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436 IEEE TRANSACTIONS ON INDUSTRY APPLICATIONS, VOL. 32, NO. 2, MARCHIAPRIL 1996

ortance ofents when Specifying Intes for Low-Voltage EJames J. Toth 111, Member, IEEE

Abstract-In a number of plants, personnel responsible for themaintenance and design of the electrical systems have done anexcellent job at sizing equipment for connected loads; however,on more than one occasion, those same personnel appeared to beunaware of the interrupting rating for low voltage equipment.This sometimes resulted in electrical systems that posed hazardsto personnel and equipment due to the incorrect application ofthe protective equipment. Several aspects of proper equipmentselection based on interrupting ratings will be addressed.

I. INTRODUCTIONLECTRICAL equipment generally has three items listedon its nameplate that are important for the proper ap-plication of the equipment: voltage, current rating (or triprating), and interrupting rating. T hese sam e three items shouldalso be used when specifying electrical equipment. On manyoccasions, plants have been found to have equipment with

an interrupting rating less than the calculated available faultcurrent. This is not to say other problems have not been found(most trip rating problems have been observed with adjustabletrip units), but the w rong interrupting rating has occurred oftenenough to cause concern. What is worse, however, is thereason it occurs: In most cases, people were not aware ofan interrupting rating and in other cases people did not thinkabout the interrupting rating.When asked how the wrong equipment was installed ina panel, switchboard, or motor control center, the followingresponses are generally given.

* When a breakedfuse had to be replaced, this was thefirst one that could be found with the proper trip/amprating from the spares in the maintenance department.Interrupting rating? I didnt know that.e This panel was ordered the same as the others.* Interrupting rating? An interrupting rating wasnt spec-Electrical personnel have found it quite easy to choose the

correct voltage rating for equipment, and most experiencedpersonnel have had little trouble choosing the correct currentrating (or trip rating) based on the con nected load. H owever, aPaper PID 95-53, approved by the Pulp and Paper Industry Committee ofthe IEEE Industry Applications Society for presentation at the 1994 IEEEPulp and Paper Industry Conference, Nashville, TN, June 20 24. Manuscriptreleased for publication September 6, 1995.The author was with the Installation Service Engineering group ofGeneral Electric Company in Charlotte, NC USA. He is now with ABBService Inc., Matthews, NC 28105 USA.

ified.

Publisher Item Identifier S 0093-9994(96)01596-4.

TABLE IC UR R ENTNTERRUPTINGATINGSMS Symmetrical Amps7,500 25,000 65,00010,000 30,000 85,00014,000 35,000 100,00018,000 42,000 125,00022,000 50,000 150,000200,000

number of personnel appeared to be unaware of the interrupt-ing rating on equipment. P erhaps it has been because the basisfor choosing an interrupting rating, the available fault current,has not been readily known.Although fuses have a small number of standard interruptingratings (10 kA 50 kA, 100 kA, and 200 kA , molded-casecircuit breakers have sixteen standard ratings [11 from 7500 Ato 200 kA (see Table I). Motor control centers have seventeenstandard ratings [2] from 5000 A to 200 kA (same as Table Iexcept add 5000 A). Low-voltage power circuit breakers havefourteen standard ratings [3] from 14 kA to 200 kA (sameas Table I except omit 7500 A and 10 kA, and replace 125kA with 130 kA). The large number of av ailable interruptingratings could mak e it extremely difficult to choose the correctinterrupting rating without know ing the available fault current.In addition to knowing the available fault current, thecalculated X / R ratio of the fault current should also beknown to properly specify the low voltage equipment. Finally,whether the protective devices are to be fully rated or seriesrated would also have an impact on the selection process.These final two points will be addressed separately in thispaper.

A number of examples were used to illustrate the impactof certain circuit parameters on the available fault current.This was not meant, however, to serve as a replacement fora good, thorough, fault analysis. The intent was to emphasizethe wide range of potential fault currents and how variouscircuit parameters (e.g., transformer size, available fault MVA,cable size, cable length, and m otor contribution) affect the faultcurrent and the eventual selection of the proper interruptingrating of equipment.

11. SAMPLE YSTEMIn order to simplify the analysis, the sample system shown inFig. 1included the following: a service transformer and cable

0093-9994/96$05,00 996 IEEE

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TOTH: THE IMPORTANCE OF KNOWING AVAILABLE FAULT CURRENTS WHEN SPECIFYING INTERRUPTING RATINGS FOR LOW-VOLTAGE EQUIPMENT 437

TABLE 11-AS UMMAR YF 480 VOLTCALCULATEDAULT URRENTSI ) ND X /R RATIOSOR 1000-kVA TRANSFORM ER.AULT URRENTS I ) N RMS SYMMETRICAL kA1000 kVA TransformerHigh Side Fault MVA50MVA IOOMVA 250MVA 500MVA 1OOOMVA InfBusFault Locatiton I xm XR xm x/R m 1 x/RTransform er Secondary 15.5 6.1 17.8 5.7 19.6 5 5 20.2 5.4 20.6 5.4 20.9 5.310 feet 15.2 5.1 17.3 4.8 19.0 4.5 19.6 4.4 19.9 4.4 20.2 4.4

