IEC_61363 FAULT StudyThis chapter examines the short-circuit current calculation procedures used in the IEC_61363 Short Circuit Study. The IEC_61363 Study follows the specifications of the International Electrotechncal Commission (IEC) International Standard 61363: Electrical installations of ships and mobile and fixed offshore units – Procedures for calculating short-circuit currents in three-phase a.c. This guide includes: Engineering Methodology • Terminology and Symbols • Assumptions and Equations • PTW Applied Methodology • Examples IEC_61363 FAULT STUDY 1.1 What is the IEC_61363 Study? 2 1.2 Engineering Methodology 2 1.3 PTW Applied Methodology 17 1.4 Application Example 24 I N T H I S C H A P T E R • SKM Power*Tools for Windows IEC_FAULT 2 Reference Manual 1.1 What is the IEC_61363 Study? The IEC_61363 Short Circuit Study (referred to hereafter as IEC363) models the current that flows in the power system under abnormal conditions and determines the prospective fault currents in an electrical power system. These currents must be calculated in order to adequately specify electrical apparatus withstand and interrupting ratings. The Study results are also used to selectively coordinate time current characteristics of electrical protective devices. IEC363 represents conditions that may affect typical marine or offshore installations more significantly than land-based systems, including more emphasis on generator and motor decay. 1.2 Engineering Methodology IEC Standard 61363 describes a detailed method for calculating three-phase short circuit duties for marine or offshore installation. The Standard contains 9 chapters. Individual paragraphs are referred to as articles or clauses, and sub-paragraphs are referred to as subclauses. 1.2.1 IEC Standard 61363 The IEC 61363 standard outlines procedures for calculating short-circuit currents that may occur on a marine or offshore a.c. electrical installation. The calculation methods are intended for use on unmeshed three-phase a.c. systems operating at 50 Hz or 60 Hz; having any system voltage specified in IEC 60092-201 table 2; having one or more different voltage levels; comprising generators, motors, transformers, reactors, cables and converter units; having their neutral point connected to the ship’s hull through an impedance (designed to limit the short-circuit current flowing to the ship’s hull; or having their neutral point isolated from the ship’s hull. The IEC 61363 standard is intended for three-phase symmetrical short circuit conditions over the first 100 ms of the fault. The effects of voltage regulators are not considered. The primary reasons for performing the IEC 61363 short circuit calculations include: 1) obtain the short-circuit current magnitude at each point in the power system; 2) compare the calculated fault current to the ratings of installed equipment to verify the equipment ratings are adequate to handle the short circuit current; 3) support proper selection of circuit protection equipment. Note that marine and offshore electrical systems typically have large generating capacities confined in a small area resulting in high short circuit values with low power factors. Special attention is required if the calculated power factor during fault conditions is below the power factor used to test the circuit breakers. 3/26/2006 of two or more points in a circuit which are normally at different voltages. [61363-1 IEC:1998] Prospective current Short-circuit current that would flow in the circuit if each pole of the device were replaced by a conductor of negligible impedance.m. [61363-1 IEC:1998] Steady-state short-circuit current Ik d in the direct axis r. symmetrical component of a prospective short-circuit current applicable at the instant of short circuit if the impedance remains at zero-time value.) component of the short-circuit current Idc Component of current in a circuit immediately after it has been suddenly short-circuited.s. value of the a.c. all components of fundamental and higher frequencies being excluded. value of the short-circuit symmetrical current flowing through a circuit with generators witch remains after the decay of the transient phenomena.m. [61363-1 IEC:1998] Transient short-circuit current Ik d in the direct axis r.2.s. [61363-1 IEC:1998] Current Theoretical maximum Peak at 1/2 cycle Top envelope idc Asymmetrical values including motor contributions 2 ip 2 I"k Decaying (aperiodic) component (DC decay) i dc 2 2I k Time Bottom