines, the This type of fault brings the zero-sequence impedance of the machines is such that Isc2 is close to Isc . impedance Zo into play. 3 Except when rotating machines are involved Phase-to-neutral short-circuit clear of earth (reduced zero-sequence impedance), the short- circuit current Isco 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 Isco may be necessary, depending V = U / 3 on the neutral system (system earthing The short-circuit current Isc is: arrangement), in view of defining the setting 1 thresholds for the zero-sequence (HV) or earth- U / 3 fault (LV) protection devices. Ιsc = 1 Figure 12 shows the various short-circuit currents. Zsc + ZLn

2.2 Determining the various short-circuit impedances

This method involves determining the shortcircuit Rup / Zup ≈ 0.1 at 150 kV. currents on the basis of the impedance = 22 represented by the “circuit” through which the As, Xup Za - Ra , short-circuit current flows. This impedance may 2 Xup  Rup be calculated after separately summing the = 1 -   various resistances and reactances in the fault Zup  Zpu  loop, from (and including) the power source to 2 Therefore, for 20 kV, the fault location. Xup 2 (The circled numbers X may be used to come = 1 - ( 0.2) = 0.980 Zup back to important information while reading the Xup = 0.980 Zup at 20kV, example at the end of this section.) hence the approximation Xup ≈ Zup. Network impedances c Internal transformer impedance c Upstream network impedance The impedance may be calculated on the basis Generally speaking, points upstream of the of the short-circuit voltage usc expressed as a power source are not taken into account. percentage: Available data on the upstream network is u U2 3 Z = sc , therefore limited to that supplied by the power T 100 Sn distributor, i.e. only the short-circuit power Ssc in U = no-load phase-to-phase voltage of the MVA. transformer; The equivalent impedance of the upstream Sn = transformer kVA rating; network is: u sc = voltage that must be applied to the U2 100 1 Zup = primary winding of the transformer for the rated Ssc current to flow through the secondary winding, where U is the no-load phase-to-phase voltage when the LV secondary terminals are of the network. shortcircuited. The upstream resistance and reactance may be For public distribution MV / LV transformers, the deduced from Rup / Zup (for HV) by: values of usc have been set by the European Rup / Zup ≈ 0.3 at 6 kV; Harmonisation document HD 428-1S1 issued in Rup / Zup ≈ 0.2 at 20 kV; October 1992 (see Fig. 13 ) .

Rating (kVA) of the MV / 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 n° 158 / p.13 Note that the accuracy of values has a direct The relative error is: 2 influence on the calculation of Isc in that an error U ∆Ιsc ΙΙ' sc - sc Zpu of x % for usc produces an equivalent error (x %) === Ssc for Z . Ιsc Ιsc Z u U2 T T sc 100 Sn 4 In general, RT << XT , in the order of 0.2 XT, ∆Ι sc = 100 Sn and the internal transformer impedance may be i.e. : Ι sc usc Ssc considered comparable to reactance XT. For low power levels, however, calculation of ZT is Figure 14 indicates the level of conservative required because the ratio RT / XT is higher. error in the calculation of Isc, due to the fact that The resistance is calculated using the joule the upstream impedance is neglected. The figure losses (W) in the windings: demonstrates clearly that it is possible to neglect W Ι 2 ⇒= the upstream impedance for networks where the W = 3 RTT n R 2 3 Ι n 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%. c v When n identically-rated transformers are Line impedance connected in parallel, their internal impedance The line 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 line. 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 rectifier units have U values of up to 10 to 12% ρ sc R = where in order to limit short-circuit currents. L A When the impedance upstream of the S = cross-sectional area of the conductor; transformer and the transformer internal ρ = conductor resistivity, however the value used impedance are taken into account, the varies, depending on the calculated short-circuit shortcircuit current may be expressed as: current (minimum or maximum). U Ιsc = + 6 The table in Figure 15 provides values for 3(Zup ZT ) 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. 2 (RL < 0.15 mΩ / m when A > 150 mm ). The upstream network impedance may be v The reactance per unit length of overhead neglected, in which case the new current value lines, cables and busbars may be calculated as is: U   d Ι'sc = XL == ω  15.7 + 144.44 Log   L   r  3Z T

∆Isc/Isc (%)

12 Ssc = 250 MVA 10

Ssc = 500 MVA 5

0 500 1,000 1,500 2,000 Sn (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 n° 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 NB : Above, Log = decimal logarithm. For overhead lines, the reactance increases ( ) and spaced, single-conductor cables slightly in proportion to the distance between ( ) ; For “sandwiched-phase” busbars  d (e.g. Canalis - Telemecanique), the reactance is conductors (Log   ), and therefore in  t  considerably lower. proportion to the operating voltage. Notes : 7 the following average values are to be used: v The impedance of the short lines between the distribution point and the HV / LV transformer X=0.3 Ω / km (LV lines); may be neglected. This assumption gives a Ω X=0.4 / km (MV or HV lines). conservative error concerning the short-circuit Figure 16 shows the various reactance values for current. The error increases in proportion to the conductors in LV applications, depending on the transformer rating wiring system (practical values drawn from French v The cable capacitance with respect to the earth standards, also used in other European countries). (common mode), which is 10 to 20 times greater The following average values are to be used: than that between the lines, must be taken into - 0.08 mΩ /m for a three-phase cable ( ), account for earth faults. Generally speaking, the capacitance of a HV three-phase cable with a and, for HV applications, between 0.1 and cross-sectional area of 120 mm2 is in the order 0.15 mΩ / m.