20 feet 14.8 4.4 16.8 4.1 18.4 3.9 18.9 3.8 19.2 3.8 19.5 3.7225A Panel 50 feet 13.7 3.2 15.4 3.0 16.7 2.8 17.1 2.7 17.4 2.7 17.6 2.7(1&4/0 AWG) 100 feet 12.1 2.3 13.4 2.1 14.3 2.0 14.6 2.0 14.8 1.9 15.0 1.9200 feet 9.6 1.6 10.4 1.5 10.9 1.4 11.0 1.4 11.1 1.4 11.2 1.3500 feet 5.7 1.0 6.0 1.0 6.1 0.9 6.2 0.9 6.2 0.9 6.2 0.910 feet 15.2 5.6 17.4 5.3 19.1 5.1 19.7 5.0 20.0 4.9 20.4 4.920 feet 14.9 5.3 17.0 4.9 18.6 4.7 19.2 4.6 19.5 4.6 19.8 4.6400A Panel 50feet 14.1 4.5 15.9 4.2 17.3 4.0 17.8 3.9 18.1 3.9 18.3 3.8(1-500kcmil) 100 feet 12.8 3.7 14.4 3.4 15.4 3.2 15.8 3.2 16.0 3 .2 16 .3 3.1200 feet 10.9 2.9 11.9 2.7 12.7 2.5 12.9 2 5 13.1 2.5 13.2 2.4500 feet 7.4 2.0 7.8 1.9 8.1 1.8 8.2 1.8 8.3 1.8 8.3 1.810 feet 15.4 5.7 17.6 5.4 19.3 5.2 19.9 5.1 20.3 5.0 20.6 5.020 feet 15.2 5.4 17.4 5.1 19.0 4.9 19.7 4.8 20.0 4.8 20.3 4.7600A Panel 50 feet 14.7 4.8 16.8 4.4 18.3 4.2 18.8 4.1 19.1 4.1 19.4 4.1(2-350 kcmjl) 100 feet 14.0 4.0 15.8 3.7 17.1 3.5 17.6 3.4 17.8 3.4 18.1 3.4200 feet 12.6 3.1 14.1 2.9 15.1 2.7 15.4 2.7 15.6 2.6 15.8 2.6500 feet 9.6 2.1 10.4 1.9 11.0 1.8 11.1 1.8 11.2 1.8 11.3 1.810 feet 15.4 5.8 17.6 5 5 19.3 5.3 20.0 5 2 20.3 5.1 20.6 5.120 feet 15.2 5.6 17.4 5.3 19.1 5.1 19.7 5.0 20.0 4.9 20.4 4.9800A Panel 50 feet 14.8 5.1 16.8 4.8 18.4 4.6 18.9 4.5 19.2 4.4 19.6 4.4(2-500 kcmil) 100 feet 14.1 4.5 15.9 4.2 17.3 4.0 17.8 3.9 18.1 3.9 18.3 3.8200 feet 12.8 3.7 14.4 3.4 15.4 3.2 15.8 3.2 16.0 3.2 16.3 3.1500 feet 10.1 2.6 11.0 2.4 11.6 2.3 11.8 2.3 11.9 2.3 12.0 2.310 feet 15.5 5.9 17.7 5 5 19.4 5.3 20.1 5.2 20.4 5.2 20.8 5.120 feet 15.4 5.7 17.6 5.4 19.3 5.2 19.9 5.1 20.3 5.0 20.6 5.01200A Panel 50 feet 15.1 5.3 17.3 5.0 18.9 4.8 19.5 4.7 1 9.8 4.6 20.2 4.6(4-350 kcmil) 100 feet 14.7 4.8 16.8 4.4 18.3 4.2 18.8 4.1 19.1 4.1 19.4 4.1200 feet 14.0 4.0 15.8 3.7 17.1 3.5 17.6 3.4 17.8 3.4 18.1 3.4500 feet 12.0 2.8 13.3 2.6 14.2 2.5 14.5 2.4 14.7 2.4 14.9 2.4

AVAILABLEOTES:1.) AVAILABLE FAULT MVA: 50, 100.250 . 500. 1000, AND INFINITE BUS.2 ) TRANSFORMER KVA RATINGS3 CABLE LENGTHS' 10, 20. 50 ,

1000. 1500 , 2000. AND 2500100. 200. AND 500 FEET @ 5 7 5 9 . 2

4 8 0 V

2 2 5 A ( I O O A 1 6 0 0 A I B O O A 1 1 2 0 0 APANEL PANEL PANEL PANEL PANEL

Fig. 1. Sample system.runs (of various conductor sizes and lengths) to downstreampanelboards or switchboards. (For convenience, they will bereferred to as panels.) The transformer could h ave kVA ratingsof 1000, 1500, 2000, and 25100. A standard 5.75% transformerimpedance was used. The high-side of the transformer hadan available fault MVA of 50, 100, 250, 500, 1000, and aninfinite bus. The panels had a nominal current rating of 225A, 400 A, 600 A, 800 A, and 120 0 A. It was assumed thesepanels were fed by the following cable runs. 14410 AWG/+,1-500 kcm il/$, 2-350 k cmil/$, 2-500 kc mil/$ , and 4-350kcm il/$, respectively. The: cable lengths being con sideredwere 10ft, 20 ft, 50 ft, 100ft, 200 ft, and 500 ft. For simplicity,no motor contribution was included in the first set of exam ples.The effect of motor contribution will be discussed later.

111. FAULT ALCULATIONESULTSThe results of the initial fault calculations for the sample

system described above are summarized in Tables 11-A to 11-D.Table 11-A included the results for the 1000-kVA transformer.Tables 11-B, 11-C, and 11-D included the results for the 1500-kVA, 2000-kVA, and 2500-kVA transformers, respectively.Each table entry included a pair of calculated numbers: therms symmetrical fault current (in kA) and the X I R ratio forthe fault impedance. The columns of the table represented theavailable high side fault MVA for the transformer (50 MVAto 1000 MVA, and an infinite bus), and the rows representedthe various fault locations (including panel size, cable size,and cable length).