envelope Steady state value (no motor contributions) Subtransient short-circuit current Ik”d in the direct axis r. [61363-1 IEC:1998] Peak short-circuit current Ip SKM Power*Tools for Windows . [61363-1 IEC:1998] Short circuit current over-current resulting from a short circuit due to a fault or an incorrect connection in and electric circuit. value of the short-circuit current flowing through a circuit with rotating machines having an impedance equal to the transient impedance of the circuit. the aperiodic component of current. [61363-1 IEC:1998] Initial symmetrical short-circuit current Ik” r. symmetrical component of a prospective short-circuit current. [61363-1 IEC:1998] Aperiodic (d.c.s.s.m. value of the a.2 Definitions Short circuit accidental or intentional connection.s.m. if any. being neglected.c.IEC_61363 FAULT Study IEC_FAULT 3 1. [61363-1 IEC:1998] Symmetrical short-circuit current r.m. value of the short-circuit current flowing through a circuit with rotating machines having an impedance equal to the transient impedance of the circuit. by a relatively low resistance or impedance. and direct-axis primary current after the decay of the transient phenomena.IEC_FAULT 4 Reference Manual Maximum possible instantaneous value of the prospective short-circuit current [61363-1 IEC:1998] Direct-axis subtransient short-circuit time constant T”d Time required for the rapidly changing component. 0. whish is due to the direct-axis flux.d. the machine running at rated speed. the machine running at rated speed. 0. component of primary voltage. the machine running at rated speed. the machine (or equivalent machine) running at rated speed. follwing a sudden change in operating conditions. and the value of the simultaneous change in fundamental a. the machine running at rated speed.c. [61363-1 IEC:1998] Direct-axis transient open—circuit time constant T’do Time required for a slowly changing component of the open-circuit primary voltage. [61363-1 IEC:1998] Short-circuit impedance Z 3/26/2006 . which is produced by the total direct-axis primary flux.c.368 of its initial value.368 of its initial value. present during the first few cycles in the direct-axis shrot-circuit current following a sudden change in operating conditions. to decrease to 1/e i. [61363-1 IEC:1998] Direct-axis subtransient open-circuit time constant T”do Time required for the rapidly changing component present during the first few cycles of the open-circuit primary winding voltage which is due to direct-axis flux following a sudden change in operation. [61363-1 IEC:1998] Direct-axis synchronous reactance Xd Quotient of the steady-state value of that fundamental a.c. i. following a sudden change in operating conditions. the machine running at rated speed and the high decrement components during the first cycles being excluded.368 of its initial value. component of primary voltage. 0. component present in the short-circuit current.c.e. to decrease to 1/e ie. and the value of the simultaneous change in fundamental a. the machine running at rated speed. [61363-1 IEC:1998] Direct-axis transient short-circuit time constant T’d Time required for the slowly changing component of the direct-axis short-circuit primary current following a sudden change in operating conditions.368 of its initial value. to decrease to 1/e i. component of direct-axis primary current. current. [61363-1 IEC:1998] Direct-axis transient reactance X’d (saturated) Quotient of the initial value of a sudden change in that fundamental a. component of primary voltage which is produced by the total direct-axis primary flux.c.e. which is produced by the total direct-axis primary flux.e 0.e.368 of its initial value. [61363-1 IEC:1998] Stator resistance of a generator Ra Resistance of the stator of a synchronous machine. 0.c. to decrease to 1/e i. to decrease to 1/e. component of direct-axis primary current.c. [61363-1 IEC:1998] DC time constant Tdc Time required for the d. measured at d. [61363-1 IEC:1998] Direct-axis subtransient reactance X”d (saturated) Quotient of the initial value of a sudden change in that fundamental a. as would be produced by a combination of motors having different ratings and different characteristics. [61363-1 IEC:1998] Transient voltage of a rotating machine E’ r.c.m. value of the symmetrical internal voltage of a machine which is active behind the subransient impedance Z” at the moment of short circuit. [61363-1 IEC:1998] Rated value (r) Quantity value assigned. which