Short-Circuit Current Calculations
Basic Point-to-Point Calculation Procedure Step 1.
Determine the transformer full load amps (F.L.A.) from
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 approximation 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 available, 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.
either the nameplate, the following formulas or Table 1: 100 Multiplier = *% Z transformer Step 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 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 1561 listed transformers 25kVA and larger have a ± 10% impedance tolerance. Short circuit amps 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 letthrough short-circuit current. See Notes 3 and 4.
1 1 +f
M=
Calculate the available short circuit symmetrical RMS current at the point of fault. Add motor contribution, if applicable. I S.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 circuit contribution is to multiply the total motor current in amps by 4. Values of 4 to 6 are commonly accepted.
Calculation of Short-Circuit Currents When Primary Available Short-Circuit Current is Known
Use the following procedure to calculate the level of fault current at the secondary of a second, downstream transformer in a system when the level of fault current at the transformer primary is known. MAIN TRANSFORMER
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 calculated 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 amps by 4. Values of 4 to 6 are commonly accepted. Step 4. Calculate the "f" factor. 3Ø Faults 1Ø Line-to-Line (L-L) Faults See Note 5 & Table 3 1Ø Line-to-Neutral (L-N) Faults See Note 5 & Table 3
f=
Step A.
2 x L x I L-L f= C x n x EL-L 2 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 secondary terminals 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: At L-N center tapped transformer terminals, IL-N = 1.5 x IL-L at Transformer Terminals. ©2014 Eaton
IS.C. primary
1.732 x L x I 3Ø C x n x E L-L
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. E = Voltage of circuit.
IS.C. primary
H.V. UTILITY CONNECTION
Step B.
IS.C. secondary
IS.C. secondary
Calculate the "f" factor (IS.C. primary known) 3Ø Transformer (I S.C. primary and I S.C. secondary are 3Ø fault values)
f=
1Ø Transformer (I S.C. primary and I S.C. secondary are 1Ø fault values: I S.C. secondary is L-L)
f=
I S.C. primary x V primary x 1.73 (%Z) 100,000 x kVA
transformer
I S.C. primary x V primary x (%Z) 100,000 x kVA
transformer
Calculate "M" (multiplier).
1 1 +f Step C. Calculate the short-circuit current at the secondary of the transformer. (See Note under Step 3 of "Basic Point-toPoint Calculation Procedure".) M=
I S.C. secondary =
Vprimary Vsecondary
x M x I S.C. primary
237
Short-Circuit Current Calculations Three-Phase Short Circuits System A One-Line Diagram
Available Utility Infinite Assumption
1500 KVA Transformer 480V, 3Ø, 3.5%Z, 3.45% X, 0.56%R 1
Fault X1
Fault X3
Step 1. If.l. = 1500 X 1000 = 1804.3A 480 X 1.732
Step 4. f = 1.732 X 50 X 55,137 = 0.4484 22,185 X 1 X 480
Step 2. Multipler
100 = 31.746 3.5 X 0.9†
Step 5. M =
1 = 0.6904 1 + 0.4483
I =1804A f.l.
