Short circuit current calculations
- 1. 3Ø Short-Circuit Calculations 43 Copyright 2002 by Cooper Bussmann® Why Short-Circuit Calculations Several sections of the National Electrical Code® relate to proper overcurrent protection. Safe and reliable application of overcurrent protective devices based on these sections mandate that a short-circuit study and a selective coordination study be conducted. These sections include, among others: 110.9 Interrupting Rating 110.10 Component Protection 240.1 Conductor Protection 250.95 Equipment Grounding Conductor Protection 517.17 Health Care Facilities - Selective Coordination 620.62 Selective Coordination for Elevator Circuits Compliance with these code sections can best be accom-plished by conducting a short-circuit study as a start to the analysis. The protection for an electrical system should not only be safe under all service conditions but, to insure continuity of service, it should be selectively coordinated as well. A coordinated system is one where only the faulted circuit is isolated without disturbing any other part of the system. Once the short-circuit levels are determined, the engi-neer can specify proper interrupting rating requirements, selectively coordinate the system and provide component protection. See the various sections of this book for further information on each topic. Low voltage fuses have their interrupting rating expressed in terms of the symmetrical component of short-circuit current, IS. They are given an RMS symmetrical interrupting rating at a specific power factor. This means that the fuse can interrupt the asymmetrical cur-rent associated with this rating. Thus only the symmetrical compo-nent of short-circuit current need be considered to determine the necessary interrupting rating of a low voltage fuse. For listed low voltage fuses, interrupting rating equals its interrupting capacity. Low voltage molded case circuit breakers also have their inter-rupting rating expressed in terms of RMS symmetrical amperes at a specific power factor. However, it is necessary to determine a molded case circuit breaker’s interrupting capacity in order to safely apply it. See the section Interrupting Rating vs. Interrupting Capacity in this book. 110.16 now requires arc-flash hazard warning labeling on certain equipment. A flash hazard analysis is required before a worker approaches electrical parts that have not been put into a safe work condition. To determine the incident energy and flash protection boundary for a flash hazard analysis the short-circuit current is typically the first step. General Comments on Short-Circuit Calculations Sources of short-circuit current that are normally taken under con-sideration include: - Utility Generation - Local Generation - Synchronous Motors - Induction Motors - Alternate Power Sources Short-circuit calculations should be done at all critical points in the system. These would include: - Service Entrance - Transfer Switches - Panel Boards - Load Centers - Motor Control Centers - Disconnects - Motor Starters - Motor Starters Normally, short-circuit studies involve calculating a bolted 3- phase fault condition. This can be characterized as all 3-phases “bolt-ed” together to create a zero impedance connection. This establishes a “worst case” (highest current) condition that results in maximum three phase thermal and mechanical stress in the system. From this calculation, other types of fault conditions can be approximated. This “worst case” condition should be used for interrupting rating, compo-nent protection and selective coordination. However, in doing an arc-flash hazard analysis it is recommended to do the arc flash hazard analysis at the highest bolted 3 phase short-circuit condition and at the “minimum” bolted three-phase short-circuit condition. There are several variables in a distribution