are connected to the system.IEC_61363 FAULT Study IEC_FAULT 5 Quotient of the sinusoidal voltage per phase on a balanced a. in series with a passive circuit element.s. [61363-1 IEC:1998] Voltage source Active element which can be represented by an ideal voltage source independent of all currents and voltages in the circuit. [61363-1 IEC:1998] Equivalent motor Fictitious motor having characteristics which will produce the same short-circuit current at any point on an electrical installation. [61363-1 IEC:1998] Nominal system voltage Un Voltage (line-to-line) by which a system is designated and to which certain operating characteristics are referred. for a specified operating condition of a component. [61363-1 IEC:1998] Subtransient voltage of a rotating machine E” r. [61363-1 IEC:1998] Equivalent generator Fictitious generator having characteristics which will produce the same short-circuit current at any point on an electrical installation.s. generally by a manufacturer. device or equipment. as would be produced by a combination of generators having different ratings and different characteristics. [61363-1 IEC:1998] Nominal value (n) Suitable approximate quantity value used to designate or identify a component. which are connected to the system. value of the symmetrical internal voltage of a machine which is active behind the transient impedance Z’ at the moment of short circuit. [61363-1 IEC:1998] SKM Power*Tools for Windows . device or equipment. system and the same frequency component of the short-circuit current in that system.m. m.m.s.) current of the equivalent generator (r.) a.s.c.s.m.s. peak value of the short-circuit current of an asynchronous motor.) symmetrical short-circuit current of an asynchronous motor (r.s. upper envelope of the short-circuit current. component of the short-circuit current of a synchronous machine (instantaneous).m. component of the short-circuit current of an asynchronous motor and an equivalent motor (instantaneous).s.m.m.s.m. including: f E”q E’q E”M f fe fr I”* I’* I* I”M* I”kd I’kd I Iac IacM ILR idc idcM ik I* Ikd iM ip ipM Ir IrM R R* Ra RC Rdc RM RR RR* RS RS* RT t T”d T’d T”d* T’d* T”e T’e T”M 3/26/2006 phase Angle subtransient q-axis voltage of a generator (r.) asynchronous motor locked rotor current d.m.3 IEC 61363 Symbols PTW’s Reports and documentation conform to IEC 61363 notation.) transient q-axis voltage of a generator (r.m.2.s.s. peak value of the short-circuit current of a synchronous machine.s.s. resistance motor resistance rotor resistance of an asynchronous motor rotor resistance of an equivalent asynchronous motor stator resistance of an asynchronous motor stator resistance of an equivalent asynchronous motor resistance of a transformer time duration from the beginning of a short circuit subtransient time constant of a synchronous machine transient time constant of a synchronous machine subtransient time constant of an equivalent generator transient time constant of an equivalent generator subtransient time constant of a synchronous machine including the non-active components transient time constant of a synchronous machine including the non-active components subtransient time constant of a an asynchronous motor .m.c.m.) frequency lowest frequency of a shaft generator rated frequency of a network subtransient short-circuit current of the equivalent generator (r.IEC_FAULT 6 Reference Manual 1.) subtransient short-circuit current of the equivalent motor (r.m.s.m.) steady-state short-circuit current of a synchronous machine (r.c.) transient initial short-circuit current of a synchronous machine (r.m. rated current (r.) subtransient initial short-circuit current of a synchronous machine (r. component of the short-circuit current of a synchronous machine (r.c.s.) transient short-circuit current of the equivalent generator (r. d. steady-state short-circuit current of an equivalent generator (r.s.) rated current of an asynchronous motor resistance resistance of an equivalent generator stator resistance of a synchronous machine cable resistance d.) subtransient voltage of a motor (r.) upper envelope of the short-circuit current of an asynchronous motor.s.) current (r.m. time constant of an asynchronous motor including the connecting cables. The initial symmetrical short circuit current can be calculated and. IEC363 requires a time-dependent calculation divided into active and nonactive components with separate AC and DC calculations. prefault voltage (line-to-line) nominal