Step 3. Is.c. = 1804.3 X 31.746 = 57,279A Is.c. motor contribution** = 4 X 1804.3 = 7217A Itotal s.c. sym RMS = 57,279 + 7217 = 64,496A
25’ - 500kcml Cu 3 Single Conductors 6 Per Phase Magnetic Conduit
Step 6. Is.c. sym RMS = 55,137 X 0.6904 = 38,067A Is.c. motor contribution** = 4 X 1804.3 = 7217A Itotal s.c. sym RMS (X3) = 38,067 + 7217 = 45,284A
Fault X2
2
Step 4. f = 1.732 X 25 X 57,279 = 0.0388 22,185 X 6 X 480
2000A Switch KRP-C 2000SP Fuse
Step 5. M = 400A Switch LPS-RK-400SP Fuse
1 = 0.9626 1 + 0.0388
Step 6. Is.c. sym RMS = 57,279 X 0.9626 = 55,137A Is.c. motor contribution** = 4 X 1804.3 = 7217A Itotal s.c. sym RMS = 55,137 + 7217 = 62,354A
50’ - 500 kcmil Cu 3 Single Conductors Magnetic Conduit
3
Motor 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 240
System B One-Line Diagram Available Utility Infinite Assumption 1000 KVA Transformer 480V, 3Ø, 3.5%Z,
Fault X1
Fault X3
Step 1. Is.c. = 1000 X 1000 = 1202.8A 480 X 1.732
Step 4. f = 1.732 X 20 X 36,761 = 0.1161 2 X 11,424 X 480
Step 2. Multipler =
If.I.=1203A 1
30’ - 500kcml Cu 3 Single Conductors 4 Per Phase PVC Conduit
100 = 31.746 3.5 X 0.9†
Step 5. M =
= 1 = 0.8960 1 + 0.1161
Step 3. Is.c. = 1202.8 X 31.746 = 38,184A
Step 6. Is.c. sym RMS = 36,761 X 0.8960 = 32,937A
Fault X2
Fault X4
Step 4. f = 1.732 X 30 X 38,184 = 0.0387 26,706 X 4 X 480
Step A. f = 32,937 X 480 X 1.732 X (1.2 X 0.9) = 1.3144 100,000 X 225
2
1600A Switch KRP-C 1500SP Fuse
=
Step 5. M =
400A Switch LPS-RK-350SP Fuse
1 = 0.9627 1 + 0.0387
Step 6. Is.c. sym RMS = 38,184 X 0.9627 = 36,761A 20’ - 2/0 Cu 3 Single Conductors 2 Per Phase PVC Conduit
Step B. M =
1 = 0.4321 1 + 1.3144
Step C. Is.c. sym RMS = 480 X 0.4321 X 32,937 = 32,842A 208
3
225 KVA Transformer 208V, 3Ø 1.2%Z
This example assumes no motor contribution. 4
238
©2014 Eaton
Short-Circuit Current Calculations
Single-Phase Short Circuits Short circuit calculations on a single-phase center tapped transformer system require a slightly different procedure than 3Ø faults on 3Ø systems.
A B C
1. 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 the source to the transformer connection. This is compensated for in the 3Ø short circuit formula by multiplying the single conductor or single-phase impedance by 1.73.
Primary Secondary
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 condition.
Short Circuit
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 multipliers 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 system 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.
Primary Secondary Short Circuit L2
N
L1
The calculations on the following pages illustrate 1Ø fault calculations on a single-phase transformer system. Both line-to-line and line-to-neutral faults are considered.
L1
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
N
50 Feet
©2014 Eaton
Short Circuit L2
239
Short-Circuit Current Calculations Single-Phase Short Circuits System A
One-Line Diagram
Available Utility Infinite Assumption
75KVA, 1Ø Transformer. 1.22%X, 0.68%R 1.40%Z 120/240V
Line-to-Line (L-L) Fault
Line-to-Neutral (L-N) Fault
Fault X1
Fault X1
Step 1. If.l. = 75 X 1000 = 312.5A 240
Step 1. If.l. = 75 X 1000 = 312.5A 240
Step 2. Multipler =
100 = 79.37 1.4 X 0.9†
Step 3. Is.c. (L-L) = 312.5 X 79.37 = 24,802A 25’ - 500kcml Cu Magnetic Conduit 3 Single Conductors
Fault X2
Step 5. M =
100 = 79.37 1.4 X 0.9†
Step 3*. Is.c. (L-N) = 24,802 X 1.5 = 37,202A
Fault X2
Step 4. f = 2 X 25 X 24,802 = 0.2329 22,185 X 1 X 240
2
Step 2. Multipler =
1 = 0.8111 1 + 0.2329
Step 4. f = 2 X 25 X 37,202 = 0.6987 22,185 X 1 X 120 Step 5. M =
1 = 0.5887 1 + 0.6987
Step 6. Is.c. (L-L) (X2) = 24,802 X 0.8111 = 20,116
400A Switch LPN-RK-400SP Fuse
Step 6*. Is.c. (L-N) (X2) = 37,202 X 0.5887 = 21,900A
50’ - 3 AWG Cu Magnetic Conduit 3 Single Conductors
Fault X3
Fault X3
Step 4. f = 2 X 50 X 20,116 = 1.7557 4774 X 1 X 240
Step 4. f = 2 X 50 X 21,900** = 3.8323 4774 X 1 X 120
3
Step 5. M =
240
1 = 0.3629 1 + 1.7557
Step 5. M =
1 = 0.2073 1 + 3.823
Step 6. Is.c. (L-L) (X3) = 20,116 X 0.3629 = 7,300A
Step 6*. Is.c. (L-N) (X3) = 21,900 X 0.2073 = 4,540A
† In addition, UL 1561 listed transformers 25kVA and larger have a ± 10% impedance tolerance. Short circuit amps can be affected by this tolerance. Therefore, for high end worst case, multiply %Z by 0.9. 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).