As can be seen from Table 11-A for the 1000-kVA trans-former, a 225-A panel in this example had a calculatedrms symmetrical fault current between 5.7 kA and 20.2 kAdepending on the length of the cable run and the availablehigh-side available fault MVA. Similarly, the calculated faultcurrent was between 7.4 kA and 20.4 kA for the 400-A panel:between 9.6 kA and 20 .6 kA for the 600-A panel; between 10.1kA and 20.6 kA for the 800-A panel; between 12.0 kA and20.8 kA for the 1200-A panel; and between 15.5 kA and 20.9kA for the switchgear on the second ary side of the transformer.

For a brief o verview of all of the results, the available faultcurrent can vary as shown in Table 111, depending on cablesize, cable length, transformer size, and high side availablefault MVA.

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438 IEEE TRANSACTIONS ON INDUSTRY APPLICATIONS, VOL. 32, NO. 2, MARCWAPRIL 1996

TABLE 11-BSUMMARYF 480 VOLTCALCULATEDAULT URRENTSI ) ND S I R RATIOSOR 1500 kVA TRANSFORMER.AULT URRENTSI ) N RMS SYMMETRICALA

1500 kVA Transform erHigh Side Fault MVA50MV A IOOMVA 250M vA 500MVA 1OOOMVA InfBusFault Location I X / R I x / R I x / R I x / R I x / R I x / RTransform er Seconda ry 20.6 6.9 24.9 6.5 28.4 6.2 29.8 6.0 30.6 6.0 31.4 5.910 feet 20.0 5.4 23.9 4.9 27.2 4.6 28.4 4.5 29.1 4.4 29.8 4.320 feet 19.3 4.4 23.0 4.0 26.0 3.7 27.1 3.6 27.7 3.5 28.4 3.5225A Panel 50 feet 17.5 3.0 20.4 2.7 22.7 2.4 23.5 2.4 23.9 2.3 24.4 2.3(1-#4/0 AWG) 100 feet 14.9 2.1 16.9 1.8 18.3 1.7 18.8 1.6 19.1 1.6 19.4 1.6200 feet 11.3 1.4 12.2 1.3 12.9 1.2 13.1 1.2 13.3 1.1 13.4 1.1500 feet 6.2 0.9 6.5 0.9 6.6 0.8 6.7 0.8 6.7 0.8 6.7 0.810feet 20.1 6.2 24.1 5.7 27.4 5.4 28.7 5.3 29.4 5.2 30.1 5 220 feet 19.6 5.6 23.4 5.2 26.4 4.9 27.6 4.7 28.3 4.7 29.0 4.6400A Panel 50 feet(1-500 kcmil) 100 feet200 feet500 feet10 feet20 feet600A Panel SO feet(2-350 kcmil) 100 feet200 feet500 feet10 feet20 feet800A Panel 50 feet

18.2 4.5 21.4 4.1 23.9 3.8 24.8 3.7 25.4 3.1 25.9 3.616.2 3.5 18.6 3.2 20.5 3.0 21.2 2.9 21.5 2.9 21.9 2.813.2 2.7 14.7 2.4 15.8 2.3 16.2 2.2 16.4 2.2 16.6 2.28.3 1.9 8.9 1.8 9.3 1.7 9.4 1.7 9.5 1.6 9.5 1.620.3 6.3 24.5 5.9 27.9 5.6 29.2 5.5 29.9 5.4 30.7 5.320.1 5.9 24.1 5.4 27.3 5.1 28.6 5.0 29.3 4.9 30.0 4.919.2 4.9 22.9 4.5 25.8 4.2 26.9 4.0 27.5 4.0 28.2 3.918.0 3.9 21.1 3.5 23.5 3.2 24.4 3.2 24.9 3.1 25.4 3.115.8 2.9 18.1 2.6 19.8 2.4 20.4 2.3 20.8 2.3 21.1 2.311.3 1.9 12.4 1.7 13.1 1.6 13.4 1.6 13.5 1.6 13.7 1.520.4 6.5 24.5 6.1 27.9 5.8 29.3 5.6 30.0 5.6 30.7 5.520.1 6.2 24.1 5.7 27.4 5.4 28.7 5.3 29.4 5.2 30.1 5.219.3 5.4 23.0 4.9 26.0 4.6 27.1 4.5 27.8 4.5 28.4 4.4(2-500 kcmil) 100 feet 18.2 4.5 21.4 4.1 23.9 3.8 24.8 3.7 25.4 3.7 25.9 3.6200 feet 16.2 3.5 18.6 3.2 20.5 3.0 21.2 2.9 21.5 2.9 21.9 2.8500 feet 12.0 2.4 13.3 2.2 14.2 2.1 14.5 2.1 14.6 2.0 14.8 2.010 feet 20.5 6.6 24.7 6.2 28.1 5.9 29.5 5.7 30.3 5.7 31.0 5.620 feet 20.3 6.3 24.5 5.9 27.9 5.6 29.2 5.5 29.9 5.4 30.7 5.31200A Panel SO feet 19.9 5.7 23.9 5.2 27.1 4.9 28.3 4.8 29.0 4.7 29.7 4.7(4-350 kcmil) 100feet 19.2 4.9 22.9 4.5 25.8 4.2 26.9 4.0 27.5 4.0 28.2 3.9200 feet 18.0 3.9 21.1 3.5 23.5 3.2 24.4 3.2 24.9 3.1 25.4 3.1500 feet 14.8 2.6 16.8 2.3 18.3 2.2 18.8 2.1 19.1 2.1 19.4 2.1

TABLE 11 CS U M M A R Y OF 480 VOLT CALCULATED FAULTURRENTS ( I ) ND X / R RATIOS OR 2000 kVA TRANSFORMER.AULTURRENTS I) N RMS SYMMETRICALA