voltage (line-to-line) rated voltage (line-to-line) rated voltage of a motor (line-to-line) subtransient reactance of an equivalent generator reactance subtransient reactance of a synchronous machine in the d-axis transient reactance of a synchronous machine in the d-axis subtransient reactance of an asynchronous motor impedance equivalent impedance 1. prepare a system one-line diagram. the asymmetrical short circuit current at various times during the onset of the fault can be calculated. define component characteristics. time constant of an asynchronous motor d.c.c. time constant of an equivalent asynchronous motor d. SKM Power*Tools for Windows .c.c.c.2.IEC_61363 FAULT Study T”Me Tdc Tdc* Tdce TdcM TdcM* TdcMe U0 Un Ur UrM X”* X X”d X’d X”M Z Z* IEC_FAULT 7 subtransient time constant of a an equivalent asynchronous motor including connecting cables. Short-Circuit Study Procedure The general study procedure outlined in the IEC 61363 standard includes: 1. and defining a driving point voltage (assuming the effect of transformer taps on bus voltage). 4. Conventional short circuit analysis techniques do not satisfy IEC Standard 61363 methodology. time constant of a synchronous machine d.c. Active components.4 Methodology The Conventional or Comprehensive short circuit analysis procedure involves reducing the network at the short circuit location to a single Thevenin equivalent impedance. d. time constant of an equivalent generator d. are combined to form equivalent motors and generators. prepare a short-circuit summary and document study conclusions. calculate the time-dependent short-circuit currents at the major points in the system using the equations and methods described in the IEC 61363 standard. 2. 3. such as generators and motors. The equivalent motors and generators are combined with non-active components. determining the associated fault point R/X ratio calculated using complex vector algebra. given the fault location R/X ratios. time constant of a synchronous machine including the non-active components. such as cables and transformers. d. to further adjust the impedance and time constants of the equivalent components. changing both the transient and steady-state values of the resulting short-circuit current. forming the system.IEC_FAULT 8 Reference Manual 1. cables. electrical installation. it is important to understand the difference between - The short-circuit current generated by an individual piece of equipment - The short-circuit current which results when several pieces of equipment are connected in a system. When an isolated machine is being considered.. however. etc.2.c. The following assumptions are applied - All system capacitance are neglected - The short-circuit arc impedance is neglected - The short circuit occurs simultaneously in all three phases (three phase fault) - Unmeshed systems When calculating short-circuit currents. for example. operating at 50 Hz or 60 Hz. this current is limited by the impedance of the non-active components.5 IEC 61363 Assumptions IEC 61363 standard outlines procedures for calculating short-circuit currents that may occur on a marine or offshore a. transformers. In a system. 3/26/2006 . only the electrical parameters of the machine affect the short-circuit current generated. The calculation methods are for use on unmeshed three-phase alternating current systems. 33 ms 2 60 * 2 SKM Power*Tools for Windows . component " I ac (t ) = ( I kd − I kd' )e − t / Td + ( I kd' − I kd )e − t / Td + I kd " I " kd I ' kd = = E q" 0 Z d" Eq' 0 Z d' E q" 0 = = ' Ra2 + X d" 2 Eq' 0 Ra2 + X d'2 E q" 0 = 2 U0 cos φ0 + Ra I 0 + 3 2 U0 sin φ0 + X d" I 0 3 Eq' 0 = 2 U0 cos φ0 + Ra I 0 + 3 2 U0 sin φ0 + X d' I 0 3 The d.c.2.6 IEC 61363 Equations Generators Three-phase short-circuit current calculation The upper envelope of the maximum values of the three-phase short-circuit current of a generator can be calculated as ik (t ) = 2 I ac (t ) + idc (t ) The a.c.IEC_61363 FAULT Study IEC_FAULT 9 1. component " idc (t ) = 2 ( I kd − I 0 sin φ0 )e − t / Tdc The peak value T T T i p = ik ( ) = 2 I ac ( ) + idc ( ) 2 2 2 for 60 Hz system T 1000 = = 8. R R = 0.16 p.034 p.u.u.4 ms TdcM = 14.08 ms 3/26/2006 .IEC_FAULT 10 Reference Manual Effects of non-active components connected in series with Generators Impedance changes [ ] [ ] Z e" = ( Ra + R) 2 + ( X d" + X ) 2 Z e' = ( Ra + R) 2 + ( X d' + X ) 2 1/ 2 1/ 2 Time-constant changes Te" = Te' = [( R ] + R) 2 + ( X d" + X ) 2 X d' Td" ( Ra + R) 2 + ( X d" + X )( X d' + X ) X d" [ [( R ] ] + R) 2 + ( X d' + X ) 2 X d Td' ( Ra + R) 2 + ( X d' + X )( X d + X ) X d' [ Tdce = a a Tdc + X ] 2πfRa 1+ R Ra Motors General motor parameter RM = RR (rotor ) + RS ( stator ) X M" = X R (rotor ) + X S ( stator ) TM" = XR + XS ω r RR TdcM = XR + XS ω r RS General data for large motors ( > 100 kW) Z M" = 0. TM" = 18. X M" = 0. TM" = 22.u. at 60 Hz. RS = 0.15 p.73 ms at 50 Hz.021 p.u.67 ms TdcM = 11. 