* Note 5. The L-N fault current is higher than the L-L fault current at the secondary terminals of a singlephase center-tapped transformer. The short-circuit 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. **Assumes the neutral conductor and the line conductor are the same size.
©2014 Eaton
Short-Circuit Current Calculations Impedance & Reactance Data Transformers Table 1. Short-Circuit Currents Available from Various Size Transformers
(Based upon actual field nameplate data or from utility transformer worst case impedance) Voltage and Phase
Full % Short Load Impedance†† Circuit 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. Transformers constructed to ANSI standards have a ±7.5% impedance tolerance (two-winding construction).
† Fluctuations in system voltage will affect the available short-circuit current.
Impedance Data for Single-Phase Transformers Normal Range Impedance Multipliers** Suggested X/R Ratio of Percent For Line-to-Neutral kVA for Impedance (%Z)* Faults Calculation for %X for %R 1Ø 0.75 1.1 1.2–6.0 0.6 25.0 0.6 0.75 37.5 1.4 1.2–6.5 0.75 1.6 1.2–6.4 0.6 50.0 0.75 1.8 1.2–6.6 0.6 75.0 100.0 2.0 1.3–5.7 0.6 0.75 0.75 167.0 2.5 1.4–6.1 1.0 0.75 3.6 1.9–6.8 1.0 250.0 4.7 2.4–6.0 1.0 0.75 333.0 1.0 0.75 5.5 2.2–5.4 500.0 * 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. . Im p ed ance Data for Sing le-Phase and Three-Phase Transform ersSupplement † kVA 1Ø 10 15
%Z 3Ø — 1.2 1.3 — 75 1.11 1.07 150 225 1.12 300 1.11 — 1.9 333 500 1.24 500 — 2.1 †These represent actual transformer installations.
Suggested X/R Ratio for Calculation 1.1 1.1 1.5 1.5 1.5 1.5 4.7 1.5 5.5 nameplate ratings taken from field
Note: UL Listed transformers 25kVA and greater have a ±10% tolerance on their impedance nameplate.
For example, a 10% increase in system voltage will result in a 10% greater available short-circuit currents than as shown in Table 1.
©2014 Eaton
241
Short-Circuit Current Calculations
Conductors & Busways "C" Values Table 4. “C” Values for Conductors Copper AWG Three Single Conductors Conduit or kcmil Steel 15kV 5kV 600V 389 14 617 12 981 10 1551 8 1557 2389 2406 2425 6 3696 3751 3806 4 4577 4674 4774 3 5574 5736 5907 2 1 7293 7029 6759 7973 1/0 8925 8544 9390 10062 2/0 10755 11022 3/0 12844 11804 12543 13606 4/0 15082 13644 14925 250 16483 14769 16293 300 18177 15678 17385 350 19704 16366 18235 400 20566 17492 19172 500 22185 17962 600 22965 20567 18889 21387 750 24137 19923 22539 1,000 25278 Aluminum 237 14 376 12 599 10 950 8 951 1472 1476 1481 6 2319 2333 4 2346 2904 2928 2952 3 3626 3670 2 3713 4498 4575 4645 1 5493 5670 5777 1/0 6733 6968 2/0 7187 8163 8467 8826 3/0 9700 10167 4/0 10741 10849 11460 250 12122 12193 13009 300 13910 14280 13288 350 15484 14188 15355 400 16671 16828 15657 500 18756 16484 18428 600 20093 17686 19685 750 21766 1,000 23478 21235 19006
Nonmagnetic 5kV 600V 389 617 982 1555 1559 2418 2430 3789 3826 4745 4811 5926 6044 7307 7493 9034 9317 10878 11424 13048 13923 16673 15351 17121 18594 20868 18975 20526 22737 21786 24297 23277 26706 25204 28033 26453 29735 28083 31491
2407 3753 4679 5809 7109 8590 10319 12360 14347 15866 17409 18672 19731 21330 22097 23408 24887
Three-Conductor Cable Conduit Steel 5kV 600V 389 617 982 1557 1559 2431 2425 3812 3830 4785 4820 5930 5989 7365 7454 9086 9210 11045 11245 13333 13656 16392 15890 17851 18311 20617 20052 21914 22646 23372 24253 25449 26980 27975 28752 30024 31051 32689 33864
