1SDC007101G0201 - SACE, Isomax, Emax, Tmax

Technical Application PapersMV/LV transformer substations:theory and examples of short-circuit calculationIndex1 General information on MV/LVtransformer substations3 Choice of protection andcontrol devices1.1 Classic typologies ....................................... 21.2 General considerations about MV/LVtransformers................................................ 51.3 MV protection devices: observations aboutthe limits imposed by the utility companies .. 81.4 LV protection devices ................................. 82 Calculation of short-circuitcurrents2.1 Data necessary for the calculation ........... 112.2 Calculation of the short-circuit current ..... 122.3 Calculation of motor contribution ............. 152.4 Calculation of the peak current value ....... 153.1 Generalities about the main electricalparameters of the protection and controldevices...................................................... 173.2 Criteria for the circuit-breaker choice ....... 193.3 Coordination between circuit-breakersand switch-disconnectors ........................ 213.4 Coordination between automatic circuitbreakers-residualcurrent devices (RCDs) .. 223.5 Example of study of a MV/LV network...... 23Annex A:Calculation of the transformer inrush current ....... 30Annex B:Example of calculation of the short-circuitcurrent ................................................................... 32B1 Method of symmetrical components ........... 33B2 Power method .............................................. 38Glossary .............................................................. 401

Technical Application Papers1 General information on MV/LV transformer substations1.1 Classic typesAn electrical transformer substation consists of a wholeset of devices (conductors, measuring and controlapparatus and electric machines) dedicated totransforming the voltage supplied by the medium voltagedistribution grid (e.g. 15kV or 20kV), into voltage valuessuitable for supplying low voltage lines with power (400V- 690V).The electrical substations can be divided into publicsubstations and private substations:public substations: these belong to the electricity utilityand supply private users in alternating single-phase orthree-phase current (typical values of the voltage for thetwo types of power supply can be 230V and 400V). Inturn, these are divided into urban or rural typesubstations, consisting of a single reduced-size powertransformer. Urban substations are usually built usingbricks, whereas rural ones are often installed externallydirectly on the MV pylon.private substations: these can often be considered asterminal type substations, i.e. substations where the MVline ends at the point of installation of the substation itself.They belong to the user and can supply both civil users(schools, hospitals, etc.) with power and industrial userswith supply from the public MV grid. These substationsare mostly located in the same rooms of the factory theysupply and basically consist of three distinct rooms:- delivery room: where the switching apparatus of theutility is installed. This room must be of a size to allowany construction of the in-feed/output system whichthe utility has the right to realise even at a later time tosatisfy its new requirements. The take-up point is foundin the delivery room, which represents the border andconnection between the public grid and the user plant.- instrument room: where the measuring units arelocated.Both these rooms must have public road access toallow intervention by authorised personnel whether theuser is present or not.- user room: destined to contain the transformer andthe MV and LV switching apparatus which are theconcern of the user. This room must normally beadjacent to the other two rooms.Figure 1 shows the typical structure of a substation withdivision of the rooms as previously described.Figure 1: Conceptual diagram of the substation2 MV/LV transformer substations: theory and examples of short-circuit calculation

1 General information on MV/LV transformer substationsIt is normally expected that the customer use MV/LVtransformers with:- delta primary winding (∆), except when specialapplications (welding machines, actuators, etc.) areforeseen, agreeing on the choice with the utility- secondary winding with grounded star point ( ),to reduce disturbances in the network and to makethe line and phase voltage easily available.The utility prescribes and defines the criteria and methodsfor connection of normal customers (intended as thosewho are not other power producers or special users withdisturbing loads characterised, for example, byharmonics or flicker) in its official documentation.These prescriptions specifically apply to connections tothe MV grid with rated voltage of 15kV and 20kV whereas,for other MV voltage values, they can be applied forsimilarity.As an example, below we give the prescriptions providedby an Italian distribution utility regarding the power ofthe transformer which can be used. The power valuesallowed are as follows:- power not higher than 1600kVA for 15kV networks- power not higher than 2000kVA for 20kV networks.The powers indicated refer to a transformer withv k%=6%. If connection to several machines is foreseen,the size limit indicated must be applied to the wholeassembly of transformers in parallel.The limit relative to the installable power is alsoestablished and, in order not to cause unwanted trips ofthe overcurrent protection of the MV line during theputting into service operations of their own plants, thecustomer cannot install more than two transformers of asize equal to the limits indicated previously with separateLV busbars. Otherwise they will have to provide suitabledevices in their plant to avoid those transformers whichwould determine these limits being exceeded being putinto service simultaneously.When, on the other hand, substation sizing foresees theuse of transformers with overall power higher than theinstallable limit, an agreement with the distributioncompany is necessary. Technically, the use of a devicemust be provided (an undervoltage protection which deenergisesthe transformers which are in excess can beused) which prevents simultaneous energisation of thetransformers so that the magnetisation current requiredremains equal to that of the two transformers of the sizelimit allowed (e.g. 2x1600kVA at 15kV).The transformer is connected to the take-up point in thedelivery room by means of a copper connection cablewhich, regardless of the power supplied, must have aminimum cross-section of 95mm 2 . This cable is theproperty of the user and must be as short as possible.The present trend regarding management of the earthingconnection of the system is to provide the passage frominsulated neutral to earthed neutral by means ofimpedance. This modification, needed to reduce thesingle-phase earth fault currents which are continuallyon the increase due to the effect of growingly commonuse of underground or overhead cables, also impliesupgrading the protections against earth faults both bythe utility and by the customers. The intention is to limitunwanted trips as far as possible, thereby improvingservice.After having indicated what the main electrical regulationsfor a MV/LV substation are, we now analyse what themost common management methods may be in relationto the layout of the power supply transformers for asubstation supplied by a single medium voltage line.Method 1I MVMV lineI MVMV lineS MVSubstation with a single transformerWhen the plant foresees installation of an “I MV” overcurrent protectiondevice where the line which supplies the substation originates, asshown in diagram 1, this device must ensure protection of both the MVline as well as the transformer.In the case where the protection device also carries out switching andisolation functions, an interlock must be provided which allows accessto the transformer only when the power supply line of the substation hasbeen isolated.Another management method is shown in diagram 1a, which foreseesinstallation of the “S MV” switching and isolation device positionedimmediately to the supply side of the transformer and separate from theprotection device which remains installed at the beginning of the line.I LVI LVL1L2L1L2Diagram 1Diagram 1aMV/LV transformer substations: theory and examples of short-circuit calculation3

1 General information on MV/LV transformer substationsTechnical Application PapersMethod 2Substation with two transformers with one as a spare for the otherI GMVWhen the plant foresees installation of a transformer considered as aspare, the circuit-breakers on the LV side must be connected with an “I”interlock whose function is to prevent the transformers from operating inparallel.I LV1I MV1II MV2I LV2Apart from the switching and isolation device on the incoming MV line(I GMV), it is advisable to provide a switching, isolation and protection deviceon the individual MV risers of the two transformers (I MV1and I MV2) as well.In this way, with opening of the device on the supply and load side of atransformer, it is possible to guarantee isolation and access the machinewithout putting the whole substation out of service.L1 L2 L3Diagram 2Method 3I GMVSubstation with two transformers which operate in parallel on thesame busbarI LV1I MV1I MV2I LV2When the plant foresees installation of two transformers operating inparallel at the same overall power required of the plant, it is possible touse two transformers with lower rated power. Compared with themanagement method described in the two previous cases, higher shortcircuitcurrents could be generated for faults in the low voltage systemdue to reduction of the possible v k%for lower power machines.Operation in parallel of the transformers could cause greater problems inmanagement of the network. Again in this case, however, outage of amachine might require a certain flexibility in load management, ensuringthe power supply of those considered to be priority loads. Whencoordinating the protections, the fact that the overcurrent on the LV side isdivided between the two transformers must be taken into consideration.L1 L2 L3Method 4Diagram 3I LV1I MV1II GMVI MV2I LV2Substation with two transformers which operate simultaneously ontwo separate half-busbarsStarting from the previous management method, by providing a “C LV” bustieand an “I” interlock which prevents the bus-tie from being closed whenboth the incoming circuit-breakers from the transformer are closed, asubstation managed as shown in diagram 4 is made, which foresees twotransformers which individually supply the low voltage busbars, which areseparate.With the same power of the transformers installed, this managementmethod allows a lower value of the short-circuit current on the busbar. Inother words, each transformer establishes the short-circuit level for thebusbar of its competence without having to consider the contribution ofother machines. Again in this case, when a transformer is out of service,with any closure of the bus-tie you pass to a system with a single busbarsupplied by the sound transformer alone, and a load management logicmust be provided with disconnection of non-priority loads.L1C LVL2 L3 L4 L5Diagram 4L6Plant management according to diagram 4 is possible, for example byusing the Emax series of air circuit-breakers with a wire interlock(mechanical interlock) between three circuit-breakers.4 MV/LV transformer substations: theory and examples of short-circuit calculation

1 General information on MV/LV transformer substations1.2 General information about MV/LVtransformersThe transformer is the most important part of thetransformer substation. Its selection affects theconfiguration of the substation and is made on the basisof various factors.Not being a specific subject of this paper and wanting togive some general indications, it can be stated that forthe request for low powers (indicatively up to 630kVA -800kVA), a single transformer can be installed, whereasfor higher powers (indicatively up to 1000kVA - 1600kVA),the power is divided over several units in parallel.Another characteristic to take into consideration whenselecting the machine is the type of cooling system, whichcan be either in air or in oil. With reference to airconditioning the structure of the substation, in the caseof oil cooled transformers, measures must be taken, forexample those to prevent the oil spreading outside byproviding an oil collection pit as shown in Figure 2.Furthermore, the substation must have a minimum flameresistance of 60 minutes (REI 60) and ventilation onlytowards the exterior. According to the type of cooling,the transformers are identified as follows:AN cooling with natural air circulation;AF cooling with forced air circulation;ONAN cooling with natural oil and air circulation;ONAF cooling with forced oil and natural aircirculation;OFAF cooling with forced oil and air circulation.The most frequent choice is for AN and ONAN types, asit is not advisable to use machines which use fans or oilcirculators because it is rarely possible to man thesubstations.Figure 2: ONAN transformers containing more than 500 kg of oil (> 800kVA)MV/LV transformer substations: theory and examples of short-circuit calculation5

1 General information on MV/LV transformer substationsTechnical Application PapersOther important characteristics to be considered arethose referring to the electrical parameters and, inaddition to the usual quantities such as rated power, noloadsecondary rated voltage, transformation ratio, ratedshort-circuit voltage in percent v k%, they acquire greatimportance above all when the transformers arefunctioning in parallel:- the connection typology of the windings (delta/stargrounded is the most used one for the substationtransformers)- connection system (CEI group), conventionallyexpressed by a number which, multiplied by 30, givesthe delay angle of the phase voltage on the LV sidecompared with the MV side.The presence of two or more MV/LV transformers and apossible bus-tie closed on the LV busbars allows theelectricity network to be managed with the transformersin parallel.In the presence of faults, this management methodcauses an increase in the short-circuit current value onthe LV side, with a possible consequent increase in thesize of the circuit-breakers outgoing from the busbar andheavier anchoring conditions for the busbars incomparison with operation with a single transformer. Thisis due to a smaller value of the v k%which characterisesthe transformers with less power. On the other hand,when suitably managed, the parallel method has theadvantage of allowing power supply, at least to the usersconsidered as primary users, through the possible bustie,even in the case of outage of one of the transformers.The following example shows the increase in the shortcircuitcurrent value on the busbar in the case oftransformers in parallel:Supply network, short-circuit power ....... S knet=750MVAPlant secondary voltage ......................... V 2n=400VPower of the single transformer .............. S nTR=1600kVARated short-circuit voltage of thesingle transformer ................................... v k%=6%Power of the transformer providedfor the parallel ......................................... S nTR=800kVAShort-circuit voltage of thetransformer in parallel ............................. v k%=4%From these data and from quick calculations, a shortcircuitcurrent value of 37 kA is obtained on the busbarwith the single 1600kVA transformer.With two 800kVA transformers in parallel, the short-circuitcurrent on the busbar shall be about 55kA.With reference to the electricity network outlined in Figure3, the following considerations have the aim ofillustrating the management philosophy for theprotections:Figure 3I GMVG4I MV1I MV2G3I LV1G2I LV2G1L1C LVL2 L3 L4 L5L66 MV/LV transformer substations: theory and examples of short-circuit calculation

