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Arc flash

Short Circuit and Arc Flash Calculator (0702e010.xls)

IntroductionHanford Electrical Safety Program Arc Flash CalculatorThis Excel file combines arc flash equations from a proposal to NFPA 70E-2003 with simplified equations to calculate available fault current. This tool uses methods that have been shown to result in conservative estimates of fault current, which should result in conservative estimates of arc incident energy if appropriate arc fault clearing times are used. An IEEE 1584 equation is used to calculate arcing fault current. That is the fault current value that is used to determine the clearing time of the overcurrent protective device ahead of the potential arc fault location. Electrical engineering should be consulted to obtain information on fault clearing times.This spreadsheet is not intended to replace existing up to date fault studies or support of experienced electrical engineers, but is one tool that can assist engineers, planners, and electrical safety POCs in performing a flash hazard analysis. Arc flash calculation is not an exact science and caution is always necessary, including use of multiple alternate methods that are available to verify results and ensure the highest level of safety based on the best information available.According to the NFPA 70E Technical Committee on Electrical Safety Requirements for Employee Workplaces, This proposal presents the best information available to date on arc fault hazards. Public review and comments are strongly encouraged. Recent testing has enabled development of improved equations for calculating the arc flash incident energy at the arc flash boundary. While the testing and development of methods is not complete this proposal contains methods that reflect significantly more laboratory data than the existing methods and will allow improved safety.This calculator is for use only with systems operating at less than 1000 volts. If any doubts exist on use of this spreadsheet, or to perform a flash hazard analysis on systems operating at more than 1000 volts, consult with an electrical engineer or other knowledgeable person.Question/comments? Contact Electrical Safety (371-7886)

&L1.0/sp00e230.xls&R(5/07)

CalculatorShort Circuit and Arc Flash CalculatorArc-In-Box energy = cal/cm2 at specified working distanceK = constant for arc in a box (see Arc Calc Info page for equations)( kVAmps > kAXfmr impedance %:Xfmr 3-Ph ISC (Amps)=0Flash Protection Boundary:00-0.0972500.0000.00Fault Clearing Time (seconds):@0arc fault current (Amps)0Conductors per phase:Feeder(S)ingle conductors or (C)able:0Enter working distance (inches):18AL or CU:Arc-In-Box Incident Energy:0.00Conductor length:Flash Protection Boundary:0Amps > kAConductor AWG or kcmilIsc at fault (Amps) =00000.00Magnetic conduit (Y or N):"f""M"Fault Clearing Time (seconds):@0arc fault current (Amps)0.00001.0000Isc at beginning of circuit (Amps):Branch CircuitConductors per phase:0Enter working distance (inches):18(S)ingle conductors or (C)able:Arc-In-Box Incident Energy:0.00AL or CU:Flash Protection Boundary:0Conductor length:0Amps > kAConductor AWG or kcmil:Isc at fault (Amps) =0000.00Metallic conduit? (Y or N):Scroll down to enter equipment ID information"f""M"Fault Clearing Time (seconds):@0arc fault current (Amps)0.00001.0000Flash Calculation Location:Transformer ID:Panel ID:Branch Circuit ID:Other Equipment ID:

&CHESP Arc Flash Calculator&L1.0/sp00e230.xls&R(5/07)Contact Electrical Safety to obtain fault clearing timesContact Electrical Safety to obtain fault clearing timesContact Electrical Safety to obtain fault clearing times

"C" ValuesThree single conductors, 600-voltThree-conductor cable, 600-voltAWG orCopper ConductorsAWG orAluminum ConductorsAWG orCopper ConductorsAWG orAluminum ConductorskcmilConduitkcmilConduitkcmilConduitkcmilConduitSteelNonmagneticSteelNonmagneticSteelNonmagneticSteelNonmagnetic143893891423623614389389142362361261761712375375126176171237537510981981105985981098198110598598815571558895195181559155989519516242524306148014816243124336148114824380638254234523504383038374235123533476048023294829583476048023294829582590660442371337292598960872373337391729274931464546781745475791468646991/0892493171/0577758381/0920994721/0585258752/010755114232/0718673012/011244117032/0732773723/012843139233/0882691103/013656144103/0907792424/015082166734/010740111744/016391174824/01118411408250164831859325012122128622501831019779250127961323630018176208673001390914922300206172252430014916154943501970322736350154841681235022646249043501541317635400205652429640016670185054002425326915400184611958750022185267065001875521390500269803002850021394229876002296528033600200932345160028752322366002363325750750241362830375021766259767503105032404750264312903610002527831490100023477287781000338643719710002986432938244828280499192096213475284982307338213707227951

&L1.0/sp00e230.xls&R(5/07)