system that affect calculated bolted 3-phase short-circuit currents. It is important to select the variable val-ues applicable for the specific application analysis. In the Point-to- Point method presented in this section there are several adjustment factors given in Notes and footnotes that can be applied that will affect the outcomes. The variables are utility source short-circuit capabilities, motor contribution, transformer percent impedance tolerance, and voltage variance. In most situations, the utility source(s) or on-site energy sources, such as on-site generation, are the major short-circuit cur-rent contributors. In the Point-to-Point method presented in the next few pages, the steps and example assume an infinite available short-circuit current from the utility source. Generally this is a good assumption for highest worst case conditions and since the property owner has no control over the utility system and future utility changes. And in many cases a large increase in the utility available does not increase the short-circuit currents a great deal for a build-ing system on the secondary of the service transformer. However, there are cases where the actual utility medium voltage available provides a more accurate short-circuit assessment (minimum bolted short-circuit current conditions) that may be desired to assess the arc flash hazard. When there are motors in the system, motor short-circuit contribu-tion is also a very important factor that must be included in any short-cir-cuit current analysis. When a short-circuit occurs, motor contribution adds to the magnitude of the short-circuit current; running motors con-tribute 4 to 6 times their normal full load current. In addition, series rated combinations can not be used in specific situations due to motor short-circuit contributions (see the section on Series Ratings in this book). For capacitor discharge currents, which are of short time duration, certain IEEE (Institute of Electrical and Electronic Engineers) publications detail how to calculate these currents if they are substantial. Procedures and Methods To determine the fault current at any point in the system, first draw a one-line diagram showing all of the sources of short-circuit cur-rent feeding into the fault, as well as the impedances of the circuit components. To begin the study, the system components, including those of the utility system, are represented as impedances in the diagram. The impedance tables include three-phase and single-phase transformers, cable, and busway. These tables can be used if information from the manufacturers is not readily available. It must be understood that short-circuit calculations are per-formed without current-limiting devices in the system. Calculations are done as though these devices are replaced with copper bars, to deter-mine the maximum “available” short-circuit current. This is necessary to project how the system and the current-limiting devices will perform. Also, multiple current-limiting devices do not operate in series to produce a “compounding” current-limiting effect. The down-stream, or load side, fuse will operate alone under a short-circuit condition if properly coordinated. The application of the point-to-point method permits the determination of available short-circuit currents with a reasonable degree of accuracy at various points for either 3Ø or 1Ø electrical distribution systems. This method can assume unlimited primary short-circuit current (infinite bus) or it can be used with limited pri-mary available current. Table Of Contents Index Glossary