027 p.u. X M" = 0. TM" = 18.u.188 p.73 ms at 50 Hz.043 p.08 ms Three-phase short-circuit current calculation The upper envelope of the maximum values of the three-phase short-circuit current of an asynchronous motor can be calculated as i pM (t ) = 2 I acM (t ) + idcM (t ) The a.u.u.c. R R = 0. component idcM (t ) = 2 ( I M" − I rM sin φ M )e − t / TdcM The peak value T T T i p = ik ( ) = 2 I acM ( ) + idcM ( ) 2 2 2 for 60 Hz system at ½ cycle SKM Power*Tools for Windows .c.2 p.67 ms TdcM = 11.4 ms TdcM = 14. TM" = 22.IEC_61363 FAULT Study IEC_FAULT 11 General data for small motors Z M" = 0. component I acM (t ) = I M" e − t / TM " I M" = E M"' = E M" = Z M" E M" RM2 + X M2 2 2 U rM U rM cosφ M + RM I rM + sin φ M + X M' I rM 3 3 The d. at 60 Hz. RS = 0. n n " * I = I " kdi " I Mj + i i n ' * ' I kdi I = i n I k* = I kdi i M * = ( I *" − I *' ) N * = ( I *' − I k * ) 3/26/2006 .33 ms 2 60 * 2 Effects of non-active components connected in series Motors Impedance Changes RMe = RR + RS + R " X Me = XR + XS + X Time-constant changes " = TMe " X Me ω r RR TdcMe = " X Me ω r ( RS + R ) Equivalent generator I ac (t ) * = ( I *" − I *' )e − t / Td * + ( I *' − I * )e − t / Td * + I k * " ' I ac (t ) * = M * e − t / Td * + N * e − t / Td * + I k * " ' I dcM (t )* = 2 I*"e − t / Tdc* where we defined the following variables.IEC_FAULT 12 Reference Manual T 1000 = = 8. 3I*' c 3 (t ) = Z* = U0 3I* 1 2πfTdc* (t ) Z *" 1 + c32 X *' (t ) = Z *'2 − R*2 X * (t ) = Z *2 − R*2 SKM Power*Tools for Windows . 3I*" Z*' = R* (t ) = c 3 (t ) X *" (t ) . n n K (t )* = K (t ) + i Td"* (t x ) = Td' * (t x ) = Tdc* (t x ) = I "jM e − t / T "M " i " i − tx K (t x )* − I *' ln M* " − tx I (t ) − ( M *e −t x / Td * + I k * ) ln ac x * N* " − tx i (t ) ln dc x " * 2 I* Equivalent generator impedance Z*" = U0 . X *" (t ) = U0 .IEC_61363 FAULT Study IEC_FAULT 13 Equivalent generator time constant For generator: K " (t ) = ( I kd" − I kd' )e − t / Td + I kd' " For motor: K " (t ) = I M" e − t / TM " Thus. I kd == 0 " Ze Ze Ze " I ac (t ) = ( I kd − I kd' )e − t / Td + ( I kd' − I kd )e − t / Td + I kd " The d.c. component " idc (t ) = 2 ( I kd )e − t / Tdc Equivalent motor 3/26/2006 ' . component " I kd = U0 ' U U . I kd = 0' .c.IEC_FAULT 14 Reference Manual Effects of non-active components connected in series with Equivalent Generators Impedance changes [ ] [ ] [ ] Z e" = ( Ra + R) 2 + ( X d" + X ) 2 Z e' = ( Ra + R) 2 + ( X d' + X ) 2 Z e' = ( Ra + R) 2 + ( X d + X ) 2 1/ 2 1/ 2 1/ 2 Time-constant changes [( R ] + R) 2 + ( X d" + X ) 2 X d' Td" T = ( Ra + R) 2 + ( X d" + X )( X d' + X ) X d" " e [ [( R a ] ] + R) 2 + ( X d' + X ) 2 X d Td' T = ( Ra + R) 2 + ( X d' + X )( X d + X ) X d' ' e [ Tdce = a Tdc + X 1+ R ] 2πfRa Ra Three-phase short-circuit current calculation The upper envelope of the maximum values of the three-phase short-circuit current of an equivalent generator can be calculated as ik (t ) = 2 I ac (t ) + idc (t ) The a. IEC_61363 FAULT Study IEC_FAULT 15 I acM (t )* = I M" *e − t / TM * " I dcM (t )* = 2 I M" *e − t / TdcM * where we defined the following variables. n " I M" * = ∑ I Mj j Equivalent motor time constant K M" (t ) = ∑ I M" e − t / TM " K dcM (t ) = 2 ∑ I M" e − t / TdcM TM" * (t x ) = Tdc* (t x ) = − tx K " (t ) ln M" x IM* − tx K dcM (t x ) ln 2 I M" * Equivalent motor impedance RR * (t ) = c1 X M" * (t ) . c1 (t ) = Z M" * = X " M* 1 . 2πfTM" * (t ) RS * (t ) = c 2 (t ) X M" * (t ) c 2 (t ) = 1 2πfTdcM * (t ) U0 " 2 "2 or Z M * = ( RR * + RS * ) + X M * " 3I M * (t ) = Z M" * 1 + (c1 (t ) + c 2 (t )) 2 SKM Power*Tools for Windows . component " idc (t ) = 2 ( I kd )e − t / Tdc 3/26/2006 .IEC_FAULT 16 Reference Manual Effects of non-active components connected in series Equivalent Motors Impedance Changes RMe = RR + RS + R " X Me = XR + XS + X Time-constant changes " Me T " X Me = ω r RR TdcMe " X Me = ω r ( RS + R ) Three-phase short-circuit current calculation The upper envelope of the maximum values of the three-phase short-circuit current of an equivalent asynchronous motor can be calculated as i M (t ) = 2 I acM (t ) + idcM (t ) The a.c. component " I kd = U0 Z e" I ac (t ) = I kd" e − t / Td " The d.c. writes the results to the database. choose Analysis. and choose the Run button. 