Rule Resistitivity Resistivity value Concerned (*) (Ω mm2/m) conductors Copper Aluminium

Max. short-circuit current ρ0 0.01851 0.02941 PH-N Min. short-circuit current c With fuse ρ2 = 1,5 ρ0 0.028 0.044 PH-N c With breaker ρ1 = 1,25 ρ0 0.023 0.037 PH-N (**)

Fault current for TN and IT ρ1 = 1,25 ρ0 0,023 0,037 PH-N systems PE-PEN

Voltage drop ρ1 = 1,25 ρ0 0.023 0.037 PH-N

Overcurrent for thermal-stress ρ1 = 1,25 ρ0 0.023 0.037 PH, PE and PEN checks on protective conductors 2 2 (*) ρ0 = resistivity of conductors at 20°C = 0.01851 Ω mm /m for copper and 0.02941 Ω mm /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 cables core cables (triangle) cables (flat) d = 2r d = 4r

dd r Diagram

Reactance per unit length, 0.08 0.13 0.08 0.09 0.13 0.13 values recommended in UTE C 15-105 (mΩ/m) Average reactance 0.15 0.08 0.15 0.085 0.095 0.145 0.19 per unit length 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 n° 158 / p.15 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 v c The reactance or resistance of the lines may Second case: Consider a three-phase cable, at ° be neglected. 20 C, with aluminium conductors. As above, If one of the values, R or X , is low with respect the impedance ZL curve may be considered L L identical to the asymptotes, but for cable cross- to the other, it may be neglected because the 2 resulting error for impedance Z is consequently sectional areas less than 120 mm and greater L than 1,000 mm2 (curves not shown) very low. For example, if the ratio between RL and XL is 3, the error in ZL is 5.1%. Impedance of rotating machines. The curves for RL and XL (see Fig. 17 ) may be c Synchronous generators used to deduce the cable cross-sectional areas The impedances of machines are generally for which the impedance may be considered expressed as a percentage, for example: comparable to the resistance or to the reactance. xnI = (where x is the equivalent of the Examples : 100 Isc v First case: Consider a three-phase cable, at transformer u ). 20°C, with copper conductors. Their reactance sc is 0.08 mΩ / m. The RL and XL curves Consider: 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 to considered, with a margin of error less than 0.2 for LV, impedance Z may be considered 5.1%, comparable to: comparable to reactance X. Values for x are - A resistance for cable cross-sectional areas given in the table in Figure 18 for less than 74 mm2 turbogenerators with smooth rotors and for “hydraulic” generators with salient poles (low speeds). In the table, it may seem surprising to see that mΩ/m the synchronous reactance for a shortcircuit exceeds 100% (at that point in time, Isc < In) . 1 However, the short-circuit current is essentially 0.8 inductive and calls on all the reactive power that the field system, even over-excited, can supply, whereas the rated current essentially carries the active power supplied by the turbine 0.2 (cos ϕ from 0.8 to 1). Z L c Synchronous compensators and motors 0.1 The reaction of these machines during a 0.08 shortcircuit is similar to that of generators. 0.05 XL 12 They produce a current in the network that depends on their reactance in % (see Fig. 19 ). RL c Asynchronous motors 0.02 When an asynchronous motor is cut from the network, it maintains a voltage across its 0.01 terminals that disappears within a few 10 2050 100 200 500 1,000 Section S 2 hundredths of a second. When a short-circuit (en mm ) occurs across the terminals, the motor supplies a Fig. 17 : Impedance ZL of a three-phase cable, at current that disappears even more rapidly, 20 °C, with copper conductors. according to time constants in the order of:

Subtransient Transient Synchronous 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 per unit.

Cahier Technique Schneider Electric n° 158 / p.16 v 20 ms for single-cage motors up to 100 kW It follows that when calculating the maximum v 30 ms for double-cage motors and motors short-circuit current, capacitor banks do not need above 100 kW to be taken into account. v 30 to 100 ms for very large HV slipring motors However, they must nonetheless be considered (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 affect current motor is therefore a generator to which an interruption. impedance (subtransient only) of 20 to 25% is c attributed. 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 have an impedance that must be taken into not easy to foresee the average number of account, for the calculation of Isc, when such a motors running that will contribute to the fault device is located upstream of the device intended when a short-circuit occurs. Individual calculation to break the given short-circuit and remain closed of the reverse current for each motor, taking into (selective circuit breakers). account the line impedance, is therefore a 15 For LV circuit breakers, for example, a tedious and futile task. Common practice, reactance value of 0.15 mΩ is typical, while the notably in the United States, is to take into resistance is negligible. account the combined contribution to the fault For breaking devices, a distinction must be made current of all the asynchronous LV motors in an depending on the speed of opening: installation. v Certain devices open very quickly and thus 13 They are therefore thought of as a unique significantly reduce short-circuit currents. This is source, capable of supplying to the busbars a the case for fast-acting, limiting circuit breakers current equal to Istart/Ir times the sum of the and the resultant level of electrodynamic forces rated currents of all installed motors. and thermal stresses, for the part of the installation concerned, remains far below the Other impedances. theoretical maximum c Capacitors v Other devices, such as time-delayed circuit A shunt capacitor bank located near the fault breakers, do not offer this advantage location will discharge, thus increasing the c Fault arc shortcircuit current. This damped oscillatory The short-circuit current often flows through an discharge is characterised by a high initial peak arc at the fault location. The resistance of the arc value that is superposed on the initial peak of the is considerable and highly variable. The voltage shortcircuit current, even though its frequency is drop over a fault arc can range from 100 to 300 V. far greater than that of the network. For HV applications, this drop is negligible with Depending on the timing between the initiation of respect to the network voltage and the arc has the fault and the voltage wave, two extreme no effect on reducing the short-circuit current. cases must be considered: For LV applications, however, the actual fault v If the initiation of the fault coincides with zero current when an arc occurs is limited to a much voltage, the short-circuit discharge current is lower level than that calculated (bolted, solid asymmetrical, with a maximum initial amplitude fault), because the voltage is much lower. peak 16 For example, the arc resulting from a v Conversely, if the initiation of the fault shortcircuit between conductors or busbars may coincides with maximum voltage, the discharge reduce the prospective short-circuit current by current superposes itself on the initial peak of 20 to 50% and sometimes by even more than the fault current, which, because it is 50% for nominal voltages under 440 V. symmetrical, has a low value However, this phenomenon, highly favourable in It is therefore unlikely, except for very powerful the LV field and which occurs for 90% of faults, capacitor banks, that superposition will result in may not be taken into account when determining an initial peak higher than the peak current of an the breaking capacity because 10% of faults take asymmetrical fault. place during closing of a device, producing a solid

Subtransient Transient Synchronous 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 per unit.

Cahier Technique Schneider Electric n° 158 / p.17 fault without an arc. This phenomenon should, filters and inductors used to limit the short-circuit however, be taken into account for the current. calculation of the minimum short-circuit current. They must, of course, be included in calculations, as well as wound-primary type c Various impedances current transformers for which the impedance Other elements may add non-negligible values vary depending on the rating and the type impedances. This is the case for harmonics of construction.