951 1479 2342 2945 3702 4632 5766 7153 8851 10749 12343 14183 15858 17321 19503 21718 23702 26109
1476 2333 2929 3673 4580 5646 6986 8627 10387 11847 13492 14955 16234 18315 19635 21437 23482
237 376 599 952 1482 2351 2963 3734 4686 5852 7327 9077 11185 12797 14917 16795 18462 21395 23633 26432 29865
237 376 599 952 1482 2350 2961 3730 4678 5838 7301 9110 11174 12862 14923 16813 18506 21391 23451 25976 28779
15kV
951 1480 2347 2955 3719 4664 5820 7271 8981 11022 12636 14698 16490 18064 20607 23196 25790 29049
2415 3779 4726 5828 7189 8708 10500 12613 14813 16466 18319 19821 21042 23126 24897 26933 29320
Nonmagnetic 5kV 600V 389 617 982 1558 1560 2433 2428 3838 3823 4803 4833 6023 6087 7507 7579 9373 9473 11529 11703 14119 14410 17020 17483 19352 19779 22525 21938 24126 24904 26044 26916 28712 30096 31258 32154 33315 34605 35749 37197
2421 3798 4762 5958 7364 9053 11053 13462 16013 18001 20163 21982 23518 25916 27766 29735 31959
1478 2339 2941 3693 4618 5717 7109 8751 10642 12115 13973 15541 16921 19314 21349 23750 26608
237 376 599 952 1482 2353 2966 3740 4699 5876 7373 9243 11409 13236 15495 17635 19588 23018 25708 29036 32938
952 1481 2350 2959 3725 4682 5852 7329 9164 11277 13106 15300 17352 19244 22381 25244 28262 31920
1479 2344 2949 3709 4646 5771 7202 8977 10969 12661 14659 16501 18154 20978 23295 25976 29135
15kV
15kV
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 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 Copper Aluminum Copper 225 28700 23000 18700 400 38900 34700 23900 600 41000 38300 36500 800 46100 57500 49300 1000 69400 89300 62900 1200 94300 97100 76900 1350 119000 104200 90100 1600 129900 120500 101000 2000 142900 135100 134200 2500 143800 156300 180500 3000 144900 175400 204100 4000 — — 277800 Note: These values are equal to one over the impedance in a survey of industry.
242
High Impedance Aluminum Copper — 12000 — 21300 — 31300 — 44100 15600 56200 16100 69900 84000 17500 90900 19200 125000 20400 166700 21700 188700 23800 256400 — impedance per foot for ©2014 Eaton
Voltage Drop Calculations
Ratings of Conductors and Tables to Determine Volt Loss With larger loads on new installations, it is extremely important to consider volt loss, otherwise some very unsatisfactory problems are likely to be encountered.
The actual conductor used must also meet the other sizing requirements such a full-load current, ambient temperature, number in a raceway, etc.
How to Figure Volt Loss
Multiply distance (length in feet of one wire) by the current (expressed in amps) by the figure shown in table for the kind of current and the size of wire to be used, by one over the number of conductors per phase. Then, put a decimal point in front of the last 6 digits–you have the volt loss to be Example – 6 AWG copper wire, one per phase, in 180 feet of steel conduit–3 phase, 40 amp load at 80% power factor. Multiply feet by amperes: 180 x 40 = 7200 Multiply this number by number from table for 6 AWG wire threephase at 80% power factor: 7200 x 745† = 5364000 1 1 x 5364000 = 5364000 Multiply by x 5364000 = #/phase 1 Place decimal point 6 places to left: This gives volt loss to be expected: 5.364V (For a 240V circuit the % voltage drop is 5.364 x 100 or 2.23%). 240 Table A and B take into consideration reactance on AC circuits as well as resistance of the wire. Remember on short runs to check to see that the size and type of wire indicated has sufficient ampacity. expected on that circuit.