1 General information on MV/LV transformer substationsG1 Fault on one of the LV usersRegardless of the presence or absence of the bus-tie:with appropriate selection of the protection devices andaccording to normal LV selectivity prescriptions, it ispossible to discriminate the fault and ensure servicecontinuity with opening just of the L1 circuit-breaker.G2 Fault on the LV busbarWithout bus-tie:the fault is extinguished by the two general LV side circuitbreakers(I LV1and I LV2) of the transformers, causing completeoutage of the plant. The transformers remain noloadsupplied. To prevent opening of the I MV. circuitbreakers,obtaining MV/LV selectivity is again importantin this case.With bus-tie:the CLV bus-tie must open, with consequent separationof the busbars and complete elimination of the fault bymeans of the main I LV1circuit-breaker opening. The actionof the bus-tie allows power supply to be maintained tothe half-busbar unaffected by the fault. The action of theLV devices (I LV1– C LV– I LV2), which are all affected by thefault, may be co-ordinated by using devices for whichthe directional zone selectivity is implemented, such asfor example protection releases PR123 for the Emaxseries and PR333 for the Emax circuit-breaker type X1.G3 Fault on the LV bus riser of the transformerWithout bus-tie:The fault current affects the two transformers and it maybe such as to cause opening of the two devices I MVandI LVof the transformers. The consequence would be tohave all the plant disconnected. In this case it becomesimportant to study and implement a dedicated managementlogic (for example directional selectivity) whichallows I LV1and I MV1opening in order to isolate only thetransformer affected by the fault. Also a logic for thedisconnection of non-priority loads should be foreseen,since the plant is functioning with one transformer only.With bus-tie:the management logic remains the same and it couldpossibly foresee also the bus-tie opening.G4 Fault on the MV bus riser of the transformerWithout bus-tie:the management logic must allow immediate opening ofthe I MV1circuit-breaker affected by the full fault current(I MV2shall see a lower current limited by the impedanceof the two transformers) and, if the plant managementforesees pulling, the opening of the I LV1circuit-breakerwith isolation of the fault point will follow with servicecontinuity of the whole plant ensured by power supplythrough the other transformer. Also a logic for thedisconnection of non-priority loads should be foreseen,since the plant is functioning with one transformer only.With bus-tie:the management logic remains the same, and the bustiewould have only the function of separating the busbarsby eliminating that of competence of the excludedtransformer.After an analysis of the fault handling modalities, whichunder some circumstances result to be quite complexdue to the double supply of the transformers in parallel,the minimum requirements to have two transformersoperating in parallel are examined now:a) the internal connections must belong to the samegroup (CEI group) and the transformers must have thesame transformation ratio. By complying with theseprescriptions, the two sets of voltage result to coincideand to be in phase opposition; consequently there areno vectorial differences between the secondary voltageof every single mesh and no circulation currents aregenerated. In the contrary case, circulation currentswould be generated, which could damage thetransformers also in no-load operation;b) the short-circuit voltages (v k%) must have the samevalue. Thanks to this measure, the total load current issubdivided between the two transformers in proportionto their respective rated powers. If not, the twotransformers would be differently loaded and the machinewith the lower internal voltage drop would tend to bemore loaded.MV/LV transformer substations: theory and examples of short-circuit calculation7

1 General information on MV/LV transformer substationsTechnical Application Papers1.3 MV protection devices: observationsabout the limits imposed by the utilitycompaniesThe MV distribution outgoing line supplying the usersubstation is provided with its own protections againstovercurrent and earth faults; therefore the utility companyshall not provide any protection device for the customer’splant.In order to prevent any internal faults of the MV and LVplant from affecting the distribution network service, theconsumer must install convenient protections. Theselection of the protection devices and their co-ordinationmust guarantee safety for the personnel and themachines, by ensuring at the same time also good servicereliability of the installation.Some indications are provided hereunder regarding thecharacteristics the MV/LV side protection functions musthave and the way they can interact.The protection of the utility company usually operateswith independent time tripping characteristics and thetripping threshold values communicated to the consumerrepresent the upper limit to comply with in order to avoidunwanted trips.Hereunder we give an example of the setting range ofthe protection device for the different protectionthresholds:- Overcurrent threshold (overload 51):Threshold (30÷600)A, with 15A steps (primary values)Delay time (0.05÷5)s, with 0.05s steps.- Overcurrent threshold (short-circuit 50):Threshold (30÷600)A, with 15A steps (primary values)Delay time (0.05÷5)s, with 0.05s steps.- Protection against earth faults:According to the characteristics of the user installation,the earth fault protection may be constituted either bya directional protection against earth faults combinedwith a zero-sequence overcurrent protection 67N orby a simple zero-sequence overcurrent protection 51N.For example, as regards the zero-sequence overcurrentprotection the setting ranges are the following:overcurrent threshold (0÷10) A, with 0.5A steps (primaryvalues delay time (0.05÷1)s, with 0.05 s steps.1.4 LV protection devicesLV protection devices are located on the load side of thetransfomer.The protection functions usually available on a LV deviceare the functions of protection against overload, againstshort-circuit and against earth fault.Here is a short description of these protection functionsimplemented on the micro-processor based electronicreleases :- protection against overloadidentified as function “L”, it is a protection with inverselong time-delay trip with adjustable current and time.On ABB electronic protection releases it is indicatedalso as function I1.- protection against short-circuitidentified as function “S”, against delayed short-circuit(on ABB electronic protection releases it is indicatedalso as function I2) and “I” against instantaneous shortcircuit(on ABB electronic protection releases it isindicated also as function I3).Function “S” can be with either inverse or definite timedelaytrip, with adjustable current and time. Function“I” is a protection with definite time-delay trip andadjustable current only.- protection against earth-faultidentified as function “G” can be with either inverse ordefinite time-delay trip, with adjustable current andtime. This protection can be realized on the star pointof the transformer with external toroid.The curve in yellow colour represents the behaviour ofthe circuit-breaker at current values much higher thanthe set protection I3.The diagram of Figure 4 shows an example of a time/currenttripping curve of a LV circuit-breaker on which all theabove mentioned protection functions have beenactivated.Figure 41E4s1E3s100s10s1s0.1s1E-2s0.1kA 1kA 10kAThe following example is aimed at explaining how it ispossible to operate with the information which8 MV/LV transformer substations: theory and examples of short-circuit calculation

1 General information on MV/LV transformer substationscharacterize the inverse time-delay curve withcharacteristic I 2 t constant as those available for functionsL - S – G.These results mathematically obtained may be obviouslyverified with immediacy through the course of the trippingcurves, as the time/current diagram of Figure 6 shows.With reference to the protection function “L” implementedon the release which is fitted on the moulded case circuitbreakersof Tmax series, for example a T2...160 In100(“In” indicates the size of the protection release mountedon the circuit-breaker), the possible tripping curves aretype A and type B.The curve of type A is characterized by its passingthrough the point identified as:6 x I1 with a time t1=3sThe curve of type B is characterized by its passingthrough the point identified:6 x I1 with a time t1=6sAssuming for I1 a generic setting I1=0.6xIn=0.6x100=60A,the above means that, in correspondence of 6 x I1=360A,the two setting curves shall be characterized by a trippingtime of 3 or 6 seconds (without the tolerances) as thetime/current diagram of Figure 5 shows.Figure 61E3s100s10s1s0.1kAIs=180ATime x 180A curve B=24sTime x 180A curve A=12sCurve BCurve A1kAFigure 5100s10s1s0.1sCurve A0.1kACurve B6xI1=360 ASince these are curves with I 2 t constant, the followingcondition shall be always verified:for the curve A:(6 x I1) 2 x 3 = const = I 2 tfor curve B:(6 x I1) 2 x 6 = const = I 2 tFor example, under the above conditions, it is possibleto determine the tripping time of the protection for anoverload current equal to 180A.Therefore, from the above formulas, the followingconditions may be obtained:(6 x I1) 2 x 3 = 180 2 x t A(6 x I1) 2 x 6 = 180 2 x t Bwhich respectively give:t A= 12st B= 24s1kA6 Sec3 SecFor example, should the installation requirements imposethat the assumed overload of 180A is eliminated in atime lower than 15 seconds, from the analysis carriedout it shall result that the tripping characteristic to beused and set on the protection release is defined as curveA (tripping time t1=3s for a current equal to 6 x I1).Still making reference to the condition(6 x I1) 2 x t = constto select the curve which is suitable to eliminate theoverload of 180 A in a time lower than 15 seconds, it ispossible to proceed in the reverse way, by setting up theequation:(6 x 0.6 x 100) 2 x t = const = 180 2 x 15This relationship allows the calculation of the maximumdelay of the tripping characteristic to comply with theinstallation requirements.By making the time explicit, the following value isobtained:t = 3.75sThe suitable curve shall be that with “t1” lower than “t”.Therefore the curve to be used is curve A, as resultedalso by the above analysis.The protections, above all the MV ones, are oftenidentified by alphanumeric codes such as 50 – 51N – 67,which do not find an equivalent in the typical LVnomenclature. Hereunder, we give some information toexplain the meaning of the most common codes and tocreate a correspondence, whenever possible, betweenthe indications used to identify MV protections and thoseuse for the LV ones.The Standard IEC 60617-7 is currently in force; it definesthe symbology and the relevant function of the releasestypically used in the electrical installations. For manypeople operating in the electrical field, it is common praxisto use the codification of the Standard ANSI/IEEE C37.2.MV/LV transformer substations: theory and examples of short-circuit calculation9

1 General information on MV/LV transformer substationsTechnical Application PapersBelow there is an example of correspondence betweenIEC and ANSI/IEEE symbology for some of the main MVprotection functions.50 Instantaneous overcurrent relayA device that operates with no intentional time-delaywhen the current exceeds a preset value. It can becompared with a protection “I” of a LV release.51 Time-delayed overcurrent relayA device that functions when the ac input current exceedsa predetermined value, and in which the input currentand operating time are inversely related. It can becompared with a protection “S” of a LV release.51N or 51G Time-delayed earth fault overcurrent relayDevices that operate with a definite time-delay when anearth fault occurs. In details:51N: residual current measured on the CT joint return.This device can be compared with a protection “G” of aLV release.51G: residual current measured directly either on a CTor on toroidal CT only. This device can be compared withthe protection which can be realized, for example,through an homopolar toroid operating a residual currentdevice with adjustable trip times (e.g. a RCQ) or throughthe function “G” of the protection release supplied by anexternal toroid.50N or 50G Instantaneous earth fault overcurrent relayA device that operates with no intentional time-delaywhen an earth fault occurs. In details:50N: residual current measured on the CT commonreturn. It can be compared with a protection “G” withdefinite time of a LV release.50G: residual current measured directly either only on aCT or on toroidal CT. It can be compared with a protectionwhich can be realized, for example, through anhomopolar toroid.67 Alternating current directional power relay ordirectional overcurrent relayA device that operates at a desired value of power flowingin a predetermined direction, or for overcurrent withpower flowing in a predetermined direction. It can becompared with a protection “D” of a LV release.49 Alternating current thermal relayA device that operates when the temperature of themachine or of the ac apparatus exceeds a predeterminedvalue. It can be compared with the overload protection“L” of a LV release, even though a real protection againstoverload is not provided for MV applications.Table 1ANSI/IEEECodeFunction definitionSimbology correspondingto the Standard IEC 60617-751Time-delayed overcurrent50Instantaneous overcurrent= 051NTime-delayed earth fault overcurrent50NInstantaneous earth fault overcurrent= 067Directional phase overcurrent= 067NDirectional zero-sequence overcurrent= 010 MV/LV transformer substations: theory and examples of short-circuit calculation

2 Calculation of short-circuit currents2.1 Data necessary for the calculationSome general indications regarding the typicalparameters characterizing the main components of aninstallation are given hereunder.Knowledge of the following parameters is fundamentalto carry out a thorough analysis of the installation.Distribution networks:In a MV network the rated voltage is the unique parameterusually known.To calculate the short-circuit currents it is necessary toknow the network short-circuit power, which canindicatively vary from 250MVA to 500MVA for systemsup to 30kV.When the voltage level rises, the short-circuit power canindicatively vary between 700MVA and 1500MVA.The voltage values of the MV distribution network andthe relevant short-circuit power values accepted by theStandard IEC 60076-5 are reported in Table 1.Table 1Distribution network Short-circuit apparent power Short-circuit apparent powervoltage Current European practice Current North-American practice[kV] [MVA] [MVA]7.2–12–17.5-24 500 50036 1000 150052–72.5 3000 5000Synchronous generatorThe data usually known for an electrical machine are therated voltage V nand the rated apparent power S n.For synchronous generators, as for every electricalmachine, to get a complete analysis it is necessary toevaluate also:- the behaviour under steady state conditions for ananalysis of the problems of static stability- the behaviour under transitory conditions when the loadsuddenly varies for an analysis of the problems of dinamicstability, in particular when a three-phase short-circuitoccurs.Therefore, it becomes necessary to know the values ofthe machine reactance, in particular:- as regards the first type of problem, the determiningparameter is represented by the synchronous reactance;- as regards the second type of problem, the transitoryreactance with the relevant time constants and thesubtransitory reactance.In this paper, the static and dynamic analysis of thephenomena connected to the generator shall not be dealtwith in details, but only the following items shall bestudied and determined:- the maximum current value in the initial instants of theshort-circuit, on which depend the stresses on thewindings, on the connections generator-to-transformerand on the foundations of the alternator;- the waveform of the short-circuit current, which resultsfundamental for the proper co-ordination of theprotections in the supplied network. The short-circuitcurrent in the time-current curve presents a typicalcourse: before reaching its steady state value, it getsto higher values which progressively falls.This behaviour is due to the fact that the impedance ofthe generator, which is constituted practically by thereactance only, has no definite value, but it varies instantby instant, because the magnetic flux, which it dependson, does not reach immediately the steady stateconfiguration. A different inductance value correspondsto any configuration of the flux, mainly because of thedifferent path of the magnetic lines. Besides, there is nota single circuit and a single inductance, but moreinductances (of the winding of the armature, of thewinding of the field, of the damping circuits) which aremutually coupled. To simplify, the following parametersshall be taken into consideration:subtransient reactance, direct axis X” dtransient reactance, direct axis X’ dsynchronous reactance, direct axisThe evolution of these parameters during the timeinfluences the course of the short-circuit current in thegenerator. Reactances are usually expressed in p.u. (perunit) and in percent, that is they are related to the nominalparameters of the machine.They can be determined by the following relationship:3 I n Xx %=100V nWhere:X is the real value in ohm of the considered reactance;I nis the rated current of the machine;V nis the rated voltage of the machine.The following values can be indicated as order of quantityfor the various reactances:- subtransient reactance: the values vary from 10% to20% in turbo-alternators (isotropic machines withsmooth rotor) and from 15% to 30% in machines withsalient pole rotor (anisotropic);- transient reactance: it can vary from 15% to 30% inturbo-alternators (isotropic machines with smoothrotor) and from 30% to 40% in machines with salientpole rotor (anisotropic);- synchronous reactance: the values vary from 120% to200% in turbo-alternators (isotropic machines withsmooth rotor) and from 80% to 150% in machines withsalient pole rotor (anisotropic).X dMV/LV transformer substations: theory and examples of short-circuit calculation11