2000 kVA TransformerHigh Side Fault MVA50MV A lOOMvA 250MV A 500MVA 1OOOMVA InfBusFault Location I x m I x / R I x / R I x / R I X / R I x / RTransforme r Secondiuy 24.7 7.5 31.1 7.0 36.7 6.7 39.1 6.5 40.4 6.4 41.8 6.410 feet 23.8 5.5 29.6. 4.9 34.7 4.5 36.8 4.3 37.9 4.3 39.2 4.220 feet 22.9 4.3 28.2 3.8 32.7 3.5 34.6 3.3 35.6 3.3 36.6 3.2225A Panel 50 feet 20.4 2.8 24.3 2.4 27.5 2.2 28.8 2.1 29.4 2.1 30.1 2.0(1-#4/0 AWG) 100 feet 16.9 - 1.9 19.4 1.7 21.2 1.5 21.9 1.4 22.2 1.4 22.6 1.4200 feet 12.3 1.3 13.4 1.2 14.2 1.1 14.4 1.0 14.6 1.0 14.7 1.0500 feet 6.5 0.9 6.7 0.8 6.9 0.8 7.0 0.8 7.0 0.8 7.0 0.710 feet 23.9 6.5 29.9 6.0 35.1 5.6 37.2 5.4 38.4 5.3 39.7 5.320 feet400A Panel 50 feet(1-500 kcmil) 100 feet200 feet500 feet10 feet20 feet600A Panel 50 feet(2-350 kcml) 100 feet200 feet

23.2 5.8 28.1 5.2 33.5 4.8 35.5 4.7 36.5 4.6 37.7 4.521.2 4.4 25.7 4.0 29.5 3.6 31.0 3.5 31.8 3.5 32.6 3.418.5 3.4 21.8 3.0 24.4 2.8 25.4 2.7 25.9 2.6 26.5 2.614.7 2.5 16 .6 2.3 18.0 2.1 18.5 2.0 18.8 2.0 19.1 2.08.9 1 8 9.5 1.7 10.0 1.6 10.1 1.6 10.2 1.5 10.3 1.524.3 6.7 30.4 6.2 35.8 5.8 38.1 5.7 39.3 5.6 40.6 5.523.9 6.1 29.8 5.6 34.9 5.2 37.1 5.0 38.2 4.9 39.5 4.922.7 4.9 28.0 4.4 32.4 4.0 34.3 3.9 35.3 3.8 36.3 3.721.0 3.8 25.3 3.3 28.8 3.0 30.3 2.9 31.0 2.8 31.8 2.818.0 2.7 21.0 2.4 23.4 2.2 24.3 2.1 24.7 2.1 25.2 2.0500 feet 124 1 8 1 3 7 1 6 1 4 6 1 5 1 4 9 1 4 1 5 0 1 4 1 5 2 1 41 0 fe et 2 4 3 6 9 3 0 4 6 4 3 5 9 6 1 3 8 2 5 9 3 9 4 5 8 4 07 5 720 feet 23.9 6.5 29.9 6.0 35.1 5.6 37.2 5.4 38.4 5.3 39.7 5.3800A Panel 50 feet 22.9 5.5 28.2 5.0 32.8 4.6 34.6 4.4 35.7 4.4 36.7 4.3,(2-500 kcmil) 100 feet 21.2 4.4 25.7 4.0 29.5 3.6 31.0 3.5 -31.8 3.5 32.6 3.4200 feet 18.5 3.4 21.8 3.0 24.4 2.8 25.4 2.7 25.9 2.6 26 5 2 6500 feet 1 33 2 3 1 4 8 2 1 1 59 1 9 1 63 1 9 1 65 1 9 1 67 1 810 feet 2 45 7 1 3 07 6 6 3 63 6 2 3 8 6 6 0 3 9 9 6 0 4 12 5 920 feet 24.3 6.7 30.4 6.2 35.8 5.8 38.1 5.7 39.3 5.6 40.6 5.51200A Panel 50 feet 23.7 5.9 29.4 5.3 34.5 4.9 36.6 4.8 37.7 4.7 38.9 4.6(4-350 kcmil) 100 feet 22.7 4.9 28.0 4.4 32.4 4.0 34.3 3.9 35.3 3.8 36.3 3.7200 feet 21.0 3.8 25.3 3.3 28.8 3.0 30.3 2.9 31.0 2.8 31.8 2.8500 feet 16 .8- 2 .4 19.4 2.2 21.3 2.0 22.0 1.9 22.4 1.9 22.8 1.8

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TOTH: THE IMPORTANCE OF KNOWING AVAILABLE FAULT CURRENTS WHEN SPECIFYING INTERRUPTING RATINGS FOR LOW-VOLTAGE EQUIPMENT 439

TABLE 11-DS U M M A R YF 480 V o u C ALC ULATI EDAULT URRENTSI ) ND S I R RATIOSOR 2500 kVA TRANSFORMER.AULT URRENTSI ) N RMS SYMMETRICALAP cp 500 kVA Transformer High Side Fault MV ASOMVA 1OOMVA 250M VA 5OOMVA 1OOOMVA InfBusFault Location I x / R I x / R I ~ I X m I x / R I x / RTransforme:rSecondary 28.0 7.9 36.5 7.4 44.6 7.0 48.1 6.9 50.1 6 .8 52.3 6.710 feet 26.8 5.5 34.5 4.8 41.6 4.4 44.6 4.2 46.3 4.1 48.2 4.0

