Short-Circuit Current Calculations: Basic Point-to-Point Calculation Procedure
Short-Circuit Current Calculations: Basic Point-to-Point Calculation Procedure
Short-Circuit Current Calculations: Basic Point-to-Point Calculation Procedure
Basic Point-to-Point Calculation Procedure At some distance from the terminals, depending upon wire size, the L-N fault
Step 1. Determine the transformer full load amps (F.L.A.) from 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.
Transformer impedance is determined as follows: The transformer secondary is short Step 6A. Motor short circuit contribution, if significant, may be
circuited. Voltage is increased on the primary until full load current flows in the added at all fault locations throughout the system. A
secondary. This applied voltage divided by the rated primary voltage (times 100) is the practical estimate of motor short circuit contribution is to
impedance of the transformer. multiply the total motor current in amps by 4. Values of 4
to 6 are commonly accepted.
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.
Calculation of Short-Circuit Currents When Primary
* Note 2. In addition, UL 1561 listed transformers 25kVA and larger have a ± 10%
Available Short-Circuit Current is Known
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. Use the following procedure to calculate the level of fault current at the secondary
Transformers constructed to ANSI standards have a ±7.5% impedance tolerance of a second, downstream transformer in a system when the level of fault current at
(two-winding construction). the transformer primary is known.
Step 3. Determine by formula or Table 1 the transformer let- MAIN
through short-circuit current. See Notes 3 and 4. TRANSFORMER
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
IS.C. primary IS.C. secondary
step 3 by 1.1 or 1.058 respectively. To find the lower end worst case, multiply results in
H.V. UTILITY
CONNECTION
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
IS.C. primary IS.C. secondary
multiply the total motor current in amps by 4. Values of 4 to 6 are commonly accepted. Step A. Calculate the "f" factor (IS.C. primary known)
Step 4. Calculate the "f" factor.
3Ø Faults 1.732 x L x I 3Ø 3Ø Transformer
f= I S.C. primary x V primary x 1.73 (%Z)
C x n x E L-L (I S.C. primary and
f=
I S.C. secondary are 100,000 x kVA
1Ø Line-to-Line (L-L) Faults transformer
2 x L x I L-L 3Ø fault values)
See Note 5 & Table 3 f=
C x n x EL-L
1Ø Transformer
1Ø Line-to-Neutral (L-N) Faults (I S.C. primary and
See Note 5 & Table 3 2 x L x I L-N† I S.C. primary x V primary x (%Z)
f= I S.C. secondary are f=
C x n x EL-N 100,000 x kVA
Where: 1Ø fault values: transformer
I S.C. secondary is L-L)
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. Step B. Calculate "M" (multiplier).
n = Number of conductors per phase (adjusts C value for
1
parallel runs) M=
1 +f
I = Available short-circuit current in amperes at beginning
Step C. Calculate the short-circuit current at the secondary of the
of circuit.
E = Voltage of circuit. transformer. (See Note under Step 3 of "Basic Point-to-
Point Calculation Procedure".)
† 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
Vprimary
I S.C. secondary = x M x I S.C. primary
(I) for this case in Step 4 should be adjusted at the transformer terminals as follows: At
Vsecondary
L-N center tapped transformer terminals, IL-N = 1.5 x IL-L at Transformer Terminals.
Fault X2
2
Step 5. M = 1 = 0.9626
1 + 0.0388
400A Switch
LPS-RK-400SP Fuse Step 6. Is.c. sym RMS = 57,279 X 0.9626 = 55,137A
Is.c. motor contribution** = 4 X 1804.3 = 7217A
50’ - 500 kcmil Cu
3 Single Conductors Itotal s.c. sym RMS = 55,137 + 7217 = 62,354A
Magnetic Conduit
3
Motor Contribution* M
System B
One-Line Diagram Fault X1 Fault X3
Available Utility
Infinite Assumption
Step 1. Is.c. = 1000 X 1000 = 1202.8A Step 4. f = 1.732 X 20 X 36,761 = 0.1161
1000 KVA Transformer 480 X 1.732 2 X 11,424 X 480
480V, 3Ø, 3.5%Z,
100 1 =
If.I.=1203A Step 2. Multipler = = 31.746 Step 5. M = = 0.8960
1 3.5 X 0.9† 1 + 0.1161
30’ - 500kcml Cu
3 Single Conductors Step 3. Is.c. = 1202.8 X 31.746 = 38,184A Step 6. Is.c. sym RMS = 36,761 X 0.8960 = 32,937A
4 Per Phase
PVC Conduit
2
1600A Switch Fault X2 Fault X4
KRP-C 1500SP Fuse
Step 4. f = 1.732 X 30 X 38,184 = 0.0387 Step A. f = 32,937 X 480 X 1.732 X (1.2 X 0.9) = 1.3144
26,706 X 4 X 480 100,000 X 225
=
Step 5. M = 1 = 0.9627 1
400A Switch Step B. M = = 0.4321
LPS-RK-350SP Fuse 1 + 0.0387 1 + 1.3144
3
225 KVA Transformer
208V, 3Ø
1.2%Z This example assumes no motor contribution.