1. 3.IEC_61363 FAULT Study IEC_FAULT 17 1. Select the check box next to Short Circuit and choose the IEC 61363 option button. The Short Circuit Study runs. SKM Power*Tools for Windows .3 PTW Applied Methodology PTW applies the methodology described in Section 1. 4. choose the Setup button.1 Before Running the IEC 61363 Fault Study Before running the IEC 61363 Fault Study. and creates a report.3 describes how to run the IEC_363 Study. 1.3. From the Run menu. To change the Study options.2. Choose the OK button to return to the Study dialog box. 1. To run the IEC 61363 Study 1. • Define feeder and transformer sizes.3 IEC 61363 Study Options The IEC_FAULT Study dialog box lets you select options for running the Study. and it always runs on the active project.3. you must: • Define the system topology and connections. 2. • Define fault contribution data. including explanations of the various options associated with the Study. Section 1.2 Running the IEC61363 Fault Study You can run the Study from any screen in PTW.3. Time Varying Setup The time varying setup allows you to specify times to report Iac and Idc time-dependent short circuit currents. The system frequency is set in the Project>Options>Application menu. Report and Study Options These boxes allow you to customize the breadth of the Study and its Report. System Modeling These options further customize the Study.IEC_FAULT 18 Reference Manual Following is a list of the available Study options. The default is to report the fault current at all buses. You can choose between a summary report.3. then the faulted bus(es) must be specified using the Select button. If a fault is to be reported at a single bus or selected group of buses. Synchronous Generators and Motors Synchronous generator and motor short circuit current contributions are defined in the Component Editor as shown in the following figures: 3/26/2006 . and asynchronous motors. 1. Model Transformer Tap You may model the transformer taps by selecting this check box.4 Component Modeling Fault Contribution Data Contribution data must be defined for synchronous generators. synchronous motors. System Frequency The system frequency must be defined for the time-dependent calculations. standard report or a detailed report Faulted Buses: All or Selected You can report a fault at a single bus. a group of buses or all buses. even if you enter a revised rated size in the motor’s first subview. you may need to modify the rated machine kVA and kVA base together. Also. For definitions of these values refer to section 1. MVA or kW. Since the IEC 61363 calculations are for 3-phase faults only. if you do not modify them together. zero sequence and neutral impedance values are not used. If the rated kVA base in the IEC Contribution subview is zero. PTW calculates the machine kVA and voltage base using the data you enter in the first subview of the Component Editor. Ra. then PTW calculates the equivalent kVA base from the machine rated size shown in the first subview of the Component Editor. and power factor is used to convert kW to kVA. If the rated kVA base is not zero. Td’ and Tdc. The motor rated size is in mechanical units of work (output) when entered as horsepower. Td”. the kVA base will remain SKM Power*Tools for Windows . Therefore. However values for these fields are still required since the IEC60909. Motor efficiency is used to convert horsepower to electrical units of work. PTW will not change it. ANSI and Comprehensive fault calculations use them.2. but in equivalent electrical units of work (input) when entered as electrical quantities of kVA. the negative sequence. Xd’. if the rated voltage is not zero.2. PTW will not change it.IEC_61363 FAULT Study IEC_FAULT 19 ANSI Contribution Format IEC Contribution Format The IEC 61363 calculations requires entry of the following values: Xd”. Xd. Either entry format can be used. Motors can be defined in the Component Editor as either a single motor (the default) or as multiple motors. 