20 feet 25.7 4.3 32.6 3.7 38.7 3.3 41.3 3.1 42.8 3.0 44.3 2.9225A Panel 50 feet 22.5 2.7 27.5 2.3 31.5 2.0 33.1 1.9 34.0 1.9 34.9 1.8(1-#4/0 AWG 100 feet 18.3 1.8 21.2 1.5 23.4 1.4 24.2 1.3 24.6 1.3 25.0 1.3200 feet 12.9 1.2 14.2 1.1 15.0 1.0 15.3 1.0 15.4 0.9 15.6 0.9500 feet 6.6 0.8 6.9 0.8 7.1 0.7 7.1 0.7 7.2 0.7 7.2 0 110 feet 27.0 6.7 34.8 6.1 42.1 5.6 45.3 5.5 47.0 5.4 48.9 5.320 feet 26.1 5.8 33.3 5.2 39.9 4.8 42.7 4.6 44.3 4.5 45.9 4.4400A Panel 50 feet 23.6 4.4 29.3 3.8 34.3 3.5 36.3 3.3 31.4 3.2 38.6 3.2(1-500 kcmil) 100 feet 20.3 3.3 24.3 2.9 27.6 2.6 28.9 2.5 29.5 2.4 30.2 2.4200 feet 15.8 2.4 18.0 2.1 19.6 2.0 20.3 1.9 20.6 1.9 20.9 1.9500 feet 9.3 1.7 10.0 1.6 10.4 1.5 10.6 1.5 10.7 1.5 10.7 1.510 feet 27.5 7.0 35.6 6.4 43.2 5.9 46.6 5.7 48.4 5.6 50.4 5.520 eet600A Panel 50 feet(2-350 kcmil) 100 feet200 feet500 feet10 feet20 feet800A Panel 50 feet(2-500 kcmil) 100 feet200 feet

26.9 6.3 34.7 5.6 41.9 5.2 45.1 5.0 46.8 4.9 48.7 4.825.5 4.8 32.3 4.2 38.4 3.8 41.0 3.7 42.4 3.6 43.9 3.523.3 3.6 28.1 3.1 33.4 2.8 35.3 2.7 36.3 2.6 37.4 2.619.7 2.6 23.3 2.2 26.2 2.0 27.3 1.9 27.9 1.9 28.5 1.913.1 1.7 14.6 1.5 15.6 1.4 15.9 1.4 16.1 1.3 16.3 1.327.5 7.2 35.6 6.7 43.3 6.2 46.7 6.1 48.5 6.0 50.6 5.927.0 6.7 34.8 6.1 42.1 5.6 45.3 5.5 47.0 5.4 48.9 5.325.7 5.5 3 2.6 4.9 38.8 4.5 41.5 4.3 43.0 4.2 44.6 4.123.6 4.4 29.3 3.8 34.3 3.5 36.3 3.3 37.4 3.2 38.6 3.220.3 3.3 24.3 2.9 27.6 2.6 28.9 2.5 29.5 2.4 30.2 2.4500 feet 14.1 2.2 15.9 2.0 17.1 1.8 17.6 1.8 17.8 1.8 18.1 1.710 feet 27.7 7.4 36.0 6.9 43.9 6.4 47.3 6.2 49.3 6.2 51.4 6.120 feet 27.5 7.0 35.6 6.4 43.2 5.9 46.6 5.7 48.4 5.6 50.4 5.51200A Panel 50 feet 26.7 6.0 34.3 5.3 41.3 4.9 44.3 4.7 46.0 4.6 47.8 4.5(4-350 kcmil) 100 feet 25.5 4.8 32.3 4.2 38.4 3.8 41.0 3.7 42.4 3.6 43.9 3.5200 feet 23.3 3.6 28.1 3.1 33.4 2.8 35.3 2.7 36.3 2.6 37.4 2.6500 feet 18.2 2.3 21.3 2.0 23.6 1.8 24.5 1.8 24.9 1.7 25.4 1.7

These results would require minimum standard rms sym-metrical interrupting ratings between 7500 A and 50 kA foIthe 225-A panel; between 7500 A and 50 kA for the 400-Apanel; between 10 kA and 65 kA for the 60 0-A panel; between14 kA and 65 kA for the 800-A panel; between 14 kA and 65kA for the 1200 -A panel; and between 18 kA and 65 kA for theswitchgear on the secondary side of the transformer, dependingon the size of the transformer, the high side available faultMVA, and the length of the cable run to the panel.Table IV-A shows information from a circuit breaker manu-facturer's catalog for various 480-V, 3-pole breaker interrupt-ing ratings based on break.er frame size. Table IV-B showssimilar information from a fuse manufacturer's catalog. Itshould be noted that although there are a number of standardratings available in the standards for circuit breakers, breakermanufacturers only offer a small subset of those ratings withina given frame size.

I v . ,L\SYMMETRYAnother important item that must be considered whenspecifying low voltage protective devices is fault asymmetry.Asymmetry is determined by the X / R atio of the equivalentfault impedance using X only and R only network reductions.In some cases, reference is made to power factor instead ofan X / R ratio. It should be noted that the two quantities arerelated as follows:

X / R = tan(cos-'(powcr factor)). (1)

TABLE 111R ANGE F CALCULATEDA U L T U R R ~ N T SkA)000 kVA 2500 kVAXfmr Secondary 15.5-20.9 28.0- 52.3

225A Panel 5.1- 20.2 6.6- 48.2400A Panel 1.4- 20.4 9.3- 48.9600A Panel 9.6- 20.6 13.1- 50.4800A Panel 10.1- 20.6 14.1- 50.618.2- 51 42.0- 20.8200A Panel

Asymm etry shows the amount of dc offset in the expectedfault current. The reason for the dc offset is based on thefollowing electrical principles:1) Current cannot change instantaneously in an inductor.2) Current in an inductive circuit lags the voltage based onIn order to calculate the highest possible fault current(conservative calculation), it is assumed that the fault will

occur at the worse possible time. For a purely inductive circuitX/R CO , this would be when the sinusoidal voltagecrosses zero. Initially, the fault current (Isym) starts at zero andcontinues to increase until it peaks a half cycle later when thevoltage crosses zero again. (Since the current lags the voltageby 90, the current peaks 90 after the voltage.) This wouldbe a completely offset wave and the dc offset would beI , , ~ , or an equivalent rms value of 1 3 I,,,.The need for including the asymmetry of the fault currentwhen specifying low voltage protective equipment is that theprotective devices are tested at a certain fault X / R ratio. For

the X / R ratio of the circuit impedance.