L2
50 Feet
50’ - 3 AWG Cu
Magnetic Conduit Fault X3 Fault X3
3 Single Conductors
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 * Note 5. The L-N fault current is higher than the L-L
and larger have a ± 10% impedance tolerance. fault current at the secondary terminals of a single-
Short circuit amps can be affected by this phase center-tapped transformer. The short-circuit
tolerance. Therefore, for high end worst case, current available (I) for this case in Step 4 should be
multiply %Z by 0.9. For low end of worst case, adjusted at the transformer terminals as follows: At L-N
multiply %Z by 1.1. Transformers constructed to center tapped transformer terminals, IL-N = 1.5 x IL-L at
ANSI standards have a ±7.5% impedance Transformer Terminals.
tolerance (two-winding construction). **Assumes the neutral conductor and the line conductor
are the same size.
Table 1. Short-Circuit Currents Available from X/R Ratio of Percent For Line-to-Neutral
kVA for Impedance (%Z)* Faults
Various Size Transformers
(Based upon actual field nameplate data or from utility transformer worst case
1Ø Calculation for %X for %R
25.0 1.1 1.2–6.0 0.6 0.75
impedance) 37.5 1.4 1.2–6.5 0.6 0.75
50.0 1.6 1.2–6.4 0.6 0.75
Voltage Full % Short
75.0 1.8 1.2–6.6 0.6 0.75
and Load Impedance†† Circuit
Phase kVA Amps (Nameplate) Amps† 100.0 2.0 1.3–5.7 0.6 0.75
25 104 1.5 12175 167.0 2.5 1.4–6.1 1.0 0.75
37.5 156 1.5 18018 250.0 3.6 1.9–6.8 1.0 0.75
120/240 50 208 1.5 23706 333.0 4.7 2.4–6.0 1.0 0.75
1 ph.* 75 313 1.5 34639 500.0 5.5 2.2–5.4 1.0 0.75
100 417 1.6 42472 * National standards do not specify %Z for single-phase transformers. Consult
167 696 1.6 66644 manufacturer for values to use in calculation.
45 125 1.0 13879 ** Based on rated current of the winding (one–half nameplate kVA divided by
75 208 1.0 23132 secondary line-to-neutral voltage).
112.5 312 1.11 31259
With larger loads on new installations, it is extremely important to consider volt Select number from Table A, three-phase at 80% power factor, that is nearest
loss, otherwise some very unsatisfactory problems are likely to be but not greater than 764. This number is 745 which indicates the size of wire
encountered. needed: 6 AWG.
The actual conductor used must also meet the other sizing requirements such a Line-to-Neutral
full-load current, ambient temperature, number in a raceway, etc. 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,
How to Figure Volt Loss
divide single phase value by 2.
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 Open Wiring
the number of conductors per phase. The volt loss for open wiring installations depends on the separation between
Then, put a decimal point in front of the last 6 digits–you have the volt loss to be conductors. The volt loss is approximately equal to that for conductors in
non-magnetic conduit.
Example – 6 AWG copper wire, one per phase, in 180 feet of steel
Installation in Conduit, Cable or Raceway
NEC® Tables 310.15(B)(16) through 310.15(B)(19) give allowable ampacities
conduit–3 phase, 40 amp load at 80% power factor.
power factor for the value nearest, but not above your result – you have the
31-35 87-95 .91 .94 .96
Conduit Wire Ampacity Direct Volt Loss (See explanation prior page.)
Size Type Type Type Current Three-Phase Single-Phase
T, TW RH, RHH, (60 Cycle, Lagging Power Factor.) (60 Cycle, Lagging Power Factor.)
