A_fault

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Copyright © 1997, SKM Systems Analysis, Inc.All Rights Reserved.

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Information in this document is subject to change without notice. No part of this document may be reproduced ortransmitted in any form or by any means, electronic or mechanical, without the express written consent of SKMSystems Analysis, Inc. No patent liability is assumed with respect to the use of the information contained herein.Although every precaution has been taken in the preparation of this manual, the publisher and author assume noresponsibility for errors or omissions. Neither is any liability assumed for damages resulting from the use ofinformation contained herein. For information, address SKM Systems Analysis, Inc., PO Box 3376, ManhattanBeach, CA 90266-1376, USA.

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1.1 What is the A_FAULT Study? .........................................................................1-2

1.2 Engineering Methodology.................................................................................1-31.2.1 ANSI Standard C37.........................................................................................1-31.2.2 Comparing the ANSI and IEC Short Circuit Standards ..................................1-31.2.3 Withstand, Closing and Latching, and Momentary Rating .............................1-41.2.4 Interrupting Ratings.........................................................................................1-71.2.5 Adjusting Machine Contributions ...................................................................1-7

1.3 PTW Applied Methodology ..............................................................................1-81.3.1 Before Running the A_FAULT Study ............................................................1-81.3.2 Running the A_FAULT Study ........................................................................1-81.3.3 A_FAULT Study Options ...............................................................................1-9

Fault Type................................................................................................................1-9Faulted Bus..............................................................................................................1-9Calculation Models..................................................................................................1-9

Transformer Tap................................................................................................1-10Pre-fault Voltage (pu)........................................................................................1-10

Low Voltage ..........................................................................................................1-10Momentary and Interrupting..................................................................................1-10

Solution Method ................................................................................................1-10NACD Option....................................................................................................1-11

1.3.4 Component Modeling ....................................................................................1-11Feeder Data............................................................................................................1-11Transformer Data...................................................................................................1-12Three-Winding Transformers ................................................................................1-12Contribution Data ..................................................................................................1-14

1.3.5 Low Voltage Duty Report .............................................................................1-151.3.6 Momentary Duty Report................................................................................1-161.3.7 Interrupting Duty Report ...............................................................................1-18

Local and Remote Fault Contributions From Generators .....................................1-18ANSI C37.5 Considerations (Total Rated Basis) ..................................................1-18ANSI C37.010 Considerations (Symmetrical Rating Basis).................................1-19Using the NACD Options......................................................................................1-19

1.3.8 Modeling ANSI Decrement Curves ..............................................................1-20

1.4 Application Examples .....................................................................................1-261.4.1 Induction Motor ac decrement Factors..........................................................1-26

Case 1 ....................................................................................................................1-26Case 2 ....................................................................................................................1-28Case 3 ....................................................................................................................1-29

1.4.2 Modeling Transformers with Taps ................................................................1-30

A_FAULT ii Reference Manual

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1.4.3 Calculating Interrupting Duties..................................................................... 1-32Case 1.................................................................................................................... 1-32Case 2.................................................................................................................... 1-35

1.4.4 Example from Plant....................................................................................... 1-39

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This chapter examines the short-circuit current calculation procedures used in theA_FAULT Short Circuit Study. The chapter includes a systematic methodology andapplies the methodology to numerous practical examples. You can also run aComprehensive Short Circuit Study (in PTW-DAPPER) or an IEC Short Circuit Study (inIEC_FAULT). The IEC_FAULT Short Circuit Study and the Comprehensive ShortCircuit Study chapters discuss the Short Circuit Methodology applied by each Study, andthe standards followed by each; the IEC_FAULT Study is based on the IEC Standard 909,while the Comprehensive Short Circuit Study is based on Thevenin equivalent circuitrepresentation and Ohm’s law.

The A_FAULT Study follows the specifications of the American National StandardsInstitute (ANSI) C37.010, C37.5, and C37.13, and IEEE Standard 141, also known as theIEEE Red Book.

This chapter discusses:

• Engineering Methodology.

• PTW Applied Methodology.

• Examples.

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CH

AP

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What is the A_FAULT Study? ....................................................................A_FAULT 1-2Engineering Methodology............................................................................A_FAULT 1-3PTW Applied Methodology.........................................................................A_FAULT 1-8Application Examples ................................................................................A_FAULT 1-26

A_FAULT 1-2 Reference Manual

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The A_FAULT Short Circuit Study (referred to hereafter as A_FAULT) models thecurrent that flows in the power system under abnormal conditions and determines theprospective fault currents in an electrical power system. These currents must be calculatedin order to adequately specify electrical apparatus withstand and interrupting ratings. TheStudy results are also used to selectively coordinate time current characteristics ofelectrical protective devices.

Electrical apparatus manufactured in North America is predominantly tested and ratedagainst the ANSI, UL and NEMA equipment Standards; these Standards outline thepreferred method for calculating fault duties when specifying North American equipment.Equipment must withstand the thermal and mechanical stresses of short-circuit currents asdescribed in the ANSI Standards. Both rms and peak short circuit withstand andinterrupting duties (referred to as making and breaking short-circuit current duties,respectively) must be calculated and then compared to the protective device and electricalapparatus ratings.

Define System Data

Define system topology and connect ionsDefine feeder and transformer sizesDefine fault contr ibution data

Run A_FAULT Study

Study Setup

Cable LibraryTransformer LibraryStudy Setup

Saved in DatabaseThree-phase faul t currentsUnbalanced faul t currentsCalculated ANSI fault currents

Reports

Used by Time CurrentCoord inat ion (CAPTOR)

Datablocks

A_FAULT Study A_FAULT 1-3


(QJLQHHULQJ0HWKRGRORJ\Most often, North American engineers think of the American National Standards Institute(ANSI) series of C37 equipment rating Standards when specifying electrical apparatuswithstand capabilities, and in the case of protective devices, equipment contact breaking orinterrupting ratings. The IEEE applies the C37 series of ANSI Standards. The C37 seriesStandards deal with circuit interruption. Also, the IEEE has published the IEEE Red Book,IEEE Standard 141 as an authoritative application guide for the selection of equipment towithstand and interrupt short-circuit currents in an industrial/commercial power system.

The ANSI C37 series Standards must be used in conjunction with other engineeringStandards, such as Underwriters Laboratory (UL) Standard 489, and the NationalElectrical Manufacturers (NEMA) Standard AB-1. This following section outlines the keyprinciples in these Standards.

$16,6WDQGDUG&Short-circuit current testing procedures and associated equipment ratings are based first onthe ability to withstand the maximum thermal and mechanical stresses during a fault. Boththermal and mechanical stresses are proportional to the magnitude of the square of thecurrent flowing through the equipment during the fault condition. Short-circuit currentsare the largest during the first cycle of the fault, and high-magnitude short-circuit currentscan be sustained for several cycles, even up to 30 cycles in extreme situations. Currentcarrying parts of electrical apparatus must be sized to withstand, or in the case of switchingand protective devices, close in and latch onto the faulted circuit. Besides the thermalheating effects associated with large short-circuit currents, electrical apparatus internalcomponents and wiring must be mechanically braced to withstand the deleterious magneticstresses that can cause current-carrying parts to be stretched or repulsed from one anotherduring a fault condition.

Further, equipment designed to operate under fault conditionsthat is, designed to opencircuit breaker contacts and extinguish the associated arcing currents, or to separate fuselinks and control the associated arcing currentsmust be tested, and rated for such duty.This second short-circuit current rating is known as an equipment interrupting rating.

Low-voltage devices often combine the withstand and interrupting ratings into a singlerating. On the other hand, high-voltage circuit breakers may have intentionally differentwithstand and interrupting ratings, and may be specifically rated for delayed contactseparation. Engineers must compare the manufacturer’s published withstand andinterrupting equipment ratings against the calculated short-circuit current duties generatedby hand calculations or computer methods.

You may be most familiar with the ANSI, UL and NEMA Standards. However,equipment manufactured in Europe is predominantly manufactured to the IEC group ofStandards. The following section highlights the significant differences between the ANSIand IEC Standards as they pertain to short-circuit calculations.

&RPSDULQJWKH$16,DQG,(&6KRUW&LUFXLW6WDQGDUGVThere are three significant differences between the IEC methodology and ANSImethodology.


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The first major difference involves calculating the dc decay component. ANSI requirescalculation of a single Thevenin equivalent fault point X/R ratio, based on separatelyderived R and X values at the fault point. From that X/R ratio, a single equivalent dc decaycan be determined for multiple sources at the fault location. The IEC Standard uses aunique R/X ratio, calculated from the complex form of the R and X values at the faultlocation for each contribution, and uses this unique ratio for calculating the asymmetricalfault currents from each machine to the fault point. It could be argued that the IECStandard is current based, while the ANSI Standard is impedance based.

The second major difference also involves the dc decay component. Both Standardsrecognize that calculating the dc decay (the transient solution to the short-circuit currentcalculation) must be uniquely accomplished when parallel or meshed paths are involved.Both Standards consider the nature of meshed or parallel paths when concerned with the dcdecay; however, the two Standards use completely different procedures for calculating thedc decay current component when meshed or parallel paths are involved.

The third major difference involves the ac decrement component. The ANSI methodglobally adjusts the machine subtransient impedances when considering different momentsof time during the fault. The IEC method modifies the prospective short-circuit currentsavailable from each machine based on the transfer impedance between the active sourceand the specific fault location in question and the defined contact breaking time. Clearly,the IEC methodology is more computationally intensive than the ANSI methodology.

Both short circuit methodologies can be considered as quasi-steady-state solutions to theshort-circuit current problem, and both Standards acknowledge that a more dynamicsolution method might yield more accurate results. They do, however, claim sufficientaccuracy for specifying electrical equipment.

The results from IEC and ANSI calculations cannot be directly compared. While bothcalculate a withstand duty, the IEC and ANSI methodologies are fundamentally different.In the sample project, the ANSI closing and latching duty can, at times, be larger than theIEC peak current duty. However, in other sections of the project, the opposite is true. Asimilar disparity can be found between the IEC’s breaking current and the ANSI’ssymmetrical current interrupting duty. Thus, it can be concluded that when equipment israted in accordance with IEC Standards, then the IEC methodology must be used tocalculate the fault duties; and when equipment is rated in accordance with the ANSIStandards, then the ANSI methodology must be used to calculate the fault duties.

:LWKVWDQG&ORVLQJDQG/DWFKLQJDQG0RPHQWDU\5DWLQJElectrical apparatus is usually rated to allow a continuous and short-time overload currentto be sustained for significant periods of time. Ratings are usually based on the ability of

the device to reject the Julian ( i t2 ) heating that occurs during the range of practicaloperating conditions.

Equipment withstand ratings are based on the highest short-circuit current expected; this iscalled the prospective short-circuit current. The prospective short-circuit current usuallyoccurs during the first half-cycle of the fault. Withstand currents are composed of asymmetrical ac current and most likely, an aperiodic current component known as the dcdecay component. Withstand ratings may be published as equivalent first-cycle rmscurrents, or more accurately, as peak or crest currents for the first half-cycle of the fault.



As noted, the asymmetrical peak fault current consists of both ac and dc components, andis a function of time. As shown in the following figure, the theoretical asymmetrical peak

to peak current is 2 2 multiplied by the initial symmetrical rms current. The initialsymmetrical rms current ′′I k is the ratio of the driving point voltage at the bus to the

Thevenin equivalent impedance. The asymmetrical nature of a fault current is best shownby the following graph:

Current

Theoretical maximumPeak at 1/2 cycle

(DC decay)

Bottom envelope

Top envelope

Decaying (aperiodic) component i

Time

2

2 I"

i

k

p2 2 Ik

dc Asymmetrical valuesincluding motor contributions

Steady state value(no motor contributions)

dci

The first half-cycle asymmetrical peak current is the sum of the de decay and ac decrementcomponents. This can be expressed in equation form as:

I = 2 I 2 I easymmetrical peak k k

2c

XR′′ + ′′

− π

where

Iasymmetrical peak asymmetrical peak fault current;′′I k initial symmetrical rms fault current;

Idc dc decaying component of fault current;Ik steady-state fault current;c time in cycles into the fault.

Besides a withstand rating, electrical protective devices are rated on both their ability towithstand a first-cycle short-circuit current, and to close and latch into a faulted circuit. Inolder, high-voltage circuit breakers, this rating is known as the momentary rating. Theclosing and latching current rating is also associated with the making current of aswitching device.