2.3 Relationships between impedances at the different voltage levels in an installation

Impedances as a function of the voltage expressed in %. The short-circuit power Ssc at a given point in 1 x Z = the network is defined by: MR Sn 100 U2 c For the system as a whole, after having Ssc = U Ι 3 = Zsc calculated all the relative impedances, the This means of expressing the short-circuit power shortcircuit power may be expressed as: implies that Ssc is invariable at a given point in = 1 the network, whatever the voltage. And the Ssc Σ from which it is possible to deduce equation ZR the fault current Isc at a point with a voltage U: Ι = U sc3 implies that all impedances 3 Zsc Ssc 1 Ιsc = = must be calculated with respect to the voltage at Σ 3 U 3 U ZR the fault location, which leads to certain complications that often produce errors in ΣZR is the composed vector sum of all the calculations for networks with two or more impedances related to elements upstream of the voltage levels. For example, the impedance of a fault. It is therefore the relative impedance of the HV line must be multiplied by the square of the upstream network as seen from a point at U reciprocal of the transformation ratio, when voltage. calculating a fault on the LV side of the Hence, Ssc is the short-circuit power, in VA, at a transformer: point where voltage is U. For example, if we consider the simplified   2 UBT diagram of Figure 20 : ZZ =   17 BT HT  U  HT U 2 At point A, Ssc = LV A simple means of avoiding these difficulties is   2 the relative impedance method proposed by ULV + ZT   ZL H. Rich.  UHV  Calculation of the relative impedances 1 Hence, Ssc = This is a calculation method used to establish a ZT + ZL 2 2 relationship between the impedances at the UHV ULV different voltage levels in an electrical installation. This method proposes dividing the impedances (in ohms) by the square of the network line-toline voltage (in volts) at the point where the impedances exist. The impedances therefore

become relative (ZR). UHT c For overhead lines and cables, the relative resistances and reactances are defined as: Z ==R X T R and X with R and X in CR U2 CR U2 ohms and U in volts. UBT c For transformers, the impedance is expressed on the basis of their short-circuit voltages u and sc Z their kVA rating Sn: C

1 u Z = sc A TR Sn 100 c For rotating machines, the equation is Fig. 20 : Calculating Ssc at point A. identical, with x representing the impedance

Cahier Technique Schneider Electric n° 158 / p.18 2.4 Calculation example (with the impedances of the power sources, the upstream network and the power supply transformers as well as those of the electrical lines)

Problem The Isc3 and ip values must be calculated at the Consider a 20 kV network that supplies a HV / various fault locations indicated in the network diagram (see Fig. 21 ), that is: LV substation via a 2 km overhead line, and a c Point A on the HV busbars, with a negligible 1 MVA generator that supplies in parallel the impedance busbars of the same substation. Two 1,000 kVA c Point B on the LV busbars, at a distance of parallel-connected transformers supply the LV 10 meters from the transformers busbars which in turn supply 20 outgoers to c Point C on the busbars of an LV subdistribution 20 motors, including the one supplying motor M. board All motors are rated 50 kW, all connection cables c Point D at the terminals of motor M are identical and all motors are running when the Then the reverse current of the motors must be fault occurs. calculated at C and B, then at D and A.

Upstream network U1 = 20 kV Ssc = 500 MVA Overhead line 3 cables, 50 mm2, copper 3L length = 2 km G Generator 1 MVA A xsubt = 15%

2 transformers 1,000 kVA secondary winding 237/410 V usc = 5%

Main LV switchboard 10 m 3 bars, 400 mm2/ph, copper B length = 10 m

Cable 1 3 single-core cables, 400 mm2, aluminium, spaced, laid flat, 3L length = 80 m

LV sub-distribution board C neglecting the length of the busbars

Cable 2 3 single-core cables 35 mm2, 3L copper 3-phase, length = 30 m

Motor D 50 kW (efficiency = 0.9 ; cos ϕ = 0.8) x = 25% M

Fig. 21 : Diagram for calculation of Isc3 and ip at points A, B, C and D.

Cahier Technique Schneider Electric n° 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 Calculation 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 = 098. Zup 2 0.78

Rup=≈02.. Zup 02 Xup 0.15

= 2. overhead line Xco 04. x 2 7 0.8 (50 mm2) = 2,000 Rco 0.018 x 6 0.72 50

3 2 15 ()20 x 10 3. generator Xx = 10 60 G 100 106 = 11 RGG 01. X 6 ↑ 20 kV X (mΩ)R (mΩ) Fault A 1 5 4102 4. transformers Zxx = 3 5 T 2 100 106 ZT on LV side ≈ XZTT 4.2 = RTT 0.2 X 4 0.84

410 V↓ = 5. circuit-breaker Xcb 015. 15 0.15

= -3 6. busbars XB 0.15 x 10 x 10 9 1.5 (one 400 mm2 bar per phase) 10 Rx = 0.023 B 400 6 0.57 Fault B = 7. circuit-breaker Xcb 015. 0.15 = −3 8. cable 1 Xc1 015. x 10 x 80 12 (one 400 mm2 cable per = 80 phase) Rc1 0.036 x 6 7.2 400 Fault C = 9. circuit-breaker Xcb 015. 0.15

= −3 8 10. cable 2 Xc2 009. x 10 x 30 2.7 (35 mm2) 30 Rc = 0.023 x 2 35 19.3 Fault D 25 4102 11. motor 50 kW Xm= x 12 605 100 (/.50 0 9 x 08 .) 103

Rm = 0.2 Xm 121

Fig. 22 : Impedance calculation.