2 Calculation of short-circuit currentsTechnical Application PapersTransformerA MV/LV transformer with delta primary winding (∆) andsecondary winding with grounded star point ( ).The electrical parameters which are usually known andwhich characterize the machine are:- rated apparent power S n[kVA]- primary rated voltage V 1n[V]- secondary rated voltage V 2n[V]- short-circuit voltage in percent v k%(typical values are 4% and 6%)With these data it is possible to determine the primaryand secondary rated currents and the currents undershort-circuit conditions.The typical values of the short-circuit voltage v k%inrelation to the rated power of the transformers arereported in Table 2 (reference Standard IEC 60076-5).Table 2Rated apparent powerShort-circuit voltage2.2 Calculation of the short-circuit currentWith reference to the electrical network schematised inFigure 1, a short-circuit is assumed on the clamps of theload. The network can be studied and represented byusing the parameters “resistances” and “reactances” ofeach electrical component.The resistance and reactance values must be all relatedto the same voltage value assumed as reference valuefor the calculation of the short-circuit current.The passage from the impedance values Z 1, related to ahigher voltage (V 1), to the values Z 2, related to a lowervoltage (V 2), occurs through the transformation ratio:K = V 1V 2in accordance with the following relationship: Z 2 = Z 1K 2Figure 1netDistribution networkS n[kVA] v k%≤ 630 4630 < S n≤ 1250 51250 < S n≤ 2500 62500 < S n≤ 6300 76300 < S n≤ 25000 8TransformerCableThe operating capacitance under overload conditionsdepends on the constructional characteristics of eachsingle transformer. As general information, the operatingcapacitance of oil transformers under overload conditionscan be considered as shown in the Standard ANSI C57.92and according to the values shown in Table 3.Table 3Multiple of the rated currentof the transformerTime [s]25 211.3 106.3 304.75 603 3002 1800Asynchronous motorThe data usually known for an asynchronous motor arethe rated active power in kW, the rated voltage V nandthe rated current I n. Among the ratings also the efficiencyvalue and the power factor are available.In case of short-circuit, the asynchronous motor functionsas a generator to which a subtransient reactance from20% to 25% is assigned. This means that a current equalto 4-5 times the rated current is assumed as contributionto the short-circuit.The structure of the electrical network taken intoconsideration can be represented through elements inseries; thus an equivalent circuit is obtained as that shownin Figure 2, which allows to calculate the equivalentimpedance seen from the fault point.Figure 2R knetLoad LFaultX knet R TR X TR R C X CV EQAt the short-circuit point, an equivalent voltage source(V EQ) is positioned, with valueV EQ= c V n3The factor “c” depends on the system voltage and takesinto account the influence of the loads and of the variationin the network voltage.On the basis of these considerations, it is possible todetermine the resistance and reactance valuescharacterizing the elements which constitute theinstallation.12 MV/LV transformer substations: theory and examples of short-circuit calculation

2 Calculation of short-circuit currentsSupply network (net)In the most cases, the installation results to be suppliedby a medium voltage distribution network, whose supplyvoltage value V netand initial short-circuit current I knetcanbe easily found.On the basis of these data and of a correction factor forthe change of voltage caused by the short-circuit it ispossible to calculate the short-circuit direct impedanceof the network through the following formula:Z knet= c V net3 I knetFor the calculation of the parameters network resistanceand network reactance, the following relationships canbe used:X knet= 0.995 Z knetR knet= 0.1 X knetIf the short-circuit apparent power A knetfor the distributionnetwork were known, it would be also possible todetermine the impedance representing the networkthrough the following relationship:Z knet= V2 netS knetTransformerThe impedance of the machine can be calculated withthe nominal parameters of the machine itself (ratedvoltage V 2n; apparent power S nTR; percentage voltage dropv k%) by using the following formula:Z TR= V2 2n v k%100 S nTRThe resistive component can be calculated with the valueof the total losses P PTRrelated to the rated current inaccordance with the following relationship:R TR=P PTR3 I 2 2nThe reactive component can be determined by theclassical relationshipX TR= ( Z TR 2 – R TR 2 )Overhead cables and linesThe impedance value of these connection elementsdepends on different factors (constructional techniques,temperature, etc....) which influence the line resistanceand the line reactance. These two parameters expressedper unit of length are given by the manufacturer of thecable.The impedance is generally expressed by the followingformula:Z c= L (r c+ x c)The resistance values are generally given for a referencetemperature of 20°C; for different operating temperaturesθ with the following formula it is possible to calculate therelevant resistance value:r θ= [ 1+ (α – 20)] r 20where:α is the temperature coefficient which depends on thetype of material (for copper it is 3.95x10 -3 ).Calculation of the short-circuit currentDetermination of the short-circuit resistance andreactance values of the main elements of a circuit allowthe short-circuit currents of the installation to becalculated.With reference to Figure 2 and applying the reductionmodality for elements in series, the following values canbe determined :- the short-circuit total resistance R Tk= Σ R- the short-circuit total reactance X Tk= Σ XOnce these two parameters are known, it is possible odetermine the short-circuit total impedance value Z TkZ Tk= ( R Tk 2 + X Tk 2 )Once determined the equivalent impedance seen fromthe fault point, it is possible to proceed with thecalculation of the three-phase short-circuit current:Value of the three-phase symmetrical short-circuit currentI k3F=c V n3 Z TkZ LThis is generally considered as the fault which generatesthe highest currents (except for particular conditions).When there are no rotary machines, or when their actionhas decreased, this value represents also the steady stateshort-circuit current and is taken as reference todetermine the breaking capacity of the protection device.Z LZ LZ NI k3FMV/LV transformer substations: theory and examples of short-circuit calculation13

2 Calculation of short-circuit currentsTechnical Application PapersAn example of calculation of the three-phase short-circuit current using the above described relationship is givenhereunder.Example:With reference to the schematized network, the electricalparameters of the different components are:netMV CableTransformerMV/LVMV cableR CMV 400V= RCMT 360 10 -3= = 0.000144ΩK 2 50 2X CMV 400V= XCMT 335 10 -3= = 0.000134ΩK 2 50 2TransformerV 2Z TR= v % 2n k 400 2 4= = 0.016Ω100 S nTR 100 400 10 3p k%S nTR3P PTR= = 10 3 = 12kW100 100 400SI 2n= nTR 400 10 3= = 577A3 V 2n 3 400P PTR12000R TR= = = 0.012Ω3 I 2 2n3 577 2LV CableX TR= ( Z 2 TR–R TR2) = ( 0.016 2 – 0.012 2 ) = 0.0106ΩShort-circuit power and current of the supply networkS knet= 500MVA I knet= 14.4kARated voltage of the supply network V net= 20kVMV cable:Resistance R CMV= 360mΩReactance X CMV= 335mΩRated power of the transformer S nTR= 400kVASecondary rated voltage of the transformer V 2n= 400VShort-circuit test for the transformer: v k%=4%; p k%= 3%LV cable with length L = 5m:Resistance R CLV= 0.388mΩReactance X CLV= 0.395mΩMaking reference to the previous relationship, thecalculation of the total impedance of the differentelements is carried out in order to determine the threephaseshort-circuit current at the given point.Since the fault is on the LV side, all the parametersdetermined for the MV section of the network shall berelated to the secondary rated voltage by applying thecoefficientK =20000400Supply network1.1Z knet= c V net 20000== 0.88Ω3 I knet 3 14.4 10 3Z knet 400V= Z knet= 0.88 = 0.00035ΩK 2 50 2X knet 400V= 0.995 Z knet 400V= 0.000348Ω=50LV cableR CLV= 0.388mΩX CLV= 0.395mΩThe total short-circuit resistance value is given by: R Tk = Σ RR Tk=R knet 400V+R CMV 400V+R TR+R CLVR Tk= 0.0000348 + 0.000144 + 0.012 + 0.000388 = 0.01256ΩThe total short-circuit reactance value is given by: X Tk = Σ XX Tk=X knet 400V+X CMV 400V+X TR+X CLVX Tk= 0.000348 + 0.000134 + 0.0106 + 0.000395 = 0.01147ΩValue of the three-phase symmetricalshort-circuit currentCalculating the value of the total short-circuit impedanceZ Tk= ( R 2 Tk+X Tk2) = ( 0.01256 2 + 0.01147 2 )= 0.017Ωand assuming the factor c (1) = 1.1 the short-circuit current value is:c V 2n 1.1 400I k3F= = = 14943A = 14.95kA3 3 0.017Z TkFor more detailed information and considerations about shortcircuitcurrent calculation, see the “Annex B” of this paper.(1)The voltage factor “c” is necessary in order to simulate the effect ofsome phenomena which are not explicitly considered in the calculation,such as for example :- the voltage changes in time- the changes of transformer taps- the subtransient phenomena of the rotary machines (generators and motors).R knet 400V= 0.1 X knet 400V= 0.0000348Ω14 MV/LV transformer substations: theory and examples of short-circuit calculation

2 Calculation of short-circuit currents2.3 Calculation of motor contributionIn case of short-circuit, the motor begins to function asa generator and feeds the fault for a limited timecorresponding to the time necessary to eliminate theenergy which is stored in the magnetic circuit of the motor.By an electrical representation of the motor with itssubtransient reactance “X”, it is possible to calculate thenumerical value of the motor contribution. This datum isoften difficult to find; therefore the general rule is toconsider motor contribution as a multiple of the ratedcurrent of the motor. The typical values of the multiplyingfactor vary from 4 to 6 times.For a LV motor, with reference to the length of time, theeffect of the contribution to the short-circuit currentresults to be negligible already after the first periods fromthe start of the short-circuit. The Standard IEC 60909 orCEI 11-28 prescribes the minimum criteria for taking intoconsideration the phenomenon; it shall be:( Σ I > I knM100 )where:ΣI nMrepresents the sum of the rated currents of themotors directly connected to the network where theshort-circuit has occurred. I kis the three-phase shortcircuitcurrent determined without motor contribution.2.4 Calculation of the peak current valueThe short-circuit current “I k” may be considered asformed by two components:• a symmetrical component “i s” with sinusoidalwaveform and precisely symmetrical with respectto the x-axis of times. This component is expressedby the following relationship:i s= 2 I ksen (ω t – ϕ k)• the unidirectional component “i u” with exponentialcurve due to the presence of an inductivecomponent. This component is characterized by atime constant τ=L/R (“R” indicates the resistanceand “L” the inductance of the circuit upstream thefault point) and dies out after 3 to 6 times τ.i u= 2 I ksenϕ ke L R tThe unidirectional component during the transient periodmakes that the asymmetrical short-circuit current ischaracterized by a maximum value called peak value,which results to be higher than the value to be due to apurely sinusoidal quantity. Generally speaking it ispossible to state that, if considering the r.m.s. value ofthe symmetrical component of the short-circuit currentI k, the value of the first current peak may vary from toFigure 330000250002000015000100005000[ms]00 10 20 30 40 50 60 70 80 90 100-5000-10000-15000-20000[A]i si uI k2 I ka 2 2 I k.After the transient period has elapsed, the short-circuitcurrent practically becomes symmetrical. The currentcurves are shown in Figure 3.As known, the performances of a circuit-breaker undershort-circuit conditions, making reference to theoperating voltage of the device, are mainly defined bythe following parameters:Icu = breaking capacityIcm = making capacityThe breaking capacity Icu is defined with reference tothe r.m.s. value of the symmetrical component of theshort-circuit current. It is possible to say that the r.m.s.value of a sinusoidal current represents that direct currentvalue which, in an equal time, produces the same thermaleffects. The sinusoidal quantities are generally expressedthrough their r.m.s. value. As r.m.s. value it is possible toconsider that short-circuit current value which can benormally calculated by the classical relationship:I k =V(R 2 + X 2 )The making capacity Icm is defined with reference to themaximum peak value of the prospective short-circuitcurrent.MV/LV transformer substations: theory and examples of short-circuit calculation15