Low-voltage devices are usually rated to withstand and interrupt a specified symmetricalrms current. Manufacturers test their protective devices against test asymmetricalwaveshapes, as noted in the following table of low-voltage test power factors andasymmetrical current withstand capabilities:

Protective DeviceTest

PF (%)TestX/R

Tested AsymmetricalWithstand Capability

LV Power Circuit Breaker 15 6.6 1.62

LV Fuse 20 4.9 1.53

Molded Case Circuit Breaker GT 20 KA AIC 20 4.9 1.53


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Protective DeviceTest

PF (%)TestX/R

Tested AsymmetricalWithstand Capability

Molded Case Circuit Breaker Between 10-20KA AIC

30 3.2 1.38

Molded Case Circuit Breaker LT 10 KA AIC 50 1.7 1.15

When low-voltage protective devices are applied in a system with an X/R greater than thetest X/R ratios listed above, then special care must be taken when specifying the device’ssymmetrical rating. In high X/R ratio situations, the symmetrical rating may not beexceeded by the calculated short circuit symmetrical duty; however, there is a possibilitythat the protective device’s tested asymmetrical withstand value could be exceeded. Low-voltage equipment is rated in symmetrical currents, but tested to a maximum asymmetricalrms 1/2 cycle test current based on the following formula:

I I 1 2easym rms cycle rms symm

-2

12

XR= × +π

When low-voltage equipment is specified in a circuit with a calculated X/R ratio greaterthan that for which the equipment was tested, you must ensure that both the symmetricaland asymmetrical half-cycle rated currents are not exceeded.

To determine the low-voltage withstand (and interrupting) duty when the system X/R ratiois greater than the test X/R for which the device is rated, the Standard calls for using thefollowing formula:

I I1 2e

1 2e

symm LVF rms symm

-2

system

-2

test

XR

XR

= × +

+

π

π

Special care must be used when calculating system X/R ratios. The ANSI C37 seriesStandards are explicit that the fault location X/R ratio must be calculated using the methodof separately-derived calculation. C37.13 and C37.010 require that the reactance at thefault point must be calculated ignoring resistance, and the resistance at the fault pointcalculated ignoring reactance. Systems with looped paths, or multiple short-circuit currentcontributions can exhibit a different X/R ratio when calculated using the separately-derived method versus the conventional method of solving for the Thevenin equivalentimpedance, and then calculating the X/R ratio from knowing the angle between theresistance and reactance of the Thevenin equivalent complex impedance.

High-voltage fuses may be rated to withstand either an rms symmetrical or asymmetricalrms current. High-voltage fuses may be tested with a test X/R of either 15 or 25. Themanufacturer must be consulted to confirm the test X/R if only the symmetrical value ispublished.

High-voltage circuit breakers manufactured since 1987 have a preferred closing andlatching rating expressed in the peak (crest) current. Breakers manufactured before 1987are rated in rms amperes symmetrical, but tested to withstand an asymmetrical rms currentof 160% of the published symmetrical rms current.

When a three-phase bolted fault occurs, it is assumed that the fault occurs such that themaximum asymmetric current occurs on Phase A. The other two phases are, respectively,120° and 240° delayed from Phase A. Thus, the complex form of the maximum fault



current in Phase A will be different than that in Phase B or C of a three-phase bolted fault.The Phase A half-cycle asymmetrical fault current is:

I I 1 2easym rms cycle rms symm12

-2X

R= × +π

But the average fault current due to each of the three individual phase currents of a three-phase fault is:

I I 1 2e 2 1 easym rms cycle ave rms symm12

-2X

R

-2X

R= × + + +

1

305

π π

.

It should be noted that the above equations are conservative, in that these equations arebased on the maximum fault current occurring at exactly one half-cycle into the fault. Itcan be shown that this is only true in a purely-inductive circuit when the fault occurs at avoltage zero.

,QWHUUXSWLQJ5DWLQJVAs stated earlier, low-voltage protective devicesboth fuses and circuit breakersarerated with a single interrupting rating. This is a symmetrical rating, usually in rms current.Also, the manufacturer tests the low-voltage protective device in accordance with NEMAStandard AB-1 to assure that the protective device will withstand, and operate under, aspecific (maximum) asymmetrical conditions.

Medium- and high-voltage circuit breakers are rated either on a Total Current orSymmetrical Interrupting Current Rating. These ratings are usually specified along with amaximum and minimum operating voltage, and a preferred contact opening time in cycles.

Circuit Breakers manufactured prior to 1964 base their short-circuit current interruptingrating on ANSI Standard C37.5. This was known as the Total Current basis, andconsidered both the ac decrement and dc decay characteristics of the calculated short-circuit current. Circuit breakers manufactured after 1964 are rated in accordance withANSI Standard C37.010, known as the Symmetrical Current basis. The primary differencebetween the Standards is the specific multiplying factor used to adjust the initialsymmetrical rms current at the expected breaker opening time. These multiplying factorsare based on a series of figures in each Standard, and are influenced by the amount ofpower system generation and the electrical distance between the power generation and thefault location. The generation may be defined as local or remote. Remote generation hasno ac decrement (NACD). As such, the Standards define local generation when:

Equivalent System Impedance to Fault Excluding X gen

X gen d

d

′′′′

<15.

Otherwise, the generator is considered remote to the fault.

$GMXVWLQJ0DFKLQH&RQWULEXWLRQVThe ANSI short-circuit current Standards have evolved since the 1940s due to the impactof increased motor contributions and the effect of their associated ac decrement. Theconcept of remote and local generation was introduced in the 1950s, and remains animportant element of the Standard methodology today. The definition of remote and localis based on the electrical distance between the generation source and the fault location.


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Local generation and motors have an ac decrement; their symmetrical wave shapedecreases with the onset of the fault. For simplicity in the ANSI Standards, the timevarying ac decrement is modeled as a constant voltage source behind a time-varyingimpedance. At the initial instant of the fault this impedance is known as the sub-transientimpedance, but this changes to the transient impedance two to three cycles into the fault.The steady-state currents are modeled as the machine’s synchronous impedance. In largemachines, the resistance component of the machine’s internal impedance is ignored.

Within the C37 series Standards, there is a set of rotating-machine reactance multipliers tobe used to model the time varying nature of the machine reactance. This set is shown inthe following table:

Type of rotating machine Low-voltagenetwork

First-cyclenetwork

Interruptingnetwork

All turbine generators; allhydrogenerators with amortisseurwindings; all condensers

1.0 Xd′′ 1.0 Xd′′ 1.0 Xd′′

Hydrogenerators without amortisseurwindings

0.75 Xd′′ 0.75 Xd′ 0.75 Xd′

All synchronous motors 1.0 Xd′′ 1.0 Xd′′ 1.5 Xd′′

Induction motors above 1000 hp at 1800r/min or less

1.0 Xd′′ 1.0 Xd′′ 1.5 Xd′′

Induction motors above 250 hp at 3600r/min

1.0 Xd′′ 1.0 Xd′′ 1.5 Xd′′

All other induction motors 50 hp andabove at 1800 r/min

1.0 Xd′′ 1.2 Xd′′ 3.0 Xd′′

All induction motors smaller than 50 hp 1.0 Xd′′ neglect neglect

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PTW applies the methodology described in Section 1.2. Section 1.3 describes how to runthe A_FAULT Study, including explanations of the various options associated with theStudy.

%HIRUH5XQQLQJWKH$B)$8/76WXG\Before running the A_FAULT Study, you must:

• Define the system topology and connections.

• Define feeder and transformer sizes.

• Define fault contribution data.

5XQQLQJWKH$B)$8/76WXG\You can run the Study from any screen in PTW, and it always runs on the active project.



X To run the A_FAULT Study

1. From the Run menu, choose Analysis.

2. Select the check box next to Short Circuit and choose the A_FAULT option button.

3. To change the Study options, choose the Setup button.

4. Choose the OK button to return to the Study dialog box, and choose the Run button.

The Short Circuit Study runs, writes the results to the database, and creates a report.

$B)$8/76WXG\2SWLRQVThe A_FAULT Study dialog box lets you select options for running the Study.

Following is a list of the available Study options.

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There are two options: Three Phase only, and 3 Phase-Unbalanced. The default is to reportonly the Three-Phase Study results. The 3 Phase-Unbalanced option allows you to studyboth the three-phase and the unbalanced fault networks (single-line-to-ground, line-to-line,and line-to-line-to-ground) in an abbreviated Report format.

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You can fault all buses or choose a specific bus to fault. If a fault is to be studied at asingle bus, then the faulted bus must be specified. The default is to study the fault currentsat all buses. If a single bus is faulted, then you can display the three-phase branch faultcontribution one branch away from the faulted location on a datablock.

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These options further customize the Study.


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You may model the primary transformer taps by selecting this check box. Secondary taps,if modeled, are ignored in the A_FAULT calculations. No adjustment in pre-fault voltage,associated with the transformer tap change, is automatically accomplished.

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The driving point voltage at the faulted bus or buses is defined in this box. The defaultdriving point voltage is 1.0 pu. The driving point voltage is not affected by the Swing Busper unit voltage, nor is it adjusted for transformer taps or the results from the Load FlowStudy.

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If you select the Low Voltage Duty check box, PTW will calculate the initial symmetricalshort-circuit currents at each specified bus, plus the half-cycle duty with no ac decrement.Also, the Three-Phase Report calculates the low-voltage protective device symmetricalduty required, based on the impact of the device test power factor. In the UnbalancedReport, the single-phase, and average three-phase (average Phase A plus Phase B and C)asymmetrical duty is calculated. All motor and generator contributions are included in theStudy results, using the user-specified subtransient reactances entered in the ANSIContribution subviews for Motor and Generator components.

The Study Report format may be selected as Complete or Summary. The Complete Reportformat is an extensive output format that includes a calculation of the branch contributionsto the faulted bus, whereas the Summary report format only includes a list of the three-phase and single-line-to-ground short-circuit current duties and the associated X/R ratios.

Because the low-voltage report calculates all motor contribution, and does not modelmotor ac decrement factors, this report format most closely resembles the ComprehensiveShort Circuit Study results. Therefore, all bus fault currents are reported, regardless ofvoltage.

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If you select the check box next to either or both of the Momentary and InterruptingDuties, PTW will calculate the initial symmetrical short-circuit current at each specifiedbus, modeling the reduction in motor fault current contribution (that is, considering motorac decrement) based on motor type, rated size, and speed (pole pairs). In the Three-PhaseReport, both the rms and peak (crest) asymmetrical current duty are calculated, based onthe 1.6 or 2.7 times test factor, and the asymmetrical rms and peak short-circuit currentduty based on the calculated X/R ratio at the fault location. In the Interrupting Report,each generator contribution is evaluated at each fault location to determine if itscontribution is considered remote or local to the fault location.

The Study Report format may be selected as Complete or Summary. The Complete Reportformat is an extensive output format that includes a calculation of the three-phase branchcontributions to the faulted bus, whereas the Summary only includes a list of the bus faultcurrents and the associated X/R ratios.

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You have the option of specifying an E/X or E/Z solution method. The Standards permitthe momentary fault currents to be calculated based on a value of E/X (voltage/reactance =current) which is conservative, or calculated based on a value of E/Z (voltage/impedance =current) which generally results in a smaller short-circuit current duty. The defaultsolution methodology is E/Z.



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For the Interrupting Duty Report, you can choose one of three operating modes, whichrepresent the most common interpretation of the ANSI Standards: All Remote,Predominant, and Interpolated. In All Remote mode, the NACD factor is 1.0; allgeneration is assumed to be remote; and the no ac decrement curves (dc decay curve only)are used. This is the most conservative solution. In Predominant mode, if the NACDfactor is greater than or equal to 0.5, then the dc decay only curve is used. If the NACDfactor is less than 0.5, the curves which model both ac decrement and dc decay are used.In Interpolated mode, interpolation between the dc decay curve and the ac decrement/dcdecay curves is used based on the percentage of generator contribution that is local andremote.

Note: The ANSI C37 Standards do not specifically recognize the interpolation methodof calculation. Technical papers state that the interpolation of the data between the dcdecay only and the ac decrement/dc decay curves of the Standard may provide morerepresentative results. The interpolation is based on the X/R ratio and the NACD factor.

&RPSRQHQW0RGHOLQJA_FAULT, IEC_FAULT, and the Comprehensive Short Circuit Study model electricalpassive devices identically. However, there are some key differences that should be noted.

One major difference is that the Comprehensive Study models secondary transformer taps,whereas A_FAULT and IEC_FAULT model only primary transformer taps. Anothermajor difference deals with the ac decrement for motors. A_FAULT modifies the motorsubtransient reactance using factors from the table in Section 1.2.5, “Contributions,”whereas the Comprehensive Study does not alter the machine subtransient reactances forac decrement. IEC_FAULT uses a series of special equations for determining machine acdecrement. Finally, the basis of the calculation of the fault location X/R is different.A_FAULT uses the method of separately-derived R and X, whereas IEC_FAULT and theComprehensive Short Circuit Study use the complex (vector) solution of the impedance todetermine the X/R ratio.