Cahier Technique Schneider Electric n° 158 / p.20 =+ -3 = Ω I - Fault at A (HV busbars) RRC ( B 72.) 10 90. m Elements concerned: 1, 2, 3. These values make clear the importance of Isc The “network + overhead line” impedance is limitation due to the cables. parallel to that of the generator, however the latter is much greater and may be neglected: =+=2 2 Ω ZRXCC C 20 . 7 m =+≈Ω XA 078. 08. 158. Ι =≈410 R0. =+ 015. 072. ≈ 87 Ω C − 11,400 A A 320710 xx . 3 ZRX =+≈ 2 2 180. Ω hence R AAA C = 048. hence κ =1.25 on the curve in 3 XC Ι =≈20 x 10 6,415 A figure 9 and therefore the peak ipC is equal to: A 3180 x . 1.25 x 2 x 11 , 400 ≈ 20,200 A IA is the “steady-state Isc” and for the purposes of calculating the peak asymmetrical I : pA IV - Fault at D (LV motor) R A = 055. hence κ =1.2 on the curve in [Elements concerned: XA (1, 2, 3) + (4, 5, 6) + (7, 8) + (9, 10)] figure 9 and therefore ipA is equal to: The reactances and the resistances of the circuit breaker and the cables must be added to X and R . 1.2 x 2 x 6,415 = 10,887 A. C C =+ + -3 = Ω XXD ( C 015. 27.) 10 2152 . m II - Fault at B (main LV switchboard busbars) and [Elements concerned: (1, 2, 3) + (4, 5, 6)] RR=+( 19 2) 10-3 = 28 2 mΩ The reactances X and resistances R calculated D C . . for the HV section must be recalculated for the =+=2 2 Ω LV network via multiplication by the square of ZRXDD D 35 . 5 m the voltage ratio 17 , i.e.: Ι =≈410 2 D 6,700 A (410/, 20 000) = 0 . 42 10-3 hence 335510 xx . -3 = ( ) ++ + -3 R XXBA [] 0.42 42. 015. 15. 10 D = 131. hence κ ≈ 1.04 on the curve in X = Ω D XB 651. m and figure 9 and therefore the peak ipD is equal to: RR = ( 0.42) ++ 084.. 057 10-3 BA[]1.04 x 2 x 6,700 ≈ 9,900 A = Ω RB 177. m 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 V - Reverse currents of the motors impedance of the 10 meter long, LV busbars. It is often faster to simply consider the motors as =+=2 2 Ω ZRXBB B 6.75 m independent generators, injecting into the fault a “reverse current” that is superimposed on the 410 Ι =≈ 35,070 A network fault current. B 3 x 6.75 x 10-3 c Fault at C R B = 027. hence κ = 1.46 on the curve in The current produced by the motor may be XB calculated on the basis of the “motor + cable” impedance: figure 9 and therefore the peak ipB is equal to: ≈ =+( ) −3 ≈ Ω 1.46 x 2 x 35 , 070 72,400 A . XM 605 2.7 10 608 m What is more, if the fault arc is taken into =+( ) -3 ≈ Ω RM 121 19. 3 10 140 m account (see § c fault arc section 16 ), I is = Ω B ZM 624 m hence reduced to a maximum value of 28,000 A and a Ι =≈410 minimum value of 17,500 A . M − 379 A 3 xx 624 10 3 III - Fault at C (busbars of LV sub-distribution For the 20 motors board) Ι = MC 7,580 A. [Elements concerned: (1, 2, 3) + (4, 5, 6) + (7, 8)] Instead of making the above calculations, it is The reactances and the resistances of the circuit possible (see 13 ) to estimate the current breaker and the cables must be added to XB and RB. =+ + -3 = Ω injected by all the motors as being equal to XXC ( B 015. 12) 10 1867 . m (Istart / Ir) times their rated current (98 A), i.e. and (4.8 x 98) x 20 = 9,400 A.

Cahier Technique Schneider Electric n° 158 / p.21 This estimate therefore provides conservative switchboard increases from 35,070 A to protection with respect to IMC : 7,580 A. 42,510 A and the peak ipB from 72,400 A to On the basis of R / X = 0.23 ⇒ κ = 1.51 and 88,200 A. However, as mentioned above, if the fault arc is i =××151., 2 7580 =16,200 A . pMC taken into account, IB is reduced between 21.3 Consequently, the short-circuit current to 34 kA. (subtransient) on the LV busbars increases from c Fault at A (HV side) 11,400 A to 19,000 A and i from 20,200 A to pC Rather than calculating the equivalent 36,400 A. impedances, it is easier to estimate c Fault at D (conservatively) the reverse current of the The impedance to be taken into account is 1 / 19th motors at A by multiplying the value at B by the of Z (19 parallel motors), plus that of the cable. M LV / HV transformation value 17 , i.e.:  608  X =+  2.7 10-3 = 34.7 mΩ 410 MD  19  7,440 × = 152.5 A 20× 103  140  R =+  19.3 10-3 ≈ 26.7 mΩ This figure, compared to the 6,415 A calculated MD   19 previously, is negligible = Ω ZMD 43. 8 m hence Rough calculation of the fault at D This calculation makes use of all the Ι = 410 = MD −3 5,400 A 343810××. approximations mentioned above (notably 15

giving a total at D of: and 16 . 6,700 + 5,400 = 12,100 A rms, and ipD ≈ 18,450 A. ΣX = 4.2 + 1.5 + 12 c Fault at B ΣX = 17.7 mΩ = X' As for the fault at C, the current produced by the D Σ Ω = motor may be calculated on the basis of the R = 7.2 + 19.3 = 26.5 m R'D “motor + cable” impedance: =++( ) -3 = Ω =+≈2 2 Ω XM 605 2. 7 12 10 620 m Z'DD R' X'D 31 . 9 m =++( ) -3 ≈Ω RM 121 19.3 7.2 10 147.5 m Ι =≈410 'D -3 7,430 A = Ω 331910 xx . ZM 637 m hence 410 I = ≈ 372 A hence the peak ip'D : M ××−3 3 637 10 2 x7 ,430 ≈ 10,500 A .

For the 20 motors IMB =7,440 A. To find the peak asymmetrical ipDtotal, the above Again, it is possible to estimate the current value must be increased by the contribution of injected by all the motors as being equal to the energised motors at the time of the fault A 4.8 times their rated current (98 A), i.e. 9,400 A. 13 i.e. 4.8 times their rated current of 98 A: The approximation again overestimates the real value of IMB. 10,500+( 4.8×× 98 2 × 20) = 23,800 A Using the fact that R / X = 0.24 = κ = 1.5 Compared to the figure obtained by the full i =××15., 2 7440 =15,800 A . pMB calculation (18,450 A), the approximate method Taking the motors into account, the short-circuit allows a quick evaluation with an error remaining current (subtransient) on the main LV on the side of safety.