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440 IEEE TRANSACTIONS ON INDUSTRY APPLICATIONS, VOL. 32, NO 2, MARCH/APRIL 1996

TABLE IV-A480-V, 3-POLE, MOLDED-CASEIRCUIT BREAKERINTERRUPTING RATINGSROM MANUFACTURERSATALOGTABLE VUL 4 8 9 R EQUI R EDOWER AC TORF TESTCIRCUITS FOR MOLDED-CASEIRCUIT BREAKERS

Int Rating Int RatingFrame Breaker (kA) Frame Breaker1 5 0 A F Type 14 600A F Type 1 30Type 2 25 Type 2 35

Type 3 65 Type 3 wISST 65Type 4 150 Type 4 w/CLF 200

Test Circuit in Amps Power Factor X R10,000 orLess 0.45-0 50 1.98-1.7310,001 to 20,000 0.25-0.30 3.87-3.17over 20,000 0.15-0.20 6.59-4.89

Tr p e 5 WICLF 2002 2 5 A F Ty pe 1 22 800AF Type 1 30Tybe 2 25 Type 2 35Type 3 65 Ty pe3 w1SST 50Type 4 150 Type 4wIS ST 65

4 0 0 A F T y p e lType 5 WICLF 2003 0 1 2 0 0 AF Ty p e l 3 0Type 2 35 Type 2 35Type 3 65 Type 3wlS ST 50Type 4 150 Type 4wlS ST 65Type 5 w/CLF 200

wlCL F means with current limiting fuses.w/SST m eans with solid-state trip unit.where:

TABLE IV-B480-V (600-V CLASS) USE NTERRUPTINGRATINGSROM MANUFACTURERSATALOGInterruptingFuse Class Amp Rating Rating @A

Class K5 1-600 amps 50Class H 1-600 amps 10Class RK1 1-600 ~ P S 200Class RK5 1-600 amps 200Class J 1-600 amps 200Class L 601-6000 amps 200

examp le, Table V shows the power facto r of the test circuit formolded -case circuit breakers as specified by [11, and Table VIshows the powe r factor of the test circuit for low voltage powercircuit breakers as specified by [4]. The X / R atio shown inTable V was calculated from the given pow er factor using (1).The fact that the circuit breakers are tested at a particularvalue of power factor ( X / R ) oes not preclude the use ofthe circuit breakers in electrical systems where the X / R atioexceeds those test values. In order to account for the higherthan test value X / R ratio, [4] gives multiplying factors forfused and unfused low voltage power circuit breakers. Themultiplying factors are shown in Table VII.When specifying a circuit breaker, it is desired to meet thefollowing condition:

( I syrn s rating) > M F ) ( I s y n s du ty ) (2)where

M F Multiplying factor;Isymc ratingIsyrnC d u tyNow one can determine the appropriate multiplying factorfor low voltage power circuit breakers (using Table VI1 forunfused and fused breakers) and molded-case circuit breakers(by deriving a similar table). Once the approp riate multiplyingfactor is known, the calculated symmetrical fault duty can bemultiplied by that factor, and then a direct comparison can be

Symmetrical short circuit rating;Symmetrical short circuit duty.

TABLE VIANSMEEE C37.13-1981 REQUIREDOWER AC TORF TESTCIRCUITS FOR LOW-VOLTAGEOWER CIRCUIT BREAKERSBreaker Type Power Factor X RUnfused 0.15 6.59Fused 5 0.20 4.89

TABLE VI1ANSI/IEEE (237.13-198 SELECTIONF MULTIPLYINGFACTOROR LOW-VOLTAGEO W E R I R C U I TR E A K E R SPower Factor XlR Multiplying Factor for Short-Circuit Current

Unfused Fused20 4.9 1.00 1 0 0I5 6 .6 1.00 1 0 7I 8 27 1 04 1.1110 9 95 1 .07 11 58.5 1172 1.09 1.187 1 4 2 5 1 1 1 1 2 15 20 .0 1 .15 1 26

made between this m odified fault current just c alculated andthe symmetrical interrupting rating of the circuit breaker.