3/26/2006 . along with the power factor. even if you change the rated size on the first subview of the Component Editor. The Component Editor ANSI and IEC contribution data boxes are shown in the following figures: ANSI Format IEC Format The fields added specifically for the IEC 61363 calculations are the ratio of Stator Resistance to Rotor Resistance. Asynchronous Induction Motors Asynchronous motor short circuit currents should be modeled in IEC 61363 calculations. and Tdc. In order to fully model a synchronous machine. Td”.IEC_FAULT 20 Reference Manual unchanged. PTW will calculate the power for multiple motors modeled at the bus. the rated size of the machine must be defined. TM" = 18.u. SKM Power*Tools for Windows .67 ms TdcM = 11.73 ms at 50 Hz.4 ms TdcM = 14. Asynchronous motors are modeled as deltaconnected. at 60 Hz. but is in equivalent electrical units of work (input) when entered as electrical quantities of kVA.u. combined with the rated kW of asynchronous machines.043 p.188 p. is used to calculate the breaking current duty.4 ms TdcM = 14. R R = 0. X M" = 0.u. R R = 0. X M" = 0.021 p.16 p. No shunt capacitance is modeled in IEC 61363. PTW will model the MW/pp of each of the individual motors that comprise the group.u.15 p. If multiple motors are modeled in a single motor object.u.IEC_61363 FAULT Study IEC_FAULT 21 The motor rated size is in mechanical units of work (output) when entered as horsepower. then PTW calculates the equivalent kVA base using the machine rated size as defined in the first subview of the Component Editor. TM" = 22.034 p. RS = 0.08 ms Cables and Transformers Cables and transformers are modeled with a series resistance and reactance as positive sequence components. MVA or kW. the following typical data can be used for the IEC 61363 calculations: Large motors ( > 100 kW) Z M" = 0.2 p. If the rated kVA base is zero.u.67 ms TdcM = 11. TM" = 22. at 60 Hz.u. TM" = 18.027 p. If specific motor data is not available. Motor efficiency is used to convert horsepower to electrical units of work.08 ms Small motors (<100kW) Z M" = 0. RS = 0. and power factor is used to convert kW to kVA. PTW assumes that the negative-sequence impedance is equal to the positive-sequence value.73 ms at 50 Hz.u. The number of pole pairs. A negative primary tap raises the secondary voltage. negative.and zero-sequence impedances from the following per-unit equations: Z1 = Z 2 1. The Study will attempt to run to completion even if fatal errors are detected. Essentially. using the equation: kVA 3Φ = 3 × I fsle × kVLL However. . The driving point voltages are defined by the generators and are not modified by the transformer tap settings. as compared to the primary and secondary bus system nominal voltages.0 Z1 = I f3Φ b g 3 × 1.3.5 Error Messages PTW examines the entered data for the IEC 61363 Study. If PTW finds missing or incomplete information. PTW uses the three-phase fault data and the single-line-to-earth fault data to calculate the positive-. Zero sequence impedance values are not used in the IEC 61363 calculations and therefore transformer earthing impedance is also not used.0 Z1 + Z 2 + Z 0 I fsle = b Z0 = 3 − Z1 − Z 2 I fsle g Utilities often report available single-line-to-earth fault duties on an equivalent threephase rating apparent power basis. in order to identify any other errors. Transformer off-nominal voltage ratios. PTW will create a fictitious primary and/or secondary tap to ensure that the voltage ratios are properly matched. Taps will only be considered if the IEC 61363 Study Setup dialog box is set to model them. Transformer taps may be modeled. It involves the entry of single-line-to-easrth short circuit contribution data. A somewhat common error is: The calculated zero sequence impedance is negative. The Study Messages dialog box will report both fatal and warning messages. the actual apparent power of a single-line-to-ground fault is: kVA1Φ = I fsle × kV 3 where kV 3/26/2006 line-to-line voltage. it sends an error message to the Study Message dialog box. The error messages are shared between all fault studies even though each has slightly different data requirements. 1. are modeled when the Model Transformer Taps check box is selected in the Study setup dialog box.IEC_FAULT 22 Reference Manual Transformers also are modeled with a positive sequence impedance value. 5 Cycles.77 Idc(A) 22907.28 6133.52 7935. 1.0 5. 5 Cycles. T4 • Iac for each branch and source feeding the fault at 1/2 Cycle. The actual apparent power to be entered into PTW is the utility equivalent single-line-to-earth duty divided by three.5 3.11 SKM Power*Tools for Windows . 