Cahier Technique Schneider Electric n° 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 “cyclical” impedances R and X are, normally c The network includes rotating machines and/or speaking, no longer useable. This calculation special transformers (Yyn connection, for 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 2π j 1 3 a rectilinear alternating field with a sinusoidal a = e3 = - + j between I1, I2, amplitude is equivalent to two rotating fields 2 2 turning in the opposite direction, the definition of symmetrical components is based on the and I3. equivalence between an unbalanced threephase This principle, applied to a current system, is system and the sum of three balanced three- confirmed by a graphical representation phase systems, namely the positivesequence, (see fig. 23). For example, the graphical addition negative-sequence and zerosequence of the vectors produces, for, the following result: (see Fig. 23 ). ΙΙΙΙ21 =++ aa2 11 The superposition principle may then be used to (1) (2) (3) . calculate the fault currents. In the description below, the system is defined Currents Ι1and Ι3 may be expressed in the using current Ι1 as the rotation reference, same manner, hence the system: where: ΙΙ11 =+ a Ι 1 + Ι 1 c Ι (1) (2) (0) 1(1) is the positive-sequence component c Ι ΙΙΙΙ21 =++ aa2 11 1(2) is the negative-sequence component (1) (2) (0) c Ι1 is the zero-sequence component 2 (0) ΙΙ31 = a + a ΙΙ1 + 1 . and by using the following operator (1) (2) (0)

Positive-sequence Negative-sequence Zero-sequence 3 I (1) I1(0) I2(2) I3 I1 I1(1) I1(2) I2(0) ωt + + = I2 ωt I3(0) 3 ω I (2) ω I2(1) t t

Geometric construction of I1 Geometric construction of I2 Geometric construction of I3 I1 I2 I1(1) I1(1) I1(0) 2 a2 I1 a I1(2) 1 1 1 (1) I (1) I (2) I (0) a I1(2) I1 I1(1) I1 (2) (2) I3

Fig. 23 : Graphical construction of the sum of three balanced three-phase systems (positive-sequence, negative-sequence and zero-sequence).

Cahier Technique Schneider Electric n° 158 / p.23 These symmetrical current components are related to the symmetrical voltage components by the corresponding impedances: Elements Z(0) V V V Transformer Z = (1) , Z = (2) and Z = (0) (seen from secondary winding) (1) ΙΙ(2) (0) Ι (1) (2) (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 Dzn or Yzn 0.1 to 0.2 X(1) note that Z ≈ Z , except for rotating machines, (2) (1) 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 Line ≈ 3Z presentation of this method for calculating solid (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 method implementing symmetrical components, that may be used by engineers not specialised in Rated Voltage factor c the field. voltage for calculation of The method is applicable to electrical networks Un Isc max. Isc min. with a nominal voltage of less than 550 kV and LV (100 to 1000 V) the standard explains the calculation of minimum If tolerance + 6% 1.05 0.95 and maximum short-circuit currents. The former is required in view of calibrating If tolerance + 10% 1.1 0.95 overcurrent protection devices and the latter is MV and HV used to determine the rated characteristics for 1 to 550 kV 1.1 1 the electrical equipment. Fig. 25 : Values for voltage factor c (see IEC 60909). Procedure 1- Calculate the equivalent voltage at the fault Ib, rms value of the symmetrical short-circuit location, equal to c Un / 3 where c is a breaking current, voltage factor required in the calculation to idc, aperiodic component, account for: Ik, rms value of the steady-state short-circuit c Voltage variations in space and in time current. c Possible changes in transformer tappings c Subtransient behaviour of generators and Effect of the distance separating the fault motors from the generator Depending on the required calculations and the When using this method, two different given voltage levels, the standardised voltage possibilities must always be considered: levels are indicated in Figure 25 . c The short-circuit is far from the generator, the 2- Determine and add up the equivalent situation in networks where the short-circuit positivesequence, negative-sequence and currents do not have a damped, alternating zerosequence impedances upstream of the fault component location. This is generally the case in LV networks, except 3- Calculate the initial short-circuit current using when high-power loads are supplied by special the symmetrical components. Practically HV substations; speaking and depending on the type of fault, the c The short-circuit is near the generator equations required for the calculation of the Isc (see fig. 11), the situation in networks where the are indicated in the table in Figure 26 . short-circuit currents do have a damped, 4- Once the rms value of the initial short-circuit alternating component. This generally occurs in current (I"k) is known, it is possible to calculate HV systems, but may occur in LV systems when, the other values: for example, an emergency generator supplies ip, peak value, priority outgoers.

Cahier Technique Schneider Electric n° 158 / p.24 Type I"k Fault occuring of short-circuit General situation far from rotating machines cUn cUn Three-phase (any Ze) " = " = Ik3 Ik3 3 Z()1 3 Z()1

In both cases, the short-circuit current depends only on Z(1). which is generally replaced by Zk 2 + 2 the short-circuit impedance at the fault location, defined by Zkk = R Xk 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. cUn Phase-to-phase clear of earth (Ze = ∞) I" = I" = cUn k2 + k2 ZZ()12 ( ) 2 Z()1

" cUn 3 " cUn 3 Phase-to-earth I = I = k1 ++ k1 + ZZ()12 ( ) Z ( 0 ) 2 ZZ()10 ( )

cUn 3 Z Phase-to-phase-to-earth I" = i I" = cUn 3 kE2E ++ kE2E + (Zsc between phases = 0) ZZ()12 ( ) Z() 2 Z () 0 ZZ() 10 ( ) ZZ()102 ( ) (see fig. 5c)  Z  cUn Z− aZ cUn  ()0  − a " = ()02 () I  Z()1  k2EL2 ZZ ++ Z Z ZZ I" = ()12 ( ) () 2 () 0() 10 ( ) k2EL2 + ZZ()102 ( )

  Z()0 2 cUn Z− a2 Z cUn   − a " ()0 ()2  Z  I = " ()1 k2EL3 ++ I = ZZ()12 ( ) Z() 2 Z () 0 ZZ() 10 ( ) k2EL3 + ZZ()102 ( )

Symbol used in this table: c phase-to-phase rms voltage of the three-pase network = Un c short-circuit impedance = Zsc c c modulus of the short-circuit current = I"k earth impedance = Ze. c symmetrical impedances = Z(1) , Z(2) , Z(0)

Fig. 26 : Short-circuit values depending on the impedances of the given network (see IEC 60909).