How to Select Size of Wire
Multiply distance (length in feet of one wire) by the current (expressed in amps), by one over the number of conductors per phase. Divide that figure into the permissible volt loss multiplied by 1,000,000.
Example – Copper in 180 feet of steel conduit–3 phase, 40 amp Ioad at 80% power factor–maximum volt loss permitted from local code equals 5.5 volts. Multiply feet by amperes by 1 1 180 x 40 x = 7200. #/phase 1 Divide permissible volt loss multiplied by 1,000,000 by this 5.5 x 1,000,000 = 764. number: 7200
Look under the column in Table A and B applying to the type of current and power factor for the value nearest, but not above your result – you have the size of wire needed.
† Value from Table A ©2014 Eaton
Select number from Table A, three-phase at 80% power factor, that is nearest but not greater than 764. This number is 745 which indicates the size of wire needed: 6 AWG.
Line-to-Neutral
For line to neutral voltage drop on a 3 phase system, divide the three phase value by 1.73. For line to neutral voltage drop on a single phase system, divide single phase value by 2.
Open Wiring
The volt loss for open wiring installations depends on the separation between conductors. The volt loss is approximately equal to that for conductors in non-magnetic conduit.
Installation in Conduit, Cable or Raceway
NEC® Tables 310.15(B)(16) through 310.15(B)(19) give allowable ampacities (current-carrying capacities) for not more than three current carrying conductors in a conduit, cable, or raceway. Where the number of current carrying conductors exceeds three the allowable ampacity of each conductor must be reduced as shown in the following tables: Installation in Conduit, Cable or Raceway per 310.15(B)(2)(a) The Number of Conductors In One Conduit, Raceway Or Cable 4 to 6 7 to 9 10 to 20 21 to 30 31 to 40 41 and over
Percentage of Values In Tables 310.16 And 310.18 80% 70% 50% 45% 40% 35%
Conditions Causing Higher Volt Loss
The voltage loss is increased when a conductor is operated at a higher temperature because the resistance increases.
Room Temperature Affects Ratings
The ampacities (carrying capacities) of conductors are based on a room temperature of either 30°C or 40ºC. For derating based upon 30ºC ambient, if room temperature is higher, the ampacities are reduced by using the following multipliers; (for 0-2000 volt, insulated conductors not more than 3 conductors in raceway or direct buried, Table 310.15(B)(2)(a)). For room temperatures based upon a 40ºC ambient, see Table 310.15(B)(2)(b). Room Temperature Affects Ratings Room Temperature °C 31-35 36-40 41-45 46-50 51-55 56-60 61-70 71-80
TW °F 87-95 96-104 105-113 114-122 123-131 132-140 141-158 159-176
Ampacity Multiplier THHN, XHHW* THW, THWN (75°C Wire) (60°C Wire) .94 .91 .88 .82 .82 .71 .75 .58 .67 .41 .58 – .33 – – –
(90°C Wire) .96 .91 .87 .82 .76 .71 .58 .41
243
Voltage Drop Calculations
Table A — Copper Conductors — Ratings & Volt Loss† Conduit
Steel Conduit
NonMagnetic Conduit (Lead Covered Cables or Installation in Fibre or Other NonMagnetic Conduit, Etc.)