2 Calculation of short-circuit currentsTechnical Application PapersSince each element with an impedance modifies theshort-circuit current on the load side, and since a circuitbreakeris an element with an impedance of its own, theprospective current is defined as the current flowing whenthe protection device is replaced by an element with nullimpedance.or through the following diagrams which show the valueof “k” as a function of the parameter “R/X” or “X/R”.a)2.0The product Standard IEC 60947-2 gives a table allowingto pass from the r.m.s. value of the short-circuit currentto its relevant peak value, through a multiplicativecoefficient linked also to the power factor of theinstallation. This table is the necessary reference todetermine the Icu and Icm values of the various circuitbreakers.When passing from the characteristics of the circuitbreakersto those of the installation, whereas calculatingthe r.m.s. value of the symmerical component of thecurrent results immediate, determining the relevant peakvalue could be less immediate. The necessaryparameters, such as the short circuit power factor or theratio between the resistance and the inductance of thecircuit on the load side of the fault point, are not alwaysavailable.The Standard IEC 60909 gives some useful informationfor the calculation of the peak current and in particularreports the following relationship:i p = k 2 I kwhere the value of “k” can be evaluated with thefollowing approximate formula:k = 1.02 + 0.98 e-3 RXkb)k1.81.61.41.21.02.01.81.61.41.20 0.2 0.4 0.6 0.8 1.0 1.2R/X1.00.5 1 2 5 10 20 50 100 200X/RExample:Assuming an r.m.s. value of the symmetrical component of the three-phase short-circuit current I k=33kA and a peakvalue under short-circuit conditions (cosϕ k=0.15), it is possible to see how to proceed in order to determine the peakvalue:from the value of cosϕ kit is possible to make the ratio X/R explicit through the tangent calculation.After calculating the ratio X/R = 6.6, through the graph or the formula it is possible to find the value of k = 1.64, whichgives a peak value Ip=76.6kA in correspondence with the three-phase short-circuit current I k=33kA.Considering the need to choose a protection device for an installation at 400V rated voltage, with reference to thethree-phase short circuit current only, a circuit-breaker with breaking capacity Icu=36kA could be used, to which amaking capacity Icm=75.6kA would correspond, in compliance with the Standard IEC 60947-2. Such making capacityresults to be lower than the peak value which can be made in the installation considered; thus the choice results tobe incorrect and forces the use of a circuit-breaker version with higher breaking capacity (for example 50 kA) andconsequently Icm greater and suitable for the peak value of the installation.From the example above it is possible to see how at first a circuit-breaker, version “N” (that is with 36 kA breakingcapacity) would have been chosen incorrectly; on the contrary the considerations regarding the peak value shalllead to use a circuit-breaker version “S” or “H”.16 MV/LV transformer substations: theory and examples of short-circuit calculation

3 Choice of protection and control devices3.1 Generalities about the main electricalparameters of protection and controldevicesGenerally speaking, when it is necessary to analyse andselect a protection and control device such as a circuitbreaker,some electrical parameters characterizing thedevice itself shall be evaluated, for example rated currentand breaking capacity.Hereunder a brief description of these parameters isgiven, relating them with the electrical quantities of theinstallation.Rated operational voltage Ue: it is the value of voltagewhich determines the application limit of an equipmentand to which all the other parameters typical of theequipment are referred to. It is generally expressed asthe voltage between phases.Rated uninterrupted current Iu: it is the value of currentwhich the device is able to carry for an indefinite time(weeks, months, or even years). This parameter is usedto define the size of the circuit-breaker.Rated current In: it is the value of current whichcharacterizes the protection release installed on boardof the circuit-breaker and determines, based on thesettings available for the release, the protectivecharacteristic of the circuit-breaker itself. Such currentis often related to the rated current of the load protectedby the circuit-breaker.Rated ultimate short-circuit breaking capacity Icu: it isthe r.m.s. value of the symmetrical component of theshort-circuit current which the circuit-breaker is able tomake. Such value is established through a clearly definedtest cycle (O-t-CO) and specified test modalitiesdescribed in the product standard IEC 60947-2. Thecircuit-breakers are classified according to their performancelevels identified with letters (“N” “S” “H” “L” etc.)referred to their breaking capacity.Rated service short-circuit breaking capacity Ics: it is ther.m.s. value of the symmetrical component of the shortcircuitcurrent which the circuit-breaker is able to make.Such value is established through a clearly defined testcycle (O-t-CO-t-CO) and specified test modalitiesdescribed in the product standard IEC 60947-2.It is expressed as a percentage 25% - 50% - 75% -100% of the rated ultimate short-circuit breakingcapacity, for example it could be Ics = 75 % Icu.The value of the breaking capacity must be put intorelation with the short-circuit current value at theinstallation point of the circuit-breaker itself and therelationship Icu>I kor Ics>I kmust be verified.Rated short-circuit making capacity Icm: it is themaximum prospective peak current which the circuitbreakermust be able to make. In alternate current, therated making capacity of a circuit-breaker under shortcircuitconditions shall not be lower than its rated ultimateshort-circuit breaking capacity multiplied by thefactor “n”, thus being Icm=n x Icu.Such value of Icm shall be put into relation with the peakvalue of the current measured in the installation point ofthe circuit-breaker and the relationship Icm>i pmust beverified.Table 1 below shows the values of coefficient “n” asspecified in the product Standard IEC 60947-2.Table 1BreakingPowercapacity I cufactor n4.5 ≤ I cu≤ 6 0.7 1.56 < I cu≤ 10 0.5 1.710 < I cu≤ 20 0.3 220 < I cu≤ 50 0.25 2.150 < I cu0.2 2.2Rated short-time withstand current Icw: it is the r.m.s.value of the alternate current component which thecircuit-breaker is able to withstand without damages fora determined time, preferred values being 1s and 3s.MV/LV transformer substations: theory and examples of short-circuit calculation17

3 Choice of protection and control devicesTechnical Application PapersMoulded-case circuit-breakersfamilycircuit breakerrated service current (Ue)rated uninterrupted current (Iu)rated ultimate short-circuit breaking capacity (Icu)(AC) 50-60 Hz 220/230V(AC) 50-60 Hz 380/415V(AC) 50-60 Hz 440V(AC) 50-60 Hz 500V(AC) 50-60 Hz 690Vrated service short-circuit breaking capacity (Ics)(AC) 50-60 Hz 220/230V(AC) 50-60 Hz 380/415V(AC) 50-60 Hz 440V(AC) 50-60 Hz 500V(AC) 50-60 Hz 690Vrated short-circuit making capacity (Icm)(AC) 50-60 Hz 220/230V(AC) 50-60 Hz 380/415V(AC) 50-60 Hz 440V(AC) 50-60 Hz 500V(AC) 50-60 Hz 690VB25161083100%100%100%100%100%52.5321713.64.3T1690160C40251510475%100%75%75%75%8452.530175.9N50362215675%75%50%50%50%10575.646.2309.2N653630256100%100%100%100%100%14375.66352.59.2S855045307100%100%100%100%100%18710594.56311.9T2690160H1007055368100%100%100%100%100%22015412175.613.6L12085755010100%75% (1)75%75%75%26418716510517N50362520575%75%75%75%75%10575.652.5407.7T3690250TmaxS85504030850%50% (2)50%50%50%187105846313.6(1) 70kA (2) 27kA (3) 75% for T5 630 (4) 50% for T5 630 (5) only for T7 800/1000/1250 AAir circuit-breakersfamilycircuit breakerrated service current (Ue)performance levelrated uninterrupted current (Iu)B630800100012501600X1690N630800100012501600L63080010001250B800100012501600E1690N800100012501600B16002000N1000125016002000E2690S8001000125016002000EmaxL12501600rated ultimate short-circuit breaking capacity (Icu)(AC) 50-60 Hz 220/230/380/415 V(AC) 50-60 Hz 440V(AC) 50-60 Hz 500/525V(AC) 50-60 Hz 660/690Vrated service short-circuit breaking capacity (Ics)(AC) 50-60 Hz 220/230/380/415 V(AC) 50-60 Hz 440V(AC) 50-60 Hz 500/525V(AC) 50-60 Hz 660/690Vrated short-circuit making capacity (Icm)(AC) 50-60 Hz 220/230/380/415 V(AC) 50-60 Hz 440V(AC) 50-60 Hz 500/525V(AC) 50-60 Hz 660/690Vrated short-time withstand current (Icw)(1) the performance at 600V is 100kA.(1s)(3s)424242424242424288.288.288.288.242656555555050424214314312112142150130100601501301004533028622013215424242424242424288.288.275.675.64236505050505050505010510575.675.65036424242424242424288.288.2848442426565555565655555143143121121554285856565858565651871871431436542130110858513011065652862421871871018 MV/LV transformer substations: theory and examples of short-circuit calculation

3 Choice of protection and control devicesN7036302520S8550403025T4690250/320H10070655040L2001201008570V30020018015080N7036302520S8550403025T5690400/630H10070655040L2001201008570V30020018015080N7036302520T6690630/800/1000S8550453522H10070505025L200100806530S8550504030T7690800/1000/1250/1600H10070655042L2001201008550V (5)20015013010060100%100%100%100%100%100%100%100%100%100%100%100%100%100%100%100%100%100%100%100%100%100%100%100%100%100%100%100%100%100%100%100%100%100%100%100%100%100%100%100% (3)100%100%100%100% (3)100% (4)100%100%100%100% (4)100% (4)100%100%100%100%75%100%100%100%100%75%100%100%100%100%75%75%75%75%75%75%100%100%100%100%100%100%100%100%100%75%100%100%100%75%75%100%100%100%100%75%15475.66352.540187105846352.52201541431058444026422018715466044039633017615475.66352.540187105846352.52201541431058444026422018715466044039633017615475.66352.54018710594.573.548.42201541051055544022017614366187105105846322015414310588,2440264220187105440330286220132N2500320065656565656565651431431431436565S10001250160020002500320075757575757575751651651651657565E3690H80010001250160020002500320010010010085 (1)858585852202201871877565V8001250160020002500320013013010010010010085852862862202208565L200025001301108585130110656528624218718715S400075757575757575751651651651657575E4690H3200400010010010085 (1)1001001008522022022018710075V3200400015015013010015015013010033033028622010075H40005000630010010010010010010010010022022022022010085E6690V3200400050006300150150130100125125100100330330286220100853.2 Criteria for the circuit-breaker choiceThe various choice criteria for a circuit-breaker impose,in addition to a verification of the typical electricalparameters of the circuit-breaker (voltage – current –breaking capacity etc.), also the verification of the circuitbreakerability to protect the devices which it has beenassigned to.Below there is a brief analysis of the verification modalitiesto be followed in order to obtain the protection of thedevices which are most commonly used in an installation.Protection of the feedersThe cable shall be protected against overload and shortcircuit.As regards protection against overload, the followingcondition shall be verified I B≤ I 1≤ I Zwhere:I Bis the load current,I 1is the overload tripping threshold (function “L”) set onthe protection release;I Zis the continuous current carrying capacity of the cable.As regards protection against short-circuit, the followingcondition shall be verified K 2 S 2 ≥ I 2 twhere:K 2 S 2 is the specific energy which can be withstand bythe cable and which results to be a function of the crosssection S and of a constant K, which is equal to 115 forPVC insulated cables and 143 for EPR insulated cables.I 2 t is the specific let-through energy of the circuit-breakerin correspondence with the maximum short-circuitcurrent of the installation.MV/LV transformer substations: theory and examples of short-circuit calculation19

3 Choice of protection and control devicesTechnical Application PapersMaximum protected lengthFor the secondary circuit of type TN-S on the LV side,the Standard IEC 60364 gives some indications for anapproximate calculation to evaluate the minimum shortcircuitcurrent at end of cable. This Standard assumesthat the minimum fault current condition occurs in caseof a phase-to-neutral fault at end of the conductor.The established difference depends on whether theneutral conductor is distributed or not, and the calculationformulas are as follows:TN-S neutral conductor not-distributedI kmin = 0.8 V S F1.5 ρ 2 LTN-S neutral conductor distributed0.8 VI kmin =0 S F1.5 ρ (1 + m) Lwhere:0.8 – 1.5 – 2 characteristic constants of the formulaunder considerationVphase-to-phase voltage of the systemV 0phase-to-neutral voltage of the systemS Fcross section of the phase conductorρresistivity of the conductive material of thecablemratio between the resistance of the neutralconductor and that of the phase conductor.In the quite common case in which phaseand neutral conductors are made of thesame material, “m” becomes the ratiobetween the phase and the neutral crosssectionsLcable length in meters [m]I kminminimum short-circuit current at end ofcable.If, in the formulas above, the value I kminis replaced by thetripping threshold I3 Maxinclusive of higher tolerance ofthe used circuit-breaker and the formula is solved bymaking the length explicit, the result obtained indicativelygives the value of the maximum cable length which resultsto be protected by the magnetic threshold setting on theprotection device.The formulas thus obtained are:L Max =L Max =0.8 V 0 S F1.5 ρ (1 + m) I3 Max0.8 V S F1.5 ρ 2 I3 MaxProtection against indirect contactProtection against indirect contact consists in protectinghuman beings against the risks deriving from touchingexposed conductive parts usually not live, but withvoltage presence due to a failure of the main insulation.Protection by automatic disconnection of the supply isrequired when, due to a fault, contact voltages can occuron the metallic frame for a time and value such as to bedangerous for human beings.The measures to obtain protection against indirectcontact for LV installations are prescribed by the StandardCEI 64-8, whereas for MV installations the referenceStandard is CEI 11-1.For a verification of protection in LV systems, theStandard gives some prescriptions which differ basedon the various distribution systems and refer to the faultloop impedance, to the voltage, to the current whichcauses the trip of the protection device and to the timeby which the device trips.In MV systems, the problem of protection against indirectcontact occurs whenever the user plant has its owntransformation substation. In compliance with theStandard CEI 11-1, the ground current I gcan becalculated through the relationshipI g= V . (0.003 . L1 + 0.2 . L2)where L1 represents the extension of the overhead lineand L2 that of the cable.The value of the current to earth is difficult to evaluate,therefore it has to be asked and assigned by themanufacturer.The Standard gives the maximum value which the stepvoltage and the touch voltage can reach based on thefault elimination time.Protection of generatorsWith reference to the typical representation of the shortcircuitcurrent of a generator, for a good protection ofthe rotary machine the protection device shall have thefollowing characteristics:- setting of the overload protection L equal or higherthanthe rated current of the generator;- tripping of the short-circuit protection (instantaneous Ior delayed S) in the very first instant of the short-circuit;- protection related to the overcurrent withstandcapability of the machine which, according to theStandard IEC 60034-1 is given by the point 1.5xI nGfor30s where I nGis the rated current of the generator.Protection of transformersA LV/LV transformer is now taken into consideration inorder to analyze the characteristics which the protectiondevices must have when located upstream ordownstream the transformer.As regards the circuit-breaker upstream, it is necessaryto make reference to the magnetizing curve of themachine; its curve shall have no intersection with thecircuit-breaker tripping curve. The breaking capacity mustbe adequate to the short-circuit current of the networkupstream the transformer.The downstream circuit-breaker shall have a trippingcharacteristic such as to guarantee protection against20 MV/LV transformer substations: theory and examples of short-circuit calculation