Isc InfoSource (except as noted):Bulletin EPR-1, Electrical Plan ReviewCooper Bussman, May 2000Pages 8 - 10http://www.bussmann.com/library/docs/EPR_Booklet.pdfXfmr FLA =(KVA * 1000) / (EL-L * 1.732)*3-Ph ISC at xfmr =(((KVA / 1000) * 106) / (1.732 * EL-L)) * 100 / Z%)(*NFPA 70E, 2000 Edition, Appendix B, Section B-2-1)"f" factor =(1.73 * L * IL-L-L) / (C * EL-L)"M" =1/(1 + f)3-Ph Isc at fault =ISC at xfmr * Mwhere:EL-L = phase-to-phase voltageZ = transformer nameplate impedance, in percentL = length of conductor to the faultIL-L-L = available 3-phase short circuit current at beginning of circuitC = constant from "C" Values sheet (multiply by number of conductors per phase for parallel runs)"f" factor = calculated variable from source document formulaM = calculated variable from source document formula

&L1.0/sp00e230.xls&R(5/07)http://www.bussmann.com/library/docs/EPR_Booklet.pdf

Arc Calc InfoReferences:The Other Electrical Hazard: Electrical Arc Blast Burns, R. Lee, IEEE Trans. Industrial Applications, Vol 1A-18. No. 3, Page 246, May/June 1982.The Use of Low Voltage Current Limiting Fuses to Reduce Arc Flash Energy, T. Neal, V. Saporita, T. Macalady, R. Doughty, K. Borgwald, Record of Conference Papers IEEE PCIC-99-36.Predicting Incident Energy to Better Manage the Electric Arc Hazard on 600 V Power Distribution Systems, R. L. Doughty, T. E. Testing Update on Protective Clothing & Equipment For Electric Arc Exposure, R. Doughty, T. Neal, T. Dear, A. Bingham, Record of Conference Papers IEEE PCIC-97-35.Testing Update on Protective Clothing & Equipment For Electric Arc Exposure, R. Doughty, T. Neal, T. Dear, A. Bingham, Record of Conference Papers IEEE PCIC-97-35.IEEE Std. 1584TM-2002, IEEE Guide for Performing Arc-Flash Hazard Calculations, IEEE Industry Applications SocietyProposal 70E-157a - (Annex XXX), Log #CP8, submitted and accepted by the Technical Committee on Electrical Safety Requirements for Employee WorkplacesTC substantiation statement: This proposal presents the best information available to date on arc fault hazards. Public review and comments are strongly encouraged. Recent testing has enabled development of improved equations for calculating the arc flash incident energy at the arc flash boundary. While the testing and development of methods is not complete this proposal contains methods that reflect significantly more laboratory data than the existing methods and will allow improved safety.Equations:B-2 Basic Equations for Calculating Incident Energy and Flash Protection Boundary Distances of Equipment.The following equations can be used to predict the incident energy and flash protection boundary distances produced by a three-phase arc and the flash protection boundary distance for that arc, based on the voltage range. The parameters required to make the calculation are:(a) The maximum bolted fault three-phase short circuit current available at the equipment,(b) The total protective device clearing time (upstream of the prospective arc location) at the arcing current,(c) The distance of the worker from the arc for the task to be performed.Voltage Range:Calculation:Equation:Vo < 1000 Volts*Ialg Ia = K + 0.662 lg Ibf + 0.0966 V + 0.000526 G + 0.5588 V (lg Ibf) 0.00304 G (lg Ibf)EiEi = 416 Ia t D 1.6DbDb = (416 Ia t / 1.2)0.625* Ia equation from IEEE 1584-2002 where:lgis the log10where:Iais arcing current (kA)Vo is the open circuit voltage of the system,Kis 0.097 for box congurationsIa is the arcing current in kA,Ibfis bolted fault current for three-phase faults (symmetrical RMS) (kA)Ib is the bolted fault current (from 0.6 to 106 kA),Vis system voltage (kV)Ei is the incident energy in cal/cm 2,Gis the gap between conductors, (calculator uses 25mm = 1 inch)D is the distance of the worker from the arc in inches (18 inches or more),t is the time of arc exposure in seconds, andDb is the boundary distance in inches from the arc (distance where incident energy is 1.2 cal/cm 2).Other Information:Calculation of Incident Energy Exposure for Open Air Arcs.The incident energy from open arcs can be better calculated through calculation programs that are commercially available in the marketplace. Most equipment incident energy values would be of the arc-in-box type, since a majority of work on voltages up through 15000 volts is in motor control cabinets, pad-mount switches, and other enclosures.

&L1.0/sp00e230.xls&R(5/07)

Typical CalculationsTypical Flash Protection Boundaries for 3-Phase, 480-volt systems(Arc-in-Box)ClearingCalculated FlashArc Incident Energy3-Phase BoltedTimeProtection Boundary (inches)cal/cm2 at 18 InchesFault Current (kA)(sec)Arc in a BoxArc in a Box500.05526.630.108113.260.2012526.52480.05516.440.107912.890.2012225.77460.05506.25460000.107812.500.2012025.00440.05496.050.107612.100.2011824.20420.05485.840.107511.690.2011523.37400.05475.630.107311.260.2011322.52380.05465.410.107110.820.2011021.64360.05455.180.106910.370.2010720.7