The main differences between these two cases v The resistances per unit length RL of lines are: (overhead lines, cables, phase and neutral c For short-circuits far from the generator conductors) should be calculated for a temperature of 20 °C v The initial (I"k), steady-state (Ik) and breaking c Calculation of the minimum short-circuit (Ib) short-circuit currents are equal (I"k = Ik = Ib) currents requires v The positive-sequence (Z(1)) and negative v Applying the voltage factor c corresponding to sequence (Z(2)) impedances are equal (Z(1) = Z(2)) the minimum permissible voltage on the network Note however that asynchronous motors may also v add to a short-circuit, accounting for up to 30% of Selecting the network configuration, and in the network Isc for the first 30 milliseconds, in some cases the minimum contribution from which case I"k = Ik = Ib no longer holds true. sources and network feeders, which result in the lowest short-circuit current at the fault location Conditions to consider when calculating the v maximum and minimum short-circuit Taking into account the impedance of the currents busbars, the current transformers, etc. v c Calculation of the maximum short-circuit Considering resistances RL at the highest currents must take into account the following foreseeable temperature  0.004  points R =+ 1 (θ - 20 °C) x R L  ° e  L20 v Application of the correct voltage factor c  C  ° θ corresponding to calculation of the maximum where RL20 is the resistance at 20 C; e is the short-circuit currents permissible temperature (°C) for the v Among the assumptions and approximations conductor at the end of the short-circuit. 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 n° 158 / p.25 Impedance correction factors The impedance of a power station unit with an Impedance-correction factors were included in on-load tap-changer is calculated by: IEC 60909 to meet requirements in terms of ZKtZZ=+( 2 ) technical accuracy and simplicity when SSr G THV calculating short-circuit currents. The various with the correction factor: factors, presented here, must be applied to the U2 U2 c short-circuit impedances of certain elements in K =⋅nQ rTLV ⋅ max S 2 2 +−'' ϕ the distribution system. UrQ UrTHV 1 xxdTsin rG c Factor KT for distribution transformers with two or three windings = UrTHV and tr = UrTLV ZKZTK T T ZS is used to calculate the short-circuit current = Cmax KT 095. for a fault outside the power station unit with an 1+ 0.6xT on-load tap-changer. where xT is the relative reactance of the The impedance of a power station unit without transformer: an on-load tap-changer is calculated by: 2 = SrT ZKtZZ=+( ) xXTT SO SO r G THV U2 rT with the correction factor: and c is the voltage factor related to the max U U c nominal voltage of the network connected to the K = nQ ⋅⋅±rTLV (1 p ) max SO + T + '' ϕ low-voltage side of the network transformer. UprG(1 G) UrTHV 1 xdsin rG The impedance correction factor must also be ZSO is used to calculate the short-circuit current applied to the transformer negative-sequence for a fault outside the power station unit without and zero-sequence impedances when an on-load tap-changer. calculating unbalanced short-circuit currents. c Factors KG,S, KT,S or KG,SO, KT,SO are used Impedances ZN between the transformer when calculating the partial short-circuit currents starpoints and earth must be introduced as 3ZN in the zero-sequence system without a for a short-circuit between the generator and the correction factor. transformer (with or without an on-load tap- changer) of a power station unit c Factors KG and KS or KSO are introduced when calculating the short-circuit impedances of v Power station units with an on-load tap- generators and power station units (with or changer without on-load tap-changers) '' = cUrG The subtransient impedance in the positive- IkG sequence network must be calculated by: 3KZGS, G == +'' where: ZKZKRjXGK G G G( G d) c with R representing the stator resistance of a K = max G GS, + '' ϕ synchronous machine and the correction factor 1 xdsin rG U c = cmax K =⋅n max KTS, G + '' ϕ 1− x sin ϕ UrG 1 xdsin rG T rG v It is advised to use the following values for RGf Power station units without an on-load tap- (fictitious resistance of the stator of a changer synchronous machine) when calculating the '' = cUrG peak short-circuit current. IkG 3KZGSO, G RX= 005. '' for generators with Gf d where: UrG > 1kV et SrG u 100 MVA = '' 1 c RXGf 007. d for generators with K = ⋅ max GSO, 1+ p 1+ x'' sin ϕ UrG > 1kV et SrG < 100 MVA G d rG = '' 1 c RXGf 015. d for generators with K = ⋅ max TSO, 11+ p − x sin ϕ UrG i 1000 V G T rG

Cahier Technique Schneider Electric n° 158 / p.26 3.4 Equations for the various currents

Initial short-circuit current (I"k) which expresses the influence of the subtransient and transient reactances, with Ir as The different initial short-circuit currents I"k are calculated using the equations in the table in the rated current of the generator. figure 26. Steady-state short-circuit current Ik The amplitude of the steady-state short-circuit Peak short-circuit current ip current I depends on generator saturation Peak value i of the short-circuit current In no k p influences and calculation is therefore less meshed systems, the peak value ip of the short- accurate than for the initial symmetrical curren I" . circuit current may be calculated for all types of k The proposed calculation methods produce a faults using the equation: sufficiently accurate estimate of the upper and = κΙ" ip 2 k where lower limits, depending on whether the short- circuit is supplied by a generator or a I"k = is the initial short-circuit current, synchronous machine. κ = is a factor depending on the R / X and can c The maximum steady-state short-circuit be calculated approximately using the following current, with the synchronous generator at its equation (see fig.9) : highest excitation, may be calculated by: R λ -3 Ikmax = max Ir κ = 1.02 + 0.98 e X c The minimum steady-state short-circuit current

Short-circuit breaking current Ib is calculated under no-load, constant (minimum) excitation conditions for the synchronous Calculation of the short-circuit breaking current I b generator and using the equation: is required only when the fault is near the generator and protection is ensured by Ikmin = λmin Ir timedelayed circuit breakers. Note that this λ is a factor defined by the saturated current is used to determine the breaking synchronous reactance Xd sat. capacity of these circuit breakers. The λmax and λmin values are indicated on next This current may be calculated with a fair degree the page in Figure 28 for turbo-generators and of accuracy using the following equation: in Figure 29 for machines with salient poles Ib = µ . I"k where: (series 1 in IEC 60909). where µ = is a factor defined by the minimum time delay tmin and the I"k / Ir ratio (see Fig. 27 )

µ

1.0

Minimum the delay tmin 0.02 s 0.9

0.05 s 0.8 0.1 s

0.7 > 0.25 s

0.6

0.5 0 1 23456789

" Three-phase short-circuit I k / Ir

Fig. 27 : Factor µ used to calculate the short-circuit breaking current Ib (see IEC 60909).