3 Choice of protection and control devicesan extreme thermal overload capacity of the machine incase of short-circuit. The Standard IEC 60076-5 indicatesas a limit to the thermal stresses due to a short-circuit(overload threshold) the short-circuit current value letthroughby the transformer for 2s.This condition shall be verified also by the upstreamcircuit-breaker in case of a short-circuit on the secondaryside not affecting the downstream circuit-breaker. Forthis analysis the current referred to the primary side mustbe obviously considered, being this the current really seenby the upstream circuit-breaker.Generally, the analysis of the behaviour of thedownstream and upstream circuit-breakers for a fault onthe secondary side shall be carried out on the basis ofthe real currents affecting the two apparatus; as a matterof fact, the short-circuit current due to a fault on thesecondary side shall be related to the primary throughthe transformation ratio.Motor co-ordinationGoing into the details of the considerations related tothe study of the motor co-ordination is quite complicatedand it is not specific subject of this paper. Generallyspeaking, the circuit-breaker for motor protection is onlyof magnetic type; the magnetic threshold shall have sucha value to allow the inrush current to be absorbed withoutunwanted trips and besides, in case of short-circuits onthe motor, to protect the contactor (switching device) andthe external thermal release (overload protection).ABB offers some co-ordination tables (available on thewebsite http://www.abb.com/lowvoltage) for variousmotor powers and for various start-up typologies relatedto the voltage and the short-circuit current of the plant.3.3 Co-ordination between circuit-breakersand switch-disconnectorsDisconnection must guarantee putting out of service ofthe whole installation or of part of it, by separating it safelyfrom any power supply, thus guaranteeing safety for thehuman beings operating on it. Disconnection shall becarried out with devices which open all the poles in asingle operation. As regards MV applications, theincoming line in the substation can be provided with aline switch-disconnector and an earth-connected switchdisconnector,which are interlocked one to the other; theyare used, in case of maintenance operations, to put toearth automatically the upstream line when the lineswitch-disconnector is opened.On the LV side, the switch-disconnector could be theincoming element into a secondary switchboard, asrepresented in Figure 1. The disconnector is a switchingequipment, which in its open position guarantees acertain isolation distance between the contacts. Ingeneral, the switch-disconnector is suitable for openingor closing circuits where little currents - or howevercurrents of the order of the device rated current - flow,and it is not provided with a protection release.Figure 1QLV1Since the switch-disconnector is not provided with adevice operating its opening, it is necessary that aprotection device is present, for example a circuit-breakerto safeguard the integrity of the switch-disconnector incase of short-circuit. In case of short-circuit, this meansthat the electrical phenomena affecting the switchdisconnectorand conditioned by the circuit-breakerbehaviour must be withstand also by the switchdisconnectoritself.To this purpose, ABB puts some tables at disposal; fromthese tables, making reference to the type of circuitbreakerand of switch-disconnector respectively on thesupply and on the load side, it is possible to find themaximum short-circuit current at which this combinationresults to be protected.Tables 2 and 3 (extracted by the documentation ABBSACE “Co-ordination Tables”) are shown below with anexample of how to read it.Moulded-case circuit-breaker and switch-disconnectorderived by the moulded-case circuit-breakers:Table 2UpstreamB16T1 C 25 160N 36N 36T2 S 50 160H 70L 85Circuit-breakerQLV2DownstreamIcw [kA]Ith [A]Version Icu [kA]Iu [A]T1D2CableSwitch-disconnectorT3D T4D T5D3.6 3.6 6250 320 400 630With reference to the highlighted combination betweena circuit-breaker type T2S160 on the load side and aswitch-disconnector type T1D160, the protection of thedisconnector is possible up to a short-circuit value equalto 50kA 400Vac.1601625363650708516253636507085162536365070851625363650708516253636507085MV/LV transformer substations: theory and examples of short-circuit calculation21

3 Choice of protection and control devicesTechnical Application PapersMoulded-case circuit-breaker and switch-disconnectorOT and OETLTable 3UpstreamReleaseDownstreamIcw [kA]Ith [A]Iu [A]320OETL2008250100*OETL250OETL3158 8315 350100*100*3.4 Co-ordination between automaticcircuit-breakers and residual currentdevices (RCDs)Residual current devices generally used in the terminalpart of a LV installation guarantee effective protectionagainst indirect contact, that is contact with metallic partswhich should be normally not live, and under determinedconditions also against direct contact, that is contactwith parts normally live.T5TMEL400630320-630100*/***100*/**100*100*/**100*100***100*/**With reference to the highlighted combination betweena circuit-breaker type T5..400 on the load side and aswitch-disconnector type OETL200, the protection of thedisconnector is possible up to a short-circuit value equalto 100kA 400Vac.As regards the asterisks in the Table, the following notesare valid:* Select the lower value between the Icu of the circuitbreakerand the value shownFor example, if the circuit-breaker is version N withIcu=36kA @ 400Vac, this means that the maximum shortcircuitcurrent of the installation shall be lower than 36kA(to use version “N”) and the switch-disconnector shallbe surely protected since the protection limit is 100 kA.If the circuit-breaker version is L, with Icu=200kA @400Vac, this means that the maximum short-circuitcurrent of the installation shall be 200kA and the switchdisconnectorshall be protected since the protection limitis 100kA.*** I1 = 0.7 x IFrom this note, linked to the thermal protection of theswitch-disconnector, it results that the maximum settingfor the thermal protection of the circuit-breaker is 0.7xIn.Analogous meaning has the note:** Maximum setting of the overload threshold PR2xx =1.28*Ith OTxx/OETLxxfrom which it can be understood why the maximumsetting for the overload protection of the circut-breakershall not exceed 1.28 times the current carrying capacityof the switch-disconnector.However, from a careful reading of the Standards, itresults clear that the protection function of human beingsagainst direct and indirect contacts is an auxiliary functionwhich the circuit-breaker carries out, since the electricalinstallation must be designed and built so that the safetyof human beings is guaranteed chiefly through anadequate earthing system.Therefore, the metallic frameworks of the loads must beconnected to an earthing system properly sized, so thatdangerous contact voltages are avoided in everysituation.In an installation, besides the normal protection againstoverload and short-circuit, which are usually demandedto the thermomagnetic circuit-breaker, it is a good ruleto provide also a residual current protection.In a wide sense, protection in a plant can be carried outby two separate devices (thermomagnetic circuit-breakerand residual current device); in this case, the RCD, whichis sensitive only to the earth fault current, shall be installedin series with a circuit-breaker which protects it againstthe thermal and dynamic stresses developing in theinstallation due to an overcurrent.An alternative is represented by a single device as thethermomagnetic residual current circuit-breaker whichunifies in a single device the protection functions, thusconsiderably reducing the possible risks deriving froman incorrect co-ordination between the two apparatus.22 MV/LV transformer substations: theory and examples of short-circuit calculation

3 Choice of protection and control devices3.5 Example of study of a MV/LV networkHereunder there is an example of how the analysis of aMV/LV installation can be carried out to evaluate the mainelectrical parameters of the network and to select thecircuit-breakers for the protection and the properhandling of the installation, with reference to protectionselectivity.Description of the installation characteristics:Distribution network:rated voltageV 1n=20kVshort-circuit currentI kMV=12.5kAThe scheme of the installation analyzed is :Figure 2MV line of the utility companyMV protection device of theutility companyTake-up point of the user in thedelivery roomMV connection cable from thedelivery room to the user roomSubstation transformer with the following data :primary rated voltage: V 1n=20kVsecondary rated voltage: V 2n=400Vrated power:S nTR=800kVArated short-circuit voltage: v k%=5%An installation normally provides supply for differentloads; to simplify and finalize the treatment of this subject,the following load typologies are now taken intoconsideration:a passive load L with:rated power P nL=300kWpower factor cosϕ = 0.8LV busbarProtection device MV userMV/LV transformerLV general protection devicesupplied by a cable C having the following characteristics:formation 2x(3x240)mm 2current carryingcapacity of the cable I Z=590Alength L C=15man active load M (three-phase asynchronous motor) with:rated power P nM=90kWcoefficient η x cosϕ = 0.8(efficiency per power factor)LLV loadsIn order to deal with the verification of the trippingcharacteristics of protections as best as possible, theelectrical characteristics of the different components ofthe installation under study are analyzed hereunder.MDistribution network:Starting from the power supply source, that is from the electrical network at 20kV owned by the utility company andhaving its own MV protection device usually characterized by independent time tripping curves with two steps, thefollowing hypothetical but realistic values can be assumed for the protection of the utility company at 20kV:Maximum currentProtections 51 (first threshold) 50 (second threshold)Fault elimination time < 0.5s < 0.25sCurrent setting values < 72A < 400AZero-sequence maximum currentProtections51N (first threshold)Fault elimination time < 0.12sCurrent setting values< 4AMV/LV transformer substations: theory and examples of short-circuit calculation23

3 Choice of protection and control devicesTechnical Application PapersTransformer:The second element affecting the electrical parameters of the installation, above all on the LV side, is the 800kVAtransformer, which has the following current values:- primary rated current (20kV side): I 1n=S nTR3 V 1n=800 x 10003 x 20 x 1000 = 23.1A- secondary rated current (400V side): I 2n=S nTR3 V 2n=800 x 1000= 1155A3 x 400By practical and quick formulas (for example by assuming the MV network on the load side with infinite short-circuitpower), which give results acceptable as first approximation and which are useful to evaluate the intensity of thecurrents really present in the installation and the dimensioning of the protection devices, the short-circuit currentscan be calculated:- three-phase short-circuit current on the secondary side (400V side)S nTR 1 800 x 1000 1I 2k3F= x 100 x =x 100 x = 23kAV k%3 xV 2n 53 x 400To this three-phase short-circuit current expressed as symmetric r.m.s. value, we associate also a short-circuitpower factor cosϕ k=0.35 and the relevant peak value equal to 43.6kA.- three-phase short-circuit current related to the MV side because of a fault on the LV sideS nTR 1 800 x 1000 1I 1k3F= x 100 x =x 100 x = 462A3 xV 1n 53 x 20 x 1000V k%or calculable by the relationship: I 1k3F= I 2k3FV 1nV 2n=2300020000400 = 460AThe functioning of the transformer can be represented through its magnetizing curve and through the maximumshort-circuit withstand capacity considered from the thermal point of view.The magnetizing curve of the transformer can be obtained through the formula:i inrush= k iI 1nTR2etτinrush, for further details see Annex A of this paper.The short-circuit withstand capacity considered from the thermal point of view can be expressed as indicated in theStandard IEC 60076-5 as the capacity of the transformer to withstand for 2 seconds the short-circuit current of thetransformer itself.In Figure 3 there is a representation of the time/currentdiagram with the following curves:Curve 1: tripping curve of the MV overcurrent protection;Curve 2: characteristic curve of the electrical parametersof the transformer described above.All the curves are represented at the reference voltage of400V of the LV side; as a consequence the current valuesrelated to the voltage of 20kV of the MV side must bemultiplied by the transformation ratio 20000V/400V.Figure 31E3s100s10s1sCurve 1I k x 2s0.1s1E-2sCurve 21E-3s0.1kA 1kA 10kA 100kA24 MV/LV transformer substations: theory and examples of short-circuit calculation