3 Cycles.56 A x(peak factor): 1. PTW may attempt to calculate the zero-sequence impedance as a negative value.200 kV Ipeak: 38583.30 1573. and one user-defined time.34 8743. T4 • Idc for each branch and source feeding the fault at 1/2 Cycle. IEC_363 reports: • Iac at 1/2 Cycle. Enter the single-line-to-ground fault current X/R ratio.6 Reports For each fault location.0 12. 5 Cycles. 3 Cycles. and one user-defined time. T4 • Idc at 1/2 Cycle.3. 3 Cycles. T4 • Ipeak for each bus and branch *FAULT BUS: B1 Voltage: 4. 3 Cycles.85 9065. 5 Cycles. and one user-defined time. not the zero sequence impedance X/R ratio.14 18048. If you do.0 0.27 14520.615 TIME (Cycles) 0.5 ================================================================================ Iac(A) 16893. and one user-defined time.23 9684.IEC_61363 FAULT Study IEC_FAULT 23 You cannot use the three-phase equivalent rating of a single-line-to-ground short circuit contribution. 0 Meters R 0.069 Ohms/km X 0.050 % X 6.4 Application Example 1.00 ms Td' 420.67 ms Tdc 11.092 Ohms/km B5 600 V 3500 kW X"d 0.4.18 pu Xd 2.095 Ohms/km X 0.00 ms Td' 320.01 pu Td" 26.29 pu Xd 2.110 Ohms/km X 0.75 pu Ra 0.0 Meters R 0.67 ms Tdc 11.29 pu Xd 2.0 kW (Output) X"d 0.0 kVA R 1.0 Meters R 0.0 kW (Output) X"d 0.400 % T2B5C 18.00 ms Td' 420.0 Meters R 0.75 pu Ra 0.00 ms Td' 420.17 pu X'd 0.0 Meters R 0.069 Ohms/km X 0.096 Ohms/km T7 250.17 pu X'd 0.01 pu Td" 26.092 Ohms/km 2000 kW X"d 0.092 Ohms/km 120 V BE 600 V 3/26/2006 B3 600 V .069 Ohms/km X 0.00 ms Tdc 93.00 ms G2 T2C 10.5926 B2B3C 30.780 % X 6.092 Ohms/km MB4/5 2000.150 % X 6.60 pu Ra 0.0 Meters R 0.0 kVA R 1.00 ms B1 4200 V M2 2000.67 ms Tdc 11.73 ms Rs/Rr 1.095 Ohms/km T4 2500.75 pu Ra 0.1 Sample Project in IEC 61363 Standard The following example was included in the IEC 61363 – 1996 Standard: G1 S 500 kW X"d 0.73 ms Rs/Rr 1.73 ms Rs/Rr 1.12 pu X'd 0.1500 pu Td" 18.00 ms Tdc 93.00 ms E1 E1C 10.410 % S 3500 kW X"d 0.01 pu Td" 26.29 pu Xd 2.IEC_FAULT 24 Reference Manual 1.6190 T4B2C 10.00 ms G3 T4C 10.125 Ohms/km X 0.00 ms Tdc 93.0 kVA R 1.1880 pu Td" 18.00 ms Tdc 93.095 Ohms/km 3500 kW X"d 0.00 ms Tdc 64.5926 B2 600 V MB2/3 B2BEC 46.29 pu Xd 2.17 pu X'd 0.0 Meters R 0.75 pu Ra 0.00 ms Td' 420.17 pu X'd 0.770 % S B7 1700.00 ms G4 G4C 11.0 kW (Output) X"d 0.098 Ohms/km T2 2000.1880 pu Td" 18.0 Meters R 0.01 pu Td" 26.069 Ohms/km X 0.164 Ohms/km X 0.01 pu Td" 20. 60 3454.0 0.96 2221.38 0.90 0.96 2221.57 14453.30 4496.60 3454.38 1571.0 0.94 9663. Iac(A) 3838.76 2.200 kV TIME(Cycles) 0.24 Idc(A) 4918.MB4/5 (Eq.52 8732.89 2419.24 Idc(A) 4918.21 1534.30 4496.52 1265.88 2608.69 A.03 Idc(A) 2893.21 7930.09 Idc(A) 22784.90 6119.50 A.G1 Ipeak: 9382.52 0.69 907. Motor) Ipeak: 2004.96 2221.76 2872.5 3.01 347.06 A.92 2007.76 2872.70 A.76 Idc(A) 3335.39 9036.89 2419.53 523.M2 Ipeak: 3127.) Ipeak: 5126.0 5.48 1421.60 3454.71 1206.616 TIME (Cycles) 0.92 2007. Iac(A) 1884.14 129.89 2419.92 2007.G2 Ipeak: 9382.72 SKM Power*Tools for Windows . Iac(A) 2358.52 .58 Idc(A) 1800.52 .51 775.28 TIME-DEPENDENT SHORT-CIRCUIT CURRENTS AT THE MAJOR POINTS: Bus Name:B1 Voltage: 4.G3 Ipeak: 9382.53 523.95 92. Iac(A) 3838. Iac(A) 1047.00 .70 A.95 40.0 12.33 1398.30 4496.B2 (Eq.IEC_61363 FAULT Study IEC_FAULT 25 A portion of the output report is shown: *FAULT BUS: B1 Voltage: 4.24 Idc(A) 4918.76 2872. Gen.5 ================================================================================ Iac(A) 16806.200 kV Ipeak: 38406.0 12.72 0.10 0. Iac(A) 3838.35 A x(peak factor): 1.88 2608.94 173.5 3.16 29.0 5.88 2608.54 2107.47 21.53 523.53 17965.52 .05 52.00 .47 2146.5 ================================================================================ .70 A.44 1. The contributions from G1.5cy) 1422 A DC (3cy) 41 A Ip 3128 A T2B5C B5 T4B2C MB4/5 B2 B2BEC E1 T7 B2B3C MB2/3 S E1C G4 G4C BE B7 B3 Note that the contributions from MB4/5 Equivalent and Equivalent Generator EG are slightly different in PTW than the hand calculation example shown in the IEC 61363 standard.5cy) 831 A AC (3cy) 185 A Ip 2136 A S T4 S AC (0.5cy) 1206 A AC (3cy) 129 A DC (0.5cy) 2108 A AC (3cy) 1535 A Ip 5143 A T2 AC (0.5cy) 18049 A DC (3cy) 9065 A Ip 38584 A T2C T4C M2 AC (0.5cy) 14520 A AC (3cy) 9684 A DC (0. G2.IEC_FAULT 26 Reference Manual Datablock results for the same faulted bus are displayed on the following one-line: AC (T0) 3842 A DC (T0) 4924 A Ip 9392 A G1 AC (T0) 3842 A DC (T0) 4924 A Ip 9392 A AC (T0) 3842 A DC (T0) 4924 A Ip 9392 A G2 G3 B1 AC (0. G3 and M2 match as expected. 3/26/2006 . These slight differences are due to neglecting motor pre-load condition in PTW and inconsistent rounding in the hand calculation.