Wire Size
14 12 10 8 6 4 3 2 1 0 00 000 0000 250 300 350 400 500 600 750 1000 14 12 10 8 6 4 3 2 1 0 00 000 0000 250 300 350 400 500 600 750 1000
Ampacity Type T, TW (60°C Wire)
20* 25* 30 40 55 70 85 95 110 125 145 165 195 215 240 260 280 320 335 400 455 20* 25* 30 40 55 70 85 95 110 125 145 165 195 215 240 260 280 320 335 400 455
Type RH, THWN, RHW, THW (75°C Wire) 20* 25* 35* 50 65 85 100 115 130 150 175 200 230 255 285 310 335 380 420 475 545 20* 25* 35* 50 65 85 100 115 130 150 175 200 230 255 285 310 335 380 420 475 545
Type RHH, THHN, XHHW (90°C Wire) 25* 30* 40* 55 75 95 110 130 150 170 195 225 260 290 320 350 380 430 475 535 615 25* 30* 40* 55 75 95 110 130 150 170 195 225 260 290 320 350 380 430 475 535 615
Direct Current
6140 3860 2420 1528 982 616 490 388 308 244 193 153 122 103 86 73 64 52 43 34 26 6140 3464 2420 1528 982 616 470 388 308 244 193 153 122 103 86 73 64 52 43 34 26
Volt Loss (See explanation prior page.) Three-Phase (60 Cycle, Lagging Power Factor.) 60% 70% 90% 80% 100%
Single-Phase (60 Cycle, Lagging Power Factor.) 70% 90% 80% 100%
60%
5369 3464 2078 1350 848 536 433 346 277 207 173 136 109 93 77 67 60 50 43 36 31 5369 3464 2078 1350 848 536 433 329 259 207 173 133 107 90 76 65 57 46 39 32 25
6200 4000 2400 1560 980 620 500 400 320 240 200 158 126 108 90 78 70 58 50 42 36 6200 4000 2400 1560 980 620 500 380 300 240 200 154 124 104 88 76 66 54 46 38 30
3836 2508 1540 1040 690 468 394 331 283 232 206 178 157 148 135 126 120 111 106 102 95 3812 2486 1520 1019 669 448 375 300 253 214 188 159 140 128 118 109 103 94 90 83 77
4887 3169 1918 1264 812 528 434 354 292 228 196 162 136 123 108 98 91 81 75 68 62 4876 3158 1908 1255 802 519 425 330 268 220 188 151 127 112 99 89 81 71 65 58 51
4371 2841 1728 1148 745 491 407 336 280 223 194 163 140 128 115 106 99 90 84 78 72 4355 2827 1714 1134 731 479 395 310 255 212 183 150 128 114 103 94 87 77 72 65 59
3848 2508 1532 1026 673 450 376 312 264 213 188 160 139 129 117 109 103 94 89 84 78 3830 2491 1516 1010 657 435 361 286 238 199 174 145 125 113 104 95 89 80 76 70 63
3322 2172 1334 900 597 405 341 286 245 200 178 154 136 128 117 109 104 96 92 88 82 3301 2153 1316 882 579 388 324 259 219 185 163 138 121 110 102 94 89 82 77 72 66
5643 3659 2214 1460 937 610 501 409 337 263 227 187 157 142 125 113 105 94 86 79 72 5630 3647 2203 1449 926 599 490 381 310 254 217 175 147 129 114 103 94 82 75 67 59
5047 3281 1995 1326 860 568 470 388 324 258 224 188 162 148 133 122 114 104 97 91 84 5029 3264 1980 1310 845 553 456 358 295 244 211 173 148 132 119 108 100 90 83 76 68
4444 2897 1769 1184 777 519 434 361 305 246 217 184 161 149 135 126 118 109 103 97 90 4422 2877 1751 1166 758 502 417 330 275 230 201 167 145 131 120 110 103 93 87 80 73
* The overcurrent protection for conductor types marked with an (*) shall not exceed 15 amperes for 14 AWG, 20 amperes for 12 AWG, and 30 amperes for 10 AWG copper; or 15 amperes for 12 AWG and 25 amperes for 10 AWG aluminum and copper-clad aluminum after any correction factors for ambient temperature and number of conductors have been applied. † Figures are L-L for both single-phase and three-phase. Three-phase figures are average for the three-phase.
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©2014 Eaton
Voltage Drop Calculations
Table B — Aluminum Conductors — Ratings & Volt Loss† Conduit
Steel Conduit
NonMagnetic Conduit (Lead Covered Cables or Installation in Fibre or Other NonMagnetic Conduit, Etc.)