3 Choice of protection and control devicesPassive load L- rated current of the load:I nL=P nL x 1000 300 x 1000== 541A3 xV 2n x cosϕ 3 x 400 x 0.8Active load M- rated current of the motor:I nM=P nM x 100090 x 1000== 160A3 xV 2n x η x cosϕ 3 x 400 x 0.8- short-circuit current of the motor:I kM= 4 x I nM= 4 x 160 =640AConsidering the size and the limited duration of thisphenomenon, the short-circuit current calculated withoutmotor contribution shall be used as short-circuit currentvalue at the busbar.The study of the co-ordination and of the selection ofMV and LV protections under the competence of thecustomer, can start by analyzing the characteristic andthe tripping values imposed by the utility company (curve1). These information, as already said, are usuallyreported in the supply contract agreement and definethe field of action for the setting the MV protection of theuser towards the supply side.Towards the load side, the limit for the protection MV useris given by the waveform of the magnetizing current ofthe transformer (curve 2).Considering a user installation having on the load sideof the protection device MV usera cable with a length suchas the MV/LV transformation unit results to be located ina unique room, the MV protections which can be usedcould be constituted by a maximum current protection(51) and by a maximum positive-sequence currentprotection (51 N).thereshold of the utility company, and downstream withthe LV general protection, guaranteeing also theprotection functions of its own competence.Generally, to the two thresholds previously identified forthe protection MV userthe following protection functionscan be assigned:• protection against the transformer overload, not strictlynecessary if already provided by the circuit-breaker onthe LV side or by other dedicated devices, such as forexample thermometric equipment which control thetemperature inside the machine through thermalprobes;• protection against short-circuits on the secondary ofthe transformer on the supply side of the LV circuitbreaker;• protection against short-circuits on the MV sidepertaining to the user, with instantaneous trip;• protection against overload for the cable constitutingthe connection between the take-up point from thedelivery room and the protection device MV user.Trying to comply with the above described conditions,here is an indication of the values to be set for theprotection device MV user. The selected values can be thefollowing ones and form the curve 3 represented in thediagram of Figure 4.threshold with low currents I>65A - 0.4s related to 20kVwhich corresponds with 65x20000/400=3250Athreshold with high currents I>>360A - 0.2s related to 20kVwhich corresponds with 360x20000/400=18kAFigure 4MV protection device of the user (MV user )The overcurrent protection on the MV side of the userhas usually two tripping thresholds:- one with low currents and which can be identified withthe overload protection, also indicated with I>- the other one with high currents and which can beidentified with the short-circuit protection, alsoindicated with I>>1E3s100s10s1sCurve 3Curve 1Curve 4I k x 2sThe setting values of currents and times for eachthreshold shall be set, whenever possible, at a level lowerthan the protections of the utility company; it is alsoimportant not to stay “too low with the settings” so thatthere are no intersections with the magnetizing curve ofthe transformer, so that there is no trip when thetransformer itself is put into service and so that the spacefor the positioning of the tripping curves of LV protectionsremains free.Of course, this means that the protection MV usershall beselective upstream with respect to the protection0.1s1E-2s1E-3sCurve 20.1kA 1kA 10kA 100kAPutting into relation the curves of the protection devicesand their relevant short-circuit currents, the diagram ofMV/LV transformer substations: theory and examples of short-circuit calculation25

3 Choice of protection and control devicesTechnical Application PapersFigure 5 is obtained, where curve 4 represents the shortcircuitcurrent value, on the LV side, affecting the MVdevices.Figure 51E3s100s10s1s0.1s1E-2s1E-3sCurve 3Curve 1Curve 2Curve 4I k x 2s0.1kA 1kA 10kA 100kAFrom the course of the time/current curves it results that:- the tripping curves of the device of the utility company(curve 1) and of the user (curve 2), do not overlap inthe whole current range and in particular incorrespondence with the short-circuit current on thesecondary winding side (curve 4); therefore, withoutconsidering the tripping tolerances typical of eachdevice, it is possible to state that, in the given example,selectivity is guaranteed between the two devices. Ifthere were no selectivity, the two MV circuit-breakerswould open simultaneously, and the MV circuit-breakerof the utility company would restart service through itsrapid reclosing, remaining closed because in themeantime the fault has been extinguished by theopening of the circuit-breaker MV user.- both MV circuit-breakers do no intersect themagnetizing curve of the transformer.Thanks to these considerations, the MV protection curvescan be held to be set properly and it is possible then toproceed with selection and setting of the LV devices.LV general protection deviceWith reference to the short-circuit current valuespreviously defined (I 2k3F=23kA i p=43.6kA) and to thesecondary rated current of the transformer (I 2n=1155A)the LV general circuit-breaker shall have:- a breaking capacity “Icu” related to the voltage on theLV side, greater than the r.m.s. short-circuit currentvalue on the LV busbar (Icu>I 2k);- a making capacity “Icm” higher than the peak value ofthe short-circuit current on the LV busbar(Icm>i p);- a rated uninterrupted current “Iu”, suitable for themaximum current of the installation, coinciding withthe rated current of the transformer secondary winding;- a size which, through proper settings, guaranteesselectivity with the MV protection device upstream andwith the circuit-breakers provided for the loadsdownstream.With reference to the electrical parameters thus calculated,the circuit-breaker to be used could be a moulded-casecircuit-breaker Tmax series T7S1250 PR332-LSI In1250,with Icu=50kA at 400V and Icm=105kA.Protection device for the passive load LThe selection of this device shall be made makingreference to:- short-circuit current value at the installation point; sincethe limitation imposed by the cable is negligible, theshort-circuit value of the LV busbar is considered,disregarding the limitation of the cable. ThereforeI 2k3F=23kA and Icu shall be > I 2k3F;- a rated uninterrupted current “Iu”, suitable for the loadmaximum current;- a size which, through proper settings, allows cableprotection to be obtained:- against overload I B

3 Choice of protection and control devicescombination T7S1250 PR332-LSI In1250 - T5N 630PR221DS-LS/I In630 allows total selectivity (indicatedwith “T”) to be guaranteed up to the lowest breakingcapacity between those of the circuit-breakers used,which is equal to 36 kA of T5N.- for current values between 1.05 and 1.3 times I1,the product Standard does not prescribe a definitebehaviour for the circuit-breaker, even if normallythe circuit-breaker tripping occurs without the timebeing exactly known.Table 4DownstreamT5VersionN,S,H,L,VUpstreamT6N,S,H,LT7S,H,LRelease TM EL ELTMELI u[A]400630400630I n[A]32040050063032040063080012501600800 800 1000 1250 1600Once the circuit-breaker sizes have been identified, amore detailed study shall be carried out, to define theproper settings and find a confirmation for the choicesmade.The first step is to analyze the settings of the LV maincircuit-breaker. The protection settings of these devicesare conditioned by the following factors:a)course of the curve 2, previously determined for thecircuit-breaker MV user;b) protection against transformer overload;c) search for selectivity towards the circuit-breakerdownstream.303030303030303030TTTTTTTTTTTTTTTTTTTTTAccording to this behaviour, which is accepted by theproduct Standards, if the setting of the protection releasehas a value I1 = I 2nof the transformer, the situation shallbe as follows:• I < 1.05 x I1: non-tripping guaranteed, with theconsequent 5% overload for the transformer;• 1.05 x I1 < I < 1.3 x I1: tripping time not defined, andconsequently in the worst hypothesis, the transformercould be subject to an overload up to 30% for 2 hours(even if the circuit-breaker usually trips in much shortertimes);• I > 1.3 x I1: tripping of the protection guaranteed incompliance with the times of the characteristic curve.As regards item “c”, in order to get the selectivity valuepreviously determined, it is necessary that the functionof protection against instantaneous short-circuit I3 is setin OFF.Based on these considerations, Figure 6 reports the time/current diagram showing how curve 5 and curve 3 areselective.In this diagram the settings assumed for the LV maincircuit-breaker are the following:L (overload; protection I1-t1):I1=0.925xIn=1156.25AS (delayed short-circuit; protection I2-t2):I2=2xIn=2500AI (instantaneous short-circuit; protection I3):t1=18st2=0.1sOFFIn particular, with reference to the point b), the followingconditions shall be complied with:• the trip in correspondence with the short-circuit currentfor a time lower than 2 seconds (thermal ability of thetransformer to withstand short-circuit);• the setting of the protection against overload shall bemade taking into consideration the fact that productStandards CEI EN 60947-2 and IEC60947-2 prescribefor the circuit-breaker, as tripping characteristic underoverload conditions, the following behaviour:- from the cold state, non-tripping in less than theconventional time (2 hours) shall be guaranteed forcurrent values equal to 1.05 x I1 (I1 is the currentset on the protection)- from the hot state, tripping in less than theconventional time (2 hours) shall be guaranteed forcurrent values equal to 1.3 x I1Figure 61E4s1E3s100s10s1s0.1s1E-2sCurve 5Curve 3Curve 60.1kA 1kA 10kACurve 4I k x 2sMV/LV transformer substations: theory and examples of short-circuit calculation27

3 Choice of protection and control devicesTechnical Application PapersOnce the tripping curve of the LV main device has beendefined, the possible settings for the circuit-breaker ofthe passive load are analyzed. As already said, theprotection of the relevant cable shall be verified and nointersections with the LV main device shall occur.Based on the these considerations, Figure 7 shows thetime/current diagram from which it results that the curveof the cable lies above the curve 7 of the relevant circuitbreakerand that there are no intersection points betweenthe curves of the two LV devices.If the fault is assumed to be upstream the LV protectiondevice, the setting of the current threshold of theprotection release should have an adequate value so thatthe protection MV usertrips due to such a fault.In compliance with these considerations, incorrespondence with the LV side three-phase shortcircuitvalue previously calculated, it is possible todetermine the fault current, related to the LV side,affecting the circuit-breaker on the MV side:Figure 71E4s1E3s100sCurve 7Curve 5CableCurve 4I 2kF-PE=I 2k x 1000 23 x 1000=3 3= 13.28kASince the first threshold of the protection device MV user,related to 400 V, has been set at 3250A, this means thatthe protection is able to trip due to a phase-to-earth faulton the LV side.10s1sWith reference to the MV side, through the transformationratio it results0.1s1E-2sI 2kF-PEI 1kF-PE= = 13280 = 265.6Ak 50which must be compared with the first protectionthreshold of the MV circuit-breaker set at 65A.0.1kA 1kA 10kAIn this diagram, the settings assumed for the load circuitbreakerare:L (overload; protection I1-t1):0.88xIn=554.4ACurva: 3sS (delayed short-circuit; protection I2-t2):not presentI (instantaneous short-circuit; protection I3):2.5xIn=1575AThe diagram shown in Figure 8 represents:curve 4, with the three-phase short-circuit current valueon the LV side;curve 8, with the current value related to the LV currentaffecting the MV circuit-breaker (value of curve 4, reducedby 3 );curve 3, relevant to the protection device MV userrelatedto the LV side, from which the tripping times can bederived.Protections against earth faultProtections against earth faults shall be studied now.In case no earth fault protection is present in thetransformer star point, the overcurrent protection on theMV side of the transformer meets also the protectionrequirements against phase-to-earth faults on thesecondary upstream the LV main circuit-breaker.Figure 8100s10sCurve 3Curve 8Curve 4For a typical transformer with connection ∆/Y a phaseto-earthfault occurring on the LV side in an installationarea immediately downstream the transformer causeson the MV primary side a current which results to be3 times lower than the value calculated for the threephasefault on the secondary side.1s0.1s10kA100kA28 MV/LV transformer substations: theory and examples of short-circuit calculation

3 Choice of protection and control devicesIf the zero-sequence protection is present, its trippingthreshold shall be lower than the threshold 51N definedby the utility company and declared in the electricalconnection agreement.This value has been fixed in 4A 0.12s; therefore, thetripping characteristic of the device MV usercould be setat the following values: 4A 0.05s.Thus, tripping curves as those represented in the diagramof Figure 9 are obtained. This diagram refers to a voltageof 400V. In particular, curve 9 shows the thresholdestablished by the utility company and curve 10 the positive-sequencetripping threshold.Figure 91E3s• zero-sequence current of the line, detected through atoroidal current transformer measuring the sum of thethree phase currents.These protections, used in the network with isolatedneutral, do not function in the network with the neutralearthed through an impedance. In these types of network,directional protections (67) with two separate settingthresholds must be used:• the first one detects the fault when the network ismanaged with the neutral earthed through animpedance• the second one detects the fault when the network ismanaged with the neutral isolated (situation occurringfor short periods in the year, that is during faults ormaintenance operations).100s10s1s0.1sCurve 9Curve 101E-2s1E-3s1E-2kA0.1kA1kA10kAObviously, the behaviour of the two protections shall bestudied with reference to the earth fault current given bythe utility company. Such value varies significantlyaccording to the fact whether the neutral is compensatedor isolated and, however, it shall be higher than theprotection threshold fixed by the utility company.If the state of the neutral were changed, it would benecessary to revise the protection modalities currentlyin use on the lines to detect the single-phase earth fault.The directional earth protection currently used processesthe module and phase of the electrical parameters (zerosequencevoltage and current) which appear during thefault:• zero-sequence voltage (voltage of the transformer starpoint with respect to earth), detected through the phasevoltage transformer with open delta-connectedsecondary windings, at the ends of which the sum ofthe three phase voltages is measured;MV/LV transformer substations: theory and examples of short-circuit calculation29

