Short-Circuit Current Calculations

Short-Circuit Current CalculationsBasic Point-to-Point Calculation ProcedureStep 1.Determine the transformer full load amps (F.L.A.) fromAt some distance from the terminals, depending upon wire size, the L-N faultcurrent is lower than the L-L fault current. The 1.5 multiplier is an approximationand will theoretically vary from 1.33 to 1.67. These figures are based on change inturns ratio between primary and secondary, infinite source available, zero feet fromterminals of transformer, and 1.2 x %X and 1.5 x %R for L-N vs. L-L resistance andreactance values. Begin L-N calculations at transformer secondary terminals, thenproceed point-to-point.Step 5.Calculate "M" (multiplier) or take from Table 2.either the nameplate, the following formulas or Table 1:100Multiplier *% Z transformerStep 2.Step 6.Find the transformer multiplier. See Notes 1 and 2* Note 1. Get %Z from nameplate or Table 1. Transformer impedance (Z) helps todetermine what the short circuit current will be at the transformer secondary.Transformer impedance is determined as follows: The transformer secondary is shortcircuited. Voltage is increased on the primary until full load current flows in thesecondary. This applied voltage divided by the rated primary voltage (times 100) is theimpedance of the transformer.Example: For a 480 Volt rated primary, if 9.6 volts causes secondary full load current toflow through the shorted secondary, the transformer impedance is 9.6/480 .02 2%Z.* Note 2. In addition, UL 1561 listed transformers 25kVA and larger have a 10%impedance tolerance. Short circuit amps can be affected by this tolerance. Therefore, forhigh end worst case, multiply %Z by .9. For low end of worst case, multiply %Z by 1.1.Transformers constructed to ANSI standards have a 7.5% impedance tolerance(two-winding construction).Step 3. Determine by formula or Table 1 the transformer letthrough short-circuit current. See Notes 3 and 4.11 fM Calculate the available short circuit symmetrical RMScurrent at the point of fault. Add motor contribution, ifapplicable.I S.C. sym. RMS IS.C. x MStep 6A. Motor short circuit contribution, if significant, may beadded at all fault locations throughout the system. Apractical estimate of motor short circuit contribution is tomultiply the total motor current in amps by 4. Values of 4to 6 are commonly accepted.Calculation of Short-Circuit Currents When PrimaryAvailable Short-Circuit Current is KnownUse the following procedure to calculate the level of fault current at the secondaryof a second, downstream transformer in a system when the level of fault current atthe transformer primary is known.MAINTRANSFORMERIS.C. TransformerF.L.A. x MultiplierNote 3. Utility voltages may vary 10% for power and 5.8% for 120 Volt lightingservices. Therefore, for highest short circuit conditions, multiply values as calculated instep 3 by 1.1 or 1.058 respectively. To find the lower end worst case, multiply results instep 3 by .9 or .942 respectively.Note 4. Motor short circuit contribution, if significant, may be added at all fault locationsthroughout the system. A practical estimate of motor short circuit contribution is tomultiply the total motor current in amps by 4. Values of 4 to 6 are commonly accepted.Step 4. Calculate the "f" factor.3Ø Faults1Ø Line-to-Line (L-L) FaultsSee Note 5 & Table 31Ø Line-to-Neutral (L-N) FaultsSee Note 5 & Table 3f Step A.2 x L x I L-Lf C x n x EL-L2 x L x I L-N†f C x n x EL-N† Note 5. The L-N fault current is higher than the L-L fault current at the secondaryterminals of a single-phase center-tapped transformer. The short-circuit current available(I) for this case in Step 4 should be adjusted at the transformer terminals as follows: AtL-N center tapped transformer terminals, IL-N 1.5 x IL-L at Transformer Terminals. 