When you run the A_FAULT Study, PTW checks for appropriate feeder sizes and lengths,and transformer sizes in the Library. If the data is inappropriate or missing, error andwarning messages are shown in the Study Run dialog box and included in the Report.

The following sections describe the minimum data required for A_FAULT to run.

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You must specify a cable’s positive-sequence impedance and one-way circuit length.PTW models the negative-sequence impedance as equal to the positive-sequenceimpedance. If a cable’s zero-sequence impedance is zero, the Short Circuit Study uses thepositive-sequence value. Cable positive and zero sequence impedances may be selectedfrom the Cable Library, or you can define them in the Component Editor.

If you make the cable User Defined, you can enter specific cable positive- and zero-sequence impedance in ohms per 1000 feet or ohms per 1000 meters. Cable lengths mustbe entered in the same units as the cable impedance data (feet or meters). If you switch theProgram Options from English to Metric units, PTW converts entered cable lengths andimpedances to the appropriate units. Cable impedances are unaffected by the wire circuitdescription characteristics.


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You can select predefined two-winding transformers from the Transformer Library or youcan define them yourself in the Component Editor. PTW defines two-windingtransformers by their percentage leakage positive- and zero-sequence impedance value,cooling capacity type, and the nominal kVA rating. If a transformer's zero-sequenceimpedance is zero, PTW uses the positive-sequence value. Transformers' rated voltagesmay differ from the bus nominal voltages. PTW models those off-nominal voltages asideal voltage shifters separate from any primary or secondary tap that is modeled. Awarning message appears in the Study Messages dialog box when PTW detects a mismatchbetween the bus nominal voltage and the transformer rated voltage. You can also definethe transformer impedance in the Component Editor using the transformer's resistance andreactance values in percent on the nominal or self-cooled kVA rating.

When you set the engineering standard for the PTW Project to IEC, user-definedtransformers can be defined in per unit on any kVA base, the Rated Short Circuit Voltagepercent or on Rated Ohmic voltage percent.

Transformer negative-sequence impedance always equals the positive-sequence value inthe Short Circuit Study. The primary and secondary transformer connections helpdetermine the effect of the zero-sequence Thevenin equivalent impedance.

The wye-grounded wye-grounded zero-sequence path appears as a non-shunt primary tosecondary leakage impedance. Grounding impedance may be placed on one or both of thegrounded points. PTW automatically multiplies this grounding impedance by three tocalculate the proper zero-sequence impedance on a per unit base. The wye-grounded wye-grounded transformer is modeled in shell form, and is defined as an infinite impedancewhen viewed from either connection.

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Three-winding transformers may be modeled. Off-nominal voltage and transformer tapsmay be modeled in a manner similar to two-winding transformers. All three-windingtransformer data must be user defined in the Component Editor. PTW models the three-winding transformer using conventional network reduction, and establishes a fictitiouscenter point bus. Also, PTW establishes a secondary to tertiary branch. This fictitious busand associated branch count against the total bus and branch limit in PTW.

There are two networks in the following one-line diagram. Transformer T1 is a three-winding transformer with a primary, secondary and tertiary power rating of 15 MVA, 15MVA and 5.25 MVA, respectively.

GEN 1 GEN 2

BUS 1

C1BUS 2

T2

BUS 3 BUS 4

T1

C2 C3

BUS 5 BUS 6

BUS 7

C4BUS 8

BUS 9

T3

BUS 10

C5BUS 11

27,856.53 A

22,405.02 A

11,150.80 A21,706.12 A

20,520.80 A 10,860.47 A

CENTER POINT

0.8790%

21,705.22 A

20,520.00 A

T4

BUS 12

C6BUS 13

27,856.53 A

22,405.02 A

7199.51 A

6.0214%

9.9790%

11,150.56 A

10,860.25 A



The manufacturer’s published test data for Transformer T1 is:

Test#

ImpedanceMeasured

intoWinding

WindingShort

Circuited

WindingOpen

Circuited

ShortCircuit

Voltage in %

ImpedanceBase for

Measure (MVA)

ImpedanceSymbol

1 Primary Secondary Tertiary 6.9 15 ZPS

2 Primary Tertiary Secondary 5.6 5.25 ZPT

3 Secondary Tertiary Primary 3.8 5.25 ZST

It is important to note that the preceding measurements are relative to different powerbases. In Test 1 when the tertiary circuit is open, short-circuit current flows only inprimary and secondary windings. Both of these windings have 15 MVA ratings. In thetest, the voltage across the primary winding is increased until 6.9 % rated voltage causesthe rated full current to flow in the secondary winding. By opening the tertiary circuit, nocurrent flows in this winding.

In Test 2, the tertiary winding is fully loaded based on its 5.25 MVA rating, even thoughthe primary carries only about one-third rated current on its 15.0 MVA rating. The teststopped when the 5.6 % rated voltage was applied to the primary winding and full loadcurrent was reached on the tertiary winding (corresponding to 5.25 MVA). It is critical toknow on what base the short circuit voltage takes place. The following drawing is anequivalent impedance diagram for the three-winding transformer:

Z

Z PS

Z PT

T

S

PST

You can convert this into an equivalent wye diagram using standard network reductiontechniques:

Z

Z 2

Z 1

T

S

P

3

From the network reduction diagram, we can write:

Z = Z + Z

Z = Z + Z

Z = Z + Z

PS 1 2

PT 1 3

ST 2 3


3/12/98

Network reduction yields:

Z = (Z + Z - Z )

Z = (Z + Z - Z )

Z = (Z + Z - Z )

11

2 PS PT ST

21

2 PS ST PT

31

2 PT ST PS

You can solve the equations by substituting the manufacturer’s data expressed on acommon 15 MVA base:

Z = 6.9 + 5.615

5.25- 3.8

15

5.25

= 6.0214 %

Z = 6.9 + 315

5.25- 5

15

5.25

= 0.879 %

Z =15

5.25

15

5.256

= 9.978 %

11

2

21

2

31

2

j j j

j

j j j

j

j j j

j

×

×

!

"

$#

×

×

!

"

$#

×

+ ×

−

!

"

$#

. .

. . .

8 6

56 38 9

The above values represent the two-winding transformer equivalent impedances that mustbe used in the one-line diagram on page A_FAULT 1-12.

&RQWULEXWLRQ'DWD

Fault duty contributions to the power system originate from the motor generator and utilitysource components. PTW provides default subtransient and X/R ratio values. You cancalculate the machine kVA and voltage base using the rated size and connected busnominal voltage. For example, if you enter a 50 hp motor with an 80% power factor and92% efficiency, PTW calculates the rated kVA base as:

kVA =50 hp 746

1000 WkW

0.8 pf 0.92 efficiency

= 50.7 kVA

base

Whp×

× ×

This is close to the rule-of-thumb that 1 hp is equal to 1 kVA. Of course, if you have a1000 hp synchronous motor with a unity power factor, PTW calculates the motor’s kVAbase value as 746 kVA for short-circuit current purposes. The fault contributioncalculation remains unaffected by the motor load factor.

Fault contributions can be at any bus and there may be multiple contributions located atany bus.

Important: . You may change the ANSI contribution calculated kVA base, for example tomodel the 50 hp motor as 50.0 kVA. However in PTW, once the machine ANSIcontribution kVA base is selected (or automatically calculated by PTW if the base kVAvalue is 0), it will not change. Therefore, if you enter the motor load as 50 hp, run aStudy, and then change the motor’s rating to 75 hp, the motor’s ANSI contribution basekVA will remain 50 kVA. You must change the base kVA to 75 kVA manually.

Induction motors are modeled as delta-connected, whereas synchronous motors aremodeled as wye-connected. Neutral (grounding) impedance may be modeled.



/RZ9ROWDJH'XW\5HSRUWFor low-voltage Studies, A_FAULT calculates the initial symmetrical short-circuit currentat each bus in the power system based on the contribution data derived from ANSI C37.13.Under this Standard, the synchronous and induction motors are assumed to contribute at1.0 Xd". At each bus (including the high voltage bus records), the magnitude of the faultcurrent and the source of each contribution is reported.

In the partial one-line diagram and associated report below, the initial symmetrical short-circuit current for a fault at Bus 15 is reported as 8.66 kA. Close inspection of the reportshows that 8.223 kA originates from Bus 14, and the remainder of the fault current isgenerated from the four 25-hp motors directly connected to Bus 15, and contained in MCC15-1A.

015-MCC 1A

I Symm 8.661 kAX/R 4.138I Duty 9.265 kA

C14

Sub Feed #1

014-DSB 1


018-RA


MCC 15-1A

PNL 18 RA

MCCB1

LVP1

015-MCC 1A FAULT: 8.661 KA AT -74.64 DEG ( 7.20 MVA) X/R:

VOLTAGE: 480. EQUIV. IMPEDANCE= .0085 + J .

LOW VOLTAGE POWER CIRCUIT BREAKER 8.661 KA

MOLDED CASE CIRCUIT BREAKER < 20KA 9.265 KA

MOLDED CASE CIRCUIT BREAKER > 20KA 8.661 KA

CONTRIBUTIONS: M15-4 .112 KA AN G

M15-3 .112 KA AN G

M15-1 .112 KA AN G M15-2 .112 KA AN G

014-DSB 1 8.223 KA AN G

Likewise, the bus initial symmetrical rms fault current at Bus 14 is 9.993 kA and at Bus 18is 7.19 kA.

Inspection of the A_FAULT three-phase Low-Voltage Report shows that the X/R ratio atBus 15 is 4.14, which is below the 6.6 test X/R ratio if a LV Power Circuit Breaker(LVPCB) is selected. The symmetrical bus duty for an LVPCB is 8.66 kA, the same as theinitial symmetrical rms short-circuit current bus. However, if a 14 kA AIC molded-casecircuit breaker is selected, then the prospective symmetrical rms bus duty is calculated as9.265 kA, since the 14 kA AIC breaker is tested at an X/R ratio of 3.2 in accordance withthe table in Section 1.2.5, “Adjusting Machine Contributions,” but applied in a locationwhere the X/R ratio is 4.14.

The low-voltage report includes the symmetrical rms fault current and the direction andmagnitude of all contributions at all points in the system. This provides required data fordetermining the specific fault duty through the device versus the total fault duty at the bus.

The X/R value reported at the faulted bus is calculated by separate reduction of the X andR networks. The magnitude and angle of the contributions are calculated using thecomplex network.


3/12/98

An option in the Low-Voltage Report is the examination of unbalanced fault currents.A_FAULT will automatically generate the negative- and zero-sequence networks from theinput data specifications. The fault currents for single-line-to-ground, line-to-line, anddouble-line-to-ground are reported. In addition, the maximum rms current (assumed tooccur on Phase A) at one half-cycle for the three-phase and the single-line-to-ground faultconditions is reported, as shown in the following report:

U N B A L A N C E D F A U L T R E P O R T (FOR APPLICATION OF LOW VOLTAGE BREAKERS) PRE FAULT VOLTAGE: 1.0000 MODEL TRANSFORMER TAPS: NO ============================================================================== LOCATION FAULT KA X/R EQUIVALENT (PU) ASYM. KA AT 0.5 CYCLES VOLTAGE DUTIES (RMS) FAULT IMPEDANCE * MAX. RMS AVG. RMS * ==============================================================================

015-MCC 1A 3 PHASE: 8.661 4. Z1= 13.8875 10.387 9.544 SLG DUTY: 8.468 4. Z2= 13.8875 9.907 480. VOLTS LN/LN 7.501 Z0= 14.8453 LN/LN/GND 8.673 ( 8.282 GND RETURN KA)

In the low-voltage unbalanced report above, the three-phase fault at Bus 15 is reported(8.661 kA), the three-phase and single-line-to-ground (SLG) duty X/R ratios are reported(4), the positive-, negative-, and zero-sequence impedances on a 100 MVA pu base arereported, and the asymmetrical rms three-phase and SLG values at one half-cycle arereported. The maximum rms value is based on the assumption that the maximumasymmetrical current flows on Phase A; the average symmetrical rms current is thearithmetical mean current for Phase A plus Phase B and Phase C. For an SLG fault, thereis no current assumed to flow on Phase B or Phase C, thus no three-phase average value isreported. Both the line-to-line fault current (7.501 kA) and the line-to-line-to-ground busfault current (8.673 kA) are reported. For a line-to-line-to-ground fault current, thequantity of fault current flowing on the ground (i.e., the zero-sequence path) is alsoreported (8.282 kA).