V. MOTORCONTRIBUTIONLow-voltage motors also add to the total calculated faultcurrent. The reason motors contribute current during a fault

is that during normal operation a voltage is applied to thestator windings, which induces a voltage in the rotor. Whena fault occurs, the residual voltage on the rotor causes theinduction motor to act as an induction gene rator and contributecurrent to the point of fault (where the voltage is zero). Themaximum amount of current a motor can contribute to the faultis approximately equ al to the locked-rotor current of the motor.ANSUIEEE Standard 141-1986 gives impedance multipliersfor motor reactances to take into account ac decay.)One problem that exists in most plants is estimating theamount of connected motor loads. In some cases, one mayattempt to estimate the motor load by measuring the loadcurrent. However, it should be noted that load current isnot always a good indication of the size of the motor. Forexample, a 200-hp motor operating at half load would drawapproximately the same amount of load current as a 100-hpmotor operating at full load. If both motors had the samelocked-rotor kVA code, the 200-hp motor would contributetwice as much to the fault as the 100-hp motor even thoughboth motors had the same running load current.The fault current contribution from a motor would begreatest for a fault at the motor (or motor control center if allmotors were lumped at the M C C ) , and it would decrease asthe fault got farther from the m otor. The rate at which the faultcurrent would decrease would depend on the cable impedance

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TOTH: THE IMPORTANCEOF KNOWING AVAILABLE FAULT CURRENTS WHEN SPECIFYING INTERRUPTING RATINGS FOR LOW-VOLTAGEEQUIPMENT 441

TABLE VI11SUMMARY OF 480 VOLT CALCULATED FAULTURRENTS I ) AND X / R RATIOSOR 2000-kVA TRANSFORMERED ROM AN INFINITE BUS

WITH 1000-HP hhOTOR CONTR IBUTION AT END F 4-350 kcmil FEEDER.AULT URRENTSI ) N RMS SYMMETRICALAMotor Location 4-350 kcmil Feeder Length)lofe et 20feet 5Ofeet 1OOfeet 200feet 500feetFault Locatiisn I x / R I x / R I x R I x i R I X R I x / RTransfonner Secondary 41.8 6.4 41.1 6.4 41.1 6.4 41.1 6.3 41.5 6.3 41.2 6.1

10 feet 44.3 4.0 44.3 4.0 44.2 4.0 44.2 4.0 44.1 4.0 43.8 3.920 feet225AP anel 50 feet(1-#4/0 AWG ) 100 feet200 feet500 feet10 feet20 feet400A Panel 50 feet(1-500 kcmil) 100 feet200 feet

41.0 3.0 41.0 3.0 41.0 3.0 40.9 3.0 40.8 3.0 40.6 3.032.9 1.9 32.9 1.9 32.9 1.9 32.8 1.9 32.8 1.9 32.6 1.924.0 1.3 24.0 1.3 24.0 1.3 24.0 1.3 24.0 1.3 23.9 1.315.2 1.0 15.2 1.0 15.2 1.0 15.2 1.0 15.2 1.0 15.2 1 .01.1 0.7 1.1 0.1 1.1 0.7 7.1 0.1 7.1 0.7 7.1 0.744.9 5.2 44.9 5.2 44.9 5.2 44.8 5.1 44.7 5.1 44.4 5.042.4 4.4 42.4 4.4 42.3 4.4 42.3 4.4 42.2 4.4 41.9 4.336.1 3.2 36.1 3.2 36.0 3.2 36.0 3.2 35.9 3.2 3 5.1 3.228.7 2.5 28.1 2.5 28.1 2.5 28.6 2.5 28.6 2.4 28.4 2.420.2 1.9 20.2 1.9 20.2 1.9 20.1 1.9 20.1 1.9 20.0 1.9500 feet 10.6 1.5 10.6 1.5 10.6 1.5 10.5 1.5 10.5 1.5 10.5 1.510 feet 46.2 5.4 46.2 5.4 46.2 5.4 46.1 5.4 46.0 5.3 45.1 5.220 feet600A Panel 50 feet(2-350 kcml 100 feet200 feet500 feet10 feet20 feet800A Panel 50 feet(2-500 kcmill) IO0 feet200 feet

44.1 4.8 44.1 4.1 44.1 4 1 44.6 4.7 44.5 4.1 44.2 4.640.6 3.5 40.6 3.5 40.6 3 5 40.6 3.5 40.5 3.5 40.2 3.535.1 2.6 35.0 2.6 35.0 2.6 35.0 2.6 34.9 2.6 34.7 2.621.2 1.9 21.2 1.9 27.1 1.9 21.1 1.9 21.1 1.9 26.9 1.915.8 1.3 15.8 1.3 15.8 1.3 15.8 1 .3 15.8 1.3 15.8 1.346.3 5.7 46.3 5.7 46.3 5.7 46.2 5.7 46.1 5.6 45.8 5.544.9 5.2 44.9 5.2 44.9 5.2 44.8 5.1 44.1 5.1 44.4 5.041.2 4.1 41.2 4.1 41.2 4.1 41.1 4.1 41.0 4.1 40.8 4.036.1 3.2 36.1 3.2 36.0 3.2 36.0 3.2 35.9 3.2 35.1 3.228.1 2.5 28.7 2.5 28.1 .2.5 28.6 2.5 28.6 2.4 28.4 2.4500 feet 17.5 1.8 17.5 1.8 17.5 1.8 17.5 1.8 17.5 1.8 17.4 1.810 feet 41.2 6.0 NIA NIA NIA NIA NIA NIA NIA NIA NIA NIA20 feet NIA NIA 46.6 5 1 NIA NIA NIA NIA NI A N I A NIA NIA1200 APa nd 50 feet NIA NIA NIA NI A 44.8 4.9 NIA NIA NIA NIA NIA NIA(4-350 kcmil) 100 feet NI A N I A NIA NIA NIA NIA 42.2 4.2 NIA NIA NIA NIAwith 1000hp 200 feet NI A NIA NIA NIA NIA NIA NIA NIA 31.6 3.4 NIA NIA500 feet NIA NIA NIA NI A NIA NIA NI A NIA NIA NIA 28.4 2.8

a function of the conductox size, number of conductors perphase, and length) between the motor and the point of the fault.Table VI11 shows the results of a fault calculation for a2000-kVA transformer with ,a5.75% mpedance, an infinite busfeeding the transformer, and 1000-hp of motor loads (5-100hp motors and 10-50 hp motors) connected at the 1200-Apanel with feeder lengths of 10, 20, 50, 100, 200, and 500 ft.Comparing the results of Table VIII with the results of Table11-C (for the infinite bus), it can be seen that the fault currentsincreased by anywhere from 100 A to 5100 A for the 225-Apanel; from 200 A to 5200 A for the 400-A panel; from 600A to 5600 A for the 600-PL panel; and from 700 A to 560 0A for the 800-A panel. Therefore, motor contribution shouldbe considered when fault calculations are made to specify orevaluate interrupting ratings of low-voltage equipment.