- 2. 3Ø Short-Circuit Calculations Basic Point-to-Point Calculation Procedure Step 1. Determine the transformer full load amperes (F.L.A.) from either the nameplate, the following formulas or Table 1: 3Ø Transformer IF.L.A.= KVA x 1000 EL-L x 1.732 1Ø Transformer IF.L.A.= KVA x 1000 EL-L Step 2. Find the transformer multiplier. See Notes 1 and 2 Multiplier = 100 *%Ztransformer Note 1. Get %Z from nameplate or Table 1. Transformer impedance (Z) helps to determine what the short circuit current will be at the transformer secondary. Transformer impedance is determined as follows: The transformer secondary is short circuited. Voltage is increased on the primary until full load current flows in the secondary. This applied voltage divided by the rated primary voltage (times 100) is the impedance of the transformer. Example: For a 480 Volt rated primary, if 9.6 volts causes secondary full load current to flow through the shorted secondary, the transformer impedance is 9.6/480 = .02 = 2%Z. Note 2. In addition, UL (Std. 1561) listed transformers 25KVA and larger have a ± 10% impedance tolerance. Short circuit amperes can be affected by this tolerance. Therefore, for high 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 let-through short-circuit current. See Notes 3 and 4. IS.C. = TransformerF.L.A. x Multiplier Note 3. Utility voltages may vary ±10% for power and ±5.8% for 120 Volt lighting services. Therefore, for highest short-circuit conditions, multiply values as calcu-lated in step 3 by 1.1 or 1.058 respectively. To find the lower end worst case, multiply results in step 3 by .9 or .942 respectively. Note 4. Motor short-circuit contribution, if significant, may be added at all fault locations throughout the system. A practical estimate of motor short-circuit contribution is to multiply the total motor current in amperes by 4. Values of 4 to 6 are commonly accepted. Step 4. Calculate the "f" factor. 3Ø Faults f = 1.732 x L x I3Ø C x n x EL-L 1Ø Line-to-Line (L-L) Faults 2 x L x I See Note 5 & Table 3 f = L-L C x n x EL-L 1Ø Line-to-Neutral (L-N) Faults 2 x L x IL-N† See Note 5 & Table 3 f = C x n x EL-N Where: L = length (feet) of conductor to the fault. C = constant from Table 4 of “C” values for conductors and Table 5 of “C” values for busway. n = Number of conductors per phase (adjusts C value for parallel runs) I = available short-circuit current in amperes at beginning of circuit. Note 5. The L-N fault current is higher than the L-L fault current at the sec-ondary terminals of a single-phase center-tapped transformer. The short-cir-cuit current available (I) for this case in Step 4 should be adjusted at the transformer terminals as follows: At L-N center tapped transformer terminals, IL-N = 1.5 x IL-L at Transformer Terminals. At some distance from the terminals, depending upon wire size, the L-N fault current is lower than the L-L fault current. The 1.5 multiplier is an approxima-tion and will theoretically vary from 1.33 to 1.67. These figures are based on change in turns ratio between primary and secondary, infinite source avail-able, zero feet from terminals of transformer, and 1.2 x %X and 1.5 x %R for L-N vs. L-L resistance and reactance values. Begin L-N calculations at transformer secondary terminals, then proceed point-to-point. Step 5. Calculate "M" (multiplier) or take from Table 2. M = 1 1 + f Step 6. Calculate the available short-circuit symmetrical RMS cur-rent at the point of fault. Add motor contribution, if applicable. IS.C. sym RMS = IS.C. x M Step 6A. Motor short-circuit contribution, if significant, may be added at all fault locations throughout the system. A practical estimate of motor short-cir-cuit contribution is to multiply the total motor current in amperes by 4. Values of 4 to 6 are commonly accepted. Calculation of Short-Circuit Currents at Second Transformer in System Use the following procedure to calculate the level of fault cur-rent at the secondary of a second, downstream transformer in a system when the level of fault current at the transformer primary is known. Procedure for Second Transformer in System Step A. Calculate the "f" factor (IS.C. primary known) 3Ø Transformer (IS.C. primary and f = IS.C. primary x Vprimary x 1.73 (%Z) IS.C. secondary are 100,000 x KVA transformer 3Ø fault values) 1Ø Transformer (IS.C. primary and IS.C. secondary are f = IS.C. primary x Vprimary x (%Z) 1Ø fault values: 100,000 x KVAtransformer IS.C. secondary is L-L) Step B. Calculate "M" (multiplier). M = 1 1 + f Step C. Calculate the short-circuit current at the secondary of the transformer. (See Note under Step 3 of "Basic Point-to-Point Calculation Procedure".) IS.C. secondary = Vprimary x M x IS.C. primary Vsecondary * * MAIN TRANSFORMER H.V. UTILITY CONNECTION IS.C. primary IS.C. secondary IS.C. primary IS.C. secondary † Table Of Contents 44 Copyright 2002 by Cooper Bussmann® Index Glossary
- 3. Fault X2 Step 4. f = 1.732 x 20 x 33,215 = .1049 2 x 11,424 x 480 Step 5. M = 1 = .905 1 + .1049 Step 6. IS.C.sym RMS = 33,215 x .905 = 30,059A Fault X3 Step A. f = 30,059 x 480 x 1.732 x 1.2 = 1.333 100,000 x 225 Step B. M = 1 = .4286 1 + 1.333 Step C. IS.C. sym RMS = 480 x .4286 x 30,059 = 29,731A 208 3Ø Short-Circuit Calculations 45 Copyright 2002 by Cooper Bussmann® Available Utility Infinite Assumption 1500 KVA Transformer, 480V, 3Ø, 3.5%Z, 3.45%X, .56%R If.l. =1804A 25' - 500kcmil 6 Per Phase Service Entrance Conductors in Steel Conduit 2000A Switch KRP-C-2000SP Fuse Fault X1 400A Switch LPS-RK-400SP Fuse 50' - 500 kcmil Feeder Cable in Steel Conduit Fault X2 Motor Contribution M Fault X1 Step 1. If.l. = 1500 x 1000 = 1804A 480 x 1.732 Step 2. Multiplier = 100 = 28.57 3.5 Step 3. IS.C.=1804 x 28.57 = 51,540A IS.C. motor contrib = 4 x 1,804* = 7,216A Itotal S.C. sym RMS = 51,504 + 7,216 = 58,720A Step 4. f = 1.732 x 25 x 51,540 = 0.0349 22,185 x 6 x 480 Step 5. M = 1 = .9663 1 + .0349 Step 6. IS.C.sym RMS = 51,540 x .9663 = 49,803A IS.C.motor contrib = 4 x 1,804* = 7,216A ItotalS.C. sym RMS = 49,803 + 7,216 = 57,019A (fault X1) System A One-Line Diagram System B Fault X1 Step 1. If.l. = 1000 x 1000 = 1203A 480 x 1.732 Step 2. Multiplier = 100 = 28.57 3.5 Step 3. IS.C. = 1203 x 28.57 = 34,370A Step 4. f = 1.732 x 30 x 34,370 = .0348 26,706 x 4 x 480 Step 5. M = 1 = .9664 1 + .0348 Step 6. IS.C.sym RMS = 34,370 x .9664 = 33,215A Available Utility Infinite Assumption 1000 KVA Transformer, 480V, 3Ø, 3.5%Z If.l.= 1203A 30’ - 500 kcmil 4 Per Phase Copper in PVC Conduit 1600A Switch KRP-C-1500SP Fuse Fault X1 400A Switch LPS-RK-350SP Fuse 20’ - 2/0 2 Per Phase Copper in PVC Conduit Fault X2 225 KVA transformer, 208V, 3Ø 1.2%Z Fault X3 One-Line Diagram 1 1 2 3 2 Fault X2 Step 4. Use IS.C.sym RMS @ Fault X1 to calculate “f” f = 1.732 x 50 x 49,803 = .4050 22,185 x 480 Step 5. M = 1 = .7117 1 + .4050 Step 6. IS.C.sym RMS = 49,803 x .7117 = 35,445A Isym motor contrib = 4 x 1,804* = 7,216A Itotal S.C. sym RMS (fault X2) = 35,445 + 7,216 = 42,661A *Assumes 100% motor load. If 50% of this load was from motors, IS.C. motor contrib. = 4 x 1,804 x .5 = 3608A Table Of Contents Index Glossary
- 4. 1Ø Short-Circuit Calculations L1 46 Copyright 2002 by Cooper Bussmann® Short-circuit calculations on a single-phase center tapped trans-former system require a slightly different procedure than 3Ø faults on 3Ø systems. 1. It is necessary that the proper impedance be used to represent the primary system. For 3Ø fault calculations, a single primary con-ductor impedance is only considered from the source to the trans-former connection. This is compensated for in the 3Ø short-circuit formula by multiplying the single conductor or single-phase impedance by 1.73. However, for single-phase faults, a primary conductor impedance is considered from the source to the transformer and back to the source. This is compensated in the calculations by multiplying the 3Ø primary source impedance by two. 2. The impedance of the center-tapped transformer must be adjusted for the