A fault summary is also provided. The fault summary contains the three-phase and single-line-to-ground fault data, and the fault X/R ratios. Care must be taken when selectingprotective device ratings using strictly the three-phase report. In some cases, theunbalanced fault current (SLG or LLG current) may be larger than the three-phase short-circuit current. This is unlikely in most grounded low-voltage systems that have cableswith significant lengths.

0RPHQWDU\'XW\5HSRUWThe momentary duty or the closing and latching duty is the current that flows through themedium- and high-voltage system at one half-cycle after the fault occurs. This currentexhibits a dc decay from the symmetrical fault current, and the fault contribution frommotors rated less than 50 hp is ignored. The Standard also allows for a reduction ininduction motor contribution due to ac decrement. The values for momentary current arefound by first calculating the initial symmetrical rms current using the magnitude of theavailable fault currents from the machine reactance values specified in the table in Section1.2.5, “Adjusting Machine Contributions,” then applying a 1.6 symmetrical multiplier tomatch the ANSI simplified momentary rms calculations, and then calculating themomentary rms current based on the calculated fault circuit X/R value. Additionally, thepeak (crest) values are calculated:

I = I 1 2emomentary rms symm rms

-4 c

XR× +π

Eq. 7-1



I 2 I 1 emomentary peak symm rms

2c

XR= × × +

− π

where

c 1/2 cycle.

A_FAULT solves Eq. 7-1 at a time equal to one half-cycle to calculate the momentarycurrent permitted by the Standard. The Standard also allows the simple multiplication ofthe symmetrical current by a factor of 1.6 to determine the momentary rms current, and 2.7to determine the momentary peak (crest) current. It can be shown that at X/R = 25 andone half-cycle, Eq. 7-1 reduces to 1.6 times the symmetrical rms current.

The Standard requires separate reduction of the resistance and the reactance networks.These separate values of X and R are used to calculate the X/R ratio. This X/R ratio canbe significantly different from the value obtained by the vector (complex) circuitreduction. This is particularly true when there are significant parallel branches wheresignificantly different component X/R ratios are modeled.

While the separately reduced X/R ratio is used to calculate the momentary asymmetricalduty, the complex network is used for calculation of the direction and magnitude of each ofthe fault currents to the faulted bus.

A_FAULT will also calculate the unbalanced fault conditions for Momentary DutyStudies. As with the low-voltage report, the single-line-to-ground, the line-to-line and thedouble-line-to-ground fault currents are calculated. The momentary values for both thethree-phase and for the single-line-to-ground fault currents are calculated and reported.

BLDG 115 SERVI C/L 11.409 kAX/R 10.539 kA C10 C11

TX E

026-TX G PRII C/L 11.172 kAX/R 9.423 kA

025-MTR 25I C/L 10.989 kAX/R 11.467 kA

007-TX E PRI

I C/L 18.666 kAX/R 5.236 kA

M25

SW1

MO/L#25

F3

In the above one-line diagram and the following report, the closing and latching(momentary) three-phase short-circuit current duty at the Main Building 115 ServiceEntrance is reported as 11.409 kA rms or 19.252 kA peak asymmetrical current based on1.6 and 2.7 times multiplying factor in the Standard.

T H R E E P H A S E M O M E N T A R Y D U T Y R E P O R T PRE FAULT VOLTAGE: 1.0000 MODEL TRANSFORMER TAPS: NO ==============================================================================

BLDG 115 SERV E/Z: 7.130 KA AT -82.21 DEG ( 51.38 MVA) X/R: 10.54 SYM*1.6: 11.409 KA MOMENTARY BASED ON X/R: 10.337 KA SYM*2.7: 19.252 KA CREST BASED ON X/R: 17.568 KA VOLTAGE: 4160. EQUIV. IMPEDANCE= .0456 + J .3337 OHMS CONTRIBUTIONS: 007-TX E PRI 5.139 KA ANG: -80.39


3/12/98

026-TX G PRI .393 KA ANG: -84.15 025-MTR 25 1.608 KA ANG: -87.57

This X/R ratio is below the manufacturer’s test case X/R of 25, thus the reason thecalculated duty (10.337 kA) is less than the reported initial symmetrical current 1.6×value of 11.409 kA. These asymmetrical currents are calculated as the bus total, andcontributions flow are from Bus 7 (5.139 kA), Bus 26 (0.393 kA), and Bus 25 (1.608 kA).A_FAULT stores the 1.6 times initial symmetrical rms current duty in the database, as thisis the more conservative rms current duty in most industrial and commercial powersystems.

,QWHUUXSWLQJ'XW\5HSRUWThe interrupting duty Report generates short-circuit current duties for specification oftwo-, three-, five-, and eight-cycle breakers. The calculations assume that the breakers willopen (contact parting time) in 1.5-, 2-, 3-, and 4-cycles respectively.

A_FAULT applies the tables and graphs in the C37.5 and the C37.010 Standards tocalculate the interrupting duties. These graphs, combined with the factors which reducemotor fault contributions due to ac decrement, result in proper calculation of asymmetricalinterrupting rating values as required by either the Total Current Standard (C37.5) and theSymmetrical Current Rated Standard (C37.010).

/RFDODQG5HPRWH)DXOW&RQWULEXWLRQV)URP*HQHUDWRUV

ANSI requires short-circuit current contributions from generators to be classified as eitherlocal or remote. The definition of the contributions as local or remote will determine theappropriate ANSI figures which are used for determination of the ac decrement and dcdecay in the power system.

Remote sources are treated by ANSI as having no ac decrement. (Motor contributions aredecremented in A_FAULT by use of factors multiplied by the motor reactance values.Refer to the table in Section 1.2.5, “Adjusting Machine Contributions”). Contributionsspecified as generators or utility sources are treated as either local or remote as describedin the following sections.

$16,&&RQVLGHUDWLRQV7RWDO5DWHG%DVLV

The following paragraphs refer to Figure 1-1 through Figure 1-3 in Section 1.3.8,“Modeling ANSI Decrement Curves.” These figures are A_FAULT interpretations of theANSI Figures 1, 2, and 3 in ANSI Standard C37.5, para 3.2.1

A generator short-circuit current contribution is considered remote when the per-unitreactance external to the generator is equal to or greater than 1.5 times the generator perunit subtransient reactance on a common MVA base. Once the total generationcontribution at a faulted location is known, the appropriate interrupting factors may beused from the Standards. For breakers tested in accordance with ANSI Standard C37.5(pre-1964), use Figure 1-1 through Figure 1-3 to determine.the appropriate interruptingduty.

If the bus is determined to be remote from the generator, then Figure 1-3 is used for both athree-phase and single-line-to-ground fault condition.

When generators are determined as local, then both the ac decrement and the dc decaymust be accounted for. Figure 1-1, from ANSI Standard C37.5, is used for determining



multiplying factors for three-phase fault conditions with local generation. Figure 1-2 isused if a single-line-to-ground fault condition is studied.

The Standards recognize that the generation in a system may consist of both local andremote contributions and therefore permit the interpolation between Figure 1-1 and Figure1-3 for three-phase multiplying factors, and between Figure 1-2 and Figure 1-3 for thesingle phase factors in the total current rated Standard is possible. The interpolation is afunction of the X/R ratio and the ratio of remote fault current to total bus fault current.

In A_FAULT, each generator is evaluated as local or remote based on a determination ofthe per-unit reactance external to the generator being equal to or greater than 1.5 times thegenerator per unit Xd" on a common MVA base for each fault location. Then, thecontribution of each generator to the fault is determined.

$16,&&RQVLGHUDWLRQV6\PPHWULFDO5DWLQJ%DVLV

The concepts of local, remote and interpolation methods discussed above also apply to theANSI Standard C37.010, para. 5.3.2. Three-phase fault multiplying factors, which includethe effects of both ac decrement and dc decay (local generation) are in Figure 8 in ANSIStandard C37.010. The line-to-ground fault multiplying factors which include the effectsof the ac decrement and dc decay are shown in Figure 9 of the Standard. Additionally,Figure 10 of the Standard illustrates the three-phase and the line-to-ground faultmultiplying factors which include the effects of the dc decay only (remote generation).Figures 8, 9, and 10 of ANSI Standard C30.010 are reproduced as Figure 1-4 throughFigure 1-15 in Section 1.3.8, “Modeling ANSI Decrement Curves.”

A_FAULT permits you to specify the generator contributions as All Remote, Predominant,or Interpolated. The program methodology for the ANSI Standard C37.010 is the same asdescribed for the ANSI Standard C37.5 in the previous sections.

Although the ANSI Standard C37.010 permits the use of a simplifying technique toconvert multiplying factors calculated by the ANSI Standard C37.5 to the ANSI StandardC37.010 format, the results may differ from the multiplying factors taken from the curvesin the ANSI Standard C37.010. A_FAULT resolves this by using only the publishedcurves.

8VLQJWKH1$&'2SWLRQV

This following one-line diagram illustrates how A_FAULT uses the three NACD options:

B L D G 1 1 5 S E R VI init symm 6.330 kAX/R 9.780 kAI symm current 6.330 kAI total current 6.352 kA

C10 C11

T X E

026-TX G PRII init symm 6.205 kAX/R 8.888 kAI symm current 6.205 kAI total current 6.212 kA

025-MTR 25I init symm 6.082 kAX/R 10.001 kAI symm current 6.082 kAI total current 6.107 kA

007-TX E PRI

I init symm 10.741 kAX/R 4.844 kAI symm current 10.741 kAI total current 10.741 kA

M 2 5

SW1

MO/L#25

F3


3/12/98

In the above one-line diagram and the following Report, the interrupting three-phase short-circuit current duty at the Main Building 115 Service Entrance is reported as an initialsymmetrical rms current of 6.33 kA at an X/R of 9.78.

T H R E E P H A S E I N T E R R U P T I N G D U T Y R E P O R T PRE FAULT VOLTAGE: 1.0000 MODEL TRANSFORMER TAPS: NO NACD OPTION: INTERPOLATED ==============================================================================

BLDG 115 SERV E/Z: 6.330 KA AT -81.61 DEG ( 45.61 MVA) X/R: 9.78 VOLTAGE: 4160. EQUIV. IMPEDANCE= .0554 + J .3754 OHMS CONTRIBUTIONS: 007-TX E PRI 5.088 KA ANG: -80.19 026-TX G PRI .173 KA ANG: -85.29 025-MTR 25 1.077 KA ANG: -87.74GENERATOR NAME -- AT BUS -- KA VOLTS PU LOCAL/REMOTE 001-EDISON 3.615 .97 R 008-GEN 1 .181 .93 R 020-GEN 2 .418 .90 L TOTAL REMOTE: 3.796 KA NACD RATIO: .5997

SYM2 SYM3 SYM5 SYM8 MULT. FACT: 1.000 1.000 1.000 1.002 DUTY (KA) : 6.330 6.330 6.330 6.344

TOT2 TOT3 TOT5 TOT8 MULT. FACT: 1.211 1.052 1.004 1.000 DUTY (KA) : 7.668 6.656 6.352 6.330

At the BLDG 115 SERV Bus, the total fault current due to generators is 4.2 kA, of whichthe generator 020-GEN2 at Bus 20 is considered local to the faulted bus. The total remotegeneration is 3.796 kA which is 59.97% of the total fault current (6.33 kA) available at thebus.

The output report lists the interrupting duty in accordance with both the symmetrical basisand the total current basis. If a 5-cycle breaker is selected, the Symmetrical Current basisduty is 6.33 kA, but if a 5-cycle breaker is selected which is rated on a Total Current basis,then it must be specified on a short circuit interrupting duty of 6.352 kA.

0RGHOLQJ$16,'HFUHPHQW&XUYHVThe published ANSI figures used for determining the interrupting duty decrement factorsare, at best, difficult to read. During the creation of A_FAULT, the published curves werephotographically enlarged and data points interpolated. These data points were thenentered into a graphic utility program and the interpolated points plotted. The graphicutility program permitted the output curves to be scaled to the same size as the enlargedcurves taken from the standard. The interpolated curves were then compared directly withthe published curves. The interpolated results agreed with the published curves.

The ac decrement/dc decay curves used by A_FAULT are discussed in this section. Figure1-1, Figure 1-2, and Figure 1-3 represent the ac decrement/dc decay curves used by theANSI Standard C37.5. Figure 1-4 through Figure 1-15 represent the ac decrement/dcdecay curves used by the ANSI Standard C37.010. In Figure 1-4 through Figure 1-15, theac decrement/dc decay curves corresponding to the breaker contact parting time are shown.The ac decrement/dc decay curves for contact parting at other than 1.5-, 2-, 3-, and 4-cycles are not shown as they are not required for the A_FAULT solution. Sufficientinformation in A_FAULT is provided to determine other multiplying factors from thestandards for breakers which may have slower contact parting times.