Technical Application PapersAnnex ACalculation of the transformer inrushcurrentHere are some considerations about the evaluation ofthe magnetizing current of a transformer.In the normal lay-out of a MV/LV installation, thephenomenon described below occurs at the put intoFigure 1service of the transformer and involves the protectiondevice on the MV side.By using the data shown in Tables 1 and 2 below andwith the help of the diagram of Figure 1, an approximatemethod is illustrated to define the minimum delay timenecessary to avoid unwanted trips of the protectiondevice upstream the transformer.Table 1: Oil transformert r/ τ inrush1.81.61.41.21.0S nTR[kVA]50100160250400630100016002000k i= ip inrushI 1nTR1514121212111098τ inrush[s]0.100.150.200.220.250.300.350.400.450.80.6Table 2: Cast resin transformer0.40.2Where:S nTRip inrushI 1nTRt inrush0.1 0.2 0.3 0.4 0.5 0.6 I r’ / ip inrushis the rated power of the transformers;is the inrush current of the transformers;primary rated current of the transformers;time constant of the inrush current.S nTR[kVA]200250315400-500630800-1000125016002000k i= ip inrushI 1nTR10.510.51010101010109.5τ inrush[s]0.150.180.20.250.260.30.350.40.4The diagram of Figure 1 shows the curve which separates the range of the possible tripping (on the left of the curve)of a generic protection from that of guaranteed non-tripping (on the right of the curve).t r= setting of the delay timeI r’= setting threshold (primary value)30 MV/LV transformer substations: theory and examples of short-circuit calculation

Annex AExample:Considering as example an oil transformer with ratedpower S nTR=630kVA and primary rated voltage V 1n=10kV,the calculation of the primary rated current gives a valueof I 1nTR= 36.4A.With reference to the rated power S nTRof the transformer,the values corresponding to k i= 11 and τ inrush= 0.30scan be read in the table.From the definition of k ithe maximum value of the inrushcurrent can be obtained ip inrush= 36.4 . 11 = 400ABy assuming a setting threshold for the primaryprotection I r’ = 40A it resultsI r ’= 40 =0.1ip inrush400corresponding on the curve to the valuet r=1.82τ inrushfrom which it results t r= 1.82 . 0.30 = 0.546srepresenting the minimum delay for the MV protectionto avoid unwanted trips.A verification of the magnetizing current phenomenonmust be carried out also for a LV/LV transformer and inthis case the LV circuit-breaker is involved.The foregoing considerations can be left out and throughthe following formula it is possible to trace the curve ofthe magnetizing current, making more direct theinterpretation of the way the magnetizing curve and theprotection curve of the LV circuit-breaker may interact:tτ inrushi inrush=k .i I 1nTR. e2When not explicitly specified by the manufacturer, thevarious quantities expressed in the formula can beassociated with the values previously indicated in theTables 1 and 2.The various parameters have the same meaning.A generic LV/LV transformer and the relevant LV circuitbreakeron its supply side are considered.With reference to the parameters already given, whichcorrespond to a transformer with a defined rated power,this formula allows the magnetizing curve shown in thediagram of Figure 2 to be represented.The same diagram shows also the tripping curve of thecircuit-breaker on the supply side of the transformer.It is highlighted how the setting of the magnetic protection(function “S” and “I”) must not intersect the magnetizingcurve, and how the protection function “L” is set withreference to the rated current of the transformer primary.Figure 21E4s1E3s100s10s1s0.1s1E-2s0.1kAI 1nTRCB on the primary side of the transformerInrush current1kA10kAMV/LV transformer substations: theory and examples of short-circuit calculation31

Technical Application PapersAnnex BExample of calculation of the short-circuitcurrentThe study of the short-circuit currents is one of the classicproblems plant engineers have to face; knowledge of thevalues of such currents is fundamental for the properdimensioning of lines and transformers, but above all ofprotection devices.If an accurate analysis which takes into account theelectromagnetic and electromechanical transients is notthe aim, the study of the short-circuit currents is quiteeasy from a conceptual point of view, since it is basedon a few concepts, which however have to be understoodin depth and correctly used. But this study may be morecomplex from a computational point of view, in particularwhen the network has remarkable dimensions or whenmeshed networks and asymmetric faults are dealt with.Here is an example of short-circuit current calculation inan electric network by using first an exact method basedon the theory of symmetrical components, and then anapproximate method defined as “power method”.Figure 1Plant dataHereunder the electrical data of the objects in the networkare defined:Supply network (net)V 1n= 20 kV rated voltagef= 50 Hz rated frequencyS k= 750 MVA short-circuit power of the supplynetworkcosϕ k= 0.2net– TR1 – TR2Load L– Cable C1BMain busbarpower factor under short-circuitconditionsDG– Cable C2ATransformers TR1-TR2V 1n= 20 kV primary rated voltageV 2n= 400 V secondary rated voltageS n= 1600 kVA rated powerv k%= 6 % voltage drop in percent undershort-circuit conditionsp k%= 1 % rated losses in percentGenerator GV 2n= 400 V rated voltageS n= 1250 kVA rated apparent powercosϕ nrated power factorx” d%= 14 % subtransient reactance in percent,direct axisx” q%= 20 % subtransient reactance in percent,quadrature axisx’ d%= 50 % synchronous transient reactance inpercentx d%= 500 % synchronous reactance inpercentagex 2%= 17 % negative-sequence short-circuitreactance in percentx 0%= 9 % zero-sequence reactance inpercentT” d= 40 ms subtransient time constantT’ d= 600 ms transient time constantT a= 60 ms armature time constant (that is ofthe unidirectional component)Cable C1Length L= 50mFormation: 3 x (2 x 185) +3 x (2 x 95) + G185R F1= 2.477 mΩ phase resistanceX F1= 1.850 mΩ phase reactanceR n1= 4.825 mΩ neutral resistanceX n1= 1.875 mΩ neutral reactanceR PE1= 4.656 mΩ PE resistanceX PE1= 1.850 mΩ PE reactanceCable C2Length L= 15 mFormation: 3 x (2 x 500) +3 x (2 x 300) + G500R F2= 0.2745 mΩ phase resistanceX F2= 1.162 mΩ phase reactanceR n2= 0.451 mΩ neutral resistanceX n2= 1.177 mΩ neutral reactanceR PE2= 0.517 mΩ PE resistanceX PE2= 1.162 mΩ PE reactance32 MV/LV transformer substations: theory and examples of short-circuit calculation

Annex BB1 Method of symmetrical componentsThis method is based on the principle that any set ofthree vectors may by resolved into three sets of vectors:- a balanced positive sequence set formed by threevectors of equal magnitude shifted by 120° and havingthe same phase sequence as the original system;- a balanced inverse sequence set formed by threevectors of equal magnitude shifted by 120° and havinginverse phase sequence to that of the original system;- a zero sequence set formed by three vectors of equalmagnitude in phase.Based on this principle, a generic asymmetric andunbalanced three-phase system can be reduced to theseparate study of three single-phase equivalent circuitswhich correspond respectively to the positive, negativeand zero-sequence.The sequence impedances can be found by replacingthe network components with the equivalent circuits forthat sequence. As regards positive and negativesequences, the equivalent circuits do not differ whenrotary machines are not present in the installation,whereas when there are rotary machines (asynchronousmotors and synchronous generators) the equivalentimpedances – positive and negative sequence – areconsiderably different. The impedance of the zerosequenceis also considerably different from the previousones and depends on the state of the neutral.Without going into the details of a theoretical treatment,we report below how the positive, negative and zerosequencecircuits represent the three-phase fault, thetwo-phase fault and the line-to-earth fault and therelevant formulas for the calculation of the fault current.This schematization can be useful to fully understandthe treatment.Three-phase faultLine-to-earth faultI d= E dZ dI k3=V 2nZ d 3I d=L1I dLine-to-neutral faultL2EZ ddL3I d=L1L2L3Two-phase faultPE o NE dV 2nI d=I k2=Z d + Z i Z d + Z iL1I dL2E Z ddL3V dI iZ iE d(Z dI k1(F-PE)=+ Z i + Z o )E d(Z dI k1(F-N)=+ Z i + Z o )E d3V 2nZ d + Z i + Z o(F-PE)3Z dZ iV 2nZ d + Z i + Z o(F-N)V iI dI iI oV dZ oV iV oMV/LV transformer substations: theory and examples of short-circuit calculation33

Annex BTechnical Application PapersThe installation typology represented by the single linediagram of Figure 1 may be significant of a genericindustrial plant, where a unique overall outgoing feederhas been considered for simplification. Only the passiveload has been taken into account, by considering alsoas negligible the contribution of possible motors to theshort-circuit current (complying with the condition:Σ I ≤ I nM k/100 prescribed by the Standard IEC 60909, whereI nMis the rated current of the various motors and I kis theinitial symmetrical short-circuit current on the busbarwithout motor contribution).The values of the impedances to be used in the sequencenetworks for the calculation of the fault currents can bederived from the data above. The subscripts have thefollowing meaning:- d positive sequence component;- i negative sequence component;- o zero-sequence component.Supply networkThe parameters of positive and negative sequence ofthe network impedance related to 400 V are:Generator GIn this example, only the subtransient reactance valuedetermining the greatest short-circuit current value forthe generator is considered.Real part of the expression of the impedances ofsequence o-d-i:X ’’ dR G== 9.507 . 10 -4 Ω2 . π . f . T aImaginary part of the expression of the positive sequenceimpedance :x ’’ d% V ’’X ’’ d= .2n= 0.018 Ω100 S nImaginary part of the expression of the negative sequenceimpedance:x 2 % V 2X 2 = .2n= 0.022 Ω100 S nX 2is a parameter of the machine among the data givenby the manufacturer.As an alternative, the imaginary part of the negativesequence impedance could have been calculated as theaverage value between the subtransient positivesequence reactance and that in quadrature:Z dnet= Z inet= V2 2n= 2.133 . 10 -4 ΩS kR dnet= R inet= Z . dnetcosϕ k = 4.266 . 10 -5 ΩX dnet= X inet= X . dnetsinϕ k = 2.090 . 10 -4 ΩThe zero-sequence impedance of the supply is notconsidered, since the delta windings of the transformersblock the zero-sequence component.X i =2Imaginary part of the expression of the zero-sequenceimpedance :Therefore:X o =X ’’ d +X’’ qx o % .V 2 2n100 S n= 0.0115 ΩZ dG= R G+ i . X ’’ dZ iG= R G+ i . X 2Z oG= R G+ i . X oTransformers TR1-TR2A classic type delta/star grounded transformer ( ∆/Y ),which allows to have a distribution system LV side ofTN-S type, is taken into consideration.The impedances of the various sequences (o-d-i) takethe same value:Z dTR= Z iTR= Z oTR=R TR=X TR=p k%.100Z 2V 2 2nS nv k%.100= 0.001 ΩV 2 2nS ndTR –R2 dTR = 5.916 . 10 -3 Ω= 0.006 ΩCables C1 - C2Z dC..= Z iC..= R F..+ i . X FZ o (F-N) C..= (R F..+ 3 . R N..) + i . (X F..+ 3 . X N..)zero-sequence impedance due to line-to-neutral faultZ o (F-PE) C..= (R F..+ 3 . R PE..) + i . (X F..+ 3 . X PE..)zero-sequence impedance due to line-to-earth faultHaving defined all the sequence impedances of thedifferent plant components, an analysis of the variousfault situations can be carried out.Making reference to the network schematization of Figure 1,the three points A-B-D are highlighted where the fault isassumed and where the current values for the differentfault typologies are calculated.34 MV/LV transformer substations: theory and examples of short-circuit calculation

Annex BStill with reference to the network represented in Figure1, the sequence networks with impedances in series orin parallel are drawn according to the way they are seenby an imaginary observer located at the fault point andlooking at the supply source.Fault in ABased on the above considerations, the followingsequence networks can be drawn for a fault at point A.PositivesequencenetworkZdnetZdGOnce identified the three sequence networks, thecalculation of the short-circuit currents for the differentfault typologies can be carried out:Three-phase faultSince the three-phase fault is a symmetrical fault, onlythe equivalent impedance of the positive sequencenetwork shall be considered, complying also with whatexpressed by the formula for the calculation of currents.Therefore the equivalent impedance which is obtainedby the reduction of the positive sequence network is:Z dEq.A= ((Z dTR1Z dTR2) + Z dnet) (Z dG+ Z dC2) = 4.237 . 10 -4 + i . 0.0027 Ω[“||” means “in parallel”] and the three-phase fault currentvalue is expressed by the following formula:I k3A=V 2n= 83.9 . 10 -3 ∠ - 81.15° AZdTR1ZdTR2ZdC23 . Z dEq.ABy using the current divider rule, the contributions of thesingle electrical machines (generator and transformer)to the short-circuit current on the main busbar can bedetermined. In particular, the contributions are subdividedas follows:Main busbarANegativesequencenetworkZdnetZdG12.07 kAZinetZiGZdTR135.94 kA35.94 kAZdTR2ZdC2ZiTR1ZiTR2ZiC2Main busbarA83.9 kAZero-sequencenetworkMain busbarATwo-phase faultIn this case the fault affects only two of the three phases;as a consequence it is necessary to evaluate the equivalentimpedance not only of the positive sequence network butalso that of the negative sequence network seen from thefault point A, as shown in the fault current formula.ZoGThe equivalent positive sequence impedance is:Z dEq.A= ((Z dTR1Z dTR2) + Z dnet) (Z dG+ Z dC2) = 4.237 . 10 -4 + i . 0.0027 ΩThe equivalent negative sequence impedance is:ZoTR1ZoTR2ZoC2Z iEq.A= ((Z iTR1Z iTR2) + Z inet) (Z iG+ Z iC2) = 4.367 . 10 -4 + i . 0.0028 ΩThe two-phase fault current value is therefore equal to:Main busbarAV 2nI k2A=Z dEq.A+ Z iEq.A= 71.77 . 10 -3 ∠ - 81.12° AMV/LV transformer substations: theory and examples of short-circuit calculation35