half-winding (generally line-to-neutral) fault condi-tion. The diagram at the right illustrates that during line-to-neutral faults, the full primary winding is involved but, only the half-winding on the secondary is involved. Therefore, the actual transformer reactance and resistance of the half-winding condition is different than the actual transformer reactance and resistance of the full winding condition. Thus, adjustment to the %X and %R must be made when considering line-to-neutral faults. The adjustment mul-tipliers generally used for this condition 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 Phase Transformers” Table may be used, however, calculations must be adjusted to indicate transformer KVA/2. 3. The impedance of the cable and two-pole switches on the sys-tem must be considered “both-ways” since the current flows to the fault and then returns to the source. For instance, if a line-to-line fault occurs 50 feet from a transformer, then 100 feet of cable impedance must be included in the calculation. The calculations on the following pages illustrate 1Ø fault cal-culations on a single-phase transformer system. Both line-to-line and line-to-neutral faults are considered. Note in these examples: a. The multiplier of 2 for some electrical components to account for the single-phase fault current flow, b. The half-winding transformer %X and %R multipliers for the line-to- neutral fault situation, and PRIMARY SECONDARY SHORT CIRCUIT PRIMARY SECONDARY SHORT CIRCUIT L2 N L1 SHORT CIRCUIT N L2 A B C 50 feet Table Of Contents Index Glossary
- 5. 1Ø Short-Circuit Calculations 47 Copyright 2002 by Cooper Bussmann® Line-to-Line Fault @ 240V — Fault X1 Available Utility Infinite Assumption 75KVA, 1Ø Transformer, 1.22%X, .68%R 1.40%Z 120/240V Negligible Distance 400A Switch LPN-RK-400SP Fuse 25' - 500kcmil Magnetic Conduit One-Line Diagram 1 Line-to-Neutral Fault @ 120V — Fault X1 Fault X1 Step 1. If.l. = 75 x 1000 = 240 312.5A Step 2. Multiplier = 100 = 71.43 1.40 Step 3. IS.C. = 312.5 x 71.43 = 22,322A Step 4. f = 2 x 25 x 22,322 = .2096 22,185 x 240 Step 5. M = 1 = .8267 1 + .2096 Step 6. IS.C. L-L (X1) = 22,322 x .8267 = 18,453A * Fault X1 Step 1. If.l. = 75 x 1000 = 312.5A 240 Step 2. Multiplier = 100 = 71.43 1.40 Step 3. IS.C. (L-L) = 312.5 x 71.43 = 22,322A IS.C. (L-N) = 22,322 x 1.5 = 33,483A Step 4. f = 2*x 25 x 22,322 x 1.5 = .6288 22,185 x 120 Step 5. M = 1 = .6139 1 + .6288 Step 6. IS.C. L-N (X1) = 33,483 x .6139 = 20,555A Assumes the neutral conductor and the line conductor are the same size. Available Utility Infinite Assumption 75KVA, 1Ø Transformer, 1.22% X, .68%R, 1.40%Z 120/240V Negligible Distance 400A Switch LPN-RK-400SP Fuse 25' - 500kcmil Magnetic Conduit One-Line Diagram 1 Table Of Contents Index Glossary
- 6. Short-Circuit, Impedance and Reactance Data Impedance Data for Single-Phase Transformers Suggested Normal Range Impedance Multipliers** X/R Ratio of Percent For Line-to-Neutral kVA for Impedance (%Z)* Faults 1Ø Calculation for %X for %R 25.0 1.1 1.2–6.0 0.6 0.75 37.5 1.4 1.2–6.5 0.6 0.75 50.0 1.6 1.2–6.4 0.6 0.75 75.0 1.8 1.2–6.6 0.6 0.75 100.0 2.0 1.3–5.7 0.6 0.75 167.0 2.5 1.4–6.1 1.0 0.75 250.0 3.6 1.9–6.8 1.0 0.75 333.0 4.7 2.4–6.0 1.0 0.75 500.0 5.5 2.2–5.4 1.0 0.75 National standards do not specify %Z for single-phase transformers. Consult manufacturer for values to use in calculation. Based on rated current of the winding (one–half nameplate KVA divided by secondary line-to-neutral voltage). Note: UL Listed transformers 25 KVA and greater have a ± 10% tolerance on their impedance nameplate. This table has been reprinted from IEEE Std 242-1986 (R1991), IEEE Recommended Practice for Protection and Coordination of Industrial and Commercial Power Systems, Copyright© 1986 by the Institute of Electrical and Electronics Engineers, Inc. with the permission of the IEEE