It should be noted that the ac decrement/dc decay curves published by ANSI StandardC37.010 occasionally result in multiplying factors which are not always intuitive. Forexample, careful examination of the symmetrical standard for three-phase faults with localgeneration indicates multiplying factors for a five-cycle breaker (3-cycle parting time) at a



value of X/R = 60 will result in a multiplying factor of 1.167. With the same X/R value,an eight-cycle breaker (4-cycle parting time) will have a multiplying factor of 1.180.Although the eight-cycle breaker opens under a lower asymmetrical current, the breakertakes longer for contact parting, thus a higher asymmetrical rating requirement.

Examination of the dc decay curves (remote sources) further illustrates that the intuitiveunderstanding of asymmetrical current values does not correspond to asymmetrical ratingscalculated by the ANSI Standard C37.010. For example, examination of a system with anX/R ratio of 30, the multiplying factors increase for slower operating breakers.

For all of the following drawings, the vertical axis is the X/R ratio based on the separatereduction of the R and X networks. Likewise, for all of the following drawings, thehorizontal axis represents the ANSI multiplying factors.


3/12/98

Figure 1-1. C37.5 Three-phase ac decrement/dc decay curves.

Figure 1-2. C37.5 Single-line-to-ground ac decrement/dc decay curves.



Figure 1-3. C37.5 Three-phase and single-line-to-ground DC decay curves (dc decay only).

Figure 1-4. C37.010 Three-phase ac decrement/dcdecay curve for 8-cycle breakers.

Figure 1-5. C37.010 Three-phase ac decrement/dcdecay curve for 5-cycle breakers.


3/12/98

Figure 1-6. C37.010 Three-phase ac decrement/dc

decay curve for 3-cycle breakers.Figure 1-7. C37.010 Three-phase ac decrement/dcdecay curve for 2-cycle breakers.

Figure 1-8. C37.010 Single-line-to-ground ac decrement/dc decay curve for 8-cycle breakers.






Figure 1-12. C37.010 Three-phase and single-line-to-ground dc decay curve for 8-cycle breakers (dc decay only).

Figure 1-13. Three-phase and single-line-to-ground dc decay curve for 5-cycle breakers (dc decay only).


3/12/98

Figure 1-14. Three-phase and single-line-to-ground dcdecay curve for 3- cycle breakers (dc decay only).

Figure 1-15. Three-phase and single-line-to-grounddc decay curve for 2 cycle breakers (dc decay only).

$SSOLFDWLRQ([DPSOHV

The examples that follow illustrate how the A_FAULT Study calculates the various short-circuit duties, given different project topologies. Unless otherwise specified, all per unitvalues are expressed on a 100 MVA base at the bus system nominal voltage.

,QGXFWLRQ0RWRUDFGHFUHPHQW)DFWRUVThis first example comprises three separate cases in which different combinations ofinduction motors are modeled to determine the impact of the ANSI ac decrement factor onthe ANSI fault current. The factors listed in the table in Section 1.2.5, “Adjusting MachineContributions” represent those modeled.

&DVH

In the following one-line diagram, a series of 4160 V 25 hp motors and a single 50 hpmotor are modeled as shown:



B1

C1

B2

C2

B3

M1

Size 25.0 hp# of Mtrs: 2

U1

C3

B4

M2


C4

B5

M4


M3


Note that the total motor short-circuit contribution from each of the three motor buses is 50hp. A fault is created at Bus B2, and inspection of the following output Report shows thatthe fault duty contributions from each of the three-motor branches are equal (47 A).

T H R E E P H A S E F A U L T R E P O R T (FOR APPLICATION OF LOW VOLTAGE BREAKERS) PRE FAULT VOLTAGE: 1.0000 MODEL TRANSFORMER TAPS: NO ============================================================================== B2 FAULT: 13.598 KA AT -86.14 DEG ( 97.98 MVA) X/R: 14.85 VOLTAGE: 4160. EQUIV. IMPEDANCE= .0119 + J .1762 OHMS CONTRIBUTIONS: B1 13.456 KA ANG: -86.16 B3 .047 KA ANG: -84.28 B4 .047 KA ANG: -84.28 B5 .047 KA ANG: -84.28

This is the expected result. In the Low-Voltage network Report all motors (regardless ofthe rated voltage of the motor) are modeled with a machine reactance of 1.0 times the ratedsubtransient reactance. This is compared to the medium/high-voltage momentary (orclosing and latching Report), where motors less than 50 hp are ignored. Note the outputresults below:


B2 E/Z: 13.496 KA AT -86.16 DEG ( 97.24 MVA) X/R: 14.89 SYM*1.6: 21.593 KA MOMENTARY BASED ON X/R: 20.519 KA SYM*2.7: 36.438 KA CREST BASED ON X/R: 34.541 KA VOLTAGE: 4160. EQUIV. IMPEDANCE= .0119 + J .1776 OHMS CONTRIBUTIONS: B1 13.456 KA ANG: -86.16 B5 .039 KA ANG: -84.28

Note that the motor contributions from Bus B3 and Bus B4 are not included in the outputreport. Also, note that the fault duty contribution from the 50 hp motor at Bus B5 is only83% of the value reported in the low-voltage report. This is equivalent to increasing themotor’s subtransient reactance by 1.2 times. Finally, note that the initial symmetrical rmsshort-circuit current is only 13.496 kA at Bus B2, compared to 13.598 kA in the previousReport.

The Interrupting Report is printed below:



3/12/98

B2 E/Z: 13.472 KA AT -86.16 DEG ( 97.07 MVA) X/R: 14.90 VOLTAGE: 4160. EQUIV. IMPEDANCE= .0119 + J .1779 OHMS CONTRIBUTIONS: B1 13.456 KA ANG: -86.16 B5 .016 KA ANG: -84.29

GENERATOR NAME -- AT BUS -- KA VOLTS PU LOCAL/REMOTE U1 13.456 .06 R TOTAL REMOTE: 13.456 KA NACD RATIO: .9988

The fault duty contribution from the single 50 hp motor at Bus B5 is now only 34% of thevalue reported in the Low-Voltage Report. This is equivalent to increasing the motor’ssubtransient reactance by 3.0 times and is consistent with a reduced ac decrementcomponent during interrupting. Also, observe that the initial symmetrical rms short-circuitcurrent is now 13.472 kA at Bus B2, reflecting the further reduction in motorcontributions. In each of the three Reports, the short-circuit current from the utility sourcewas 13.456 kA.

&DVH

This second example follows from the format of Case 1. The 4160 V motors have beenresized, as shown in the following one-line diagram:

B1

C1

B2

C2

B3

M1


U1

C3

B4

M2


C4

B5

M4


First the Comprehensive Short Circuit Study is run for a fault at Bus B2:

***************** F A U L T A N A L Y S I S R E P O R T ****************

FAULT TYPE: 3PH MODEL INDUCTION MOTOR CONTRIBUTION: YES MODEL TRANSFORMER TAPS: NO MODEL TRANSFORMER PHASE SHIFT: NO

B2 VOLTAGE BASE LL: 4160.0 (VOLTS) INI. SYM. RMS FAULT CURRENT: 15014.0 / -86. ( AMPS/DEG ) THEVENIN EQUIVALENT IMPEDANCE: .065 +j .922 (PU) THEVENIN IMPEDANCE X/R RATIO: 14.105

ASYM RMS INTERRUPTING AMPS 1/2 CYCLES 3 CYCLES 5 CYCLES 8 CYCLES 22675.9 16017.4 15187.5 15026.0

B2 ==== INI. SYM. RMS SYSTEM BRANCH FLOWS ( AMPS/DEG ) ======= ALL BRANCHES REPORTED AT TIME = .5 CYCLES VOLTS --PHASE A--- ---PHASE B--- ---PHASE C

M1 4160. 94.7/ -84. 94.7/ 156. 94.7/ 36. M2 4160. 283.9/ -84. 283.9/ 156. 283.9/ 36. M4 4160. 1180.1/ -84. 1180.1/ 156. 1180.1/ 36.

*****************************************************************************



The results of A_FAULT’s low-voltage report for the same fault bus (B2) are:

T H R E E P H A S E F A U L T R E P O R T (FOR APPLICATION OF LOW VOLTAGE BREAKERS) PRE FAULT VOLTAGE: 1.0000 MODEL TRANSFORMER TAPS: NO ============================================================================== B2 FAULT: 15.014 KA AT -85.94 DEG ( 108.18 MVA) X/R: 14.36 VOLTAGE: 4160. EQUIV. IMPEDANCE= .0113 + J .1596 OHMS CONTRIBUTIONS: B1 13.456 KA ANG: -86.16 B3 .095 KA ANG: -84.27 B4 .284 KA ANG: -84.22 B5 1.180 KA ANG: -84.02

Close examination of the two Reports shows that the bus initial symmetrical rms currentand each of the motor contributions are identical in magnitude. In fact, the only significantdifference between the two Reports is the reported X/R ratio. The Comprehensive Studyreported an X/R ratio of 14.11, whereas the A_FAULT study Low-Voltage Reportcalculated the X/R ratio as 14.36. Indeed, even the difference in calculated X/R ratiobetween the two methods is not particularly significant in this case.

Next, A_FAULT’s Momentary (closing and latching) Report is examined:


B2 E/Z: 14.951 KA AT -85.95 DEG ( 107.73 MVA) X/R: 14.38 SYM*1.6: 23.922 KA MOMENTARY BASED ON X/R: 22.634 KA SYM*2.7: 40.368 KA CREST BASED ON X/R: 38.137 KA VOLTAGE: 4160. EQUIV. IMPEDANCE= .0113 + J .1602 OHMS CONTRIBUTIONS: B1 13.456 KA ANG: -86.16 B3 .079 KA ANG: -84.27 B4 .237 KA ANG: -84.23 B5 1.180 KA ANG: -84.02

The 100 hp and 300 hp motors are modeled with a 1.2 times subtransient reactance acdecrement factor, whereas the 1250 hp motor does not use a multiplying factor, based onthe ac decrement factors from the table in Section 1.2.5, “Adjusting MachineContributions.”

&DVH

This third example follows from the first two cases. The 4160 V motors have been re-sized once again, as shown in the following one-line diagram:


3/12/98

B1

C1

B2

C2

B3

M1


U1

C3

B4

M2


1800 rpm 3600 rpm

In this problem each of the two induction motors is rated 300 hp, but Motor M1 is rated at1800 rpm, whereas Motor M2 is rated at 3600 rpm. The A_FAULT Momentary (closingand latching) Report predicts the following ac decrement factors for the two motors:


B2 E/Z: 13.976 KA AT -86.09 DEG ( 100.70 MVA) X/R: 14.72 SYM*1.6: 22.362 KA MOMENTARY BASED ON X/R: 21.220 KA SYM*2.7: 37.736 KA CREST BASED ON X/R: 35.732 KA VOLTAGE: 4160. EQUIV. IMPEDANCE= .0117 + J .1714 OHMS CONTRIBUTIONS: B1 13.456 KA ANG: -86.16 B3 .237 KA ANG: -84.23 B4 .284 KA ANG: -84.22

As expected, the slower motor (Motor M1 on Bus B3) has a smaller momentary short-circuit current contribution (237 A) than the faster motor attached to Bus B4 (284 A).

0RGHOLQJ7UDQVIRUPHUVZLWK7DSVIn this example the impact of modeling transformer taps in A_FAULT is investigated. Thefollowing one-line diagram posts the results of the Comprehensive Short Circuit Study,modeling the effects of transformer taps.



B1

14753.23 Amps 3P13600.55 Amps SLG C1

B2

14266.92 Amps 3P12999.28 Amps SLG

U1

G1

TX1

B3

35266.01 Amps 3P37492.43 Amps SLG C2

B4

19772.67 Amps 3P16101.20 Amps SLG

M1 L1

Because of a significant steady-state load flow voltage drop at Bus B4, a -2.5% primarytransformer tap is placed on Transformer TX1. The pre-fault no-load voltage used tomodel the short-circuit current for a fault at Bus B4 is 1.0256 % of nominal due to the-2.5% primary tap. The results of the Comprehensive Short Circuit Report are posted onthe one-line diagram. The calculated pre-fault, no-load voltage at Bus B4, given the-2.5% primary tap, is 1.0256 pu V.