VI. FULLY ATEDOR SERIES RATEDEQUIPMENTAnother question often arises when specifying low voltageequipment ratings: Should fully rated equipment or series-rated equipment be specified?In a fully rated system, the interrupting ratings of circuitbreakers are chosen in accordance with (2). This says everyprotective device is rated for the available fault current.In some cases, it may be more cost effective to consider aseries connected application. In this way, circuit breakers ina downstream panel may be chosen with a lower interruptingrating than the calculated fault current (in most cases, reducing

the cost of the circuit breakers) provided they have a listedseries connected rating that meets the criteria in (2). Forexample, in a panel with a main breaker and twenty feederbreakers, only the main breaker would need to be fully rated.The other twenty feeder breakers could have a lower individualrating provided they had an adeq uate series connected rating.

A few comments may be worth noting about series con-nected ratings.1) Series connected ratings only apply to molded-casecircuit breakers, rated 600 V and below.2) The upstream device may be a fuse or a molded-casecircuit breaker.3 If the upstream device is a fuse, it is identified ingeneric terms, by class (e.g., Class R, Class J, Class

L, etc.), with a maximum amp rating. Therefore, anyfuse manufacturer may be used.4) If the upstream device is another molded-case circuitbreaker, it is identified by typ e and maximum trip rating.Therefore, the upstream and downstream devices mustbe by the same manufacthrer.5) The upstream device does not need to be located in thesame panel. It can be located in a remote panel upstream.6) All of the fault current that flows through the down-stream molded-case circuit breaker must also flowthrough the upstream protective device. This wouldpreclude the ability to series rate protective devicesin a motor control center since it would be unlikely that

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442 IEEE TRANSACTIONS ON INDUSTRY APPLICATIONS, VOL 32, NO 2, MARCWAPRIL 1996

all of the motor contribution would flow through boththe upstream and downstream devices. This would alsoapply to other panels (besides MCCs) that fed somemotor loads.There is a case, with m otor control centers, where protective

devices appear to be series rated. Thats when the breaker ispart of a combination starter and the unit has a higher ratingthan the individual molded-case circuit breaker in the starter.This is allowed because all of the fault current flows throughall parts of the starter (the overload, the contactor, and themolded-case circuit breaker).Molded-case circuit breaker manufacturers normally publisha list of their UL recognized series connected ratings.One final warning about trying to interpolate or extrapolatea series combination that does not appear in the manufacturerspublication. The combination may not be included for one ofthe following reasons.

1) It was never tested, and it would pass the test.2) It was never tested, and it would fail the test.3) It was tested, and it failed the test.Since one may not know for which of the reasons the desiredcombination was not listed, making an assumption would notbe a wise choice. Use only what is published.

VII. CONCLUSIONSHopefully this will help clear up some of the mystery

about that often forgotten characteristic of low-voltage pro-tective equipment-the interrupting rating. Know ledge of theexistence of such a rating, and knowing the available faultcurrent (including any motor contribution and appropriatemultiplying factors), should make the selection process simple.Just remember, the correct interrupting rating is as important

as the correct voltage and trip rating. Its one more ite m to beaware of when specifying low-voltage protective equipment.REFERENCES

f 11 M olded-Case Circuit Breakers and Circuit-Breaker Enclosures, UL 489,. table 48.1, April 6, 1987.121 Motor Control Centers, UL 845, table 32.1. March 26. 198731 Low-Voltage AC and DC Power Circuit-Breakers Used in Enclosures,. UL 1066, sec 21 4, Jan 24 , 1985[ ] EEE Standard for Low-Voltage AC Power Circuit Breakers,ANSVIEEE C37 13, table 3, 1981[ 5 ] IEEE Recommended Practice for Electric Power Distribution for Indus-trial Plants IEEE Red Book), ANSIAEEE Standard 141, 1986

James J Toth (S79-M82) received theBachelor of Engineering (with Honor) degree fromStevens Institute of Technology and the Master ofEngineering in electric power engineering degreefrom Rensselaer Polytechnic Institute, Troy, NY.While enrolled in a doctoral program at GeorgiaInstitute of Technology, his research activitiesincluded optimal var compensation allocation,electromagnetic transients, and subsynchronousresonance.He was a Project Engineer with the IndustrialPower Systems Engineering group of Westinghouse Electric Corporationin Pittsburgh, PA, and a Power Systems Engineer with the ConsultingAdvisory Engineering group of Westinghouse Electric Corporation inAtlanta, GA. Later, he was a Power Systems Engineer with the InstallationService Engineering group of General Electric Company in Charlotte,NC. He is currently a Senior Power Systems Engineer with ABB ServiceInc. in Matthews (Charlotte), NC, where he is responsible for performinga variety of power systems analyses (e.g., short circuit, coordination, loadflow, power factor correction, harmonics, an d transients) in support of projectsthroughout the country. He has also taught courses ifi industrial power systemsengineering and industrial power systems coordination.Mr. Toth is a member of the Industry Applications Society and the N ationalSociety of Professional Engineers. He is a licensed Professional Engineer inthe states of North Carolina, South Carolina, Virginia, and Georgia.