Standards Department. .Impedance Data for Single-Phase and Three-Phase Transformers- Supplement† kVA Suggested 1Ø 3Ø %Z X/R Ratio for Calculation 10 — 1.2 1.1 15 — 1.3 1.1 75 1.11 1.5 150 1.07 1.5 225 1.12 1.5 300 1.11 1.5 333 — 1.9 4.7 500 1.24 1.5 500 — 2.1 5.5 †These represent actual transformer nameplate ratings taken from field installations. Note: UL Listed transformers 25KVA and greater have a ±10% tol-erance on their impedance nameplate. Table 3. Various Types of Short Circuit Currents as a Percent of Three Phase Bolted Faults (Typical). Three Phase Bolted Fault 100% Line-to-Line Bolted Fault 87% Line-to-Ground Bolted Fault 25-125%* (Use 100% near trans-former, 50% otherwise) Line-to-Neutral Bolted Fault 25-125% (Use 100% near tans-former, 50% otherwise) Three Phase Arcing Fault 89% (maximum) Line-to-Line Arcing Fault 74% (maximum) Line-to-Ground Arcing Fault (minimum) 38% (minimum) *Typically much lower but can actually exceed the Three Phase Bolted Fault if it is near the transformer terminals. Will normally be between 25% to 125% of three phase bolted fault value. * 48 Copyright 2002 by Cooper Bussmann® Table 1. Short-Circuit Currents Available from Various Size Transformers (Based Upon actual field nameplate data or from utility transformer worst case impedance) Voltage Full % Short and Load Impedance†† Circuit Phase KVA Amps (Nameplate) Amps† 25 104 1.5 12175 37.5 156 1.5 18018 120/240 50 208 1.5 23706 1 ph.* 75 313 1.5 34639 100 417 1.6 42472 167 696 1.6 66644 45 125 1.0 13879 75 208 1.0 23132 112.5 312 1.11 31259 150 416 1.07 43237 120/208 225 625 1.12 61960 3 ph.** 300 833 1.11 83357 500 1388 1.24 124364 750 2082 3.50 66091 1000 2776 3.50 88121 1500 4164 3.50 132181 2000 5552 4.00 154211 2500 6940 4.00 192764 75 90 1.00 10035 112.5 135 1.00 15053 150 181 1.20 16726 225 271 1.20 25088 300 361 1.20 33451 277/480 500 602 1.30 51463 3 ph.** 750 903 3.50 28672 1000 1204 3.50 38230 1500 1806 3.50 57345 2000 2408 4.00 66902 2500 3011 4.00 83628 Single-phase values are L-N values at transformer terminals. These figures are based on change in turns ratio between primary and secondary, 100,000 KVA primary, zero feet from terminals of transformer, 1.2 (%X) and 1.5 (%R) multipliers for L-N vs. L-L reactance and resistance values and transformer X/R ratio = 3. Three-phase short-circuit currents based on “infinite” primary. UL listed transformers 25 KVA or greater have a ±10% impedance toler-ance. Short-circuit amps shown in Table 1 reflect –10% condition. Trans-formers constructed to ANSI standards have a ±7.5% impedance tolerance (two-winding construction). Fluctuations in system voltage will affect the available short-circuit current. For example, a 10% increase in system voltage will result in a 10% greater available short-circuit currents than as shown in Table 1. * ** †† † TRANSFORMERS ** f M 0.01 0.99 0.02 0.98 0.03 0.97 0.04 0.96 0.05 0.95 0.06 0.94 0.07 0.93 0.08 0.93 0.09 0.92 0.10 0.91 0.15 0.87 0.20 0.83 0.25 0.80 0.30 0.77 0.35 0.74 0.40 0.71 f M 0.50 0.67 0.60 0.63 0.70 0.59 0.80 0.55 0.90 0.53 1.00 0.50 1.20 0.45 1.50 0.40 1.75 0.36 2.00 0.33 2.50 0.29 3.00 0.25 3.50 0.22 4.00 0.20 5.00 0.17 6.00 0.14 f M 7.00 0.13 8.00 0.11 9.00 0.10 10.00 0.09 15.00 0.06 20.00 0.05 30.00 0.03 40.00 0.02 50.00 0.02 60.00 0.02 70.00 0.01 80.00 0.01 90.00 0.01 100.00 0.01 Table 2. “M” (Multiplier) 1 1 + f M = Table Of Contents Index Glossary