Rerunning the study using A_FAULT and modeling a 1.0256 pu driving point voltageyields the same fault current magnitude at Bus B4, as compared to the ComprehensiveShort Circuit Study; this is noted in the following Report:

T H R E E P H A S E F A U L T R E P O R T (FOR APPLICATION OF LOW VOLTAGE BREAKERS) PRE FAULT VOLTAGE: 1.0256 MODEL TRANSFORMER TAPS: YES ==============================================================================

B4 FAULT: 19.772 KA AT -55.84 DEG ( 16.86 MVA) X/R: 2.59 VOLTAGE: 480. EQUIV. IMPEDANCE= .0081 + J .0119 OHMS LOW VOLTAGE POWER CIRCUIT BREAKER 19.772 KA MOLDED CASE CIRCUIT BREAKER < 20KA 19.772 KA MOLDED CASE CIRCUIT BREAKER > 20KA 19.772 KA CONTRIBUTIONS: M1 4.208 KA ANG: -84.29 B3 16.196 KA ANG: -48.7

The results between the Comprehensive Short Circuit Study and the A_FAULT Studymatch.


3/12/98

If the transformer tap is ignored, and the pre-fault voltage is reset to 1.0 pu, A_FAULTpredicts a slightly smaller fault current at Bus B4:

T H R E E P H A S E F A U L T R E P O R T (FOR APPLICATION OF LOW VOLTAGE BREAKERS) PRE FAULT VOLTAGE: 1.0 MODEL TRANSFORMER TAPS: NO ==============================================================================B4 FAULT: 19.360 KA AT -55.63 DEG ( 16.10 MVA) X/R: 2.58 VOLTAGE: 480. EQUIV. IMPEDANCE= .0081 + J .0118 OHMS LOW VOLTAGE POWER CIRCUIT BREAKER 19.360 KA MOLDED CASE CIRCUIT BREAKER < 20KA 19.360 KA MOLDED CASE CIRCUIT BREAKER > 20KA 19.360 KA CONTRIBUTIONS: M1 4.103 KA ANG: -84.29 B3 15.882 KA ANG: -48.52

This case may also be used to understand the significance of examining both the three-phase and the single-line-to-ground fault Reports. It is important to understand that thethree-phase bolted fault may not be the largest fault current available at the bus. Forexample, A_FAULT predicts that the single-line-to-ground fault current at Bus B3 will belarger than the three-phase short-circuit current. Note the following one-line results fromA_FAULT:

B1

I Symm 14.733 kA 3 PhI Symm 13.589 kA SLG C1

B2

I Symm 14.246 kA 3 PhI Symm 12.988 kA SLG

U1

G1

TX1

B3

I Symm 34.795 kA 3 PhI Symm 36.864 kA SLG C2

B4

I Symm 19.360 kA 3 PhI Symm 15.732 kA SLG

M1 L1

Close in to the delta-wye-grounded transformer, the single-line-to-ground fault current islarger than the three-phase short-circuit current.

&DOFXODWLQJ,QWHUUXSWLQJ'XWLHVThis example investigates how A_FAULT calculates the interrupting duties associatedwith medium-voltage protective devices. Two issues of importance must be examined.First, how the generator local/remove determination is made, and second, how A_FAULTuses the various ac decrement figures to determine protective device Total Current andSymmetrical Current duties.

&DVH

Consider the following one-line with the per unit impedances (expressed on a 100 MVAbase) as noted. Note that in this example resistances have been ignored:



B1

C1

B2

G1

X1 = 0.2 PU

C2

B3

C3

B4

C7

C4

B5

X1 = 0.4 PU

X1 = 0.6 PU

X1 = 0.9 PU

X1 = 0.1 PU

X1 = 1.0 PU

A portion of the A_FAULT three-phase interrupting report is reported below for theimpedances modeled above:


B4 E/Z: 8.292 KA AT -89.65 DEG ( 59.75 MVA) X/R: 165.70 VOLTAGE: 4160. EQUIV. IMPEDANCE= .0017 + J .2896 OHMS CONTRIBUTIONS: B2 4.363 KA ANG: -89.65 B3 3.929 KA ANG: -89.65

GENERATOR NAME -- AT BUS -- KA VOLTS PU LOCAL/REMOTE G1 8.292 .40 L TOTAL REMOTE: .000 KA NACD RATIO: .0000

B5 E/Z: 7.825 KA AT -89.67 DEG ( 56.38 MVA) X/R: 175.60 VOLTAGE: 4160. EQUIV. IMPEDANCE= .0017 + J .3069 OHMS CONTRIBUTIONS: B4 7.825 KA ANG: -89.67


Inspection of this partial report shows that at for a fault at either Bus B4 or Bus B5,A_FAULT has determined that the single generator at Bus B1 is considered local to thefaulted bus.

The impedance of the branch from Bus B4 to Bus B5 is now increased, as noted below:


3/12/98

B1

C1

B2

G1

X1 = 0.2 PU

C2

B3

C3

B4

C7

C4

B5

X1 = 0.4 PU

X1 = 0.6 PU

X1 = 0.9 PU

X1 = 0.95 PU

X1 = 1.0 PU


B4 E/Z: 8.292 KA AT -89.65 DEG ( 59.75 MVA) X/R: 165.70 VOLTAGE: 4160. EQUIV. IMPEDANCE= .0017 + J .2896 OHMS CONTRIBUTIONS: B2 4.363 KA ANG: -89.65 B3 3.929 KA ANG: -89.65


B5 E/Z: 5.290 KA AT -89.78 DEG ( 38.12 MVA) X/R: 259.76 VOLTAGE: 4160. EQUIV. IMPEDANCE= .0017 + J .4540 OHMS CONTRIBUTIONS: B4 5.290 KA ANG: -89.78 GENERATOR NAME -- AT BUS -- KA VOLTS PU LOCAL/REMOTE G1 5.290 .62 R TOTAL REMOTE: 5.290 KA NACD RATIO: 1.0000

As the impedance of the branch from Bus B4 to Bus B5 increased from 0.1 to 0.95 pu Ω,A_FAULT determined that when the fault occurred at Bus B5, the Generator contributionswitched from being a local contribution, where ac decrement and dc decay is considered,to a remote contribution where there is no ac decrement considered for the generator inquestion.

Para 5.3.2 of ANSI Standard C37.010 states that a generator should be considered remoteto the fault location if the per-unit reactance external to the generator is equal to or greaterthan 1.5 times the generator per-unit sub transient reactance on a common system MVAbase. In Case 1, the total system reactance (on a 100 MVA base) between Bus B1 to BusB5 is:

X pu

= 0.2 + 0.45+ 0.1 pu

= 0.75 pu

total = ++ ×+ +

+0 20 4 05 09

0 4 05 0901.

. . .

. . ..

1 6



The generator subtransient reactance is 1.0, and 1.5 times Xd”gen (1.5 pu Ω) is less thanthe system impedance less the Generator (0.75 pu), thus A_FAULT determined that whenthe fault is at Bus B5, the Generator contribution from Bus B1 is a local contribution, andac decrement is considered.

However in the second study, the branch reactance from Bus B4 to Bus B5 is increasedfrom 0.1 pu to 0.95 pu, thus making the total external system reactance between Bus B1 toBus B5 equal to 1.6 pu. This makes the total external reactance greater than 1.5 times Xd”gen, therefore A_FAULT determined in this study that when the fault is at Bus B5 theGenerator contribution from Bus B1 is a remote contribution, and no ac decrement isconsidered.

&DVH

In this second example a small industrial power system with co-generation is modeled, asshown in the following one-line diagram. The Load Flow Study data presented on theone-line diagram shows that the system load is approximately 12 MVA, and a 2 MVA co-generator supplies some of the plant load on a continuous basis:

UTIL FDR BUS13800 V1.08 % VD

UTIL-0001

8187.41 kW3046.42 kvar

C18181.50 kW2780.41 kvarTX A SEC BUS

13800 V1.26 % VD

TXA

8167.88 kW2773.68 kvar

SW-YARD BUS

4160 V0.21 % VD

C2

8096.81 kW2781.32 kvar

PLANT MAIN4160 V0.25 % VD

C3

1498.69 kW749.21 kvar

C4

4084.45 kW2315.12 kvar

C5

5508.89 kW1211.20 kvar

CO GEN BUS

4160 V0.16 % VD

G1

1500.00 kW750.00 kvar

PROCESS A BUS

4160 V0.52 % VD

TXB

4074.21 kW2807.66 kvar

PROC A MTR BUS

2400 V2.45 % VD

PROCESS B BUS

4160 V0.43 % VD

PROC B MTR BUS

4160 V0.80 % VD

M1 LOAD

2500.0 hp

PROC B MTR LD

5000.0 hp

NON MTR LOADS

2500.0 kVA

C6

3899.60 kW5.61 kvar

BLDG 115 LD2000.0 kVA

PF CAPACITOR

500.0 kVA

CAP BANK #2500.0 kVA

Next the three-phase Comprehensive Short Circuit current Report is run on the aboveproject, first ignoring the effects of transformer taps and associated change in pre-fault no-load bus voltages due to the transformer taps. From the following one-line diagram, noteparticularly the bus fault currents on the PROC A MTR BUS where there is a lumped loadof 5 MVA, split equally between motor and non-motor loads.


3/12/98

UTIL FDR BUS13543.65 Amps 3PX/R 42.18

UTIL-0001

C1

TX A SEC BUS13231.92 Amps 3PX/R 14.95

TXA

SW-YARD BUS20118.43 Amps 3PX/R 14.12

C2

PLANT MAIN20065.35 Amps 3PX/R 13.90

C 3 C4 C5

CO GEN BUS19314.84 Amps 3PX/R 8.19

G1

PROCESS A BUS19331.02 Amps 3PX/R 11.75

TXB

PROC A MTR BUS15593.57 Amps 3PX/R 11.28

PROCESS B BUS19698.84 Amps 3PX/R 10.27

PROC B MTR BUS18815.79 Amps 3PX/R 5.79

M1 LOAD PROC B MTR LDNON MTR LOADS

C6

BLDG 115 LD

PF CAPACITOR

CAP BANK #2

As noted in the Comprehensive Short Circuit Report following, the initial symmetrical rmsshort-circuit current at the PROC A MTR BUS is three-phase with a positive-sequencenetwork X/R ratio of over 11. The Comprehensive Short Circuit Study predicts anasymmetrical rms current at five cycles into the fault of 15652.9 A, and even by eightcycles into the fault there is some (slight) dc offset current present.

PROC A MTR BUS VOLTAGE BASE LL: 2400.0 (VOLTS) INI. SYM. RMS FAULT CURRENT: 15593.6 / -85. ( AMPS/DEG ) THEVENIN EQUIVALENT IMPEDANCE: .136 +j 1.537 (PU) THEVENIN IMPEDANCE X/R RATIO: 11.282

ASYM RMS INTERRUPTING AMPS 1/2 CYCLES 3 CYCLES 5 CYCLES 8 CYCLES 22843.0 16135.9 15652.9 15595.7

Next, the A_FAULT Study is run with transformer taps ignored, as shown in the results onthe following one-line:



UTIL FDR BUSI symm 13.362 kAX/R 55.568 kA

UTIL-0001

C1

TX A SEC BUSI symm 13.048 kAX/R 15.452 kA

TXA

SW-YARD BUSI symm 18.218 kAX/R 16.019 kA

C2

PLANT MAINI symm 18.160 kAX/R 15.830 kA

C3 C4 C5

CO GEN BUSI symm 17.561 kAX/R 11.960 kA

G1

PROCESS A BUSI symm 17.536 kAX/R 13.097 kA

TXB

PROC A MTR BUSI symm 13.871 kAX/R 11.885 kA

PROCESS B BUSI symm 17.818 kAX/R 11.299 kA

PROC B MTR BUSI symm 17.004 kAX/R 7.323 kA

M1 LOAD PROC B MTR LDNON MTR LOADS

C6

BLDG 115 LD

PF CAPACITOR

CAP BANK #2

The associated interrupting report for bus PROC A MTR BUS is shown below:

T H R E E P H A S E I N T E R R U P T I N G D U T Y R E P O R T PRE FAULT VOLTAGE: 1.0000 MODEL TRANSFORMER TAPS: NO NACD OPTION: INTERPOLATED

PROC A MTR BUS E/Z: 13.871 KA AT -85.05 DEG ( 57.66 MVA) X/R: 11.88 VOLTAGE: 2400. EQUIV. IMPEDANCE= .0086 + J .0995 OHMS CONTRIBUTIONS: M1 LOAD 2.114 KA ANG: -84.29 PROCESS A BUS 11.757 KA ANG: -85.19

GENERATOR NAME -- AT BUS -- KA VOLTS PU LOCAL/REMOTE UTIL-0001 9.401 .86 R G1 1.307 .59 L TOTAL REMOTE: 9.401 KA NACD RATIO: .6778



Note that the initial symmetrical rms short-circuit interrupting duty is calculated as 13.87kA. The difference between the 15593.6 A short-circuit current calculated in theComprehensive report and the 13.87 kA above is due to the modeling of motor acdecrement allowed by the Standard and used in the A_FAULT calculations. TheComprehensive Study does not explicity model the change in subtransient reactance ofmachines.