(60°C THWN, THHN, 100% 90% 80% 70% 60% 100% 90% 80% 70% 60%
Wire) RHW, XHHW
THW (90°C
(75°C Wire)
Wire)
Steel 14 20* 20* 25* 6140 5369 4887 4371 3848 3322 6200 5643 5047 4444 3836
Conduit 12 25* 25* 30* 3860 3464 3169 2841 2508 2172 4000 3659 3281 2897 2508
10 30 35* 40* 2420 2078 1918 1728 1532 1334 2400 2214 1995 1769 1540
8 40 50 55 1528 1350 1264 1148 1026 900 1560 1460 1326 1184 1040
6 55 65 75 982 848 812 745 673 597 980 937 860 777 690
4 70 85 95 616 536 528 491 450 405 620 610 568 519 468
3 85 100 110 490 433 434 407 376 341 500 501 470 434 394
2 95 115 130 388 346 354 336 312 286 400 409 388 361 331
1 110 130 150 308 277 292 280 264 245 320 337 324 305 283
0 125 150 170 244 207 228 223 213 200 240 263 258 246 232
00 145 175 195 193 173 196 194 188 178 200 227 224 217 206
000 165 200 225 153 136 162 163 160 154 158 187 188 184 178
0000 195 230 260 122 109 136 140 139 136 126 157 162 161 157
250 215 255 290 103 93 123 128 129 128 108 142 148 149 148
300 240 285 320 86 77 108 115 117 117 90 125 133 135 135
350 260 310 350 73 67 98 106 109 109 78 113 122 126 126
400 280 335 380 64 60 91 99 103 104 70 105 114 118 120
500 320 380 430 52 50 81 90 94 96 58 94 104 109 111
600 335 420 475 43 43 75 84 89 92 50 86 97 103 106
750 400 475 535 34 36 68 78 84 88 42 79 91 97 102
1000 455 545 615 26 31 62 72 78 82 36 72 84 90 95
Non- 14 20* 20* 25* 6140 5369 4876 4355 3830 3301 6200 5630 5029 4422 3812
Magnetic 12 25* 25* 30* 3464 3464 3158 2827 2491 2153 4000 3647 3264 2877 2486
Conduit 10 30 35* 40* 2420 2078 1908 1714 1516 1316 2400 2203 1980 1751 1520
(Lead 8 40 50 55 1528 1350 1255 1134 1010 882 1560 1449 1310 1166 1019
Covered 6 55 65 75 982 848 802 731 657 579 980 926 845 758 669
Cables or 4 70 85 95 616 536 519 479 435 388 620 599 553 502 448
Installation 3 85 100 110 470 433 425 395 361 324 500 490 456 417 375
in Fibre or 2 95 115 130 388 329 330 310 286 259 380 381 358 330 300
Other 1 110 130 150 308 259 268 255 238 219 300 310 295 275 253
Non- 0 125 150 170 244 207 220 212 199 185 240 254 244 230 214
Magnetic 00 145 175 195 193 173 188 183 174 163 200 217 211 201 188
Conduit, 000 165 200 225 153 133 151 150 145 138 154 175 173 167 159
Etc.) 0000 195 230 260 122 107 127 128 125 121 124 147 148 145 140
250 215 255 290 103 90 112 114 113 110 104 129 132 131 128
300 240 285 320 86 76 99 103 104 102 88 114 119 120 118
350 260 310 350 73 65 89 94 95 94 76 103 108 110 109
400 280 335 380 64 57 81 87 89 89 66 94 100 103 103
500 320 380 430 52 46 71 77 80 82 54 82 90 93 94
600 335 420 475 43 39 65 72 76 77 46 75 83 87 90
750 400 475 535 34 32 58 65 70 72 38 67 76 80 83
1000 455 545 615 26 25 51 59 63 66 30 59 68 73 77
* 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.
* 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.
Electrical Formulas
To Find Single-Phase Two-Phase Three-Phase Direct Current
kVA ≈ 1000 kVA ≈ 1000 kVA ≈ 1000
Amperes when kVA is known Not Applicable
E E≈ 2 E ≈ 1.73
HP ≈ 746 HP ≈ 746 HP ≈ 746 HP ≈ 746
Amperes when horsepower is known
E ≈ % eff. ≈ pf E ≈ 2 ≈ % eff. ≈ pf E ≈ 1.73 ≈ % eff. ≈ pf E ≈ % eff.
kW ≈ 1000 kW ≈ 1000 kW ≈ 1000 kW ≈ 1000
Amperes when kilowatts are known
E ≈ pf E ≈ 2 pf E ≈ 1.73 ≈ pf E
I ≈ E ≈ pf I ≈ E ≈ 2 ≈ pf I ≈ E ≈ 1.73 ≈ pf I≈ E
Kilowatts
1000 1000 1000 1000
I≈ E I≈ E≈ 2 I ≈ E ≈ 1.73
Kilovolt-Amperes Not Applicable
1000 1000 1000
I ≈ E % eff. ≈ pf I ≈ E ≈ 2 ≈ % eff. ≈ pf I ≈ E ≈ 1.73 ≈ % eff. ≈ pf I ≈ E ≈ % eff.
Horsepower
746 746 746 746
Watts E ≈ I ≈ pf I ≈ E ≈ 2 ≈ pf I ≈ E ≈ 1.73 ≈ pf E≈ I
Energy Efficiency = Load Horsepower ≈ 746
Load Input kVA ≈ 1000
Power Consumed = W or kW = cosθ
Power Factor = pf =
Apparent Power VA kVA
I = Amperes E = Volts kW = Kilowatts kVA = Kilovolt-Amperes
HP = Horsepower % eff. = Percent Efficiency pf = Power Factor