2014 EatonIS.C. primary1.732 x L x I 3ØC x n x E L-LWhere:L length (feet) of conductor to the fault.C constant from Table 4 of “C” values for conductors andTable 5 of “C” values for busway.n Number of conductors per phase (adjusts C value forparallel runs)I Available short-circuit current in amperes at beginningof circuit.E Voltage of circuit.IS.C. primaryH.V. UTILITYCONNECTIONStep B.IS.C. secondaryIS.C. secondaryCalculate the "f" factor (IS.C. primary known)3Ø Transformer(I S.C. primary andI S.C. secondary are3Ø fault values)f 1Ø Transformer(I S.C. primary andI S.C. secondary are1Ø fault values:I S.C. secondary is L-L)f I S.C. primary x V primary x 1.73 (%Z)100,000 x kVAtransformerI S.C. primary x V primary x (%Z)100,000 x kVAtransformerCalculate "M" (multiplier).11 fStep C. Calculate the short-circuit current at the secondary of thetransformer. (See Note under Step 3 of "Basic Point-toPoint Calculation Procedure".)M I S.C. secondary VprimaryVsecondaryx M x I S.C. primary237

Short-Circuit Current CalculationsThree-Phase Short CircuitsSystem AOne-Line DiagramAvailable UtilityInfinite Assumption1500 KVA Transformer480V, 3Ø, 3.5%Z,3.45% X, 0.56%R1Fault X1Fault X3Step 1. If.l. 1500 X 1000 1804.3A480 X 1.732Step 4. f 1.732 X 50 X 55,137 0.448422,185 X 1 X 480Step 2. Multipler100 31.7463.5 X 0.9†Step 5. M 1 0.69041 0.4483I 1804Af.l.Step 3. Is.c. 1804.3 X 31.746 57,279AIs.c. motor contribution** 4 X 1804.3 7217AItotal s.c. sym RMS 57,279 7217 64,496A25’ - 500kcml Cu3 Single Conductors6 Per PhaseMagnetic ConduitStep 6. Is.c. sym RMS 55,137 X 0.6904 38,067AIs.c. motor contribution** 4 X 1804.3 7217AItotal s.c. sym RMS (X3) 38,067 7217 45,284AFault X22Step 4. f 1.732 X 25 X 57,279 0.038822,185 X 6 X 4802000A SwitchKRP-C 2000SP FuseStep 5. M 400A SwitchLPS-RK-400SP Fuse1 0.96261 0.0388Step 6. Is.c. sym RMS 57,279 X 0.9626 55,137AIs.c. motor contribution** 4 X 1804.3 7217AItotal s.c. sym RMS 55,137 7217 62,354A50’ - 500 kcmil Cu3 Single ConductorsMagnetic Conduit3Motor Contribution*M*See note 4 on page 240.**Assumes 100% motor load. If 50% of this load was from motors. Is.c. motor contrib. 4 X 1804 X 0.5 3,608A† See note 2 on page 240System BOne-Line DiagramAvailable UtilityInfinite Assumption1000 KVA Transformer480V, 3Ø, 3.5%Z,Fault X1Fault X3Step 1. Is.c. 1000 X 1000 1202.8A480 X 1.732Step 4. f 1.732 X 20 X 36,761 0.11612 X 11,424 X 480Step 2. Multipler If.I. 1203A130’ - 500kcml Cu3 Single Conductors4 Per PhasePVC Conduit100 31.7463.5 X 0.9†Step 5. M 1 0.89601 0.1161Step 3. Is.c. 1202.8 X 31.746 38,184AStep 6. Is.c. sym RMS 36,761 X 0.8960 32,937AFault X2Fault X4Step 4. f 1.732 X 30 X 38,184 0.038726,706 X 4 X 480Step A. f 32,937 X 480 X 1.732 X (1.2 X 0.9) 1.3144100,000 X 22521600A SwitchKRP-C 1500SP Fuse Step 5. M 400A SwitchLPS-RK-350SP Fuse1 0.96271 0.0387Step 6. Is.c. sym RMS 38,184 X 0.9627 36,761A20’ - 2/0 Cu3 Single Conductors2 Per PhasePVC ConduitStep B. M 1 0.43211 1.3144Step C. Is.c. sym RMS 480 X 0.4321 X 32,937 32,842A2083225 KVA Transformer208V, 3Ø1.2%ZThis example assumes no motor contribution.4238 2014 Eaton

Short-Circuit Current CalculationsSingle-Phase Short CircuitsShort circuit calculations on a single-phase center tapped transformer systemrequire a slightly different procedure than 3Ø faults on 3Ø systems.ABC1. It is necessary that the proper impedance be used to represent the primary system.For 3Ø fault calculations, a single primary conductor impedance is used from thesource to the transformer connection. This is compensated for in the 3Ø short circuitformula by multiplying the single conductor or single-phase impedance by 1.73.PrimarySecondaryHowever, for single-phase faults, a primary conductor