3/12/98

Close comparison of the calculated X/R ratios shows that at the bus in question theComprehensive Report predicted an X/R of 11.28, whereas the A_FAULT procedure ofseparately derived R and X calculation predicted an X/R of 11.88.

Referring to the A_FAULT report, note that the total remote generation to the PROC AMTR BUS is 9.4 kA, or 67.78% of the initial symmetrical short-circuit current. TheInterrupting Duty Study setup requested that the results of the ac decrement be interpolatedbetween the ac decrement that was associated with the co-generator (local) and that faultcurrent from remote generators associated with the utility service. Thus, when calculatingthe Total Current and Symmetrical Current interrupting duty for the bus, a weightedaverage of the multiplying factors from the local and remote curves associated with therespective Standard.

It is important to note that in this project and at this specific faulted bus, the X/R ratio isnot large, thus the interrupting ratings will not differ significantly from the initialsymmetrical rms short-circuit current. Indeed, inspection of the symmetrical currentratings show multiplying factors of 1.0.

Next, the A_FAULT Study is rerun, with the NACD option set as All Remote. This yieldsa more conservative answer, as shown in the Report following:

T H R E E P H A S E I N T E R R U P T I N G D U T Y R E P O R T PRE FAULT VOLTAGE: 1.0000 MODEL TRANSFORMER TAPS: YES NACD OPTION: ALL REMOTE


GENERATOR NAME -- AT BUS -- KA VOLTS PU LOCAL/REMOTE UTIL-0001 9.401 .86 R G1 1.307 .59 R TOTAL REMOTE: 10.709 KA NACD RATIO: 1.0000



As noted above, the co-generator is now assumed to be remote, and the 8-cycleSymmetrical current multiplying factor of 1.019 is from Figure 1-12.

Also, note that the 5-cycle Total Current rating is increased to 1.03 (from 1.02 when aninterpolated solution is considered).

In this last case, the A_FAULT study is rerun, but the NACD option is set as Predominant.The results are reported below:

T H R E E P H A S E I N T E R R U P T I N G D U T Y R E P O R T PRE FAULT VOLTAGE: 1.0000 MODEL TRANSFORMER TAPS: YES NACD OPTION: PREDOMINANT


GENERATOR NAME -- AT BUS -- KA VOLTS PU LOCAL/REMOTE UTIL-0001 9.401 .86 R G1 1.307 .59 L TOTAL REMOTE: 9.401 KA NACD RATIO: 1.0000





When the NACD Option is set as Predominant, then the Study will calculate the acdecrement based on whether the short-circuit current from the generator sources is eitherpredominantly local or predominantly remote. In this case the total utility and generatorshort-circuit current contribution is over 50% remote (9.4 kA/13.871 kA is 67%), thus themultiplying factors for both the Symmetrical Current and Total Current basis match theresults for the All Remote NACD Option.

([DPSOHIURP3ODQWThe following figure is a one-line diagram for the Plant project. The Plant project isincluded on the PTW diskettes.


3/12/98

001-

UT

ILIT

Y C

OU

1

002-

TX

A P

RI

TX

A

003-

HV

SW

GR

C1

C2

C3

C4

M8

M10

004-

TX

B P

RI

005-

TX

D P

RI

006-

TX

3 P

RI

007-

TX

E P

RI

TX

E

BLD

G 1

15 S

ER

V

C10

C11

C19

026-

TX

G P

RI

025-

MT

R 2

5

029-

TX

D S

EC

TX

G

027-

DS

B 3

C13

A

L3

028-

MT

R 2

8

M 2

8 #

1&2

011-

TX

3 S

EC

012-

TX

3 T

ER

C7

013-

DS

SW

G2

C8

020-

DS

SW

G3

C9

M3

G2

021-

TX

F P

RI

TX

6

022-

DS

B 2

C12

023-

MT

R 2

3

M7

M9

L1

M4

TX

4

M25

TX

C

008-

DS

SW

G1

C5

C6G

1

009-

TX

C P

RI

010-

MT

R 1

0

TX

3

LV D

IST

RIB

M5

C14

C16

C17

015-

MC

C 1

A01

6-H

2A

017-

H1A

Sub

feed

#1

018-

RA

Sub

feed

#2A

019-

H3A

L9

L10

L2L1

1L1

2

G3

TX

L1

C21

PA

NE

L S

1

MC

C 1

5AP

AN

EL

S3

TX

3W

ND

DE

MO

NS

TR

AT

ION

PR

OJE

CT

FO

R

PO

WE

R*T

OO

LS F

OR

WIN

DO

WS

SW

BD

1

CB

3

R3

F T

X C

F5

F T

X 3

PD

-001

1

LVP

1

CB

6

R6

R G

2

CB

G2

R G

3

CB

G3

R G

1

CB

G1 MC

P5

F2

LVP

2LV

P3

CB

5

R5

R1

CB

1

CB

2

R2

CB

M8

R M

8

CB

M10

R M

10

CB

7

R7

SW

1

CA

P #

1

MC

P M

25

F M

25

SW

M25

F 4

C13

B

LVP

5

028-

MT

R 2

8 B

LVP

4

M28

#3

MC

P M

28 #

3M

CP

M28

#4

M28

#4

MC

P M

28 #

1&2



The following figure shows a portion of the Plant project, including A_FAULT results.

BLDG 115 SERV

I symm 10.560 kA (lv)I breaker Rating 0.000 kAI C/L 16.758 kA (mv)I Interrupting 9.449 kA (mv)

C10C11026-TX G PRI


025-MTR 25I symm 9.802 kA (lv)I breaker Rating 0.000 kAI C/L 15.568 kA (mv)I Interrupting 8.786 kA (mv)

F 4

TX G

027-DSB 3I symm 38.277 kA (lv)I breaker Rating 38.277 kAI C/L 0.000 kA (mv)I Interrupting 0.000 kA (mv)

L3C13 B

028-MTR 28 A


LVP5028-MTR 28 B


MCP M28 #4 MCP M28 #3

M28 #4 M28 #3

LVP4

MCP M28 #1&2

C13 A

M 28 # 1&2

SW M25F M25MCP M25

M25

ANSI SHORT CIRCUIT STUDYFAULT ALL BUSESBUILDING 115 SERVICE

A segment of the A_FAULT Report follows. First the Three-Phase Report for applicationof low-voltage protective devices is presented:

T H R E E P H A S E F A U L T R E P O R T

(FOR APPLICATION OF LOW VOLTAGE BREAKERS) PRE FAULT VOLTAGE: 1.0000 MODEL TRANSFORMER TAPS: NO ==============================================================================

BLDG 115 SERV FAULT: 10.560 KA AT -79.09 DEG ( 76.08 MVA) X/R: 7.03 VOLTAGE: 4160. EQUIV. IMPEDANCE= 0.0431 + J 0.2233 OHMS CONTRIBUTIONS: 007-TX E PRI 4.163 KA ANG: -80.74 026-TX G PRI 1.021 KA ANG: -63.58 025-MTR 25 1.368 KA ANG: -86.40 029-TX D SEC 4.057 KA ANG: -78.78

025-MTR 25 FAULT: 9.802 KA AT -77.60 DEG ( 70.63 MVA) X/R: 6.49 VOLTAGE: 4160. EQUIV. IMPEDANCE= 0.0526 + J 0.2393 OHMS CONTRIBUTIONS: M25 1.386 KA ANG: -86.82 BLDG 115 SERV 8.437 KA ANG: -76.09

026-TX G PRI FAULT: 9.517 KA AT -69.53 DEG ( 68.57 MVA) X/R: 3.09 VOLTAGE: 4160. EQUIV. IMPEDANCE= 0.0883 + J 0.2364 OHMS CONTRIBUTIONS: BLDG 115 SERV 8.459 KA ANG: -70.10 027-DSB 3 1.061 KA ANG: -64.93

027-DSB 3 FAULT: 38.277 KA AT -70.78 DEG ( 31.82 MVA) X/R: 3.69 VOLTAGE: 480. EQUIV. IMPEDANCE= 0.0024 + J 0.0068 OHMS LOW VOLTAGE POWER CIRCUIT BREAKER 38.277 KA MOLDED CASE CIRCUIT BREAKER > 20KA 38.277 KA CONTRIBUTIONS: 026-TX G PRI 26.612 KA ANG: -76.87 023-MTR 23 6.328 KA ANG: -30.29 028-MTR 28 A 3.560 KA ANG: -81.20 028-MTR 28 B 3.560 KA ANG: -81.20


3/12/98

028-MTR 28 A FAULT: 21.638 KA AT -52.40 DEG ( 17.99 MVA) X/R: 2.85 VOLTAGE: 480. EQUIV. IMPEDANCE= 0.0078 + J 0.0101 OHMS LOW VOLTAGE POWER CIRCUIT BREAKER 21.638 KA MOLDED CASE CIRCUIT BREAKER > 20KA 21.638 KA CONTRIBUTIONS: M 28 # 1&2 3.752 KA ANG: -86.63 027-DSB 3 18.656 KA ANG: -45.91

028-MTR 28 B FAULT: 21.638 KA AT -52.40 DEG ( 17.99 MVA) X/R: 2.85 VOLTAGE: 480. EQUIV. IMPEDANCE= 0.0078 + J 0.0101 OHMS LOW VOLTAGE POWER CIRCUIT BREAKER 21.638 KA MOLDED CASE CIRCUIT BREAKER > 20KA 21.638 KA CONTRIBUTIONS: M28 #3 1.876 KA ANG: -86.63 M28 #4 1.876 KA ANG: -86.63 027-DSB 3 18.656 KA ANG: -45.91

Next, the Unbalanced Fault Report for application of low-voltage protective devices:

U N B A L A N C E D F A U L T R E P O R T (FOR APPLICATION OF LOW VOLTAGE BREAKERS) PRE FAULT VOLTAGE: 1.0000 MODEL TRANSFORMER TAPS: NO ============================================================================== LOCATION FAULT KA X/R EQUIVALENT (PU) ASYM. KA AT 0.5 CYCLES VOLTAGE DUTIES (RMS) FAULT IMPEDANCE * MAX. RMS AVG. RMS * ==============================================================================

BLDG 115 SERV 3 PHASE: 10.560 7. Z1= 1.3143 14.237 12.471 SLG DUTY: 10.828 7. Z2= 1.3143 14.461 4160. VOLTS LN/LN 9.145 Z0= 1.2170 LN/LN/GND 10.619 ( 11.109 GND RETURN KA)

025-MTR 25 3 PHASE: 9.802 6. Z1= 1.4158 13.002 11.462 SLG DUTY: 9.512 6. Z2= 1.4158 12.256 4160. VOLTS LN/LN 8.489 Z0= 1.5455 LN/LN/GND 9.660 ( 9.238 GND RETURN KA)

026-TX G PRI 3 PHASE: 9.517 3. Z1= 1.4583 10.687 10.111 SLG DUTY: 9.143 3. Z2= 1.4583 9.948 4160. VOLTS LN/LN 8.242 Z0= 1.6415 LN/LN/GND 9.585 ( 8.790 GND RETURN KA)

027-DSB 3 3 PHASE: 38.277 4. Z1= 3.1424 44.697 41.551 SLG DUTY: 40.240 4. Z2= 3.1424 47.789 480. VOLTS LN/LN 33.149 Z0= 2.6891 LN/LN/GND 38.382 ( 42.385 GND RETURN KA)



028-MTR 28 A 3 PHASE: 21.638 3. Z1= 5.5587 23.913 22.790 SLG DUTY: 18.418 2. Z2= 5.5587 18.772 480. VOLTS LN/LN 18.739 Z0= 8.5031 LN/LN/GND 20.976 ( 16.009 GND RETURN KA)

028-MTR 28 B 3 PHASE: 21.638 3. Z1= 5.5587 23.913 22.790 SLG DUTY: 18.418 2. Z2= 5.5587 18.772 480. VOLTS LN/LN 18.739 Z0= 8.5031 LN/LN/GND 20.976 ( 16.009 GND RETURN KA)

The above two reports are associated with the Complete Report format. The following isthe Summary Report Format for application of low-voltage protective devices.