- 7. “C” Values for Conductors and Busway Table 4. “C” Values for Conductors Copper AWG Three Single Conductors Three-Conductor Cable or Conduit Conduit kcmil Steel Nonmagnetic Steel Nonmagnetic 600V 5kV 15kV 600V 5kV 15kV 600V 5kV 15kV 600V 5kV 15kV 14 389 - - 389 - - 389 - - 389 - - 12 617 - - 617 - - 617 - - 617 - - 10 981 - - 982 - - 982 - - 982 - - 8 1557 1551 - 1559 1555 - 1559 1557 - 1560 1558 - 6 2425 2406 2389 2430 2418 2407 2431 2425 2415 2433 2428 2421 4 3806 3751 3696 3826 3789 3753 3830 3812 3779 3838 3823 3798 3 4774 4674 4577 4811 4745 4679 4820 4785 4726 4833 4803 4762 2 5907 5736 5574 6044 5926 5809 5989 5930 5828 6087 6023 5958 1 7293 7029 6759 7493 7307 7109 7454 7365 7189 7579 7507 7364 1/0 8925 8544 7973 9317 9034 8590 9210 9086 8708 9473 9373 9053 2/0 10755 10062 9390 11424 10878 10319 11245 11045 10500 11703 11529 11053 3/0 12844 11804 11022 13923 13048 12360 13656 13333 12613 14410 14119 13462 4/0 15082 13606 12543 16673 15351 14347 16392 15890 14813 17483 17020 16013 250 16483 14925 13644 18594 17121 15866 18311 17851 16466 19779 19352 18001 300 18177 16293 14769 20868 18975 17409 20617 20052 18319 22525 21938 20163 350 19704 17385 15678 22737 20526 18672 22646 21914 19821 24904 24126 21982 400 20566 18235 16366 24297 21786 19731 24253 23372 21042 26916 26044 23518 500 22185 19172 17492 26706 23277 21330 26980 25449 23126 30096 28712 25916 600 22965 20567 17962 28033 25204 22097 28752 27975 24897 32154 31258 27766 750 24137 21387 18889 29735 26453 23408 31051 30024 26933 34605 33315 29735 1,000 25278 22539 19923 31491 28083 24887 33864 32689 29320 37197 35749 31959 Aluminum 14 237 - - 237 - - 237 - - 237 - - 12 376 - - 376 - - 376 - - 376 - - 10 599 - - 599 - - 599 - - 599 - - 8 951 950 - 952 951 - 952 951 - 952 952 - 6 1481 1476 1472 1482 1479 1476 1482 1480 1478 1482 1481 1479 4 2346 2333 2319 2350 2342 2333 2351 2347 2339 2353 2350 2344 3 2952 2928 2904 2961 2945 2929 2963 2955 2941 2966 2959 2949 2 3713 3670 3626 3730 3702 3673 3734 3719 3693 3740 3725 3709 1 4645 4575 4498 4678 4632 4580 4686 4664 4618 4699 4682 4646 1/0 5777 5670 5493 5838 5766 5646 5852 5820 5717 5876 5852 5771 2/0 7187 6968 6733 7301 7153 6986 7327 7271 7109 7373 7329 7202 3/0 8826 8467 8163 9110 8851 8627 9077 8981 8751 9243 9164 8977 4/0 10741 10167 9700 11174 10749 10387 11185 11022 10642 11409 11277 10969 250 12122 11460 10849 12862 12343 11847 12797 12636 12115 13236 13106 12661 300 13910 13009 12193 14923 14183 13492 14917 14698 13973 15495 15300 14659 350 15484 14280 13288 16813 15858 14955 16795 16490 15541 17635 17352 16501 400 16671 15355 14188 18506 17321 16234 18462 18064 16921 19588 19244 18154 500 18756 16828 15657 21391 19503 18315 21395 20607 19314 23018 22381 20978 600 20093 18428 16484 23451 21718 19635 23633 23196 21349 25708 25244 23295 750 21766 19685 17686 25976 23702 21437 26432 25790 23750 29036 28262 25976 1,000 23478 21235 19006 28779 26109 23482 29865 29049 26608 32938 31920 29135 Note: These values are equal to one over the impedance per foot and based upon resistance and reactance values found in IEEE Std 241-1990 (Gray Book), IEEE Recommended Practice for Electric Power Systems in Commerical Buildings & IEEE Std 242-1986 (Buff Book), IEEE Recommended Practice for Protection and Coordination of Industrial and Commercial Power Systems. Where resistance and reac-tance values differ or are not available, the Buff Book values have been used. The values for reactance in determining the C Value at 5 KV & 15 KV are from the Gray Book only (Values for 14-10 AWG at 5 kV 49 Copyright 2002 by Cooper Bussmann® and 14-8 AWG at 15 kV are not available and values for 3 AWG have been approximated). Table 5. “C” Values for Busway Ampacity Busway Plug-In Feeder High Impedance Copper Aluminum Copper Aluminum Copper 225 28700 23000 18700 12000 — 400 38900 34700 23900 21300 — 600 41000 38300 36500 31300 — 800 46100 57500 49300 44100 — 1000 69400 89300 62900 56200 15600 1200 94300 97100 76900 69900 16100 1350 119000 104200 90100 84000 17500 1600 129900 120500 101000 90900 19200 2000 142900 135100 134200 125000 20400 2500 143800 156300 180500 166700 21700 3000 144900 175400 204100 188700 23800 4000 — — 277800 256400 — Note: These values are equal to one over the impedance per foot for impedance in a survey of industry. Table Of Contents Index Glossary