Wire Size
12 10 8 6 4 3 2 1 0 00 000 0000 250 300 350 400 500 600 750 1000 12 10 8 6 4 3 2 1 0 00 000 0000 250 300 350 400 500 600 750 1000
Ampacity Type T, TW (60°C Wire)
20* 25 30 40 55 65 75 85 100 115 130 150 170 190 210 225 260 285 320 375 20* 25 30 40 55 65 75 85 100 115 130 150 170 190 210 225 260 285 320 375
Type RH, THWN, RHW, THW (75°C Wire) 20* 30* 40 50 65 75 90 100 120 135 155 180 205 230 250 270 310 340 385 445 20* 30* 40 50 65 75 90 100 120 135 155 180 205 230 250 270 310 340 385 445
Type RHH, THHN, XHHW (90°C Wire) 25* 35* 45 60 75 85 100 115 135 150 175 205 230 255 280 305 350 385 435 500 25* 35* 45 60 75 85 100 115 135 150 175 205 230 255 280 305 350 385 435 500
Direct Current
6360 4000 2520 1616 1016 796 638 506 402 318 259 200 169 141 121 106 85 71 56 42 6360 4000 2520 1616 1016 796 638 506 402 318 252 200 169 141 121 106 85 71 56 42
Volt Loss (See explanation two pages prior.) Three-Phase (60 Cycle, Lagging Power Factor.) 100% 90% 80% 70% 60%
Single-Phase (60 Cycle, Lagging Power Factor.) 100% 90% 80% 70%
60%
5542 3464 2251 1402 883 692 554 433 346 277 225 173 148 124 109 95 77 65 53 43 5542 3464 2251 1402 883 692 554 433 346 277 225 173 147 122 105 93 74 62 50 39
6400 4000 2600 1620 1020 800 640 500 400 320 260 200 172 144 126 110 90 76 62 50 6400 4000 2600 1620 1020 800 640 500 400 320 260 200 170 142 122 108 86 72 58 46
3948 2500 1663 1074 708 574 475 391 328 278 239 201 186 168 155 144 130 122 114 103 3926 2480 1643 1053 668 555 456 373 310 260 223 185 167 150 137 128 114 105 95 86
5039 3165 2075 1310 840 668 541 432 353 290 241 194 173 150 135 122 106 95 84 73 5029 3155 2065 1301 831 659 532 424 344 281 234 186 163 141 125 114 96 85 73 63
4504 2836 1868 1188 769 615 502 405 334 277 234 191 173 152 139 127 112 102 92 82 4490 2823 1855 1175 756 603 490 394 322 266 223 181 160 140 125 116 100 90 79 70
3963 2502 1656 1061 692 557 458 373 310 260 221 184 168 150 138 127 113 105 96 87 3946 2486 1640 1045 677 543 443 360 296 247 209 171 153 136 123 114 100 91 82 73
3419 2165 1441 930 613 497 411 338 284 241 207 174 161 145 134 125 113 106 98 89 3400 2147 1423 912 596 480 394 323 268 225 193 160 145 130 118 111 98 91 82 75
5819 3654 2396 1513 970 771 625 499 407 335 279 224 200 174 156 141 122 110 97 85 5807 3643 2385 1502 959 760 615 490 398 325 270 215 188 163 144 132 111 98 85 73
5201 3275 2158 1372 888 710 580 468 386 320 270 221 200 176 160 146 129 118 107 95 5184 3260 2142 1357 873 696 566 455 372 307 258 209 185 162 145 134 115 104 92 81
4577 2889 1912 1225 799 644 529 431 358 301 256 212 194 173 159 146 131 121 111 100 4557 2871 1894 1206 782 627 512 415 342 285 241 198 177 157 142 132 115 106 94 85
* The overcurrent protection for conductor types marked with an (*) shall not exceed 15 amperes for 14 AWG, 20 amperes for 12 AWG, and 30 amperes for 10 AWG copper; or 15 amperes for 12 AWG and 25 amperes for 10 AWG aluminum and copper-clad aluminum after any correction factors for ambient temperature and number of conductors have been applied. † Figures are L-L for both single-phase and three-phase. Three-phase figures are average for the three-phase.
©2014 Eaton
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Glossary Common Electrical Terminology Ohm
The unit of measure for electric resistance. An ohm is the amount of resistance that will allow one amp to flow under a pressure of one volt.
Semiconductor Fuses
Fuses used to protect solid-state devices. See “High Speed Fuses.”
Short-Circuit
The relationship between voltage, current, and resistance, expressed by the equation E = IR, where E is the voltage in volts, I is the current in amps, and R is the resistance in ohms.
Can be classified as an overcurrent which exceeds the normal full load current of a circuit by a factor many times (tens, hundreds or thousands greater). Also characteristic of this type of overcurrent is that it leaves the normal current carrying path of the circuit – it takes a “short cut” around the load and back to the source.