Annex BTechnical Application PapersSingle-phase faultFault in B3 . networkV 2nI k1(F-PE)A== 85.43 . 10 -3 ∠ - 80.89° AZ dEq.A+ Z iEq.A+ Z o(F-PE)Eq.AZoGAs regards the single-phase fault a distinction must bemade between:- single-phase fault to earth, then return through theprotection conductor, being a distribution system ofTN-S typeComplying with what described for the fault at point A,the three sequence networks are drawn now taking intoconsideration the impedances as seen from point B. Asit results evident in this new case, also the cable C1 is tobe considered in the sequence circuits.- line-to-neutral fault, then return through the neutralconductor.As expressed in the formulas for the calculation of the Positivefault current, it is necessary to take into consideration sequencenetworkthe contribution of the three sequence circuits.ZdnetZdGTo this purpose, it should be noted how the zerosequencenetwork is topologically different from the othersequence networks, since it is strongly influenced bythe typology of the transformer windings.Besides, the values of the zero-sequence impedancesof the cables depend on the type of single-phase fault(F-N or F-PE).ZdTR1ZdTR2ZdC2Main busbarThe equivalent positive sequence impedance is:ZdC1Z dEq.A= ((Z dTR1Z dTR2) + Z dnet) (Z dG+ Z dC2) = 4.237 . 10 -4 + i . 0.0027 ΩBThe equivalent negative sequence impedance is:Z iEq.A= ((Z iTR1Z iTR2) + Z inet) (Z iG+ Z iC2) = 4.367 . 10 -4 + i . 0.0028 Ω NegativesequenceThe equivalent zero-sequence impedance line-to-neutral is:networkZinetZiGZ o(F-N)Eq.A= ((Z oTR1Z oTR2) (Z oG+ Z o(F-N)C2) = 4.189 . 10 -4 + i . 0.0025 ΩThe equivalent zero-sequence impedance line-to-earth is:Z o(F-PE)Eq.A= ((Z oTR1Z oTR2) (Z oG+ Z o(F-PE)C2) = 4.237 . 10 -4 + i . 0.0025 ΩZiTR1ZiTR2ZiC2The value of the fault current line-to-neutral instead isequal to:Main busbar3 . VZiC12nI k1(F-N)A== 85.43 . 10 -3 ∠ - 80.92° AZ dEq.A+ Z iEq.A+ Z o(F-N)Eq.ABThe value of the fault current line-to-earth is equal to: Zero-sequenceZoTR1ZoTR2ZoC2Main busbarZoC1B36 MV/LV transformer substations: theory and examples of short-circuit calculationThrough a process and considerations analogous to theabove case, the equivalent impedances are obtained andcalculation of the short-circuit currents for the differentfault typologies can be carried out.

Annex BThree-phase faultThe equivalent positive sequence impedance derivingfrom the reduction of the relevant sequence network is:Z dEq.B= ((Z dTR1Z dTR2) + Z dnet) (Z dG+ Z dC2) + Z dC1= 0.003 + i . 0.0046 ΩThen the three-phase fault current value is equal to:V 2nI k3B== 42.66 . 10 3 ∠ - 57.59° A3 . Z dEq.BThe contributions are subdivided as follows:Fault in DAssuming a fault in D, we take into consideration thecase when the fault occurs immediately on the load sideof the transformer. In accordance with what described inthe cases above, the three sequence networks are drawnconsidering the impedances as seen from point D.PositivesequencenetworkZdnetZdnetZdG6.14 kAZdTR1ZdTR2ZdTR118.28 kA18.28 kAZdTR2ZdC2Main busbarMain busbarZdC1ZdGZdC142.66 kADBTwo-phase faultThe equivalent positive sequence impedance is:NegativesequencenetworkZinetZ dEq.B= ((Z dTR1Z dTR2) + Z dnet) (Z dG+ Z dC2) + Z dC1= 0.003 + i . 0.0046 ΩThe equivalent negative sequence impedance is:Z iEq.B= ((Z iTR1Z iTR2) + Z inet) (Z iG+ Z iC2) + Z iC1= 0.003 + i . 0.0046 ΩZiTR1ZiTR2Then the two-phase fault current value is equal to:V 2nI k2B== 36.73 . 10 3 ∠ - 57.72° AZ dEq.B+Z iEq.BSingle-phase faultThe equivalent positive sequence impedance is :Main busbarZiC1ZiGDZ dEq.B= ((Z dTR1Z dTR2) + Z dnet) (Z dG+ Z dC2) + Z dC1= 0.003 + i . 0.0046 ΩThe equivalent negative sequence impedance is:Z iEq.B= ((Z iTR1Z iTR2) + Z inet) (Z iG+ Z iC2) + Z iC1= 0.003 + i . 0.0046 ΩZero-sequencenetworkThe equivalent zero-sequence impedance line-to-neutral is:Z o(F-N)Eq.B= ((Z oTR1Z oTR2) (Z oG+ Z o(F-N)C2) + Z o(F-N)C1= 0.017 + i . 0.010 ΩZoTR1ZoTR2The equivalent zero-sequence impedance line-to-earth is:Z o(F-PE)Eq.B= ((Z oTR2Z oTR2) (Z oG+ Z o(F-PE)C2) + Z o(F-PE)C1= 0.017 + i . 0.010 ΩMain busbarThe fault current value line-to-neutral is then equal to:ZoC1ZoG3 . V 2nI k1(F-N)B=Z dEq.B+ Z iEq.B+ Z o(F-N)Eq.B= 23.02 . 10 3 ∠ - 39.60° ADwhereas the fault current value line-to-earth is equal to:3 . V 2nI k1(F-PE)B== 23.35 . 10 3 ∠ - 40.09° AZ dEq.B+ Z iEq.B+ Z o(F-PE)Eq.BThrough a process and considerations analogous to theabove ones, the equivalent impedances are obtained andcalculation of the short-circuit currents for the differentfault typologies can be carried out.MV/LV transformer substations: theory and examples of short-circuit calculation37

Annex BTechnical Application PapersThree-phase faultThe equivalent positive sequenceZ dEq.B= ((Z dTR1Z oTR2) + Z dnet+ Z dC2) (ZThen the three-phase fault currentV 2nI k3D== 65.193 . Z dEq.DThe contributions are subdividedZdnet26.21 kAZdTR1ZdTR226.21 kAMain busbar52.41 kAZdC2Two-phase faultThe equivalent positive sequenceZ dEq.D= ((Z dTR1Z dTR2) + Z dnet+ Z dC2) (Z dGThe equivalent negative sequenceZ iEq.D= ((Z iTR1Z iTR2) + Z inet+ Z iC2) (Z iGThe two-phase fault current valueV 2nI k2D== 55.46Z dEq.D+Z iEq.DSingle-phase faultThe equivalent positive sequenceZ dEq.D= ((Z dTR1Z dTR2) + Z dnet+ Z dC2) (Z dGThe equivalent negative sequenceZ iEq.D= ((Z iTR1Z iTR2) + Z inet+ Z iC2) (Z iGThe equivalent zero-sequence impedanceZ o(F-N)Eq.D= ((Z oTR1Z oTR2) + Z o(F-N)C2) (Z oGThe equivalent zero-sequence impedanceZ o(F-PE)Eq.D= ((Z oTR1Z oTR2) + Z o(F-PE)C2) (ZThe fault current value line-to-neutral3 . V 2nI k1(F-N)D=Z dEq.D+ Z iEq.D+ Z o(F-N)Eq.Dwhereas, the fault current value3 . V 2nI k1(F-PE)D=Z dEq.D+ Z iEq.D+ Z o(F-PE)Eq.Dimpedance is:dG) = 5.653 . 10 -4 + i . 0.0035value is:10 3 ∠ - 80.82° Aas follows:ZdG12.87 kAimpedance is:) = 5.653 . 10 -4 + i . 0.0035impedance is:) = 5.94 . 10 -4 + i . 0.0036 Ωis therefore equal 10 3 ∠ - 80.75° Aimpedance is:) = 5.653 . 10 -4 + i . 0.0035impedance is:) = 5.94 . 10 -4 + i . 0.0036 Ωline-to-neutral) = 9.127 . 10 -4 + i . 0.0046line-to-earth is:oG) = 9.85 . 10 -4 + i . 0.0046is therefore:= 58.03 . 10 3 ∠ - 80.01° Aline-to-earth is equal= 57.99 . 10 3 ∠ - 79.66° AB2 Power methodSupply networkS knet=750MVA is a plantTransformer TR1-TR2S nTR1S kTR1= . 100vk%S nTR2S kTR2= . 100vk%Generator GS nGS kG= . 100x’’d %Cables C1-C22V 2nS kC1=ZFC12V 2nS kTR2=ZFC2where:Z FC1= 2(R F1+ X 2 F1)Z FC2= 2(R F2+ X 2 F2)datumS kTR1= 26.67MVAS kTR2= 26.67MVAS kG= 8.93MVAS kC1= 51.75MVAS kC2= 133.95MVAZ FC1= 0.0031ΩZ FC2= 0.0012ΩΩ. DΩto:Ωis:ΩΩto:This method allows a quick but approximate evaluationof the three-phase short-circuit current in a network. It isnecessary to calculate the power short-circuits of thevarious elements constituting the network (transformers– generators – cables), before determining the total shortcircuitpower at the point where the fault current has tobe evaluated.Power fluxes due to elements operating in parallel canbe reduced by applying the formula of the resistances inseries, whereas power fluxes due to elements workingin series can be reduced by applying the formula of theresistances in parallel.Here is an example of calculation applied to the networkpreviously examined.It can be observed how, for the same typology of fault(three-phase short-circuit at points A – B – D), this“approximate” method gives results quite similar to thoseobtained by applying the method of the symmetricalcomponents.Making reference to the plant data previously reported,it is possible now to proceed with the calculation of theshort-circuit powers of the different elements of theinstallation:38 MV/LV transformer substations: theory and examples of short-circuit calculation

Annex BTaking into consideration the fault in A, the networkschematising the contribution of the short-circuit powersis the following:Taking into consideration the fault in D, the networkschematising the contribution of the short-circuit powersis:SknetSkGSknetSkTR1SkTR2SkC2SkTR1SkTR2Main busbarAMain busbarBy the reduction of the elements in series – in parallel,the following expression for the total power is obtained:S kTOT(A)= ((S kTR1+ S kTR2) // S kR) + (S kG// S kC2) = 58.16MVAI k3A=S kTOT(A)3 . V 2nfrom which it resultsI k3A= 83.95kASkC2SkGBy the reduction of the elements in series – in parallel,the following expression for the total power is obtained:S kTOT(D)= {[(S kTR1+ S kTR2) // S kR] // S kC2} + S kG= 45.23MVADTaking into consideration the fault in B, the networkschematising the contribution of the short-circuit powersis the following:I k3D=S kTOT(D)3 . V 2nfrom which it resultI k3D= 65.28kASknetSkGConsiderations about the results obtainedFrom the above example, it is evident that the use of thepower method offers the advantage of simplicity andspeed, but it could give results less precise comparedwith the method of the symmetrical components.SkTR1SkC1BSkTR2SkC2Main busbarThe most evident difference regards the three-phase faultcalculated at point B, where the presence of the cableC2, characterized by particular values for “L” and “R”,introduces a different ratio between the imaginary andthe real parts of the expressions as regards to the otherelements, thus highlighting the approximate characterof the power method.By the reduction of the elements in series – in parallel,the following expression for the total power is obtained:S kTOT(B)= [((S kTR1+ S kTR2) // S kR) + (S kG// S kC2)] // S kC1= 27.38MVAHowever, the effect of the approximation is not such asto invalidate this method, in particular if it is used to carryout preliminary calculations, as often happens.I k3B=S kTOT(B)3 . V 2nfrom which it resultsI k3B= 39.52kAMV/LV transformer substations: theory and examples of short-circuit calculation39

GlossaryTechnical Application PapersGlossaryv k%p k%V nS nI nV 1nV 2nshort-circuit voltage in percentshort-circuit power in percentrated voltagerated powerrated currentprimary rated voltagesecondary rated voltageX” dsubtransient reactance, direct axisX’ dtransient reactance, direct axisX dS kI ki pZ kX ksynchronous reactance, direct axisshort-circuit apparent powershort-circuit currentpeak currentshort-circuit impedanceshort-circuit reactanceR kshort-circuit resistanceZ… impedance of a generic elementR… resistance of a generic elementX… reactance of a generic elementi si uηcosϕa∠ba+ibsymmetrical component of the short-circuit currentunidirectional component of the short-circuit currentefficiencypower factorpolar representation: “a” is the modulus; “b” is the phase displacement anglerectangular representation: “a” is the real-part and “b” is the imaginary-partSubscripts:…L…TR…G…M…n…C…net…N… F… PE…1F-PE…1F-npassive generic loadtransformergeneratormotorratedcableplant supply networkneutralphaseprotection conductorsingle-phase to earthline-to-neutral…2 two-phase…3 three-phase…LV…MV…klow voltagemedium voltageshort-circuit condition40 MV/LV transformer substations: theory and examples of short-circuit calculation

Due to possible developments of standards as well as ofmaterials, the characteristics and dimensions specified in thisdocument may only be considered binding after confirmationby ABB SACE.1SDC007101G0201 September ’05Printed in Italy6.000 - CALDistributed by:Gross Automation1725 South Johnson RoadNew Berlin, WI 53146262-446-0000, FAX 262-446-0300