impedance is considered fromthe source to the transformer and back to the source. This is compensated in thecalculations by multiplying the 3Ø primary source impedance by two.2. The impedance of the center-tapped transformer must be adjusted for thehalf-winding (generally line-to-neutral) fault condition.ShortCircuitThe diagram at the right illustrates that during line-to-neutral faults, the full primarywinding is involved but, only the half-winding on the secondary is involved.Therefore, the actual transformer reactance and resistance of the half-windingcondition is different than the actual transformer reactance and resistance of the fullwinding condition. Thus, adjustment to the %X and %R must be made whenconsidering line-to-neutral faults. The adjustment multipliers generally used for thiscondition are as follows: 1.5 times full winding %R on full winding basis. 1.2 times full winding %X on full winding basis.Note: %R and %X multipliers given in “Impedance Data for Single PhaseTransformers” Table may be used, however, calculations must be adjusted toindicate transformer kVA/2.3. The impedance of the cable and two-pole switches on the system must beconsidered “both-ways” since the current flows to the fault and then returns to thesource. For instance, if a line-to-line fault occurs 50 feet from a transformer, then100 feet of cable impedance must be included in the calculation.PrimarySecondaryShort CircuitL2NL1The calculations on the following pages illustrate 1Ø fault calculations on asingle-phase transformer system. Both line-to-line and line-to-neutral faults areconsidered.L1Note in these examples:a. The multiplier of 2 for some electrical components to account for the single-phasefault current flow,b. The half-winding transformer %X and %R multipliers for the line-to-neutral faultsituation, andN50 Feet 2014 EatonShort CircuitL2239

Short-Circuit Current CalculationsSingle-Phase Short CircuitsSystem AOne-Line DiagramAvailable UtilityInfinite Assumption75KVA, 1Ø Transformer.1.22%X, 0.68%R1.40%Z120/240VLine-to-Line (L-L) FaultLine-to-Neutral (L-N) FaultFault X1Fault X1Step 1. If.l. 75 X 1000 312.5A240Step 1. If.l. 75 X 1000 312.5A240Step 2. Multipler 100 79.371.4 X 0.9†Step 3. Is.c. (L-L) 312.5 X 79.37 24,802A25’ - 500kcml CuMagnetic Conduit3 Single ConductorsFault X2Step 5. M 100 79.371.4 X 0.9†Step 3*. Is.c. (L-N) 24,802 X 1.5 37,202AFault X2Step 4. f 2 X 25 X 24,802 0.232922,185 X 1 X 2402Step 2. Multipler 1 0.81111 0.2329Step 4. f 2 X 25 X 37,202 0.698722,185 X 1 X 120Step 5. M 1 0.58871 0.6987Step 6. Is.c. (L-L) (X2) 24,802 X 0.8111 20,116400A SwitchLPN-RK-400SP FuseStep 6*. Is.c. (L-N) (X2) 37,202 X 0.5887 21,900A50’ - 3 AWG CuMagnetic Conduit3 Single ConductorsFault X3Fault X3Step 4. f 2 X 50 X 20,116 1.75574774 X 1 X 240Step 4. f 2 X 50 X 21,900** 3.83234774 X 1 X 1203Step 5. M 2401 0.36291 1.7557Step 5. M 1 0.20731 3.823Step 6. Is.c. (L-L) (X3) 20,116 X 0.3629 7,300AStep 6*. Is.c. (L-N) (X3) 21,900 X 0.2073 4,540A† In addition, UL 1561 listed transformers 25kVAand larger have a 10% impedance tolerance.Short circuit amps can be affected by thistolerance. Therefore, for high end worst case,multiply %Z by 0.9. For low end of worst case,multiply %Z by 1.1. Transformers constructed toANSI standards have a 7.5% impedancetolerance (two-winding construction).* Note 5. The L-N fault current is higher than the L-Lfault current at the secondary terminals of a singlephase center-tapped transformer. The short-circuitcurrent available (I) for this case in Step 4 should beadjusted at the transformer terminals as follows: At L-Ncenter tapped transformer terminals, IL-N 1.5 x IL-L atTransformer Terminals.**Assumes the neutral conductor and the line conductorare the same size. 2014 Eaton