One Time Fuses
Short-Circuit Current Rating
Overcurrent
Single-Phasing
Ohm’s Law
Generic term used to describe a Class H nonrenewable cartridge fuse, with a single element. A condition which exists on an electrical circuit when the normal load current is exceeded. Overcurrents take on two separate characteristics – overloads and shortcircuits.
Overload
Can be classified as an overcurrent which exceeds the normal full load current of a circuit. Also characteristic of this type of overcurrent is that it does not leave the normal current carrying path of the circuit – that is, it flows from the source, through the conductors, through the load, back through the conductors, to the source again.
Peak Let-Through Current, lp
The instantaneous value of peak current let-through by a current-limiting fuse, when it operates in its current-limiting range.
Renewable Fuse (600V & below)
A fuse in which the element, typically a zinc link, may be replaced after the fuse has opened, and then reused. Renewable fuses are made to Class H standards.
Resistive Load
An electrical load which is characteristic of not having any significant inrush current. When a resistive load is energized, the current rises instantly to its steady-state value, without first rising to a higher value.
RMS Current
The RMS (root-mean-square) value of any periodic current is equal to the value of the direct current which, flowing through a resistance, produces the same heating effect in the resistance as the periodic current does.
The maximum short-circuit current an electrical component can sustain without the occurrence of excessive damage when protected with an overcurrent protective device. That condition which occurs when one phase of a three-phase system opens, either in a low voltage (secondary) or high voltage (primary) distribution system. Primary or secondary single-phasing can be caused by any number of events. This condition results in unbalanced currents in polyphase motors and unless protective measures are taken, causes overheating and failure.
Threshold Current
The symmetrical RMS available current at the threshold of the current-limiting range, where the fuse becomes current-limiting when tested to the industry standard. This value can be read off of a peak let-through chart where the fuse curve intersects the A - B line. A threshold ratio is the relationship of the threshold current to the fuse’s continuous current rating.
Time-Delay Fuse
A fuse with a built-in delay that allows temporary and harmless inrush currents to pass without opening, but is so designed to open on sustained overloads and short-circuits.
Voltage Rating
The maximum open circuit voltage in which a fuse can be used, yet safely interrupt an overcurrent. Exceeding the voltage rating of a fuse impairs its ability to clear an overload or short-circuit safely.
Withstand Rating
The maximum current that an unprotected electrical component can sustain for a specified period of time without the occurrence of extensive damage.
Electrical Formulas Single-Phase
To Find
Amperes when kVA is known Amperes when horsepower is known Amperes when kilowatts are known Kilowatts Kilovolt-Amperes Horsepower Watts
I = Amperes HP = Horsepower
©2014 Eaton
Two-Phase
Three-Phase
kVA ≈ 1000 E HP ≈ 746 E ≈ % eff. ≈ pf kW ≈ 1000 E ≈ pf I ≈ E ≈ pf 1000 I≈ E 1000 I ≈ E % eff. ≈ pf 746 E ≈ I ≈ pf
kVA ≈ 1000 kVA ≈ 1000 E≈ 2 E ≈ 1.73 HP ≈ 746 HP ≈ 746 E ≈ 2 ≈ % eff. ≈ pf E ≈ 1.73 ≈ % eff. ≈ pf kW ≈ 1000 kW ≈ 1000 E ≈ 2 pf E ≈ 1.73 ≈ pf I ≈ E ≈ 2 ≈ pf I ≈ E ≈ 1.73 ≈ pf 1000 1000 I≈ E≈ 2 I ≈ E ≈ 1.73 1000 1000 I ≈ E ≈ 2 ≈ % eff. ≈ pf I ≈ E ≈ 1.73 ≈ % eff. ≈ pf 746 746 I ≈ E ≈ 2 ≈ pf I ≈ E ≈ 1.73 ≈ pf Energy Efficiency = Load Horsepower ≈ 746 Load Input kVA ≈ 1000 Power Consumed = W or kW = cosθ Power Factor = pf = kVA VA Apparent Power
E = Volts % eff. = Percent Efficiency
kW = Kilowatts pf = Power Factor
Direct Current
Not Applicable HP ≈ 746 E ≈ % eff. kW ≈ 1000 E I≈ E 1000 Not Applicable I ≈ E ≈ % eff. 746 E≈ I
kVA = Kilovolt-Amperes
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