F A U L T S T U D Y S U M M A R Y (FOR APPLICATION OF LOW VOLTAGE BREAKERS) PRE FAULT VOLTAGE: 1.0000 MODEL TRANSFORMER TAPS: NO BUS RECORD VOLTAGE A V A I L A B L E F A U L T D U T I E S (KA) NO NAME L-L 3 PHASE X/R LINE/GRND X/R ==============================================================================

BLDG 115 SERV 4160. 10.560 7.03 10.828 6.71

025-MTR 25 4160. 9.802 6.49 9.512 5.67 026-TX G PRI 4160. 9.517 3.09 9.143 2.63

027-DSB 3 480. 38.277 3.69 40.240 3.97 028-MTR 28 A 480. 21.638 2.85 18.418 1.59

028-MTR 28 B 480. 21.638 2.85 18.418 1.59

The following is the Three-Phase Medium/High Voltage Momentary, or Closing andLatching Report:


BLDG 115 SERV E/Z: 10.474 KA AT -79.07 DEG ( 75.47 MVA) X/R: 7.06 SYM*1.6: 16.758 KA MOMENTARY BASED ON X/R: 14.133 KA SYM*2.7: 28.279 KA CREST BASED ON X/R: 24.301 KA VOLTAGE: 4160. EQUIV. IMPEDANCE= 0.0435 + J 0.2252 OHMS CONTRIBUTIONS: 007-TX E PRI 4.161 KA ANG: -80.76 026-TX G PRI 0.944 KA ANG: -61.98 025-MTR 25 1.368 KA ANG: -86.40 029-TX D SEC 4.055 KA ANG: -78.80

025-MTR 25 E/Z: 9.730 KA AT -77.60 DEG ( 70.11 MVA) X/R: 6.52 SYM*1.6: 15.568 KA MOMENTARY BASED ON X/R: 12.919 KA SYM*2.7: 26.271 KA CREST BASED ON X/R: 22.258 KA VOLTAGE: 4160. EQUIV. IMPEDANCE= 0.0530 + J 0.2411 OHMS CONTRIBUTIONS: M25 1.386 KA ANG: -86.82 BLDG 115 SERV 8.365 KA ANG: -76.08


3/12/98

026-TX G PRI E/Z: 9.432 KA AT -69.41 DEG ( 67.96 MVA) X/R: 3.08 SYM*1.6: 15.092 KA MOMENTARY BASED ON X/R: 10.591 KA SYM*2.7: 25.467 KA CREST BASED ON X/R: 18.157 KA VOLTAGE: 4160. EQUIV. IMPEDANCE= 0.0896 + J 0.2384 OHMS CONTRIBUTIONS: BLDG 115 SERV 8.456 KA ANG: -70.12 027-DSB 3 0.983 KA ANG: -63.29

027-DSB 3 VOLTAGE: 480. ( SEE LOW VOLTAGE REPORT )

The following is the Unbalanced Fault Medium/High Voltage Momentary, or Closing andLatching Report:

U N B A L A N C E D M O M E N T A R Y D U T Y R E P O R T PRE FAULT VOLTAGE: 1.0000 MODEL TRANSFORMER TAPS: NO ============================================================================== LOCATION FAULT E/Z X/R EQUIVALENT MOMENTARY FAULT DUTIES VOLTAGE TYPE KA IMPEDANCE (PU) E/Z * 1.6 @ 0.5 CYCLE ==============================================================================

3 PHASE: 10.47 7.1 Z1= 1.3251 16.76 14.13 BLDG 115 S SLG DUTY: 10.77 6.7 Z2= 1.3251 17.23 14.39 4160. VOLTS LN/LN 9.07 Z0= 1.2170 LN/LN/GND 10.55 ( 11.08 GND RETURN KA)

025-MTR 25 SLG DUTY: 9.47 5.7 Z2= 1.4264 15.15 12.21 4160. VOLTS LN/LN 8.43 Z0= 1.5455 LN/LN/GND 9.60 ( 9.22 GND RETURN KA)

3 PHASE: 9.43 3.1 Z1= 1.4714 15.09 10.59 026-TX G P SLG DUTY: 9.09 2.6 Z2= 1.4714 14.55 9.89 4160. VOLTS LN/LN 8.17 Z0= 1.6415 LN/LN/GND 9.51 ( 8.77 GND RETURN KA)

3 PHASE: 9.73 6.5 Z1= 1.4264 15.57 12.92

The above two Reports are associated with the Complete Report format. The following isthe Summary Report format for application of medium/high-voltage momentary or closingand latching protective devices.

M O M E N T A R Y D U T Y S U M M A R Y R E P O R T PRE FAULT VOLTAGE: 1.0000 MODEL TRANSFORMER TAPS: NO ============================================================================== BUS RECORD VOLTAGE * 3 P H A S E * * * * SLG * * * NO NAME L-L E/Z * 1.6 X/R E/Z * 1.6 X/R ==============================================================================

BLDG 115 SERV 4160. 16.758 7.06 17.228 6.73 025-MTR 25 4160. 15.568 6.52 15.146 5.69 026-TX G PRI 4160. 15.092 3.08 14.545 2.63



The following is the Three-Phase Medium/High Voltage Interrupting Report Format.


BLDG 115 SERV E/Z: 9.449 KA AT -78.48 DEG ( 68.09 MVA) X/R: 6.76 VOLTAGE: 4160. EQUIV. IMPEDANCE= 0.0508 + J 0.2491 OHMS CONTRIBUTIONS: 007-TX E PRI 4.019 KA ANG: -80.59 026-TX G PRI 0.673 KA ANG: -53.33 025-MTR 25 0.916 KA ANG: -86.54 029-TX D SEC 3.917 KA ANG: -78.63

GENERATOR NAME -- AT BUS -- KA VOLTS PU LOCAL/REMOTE U1 5.155 0.93 R G1 0.211 0.80 L G2 0.522 0.75 L G3 0.232 0.75 L TOTAL REMOTE: 5.155 KA NACD RATIO: 0.5456



025-MTR 25 E/Z: 8.786 KA AT -77.01 DEG ( 63.31 MVA) X/R: 6.07 VOLTAGE: 4160. EQUIV. IMPEDANCE= 0.0615 + J 0.2663 OHMS CONTRIBUTIONS: M25 0.924 KA ANG: -86.82 BLDG 115 SERV 7.878 KA ANG: -75.86




026-TX G PRI E/Z: 8.558 KA AT -69.21 DEG ( 61.66 MVA) X/R: 3.10 VOLTAGE: 4160. EQUIV. IMPEDANCE= 0.0996 + J 0.2624 OHMS CONTRIBUTIONS: BLDG 115 SERV 7.875 KA ANG: -70.48 027-DSB 3 0.707 KA ANG: -54.95



3/12/98



027-DSB 3 VOLTAGE: 480. ( SEE LOW VOLTAGE REPORT )

The following is the Unbalanced Fault Medium/High-Voltage Interrupting Report format.

U N B A L A N C E D I N T E R R U P T I N G D U T Y R E P O R T PRE FAULT VOLTAGE: 1.0000 MODEL TRANSFORMER TAPS: NO NACD OPTION: INTERPOLATED ============================================================================== LOCATION FAULT E/Z X/R ANSI AC/DC INTERRUPTING TYPE KA DECREMENT FACT. DUTIES (KA) 3 PHASE SLG 3 PHASE SLG ============================================================================== BLDG 115 SERV 3 PHASE: 9.45 6.8 SYM2: 1.00 1.00 9.45 10.02 VOLTS: 4160.0 SLG: 10.02 6.6 SYM3: 1.00 1.00 9.45 10.02 NACD: 0.546 LN/LN: 8.18 SYM5: 1.00 1.00 9.45 10.02 LN/LN/GND: 9.67 SYM8: 1.00 1.00 9.45 10.02 GND RETURN: 10.67 TOT2: 1.10 1.09 10.36 10.93 Z1(PU): 1.46874 TOT3: 1.01 1.01 9.53 10.11 Z2(PU): 1.46874 TOT5: 1.00 1.00 9.45 10.03 Z0(PU): 1.21698 TOT8: 1.00 1.00 9.45 10.02

025-MTR 25 3 PHASE: 8.79 6.1 SYM2: 1.00 1.00 8.79 8.85 VOLTS: 4160.0 SLG: 8.85 5.5 SYM3: 1.00 1.00 8.79 8.85 NACD: 0.541 LN/LN: 7.61 SYM5: 1.00 1.00 8.79 8.85 LN/LN/GND: 8.79 SYM8: 1.00 1.00 8.79 8.85 GND RETURN: 8.91 TOT2: 1.07 1.05 9.40 9.28 Z1(PU): 1.57954 TOT3: 1.00 1.00 8.81 8.87 Z2(PU): 1.57954 TOT5: 1.00 1.00 8.79 8.85 Z0(PU): 1.54554 TOT8: 1.00 1.00 8.79 8.85

026-TX G PRI 3 PHASE: 8.56 3.1 SYM2: 1.00 1.00 8.56 8.53 VOLTS: 4160.0 SLG: 8.53 2.7 SYM3: 1.00 1.00 8.56 8.53 NACD: 0.542 LN/LN: 7.41 SYM5: 1.00 1.00 8.56 8.53 LN/LN/GND: 8.75 SYM8: 1.00 1.00 8.56 8.53 GND RETURN: 8.50 TOT2: 1.02 1.01 8.69 8.64 Z1(PU): 1.62180 TOT3: 1.00 1.00 8.56 8.53 Z2(PU): 1.62180 TOT5: 1.00 1.00 8.56 8.53 Z0(PU): 1.64149 TOT8: 1.00 1.00 8.56 8.53



The previous two reports are associated with the Complete Report format. The followingis the Summary Report Format for interrupting ratings for medium/high voltage protectivedevices.

I N T E R R U P T I N G D U T Y S U M M A R Y R E P O R T PRE FAULT VOLTAGE: 1.0000 MODEL TRANSFORMER TAPS: NO NACD OPTION: INTERPOLATED ============================================================================== BUS RECORD VOLTAGE NACD * 3 P H A S E * * * * S L G * * * NO NAME L-L RATIO E/Z KA X/R E/Z KA X/R ==============================================================================

BLDG 115 SERV 4160. 0.546 9.449 6.76 10.023 6.56 025-MTR 25 4160. 0.541 8.786 6.07 8.850 5.49 026-TX G PRI 4160. 0.542 8.558 3.10 8.530 2.67


3/12/98

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,QGH[

AA_FAULT Study

ANSI Considerations, 1-18ANSI decrement curves used by, 1-20before running the Study, 1-8component modeling, 1-11

contribution data, 1-14feeders, 1-11transformers, 1-12

definition of, 1-2examples, 1-26

calculating interrupting duties, 1-32induction motor ac decrement factors, 1-26modeling transformers with taps, 1-30Plant project, 1-39

methodology, 1-3, 1-8reports

interrupting-duty, 1-18low-voltage duty, 1-15momentary-duty, 1-16

running the Study, 1-8Study options, 1-9

ac decrement, 1-4American National Standards Institute. See ANSI StandardANSI Considerations, 1-18ANSI Decrement Curves, 1-20ANSI Methodology

compared to IEC methodology, 1-3ANSI Standard, 1-1, 1-2Asymmetrical Peak Fault Current, 1-5

CCircuit Breakers. See Protective DevicesClosing and Latching Rating, 1-4Component Modeling in A_FAULT Study. See A_FAULT Study

Ddc decay, 1-4dc Offset Current, 1-4

EEquipment

low-voltage test power factors, 1-5Equipment Interrupting Rating, 1-3. see also RatingEquivalent First Cycle Rms Currents, 1-4

FFault

asymmetrical peak current, 1-5half-cycle, 1-4initial symmetrical rms current, 1-5

Fuses. See Protective Devices

GGenerators

local and remote determination, 1-18

HHalf-Cycle. See FaultHigh-Voltage Protective Devices. See Protective Devices

IIEC Methodology

compared to ANSI methodology, 1-3IEEE Red Book, 1-1Interrupting Fault Duty, 1-2Interrupting-Duty Report. See A_FAULT Study

NACD options, 1-19

JJulian Heating, 1-4

LLow-Voltage Protective Devices. See Protective DevicesLow-Voltage Report. See A_FAULT Study

MMachine

rotating. See Rotating-MachineMachine Contributions

adjusting, 1-7

A_FAULT ii Reference Manual

3/12/98

MethodologyA_FAULT Study, 1-3, 1-8

Momentary Rating Rating, 1-4Momentary-Duty Report. See A_FAULT StudyMulitpliers, Reactance

rotating-machine, 1-8

NNational Electrical Manufacturers Standard. See NEMA StandardNEMA Standard, 1-3, 1-7

PProtective Devices

interrupting ratings, 1-7low-voltage test power factors, 1-5withstand, closing and latching, and momentary rating, 1-4

RRating

closing and latching, 1-4withstand, 1-4

Reports in A_FAULT Study. See A_FAULT StudyRotating-Machine

reactance multipliers, 1-8

SShort-Circuit Current, 1-3

TThevenin Equivalent Fault Point X/R Ratio, 1-4

UUL Standard, 1-3Underwriters Laboratory Standard. See UL Standard

WWithstand Fault Duty, 1-2Withstand Rating, 1-4

XX/R Ratio, 1-6

Documents

A_fault