Cahier Technique Schneider Electric n° 158 / p.27 λ λ 2.4 6.0 λ X max d sat 2.2 1.2 5.5

1.4 2.0 5.0 1.6 1.8 X 1.8 2.0 4.5 d sat 2.2 λ 1.6 4.0 max 0.6

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 λ 1.5 min λ min 0.4 1.0

0.2 0.5

0 0 012 345678 12345678 " " 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 (overexcitation = 1.3 as per IEC 60909). salient poles (overexcitation = 1.6 as per IEC 60909).

3.5 Examples of short-circuit current calculations

Problem 1. A transformer supplied by a network Supply network A 20 kV network supplies a transformer T UnQ = 20 kV connected to a set of busbars by a cable L " IkQ = 10 kA (see Fig. 30 ). It is necessary to calculate, in compliance with T (Dyn5) S = 400 kVA IEC 60909, the initial short-circuit current I"k and rT U = 20 kV the peak short-circuit current ip during a three- rTHV phase, then a phase-to-earth fault at point F1. UrTLV = 410 V U = 4% The following information is available: kr PkrT = 4.6 kW c The impedance of the connection between the R(0)T / RT = 1.0 Cable L supply and transformer T may be neglected X(0)T / XT = 0.95 l = 4 m c Cable L is made up of two parallel cables with three conductors each, where: l = 4 m; 3 x 185 mm2 Al

ZL = (0.208 + j0.068) Ω/km Un = 400 V R(0)L = 4.23RL; X(0)L = 1.21XL F1 c The short-circuit at point F1 is assumed to be Fig. 30 far from any generator

Solution: c Three-phase fault at F1 v Impedance of the supply network (LV side) 2 cU  U  11..× 20  041 2 Z =×QnQ  rTLV  = ×   = 0.534 mΩ Qt ''   ×   3 IkQ UrTHV 310 20 R Failing other information, it is assumed that Q = 01. , hence: XQ

Cahier Technique Schneider Electric n° 158 / p.28 ==Ω XZmQt0..995 Qt 0 531 == Ω RXQt01.. Qt 0053 m =+ Ω ZjmQt (0..053 0 531) c Impedance of the transformer 2 ( )2 =×uUkr rTLV =×4 410 = Ω ZTLV 3 16. 81 m 100 SrT 100 400× 10 2 2 ==UrTLV ()410 = Ω RPTLV krT 2 4,600 2 483. m SrT (400× 103)

=−=22 Ω XZRTLV TLV TLV 16. 10 m =+ Ω ZjmTLV (.483 1610 . ) ==×SrT 400 = xXTT2 16. 10 2 0.03831 UrTLV 410 The impedance correction factor can be calculated as: c 105. K = 095. max = 095. = 0.975 T + +×( ) 106. xT 10..6003831 ==+ Ω ZKZTK T TLV (471.. j 1570) m c Impedance of the cable =× + ×=+−3 Ω ZjL 05..( 0208 0 .068) 4 10( 0 . 416 jm 0 . 136) c Total impedance seen from point F1 =+ +=( + ) Ω ZZk Qt ZTK Z L 518.. 1637 m c Calculation of I"k and ip for a three-phase fault cU 105. × 400 I'' ==n = 14. 12 kA k × 3Zk 31717. R R 518. ==k =0.316 X Xk 16. 37 R −3 κ =+102.. 098e X = 14 . =×=×κ '' = ikAp 214214Ik ...12 27 96 c Phase-to-earth fault at F1 v Determining the zero-sequence impedances For transformer T (Dyn5 connection), the manufactures indicates: == RR()00TT and X () T095. X T with the impedance-correction factor KT, the zero-sequence impedance is: =+ =+ Ω ZKRjX()0 TK T( T095... T ) ( 4712 j 14 913) m For cable L: =+=+ Ω ZRXjm()0 LLL(.423 121 . )(176 . 0 .165) v Calculation of I"k and ip for a phase-to-earth fault ===+ Ω ZZ()1 ()2 ZK (518.. j 1637) m =+=+ Ω ZZ()0 ()00TK Z () L (647.. j 1508) m ++= + Ω ZZ()1 ()2 Z()0 (16.. 83 j 47 82) m The initial phase-to-earth short-circuit current can be calculated using the equation below: cU 3105. × 400 3 I'' = n = = 14. 35 kA k1 ++ ZZ()1 ()2 Z()0 50. 70

The peak short-circuit current ip1 is calculated with the factor κ obtained via the positive-sequence: =×=×κ '' = ikAp1 214214Ik1 ...35 28 41

Cahier Technique Schneider Electric n° 158 / p.29 Problem 2. A power station unit

A power station unit S comprises a generator G SrG = 250 MVA and a transformer T with an on-load tap-changer G UrG = 21 kV Ω (see Fig. 31 ). RG = 0.0025 " It is necessary to calculate, in compliance with xd = 17% IEC 60909, the initial short-circuit current I’’ as xdsat = 200% k ϕ cos rG = 0.78 well as the peak ip and steady-state Ikmax short- circuit currents and the breaking short-circuit current Ib during a three-phase fault: F2 c Outside the power station unit on the busbars at point F1 c Inside the power station unit at point F2 SrT = 250 MVA UrTHV 240 kV The following information is available: T = c The impedance of the connection between UrTLV 21 kV generator G and transformer T may be Ukr = 15% neglected PkrT = 520 kW c The voltage factor c is assumed to be 1.1 U = 220 kV c The minimum dead time tmin for calculation of nQ Ib is 0.1 s F1 c Generator G is a cylindrical rotor generator (smooth poles) Fig. 31 c All loads connected to the busbars are passive

Solution: c Three-phase fault at F1 v Impedance of the transformer

2 2 =×uUkr rTHV =×15 240 = Ω ZTHV 34. 56 100 SrT 100 250 2 2 ==UrTHV 240 =Ω RPTHV krT 2 052.. x 2 0 479 SrT 250 =−=22 Ω XZRTHV THV THV 34. 557 =+ Ω ZjTHV (.0 479 34 . 557 ) v Impedance of the generator

