Calculation of short-circuit currents (no 158) by hamada1331

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									Collection Technique ..........................................................................

Cahier technique no. 158

Calculation of short-circuit

                                                         B. De Metz-Noblat
                                                         F. Dumas
                                                         G. Thomasset
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no. 158
Calculation of short-circuit

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.

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

                             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
                             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 %.
                                                                                 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.

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 (%)

                                                                                   at transformer

 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
                                                                                                               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
                                                       θ1 > θ2

                                                                                                                            Fuse time-current
                   a5   s                                                                                                   curve
                                                                 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
                            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 )
                  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.

                                                 L3                                                       L3
                                                 L2                                                       L2
                                                 L1                                                       L1

                                                      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
                   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 ,
                                R               X                                   α = angle characterising the difference between
                                                                                    the initiation of the fault and zero voltage.
                                                                                    c The second (idc) is an aperiodic component,
                                                                                                        -       t
                                                               A                    idc = - Ι sin α e L . Its initial value depends on
                                       Zsc                                          a and its decay rate is proportional to R / L.

                                                                                    At the initiation of the short-circuit, i is equal to
                                                                                    zero by definition (the design current Is is
                                                                                    negligible), hence:
                                                                                    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


                                                       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 ).
                                                                                         The fault current can be defined by: i =     sin ω t
                                     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
                              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
                             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.



                                                         Subtrans.                  Transient                       Steady-state


                            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

                                                                                                                           U/ 3
                                                                                                 V              Ιsc 3 =
                                                                           ZL                                               Zsc


                                 Phase-to-phase fault                  ZL                            Zsc

                                                                                                 U              Ιsc 2 =
                                                                       ZL                                                   2 Zsc

                                 Phase-to-neutral fault                ZL                            Zsc

                                                                                                                             U/ 3
                                                                       ZLn                       V              Ιsc1 =
                                                                                                                           Zsc + Z Ln

                                 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
                 the fault location.                                                  = 1 - (0,2) = 0, 980

                 (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
                 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
                 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-
                 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 = 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
                            rectifier units have usc values of up to 10 to              RL =
                            12 % in order to limit short-circuit currents.                      A
                            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).
                             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 


                                                 12                                                    Psc = 250 MVA

                                                                                                       Psc = 500 MVA

                                                     500              1,000                1,500                 2,000       Pn

                            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

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-
                                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
                                                                      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.
                                                                                         12 They produce a current in the
                                                                                         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 =
                                                 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
                                                      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
                            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:
                                   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
                 Psc = 500 MVA

                 Overhead line
                 3 cables, 50 mm2, copper,                                                                                         G
                 length = 2 km                                                                     A
                 1 MVA
                 Xsubt = 15 %
                 2 transformers
                 1,000 kVA
                 secondary winding 237 / 410 V
                 usc = 5 %                                                                                             10 m
                 Main LV switchboard
                 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
                 3 three-phase cables,
                 35 mm2, copper,
                 length = 30 m                                                                                     D
                 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.

                            Section                      Calculations                                                       Results

                            (the circled numbers X indicate where explanations may be found in the preceding text)

                            20 kV↓                                                                                          X (Ω)       R (Ω)

                                                                 (           )
                            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

                                                                         (               )
                                                                 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)
                                                          RB = 0.0225 x                                       6                         ≈0
                                                                             3 x 400
                            Fault B
                            7. circuit-breaker           X cb = 0.15                                                        0.15
                            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)
                                                         Rc 2 = 0.0225 x                                                                19.2
                            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
ZA =     R2
          A   +   X2
                   A   ≈ 1.80 Ω hence
                                                      figure 9 and therefore the peak Isc is equal to:
         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
                                                      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
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) +                      ]
XB =                   4.2 + 0.15 + 1.5 10-3                = 1.06 hence k ≈ 1.05 on the curve in
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
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"
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:
                             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
                             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(2)
                                                         +               I2(0)

                                                                                              =             I2
I2(1)         ωt                                   ωt

                    Geometric construction of I1                                        Geometric construction of I2
                                       I1                                                  I1(0)
                         I1(1)   I1(2) I1(0)                                                       a2 I1(1)
                                                                                     a 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:
                                     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,
                            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.

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
                            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 +
                                                                                                               (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
                                                                                         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.



                                                                                           Minimum time delay tmin
                                                                                             0.02 s

                                                                                                0.05 s

                                                                                                0.1 s

                                                                                                > 0.25 s


                                                              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
                    2.2                                               1.2      5.5
                    2.0                                                        5.0
                    1.8                                               2.0      4.5                                               Xd sat
                    1.6                                                                                   λmax                         0.6
                    1.4                                                        3.5                                                     0.8
                    1.2                                                        3.0                                                     1.0
                    1.0                                                        2.5                                                     1.7
                    0.8                                                        2.0
                                      λmin                                     1.5
                    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
                 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.


                                                          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


                            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
                                        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)
               j 18                                            j 270
                  j7                                                                                  j 21
                                    H                                                                                  H

                               Z(1), Z(2)                                                                             Z(0)
Fig. 33.

                                    j 3.1                                                                                         Z'
                                                                                                                                      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

                         Z(1), Z(0)

                                                     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

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
                                                                                              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

                            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
                            c NF C 15-100: Installations électriques à basse
                            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

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
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