'' 2 2 '' =×=×=xdrGU 17 21 Ω Xd 0.2999 100 SrG 100 250 =+'' = + ZRjXGG d 0..0025j 0 2999 = Ω ZG 0.2999

SrG > 100 MVA, therefore RGf = 0.05 X"d, hence ZGf = 0.015 + j0.2999

U2 U2 c 2202 212 11. K =×nQ rTLV × max =×× = 0.913 S 2 2 +−'' ϕ 2 2 10+−×...17 0 15 0 6258 UrG UrTHV 1 xxdTsin rG 21 240

 240 2  ZKtZZ=+=().(..)(..)2 0 913   ×+0 0025 j 0 2999 ++ 0 479 j 34 557  SSr G THV  21   =+ ZjS 0..735 67 313 (ZSf = 2.226 + j67.313 if we consider ZGf (to calculate ip)) cU 11, × 220 I'' ==nQ =−0..023j 2 075 kS + 3ZjS 30( ..735 67 313) '' = IkS 208. kA

Cahier Technique Schneider Electric n° 158 / p.30 Based on impedance ZSf, it is possible to calculate RSf / XSf = 0.033 and κS = 1.908

The peak short-circuit current ipS is calculated by: =×κ '' ipS S2 I kS =×= ikApS 1...908 2 2 08 5 61

The short-circuit breaking current IbS is calculated by: =×µ '' IIbS kS Factor µ is a function of radio I"kG / IrG and the minimum dead time tmin. Ratio I"kG / IrG is calculated by: I''I '' U 208. 240 kG ==kS rTHV =346. IrG IrG UrTLV 6.873 21

According to figure 27 (curve at tmin = 0.1 s), µ ≈ 0.85, hence: =×= IbS 085... 208 177 kA

The maximal steady-state short-circuit current Ikmax is calculated by: ==××=λ UrTLV 21 IIkSmax rG 165.. 6873 099 . kA UrTHV 240

Factor λmax = 1.65 is obtained in figure 28 for the ratio I"kG / IrG = 3.46 and xdsat = 2.0 c Three-phase fault at F2

'' = cUrG IkG 3KZGS, G where: c 11. K = max = = 0.994 GS, + '' ϕ +×( ) 1 xdsin rG 1017..0626 cU 11. × 21 I'' ==rG = 44. 74 kA kG ×× 3KZGS, G 30..994 0 2999

The peak short-circuit current ipG is calculated by: =×κ '' ipG G2 I kG

Based on impedance ZGf, it is possible to calculate RGf / X"d = 0.05, hence κG = 1.86 =×= ikApG 186... 2 4474 117 69

The short-circuit breaking current IbG is calculated by: =×µ '' IIbG kG

Factor µ is a function of ratio I"kG / IrG and the minimum dead time tmin. Ratio I"kG / IrG is calculated by: I'' 44. 74 kG ==651. IrG 6.873

According to figure 27 (curve at tmin = 0.1 s), µ ≈ 0,71, hence: =× = IbS 071.. 4474 3177 . kA

The maximum steady-state short-circuit current Ikmax is calculated by: ==×=λ IIkGmax rG 175.. 6873 12 . 0 kA

Factor λmax = 1.75 is obtained in figure 28 for the ratio I"kG / IrG = 6.51 and xdsat = 2.0

Cahier Technique Schneider Electric n° 158 / p.31 4 Conclusion

Various methods for the calculation of short- All computer programs designed to calculate circuit currents have been developed and short-circuit currents are predominantly subsequently included in standards and in this concerned with: “Cahier Technique” publication as well. c Determining the required breaking and making capacities of switchgear and the electro- A number of these methods were initially mechanical withstand capabilities of equipment designed in such a way that short-circuit currents could be calculated by hand or using a small c Determining the settings for protection relays calculator. Over the years, the standards have and fuse ratings to ensure a high level of been revised and the methods have often been discrimination in the electrical network modified to provide greater accuracy and a Other software is used by experts specialising in better representation of reality. However, in the electrical network design, for example to study process, they have become more complicated the dynamic behaviour of electrical networks. and time-consuming, as is demonstrated by the Such computer programs can be used for recent changes in IEC 60909, where hand precise simulations of electrical phenomena over calculations are possible only for the most time and their use is now spreading to include simple cases. the entire electro-mechanical behaviour of With the development of ever more sophisticated networks and installations. computerised calculations, electrical-installation Remember, however, that all software, whatever designers have developed software meeting its degree of sophistication, is only a tool. To their particular needs. Today, a number of ensure correct results, it should be used by software packages comply with the applicable qualified professionals who have acquired the standards, for example Ecodial, a program relevant knowledge and experience. designed for low-voltage installations and marketed by Schneider Electric.

Bibliography

Standards Schneider Electric Cahiers Techniques c EC 60909: Short-circuit currents in three- c Analysis of three-phase networks in disturbed phase AC systems. operating conditions using symmetrical v Part 0: Calculation of currents. components, Cahier Technique no. 18 - v Part 1: Factors for the calculation of short- B. DE METZ-NOBLAT. circuit currents. c Neutral earthing in an industrial HV network. v Part 2: Electrical equipment. Data for short- Cahier Technique no. 62 - F. SAUTRIAU. circuit current calculations. c LV circuit-breaker breaking capacity. v Part 3: Currents during two separate Cahier Technique no. 154 - R. MOREL. simultaneous single phase line-to-earth short circuits and partial short-circuit currents flowing Other publications through earth. c Electrical Installation Guide v Part 4: Examples for the calculation of short- In English in accordance with IEC 60364: circuit currents. 2005 edition. In French in accordance with NF C15-100: c NF C 15-100: Installations électriques à basse 2004 edition. tension. Published by Schneider Electric c C 15-105: Guide pratique. Détermination des (Schneider Training Institute). sections de conducteurs et choix des dispositifs c Les réseaux d’énergie électrique (Part 2), de protection. R. PELISSIER. Published by Dunod.

Cahier Technique Schneider Electric n° 158 / p.32 ic

Schneider Electric Direction Scientifique et Technique, DTP: Axess Service Communication Technique Transl.: Cabinet Harder - Grenoble - France F-38050 Grenoble cedex 9 Editor: Schneider Electric E-mail : [email protected]

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