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Short-Circuit Current

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Overview

Calculates three-phase prospective short-circuit (fault) current at any point in a distribution system using the Bussmann point-to-point method (Cooper Bussmann SPD tables). Supports two starting methods: from transformer nameplate data or from a known source fault current. The calculation traces fault current through cables and busbars using the Bussmann f-factor and M = 1/(1 + f), then adds the motor contribution (4 × motor full-load current) to determine total available fault current.

Use this calculator when you need to:

  • Verify interrupting capacity ratings for circuit breakers and fuses
  • Determine withstand ratings for switchboards and panelboards
  • Establish input data for arc flash analysis (IEEE 1584)
  • Check protective device coordination

Input Parameters

General Parameters

Parameter Type Default Range Description
calculation_method Enum: transformer / source_fault transformer — Starting point for fault current calculation
voltage Number 480 100 – 15,000 V System line-to-line voltage at fault location

Transformer Method Parameters

Visible when calculation_method = transformer.

Parameter Type Default Range Description
transformer_kva Number 1000 10 – 50,000 kVA Transformer capacity rating
transformer_primary_v Number 4160 100 – 50,000 V Primary side line-to-line voltage
transformer_impedance_percent Number — 1 – 20% Transformer impedance; blank = use standard table lookup
primary_fault_current Number — 100 – 500,000 A Optional — available fault current at the transformer primary (Method 2, Bussmann second-transformer procedure); blank = infinite primary

Source Fault Method Parameters

Visible when calculation_method = source_fault.

Parameter Type Default Range Description
source_fault_current Number — 100 – 500,000 A Known fault current at source/beginning of system

Cables (Array)

Parameter Type Default Range Description
wire_size Enum — AWG 14 – 1000 kcmil Conductor size (AWG or kcmil)
length_ft Number — 1 – 10,000 ft Cable run length in feet
material Enum: copper / aluminum copper — Conductor material
conduit Enum: steel / nonmagnetic steel — Three single conductors in steel (magnetic) or nonmagnetic conduit — selects the C-value column
num_parallel Integer 1 1 – 10 Parallel conductors per phase

Busbars (Array)

Parameter Type Default Range Description
ampacity Enum — 225 – 4000 A Busway ampacity rating
busway_type Enum: plug_in / feeder / high_impedance plug_in — Busway construction (high-impedance: copper 1000–3000 A only; 4000 A: feeder only)
material Enum: copper / aluminum copper — Busbar material
length_ft Number — 1 – 1,000 ft Busway length in feet

Motors (Array)

Parameter Type Default Range Description
hp Enum — 0.5 – 200 HP Motor horsepower rating (full-load current from NEC Table 430.250)
quantity Integer 1 1 – 100 Number of identical motors at this location

Output Results

Output Unit Description
total_fault_current A Total available fault current including motor contribution
fault_current_without_motors A Fault current from source/transformer through cables and busbars only
motor_contribution A 4 × the total motor full-load current (NEC Table 430.250)
steps — Step-by-step calculation trace through each component

Per-Step Outputs (each component in the chain)

Output Unit Description
fault_current A Fault current at the output of this component
f_factor — f-factor for cable/busbar (dimensionless)
m_multiplier — The divisor 1 + f (Bussmann M = 1 / (1 + f))
c_value — Bussmann C value (1 / impedance per foot) from Table 2 (cable) or Table 3 (busway)
impedance_percent % Transformer impedance (transformer step only)
full_load_current A Transformer secondary FLC (transformer step only)

Formulas

Transformer Full-Load Current

\[ I_{FL} = \frac{S \times 1000}{\sqrt{3} \times V_{secondary}} \]

Where \(S\) is transformer rating in kVA and \(V_{secondary}\) is the secondary line-to-line voltage.

Transformer Short-Circuit Current (Method 1 — Infinite Utility)

\[ I_{sc} = \frac{I_{FL} \times 100}{Z\%} \]

Where \(Z\%\) is the transformer impedance percentage. This assumes an infinite source on the primary side.

Transformer Short-Circuit Current (Method 2 — Known Primary Fault)

\[ f = \frac{I_{sca(P)} \times V_P \times 1.732 \times Z\%}{100\,000 \times kVA}, \qquad M = \frac{1}{1 + f}, \qquad I_{sca(S)} = \frac{V_P}{V_S} \times M \times I_{sca(P)} \]

Bussmann SPD (2005) point-to-point method, “Procedure for second transformer in system” (Steps A–C), used when the available fault current \(I_{sca(P)}\) at the transformer primary is known. It is the source impedance and the transformer impedance added in series; as \(I_{sca(P)} \to \infty\) it tends to Method 1. The calculator reports the divisor \(1 + f\).

Cable f-Factor

\[ f = \frac{1.732 \times L \times I_{upstream}}{C \times n \times E_{L-L}} \]

Where \(L\) is the cable length in feet, \(I_{upstream}\) is the fault current at the start of the cable (A), \(C\) is the Bussmann conductor constant from Table 2 (a reciprocal-impedance figure, larger = lower impedance), \(n\) is the number of parallel conductors per phase, and \(E_{L-L}\) is the line-to-line voltage. This is the three-phase form of the Bussmann point-to-point method.

Busway f-Factor

\[ f = \frac{1.732 \times L \times I_{upstream}}{C \times E_{L-L}} \]

Same form as the cable f-factor with the busway constant \(C\) from Table 3 and no parallel-set term.

M Multiplier

\[ M = \frac{1}{1 + f} \]

Exact for every value of \(f\); no table lookup or interpolation is involved. The calculator reports the divisor \(1 + f\) and divides by it, which is the same thing.

Fault Current at Cable/Busbar End

\[ I_{sc} = I_{upstream} \times M = \frac{I_{upstream}}{1 + f} \]

Applied element by element along the path, each f using the fault current at the start of that element.

Motor Contribution

\[ I_{motor,total} = 4 \times \sum_{i} (I_{FLA,i} \times Q_i) \]

Bussmann point-to-point Note 1: “a practical estimate of motor short-circuit contribution is to multiply the total motor full-load current in amperes by 4” (values of 4 to 6 are commonly accepted). \(I_{FLA,i}\) is the full-load current of motor \(i\) from NEC Table 430.250 (Table 4 below) and \(Q_i\) the quantity.

Total Fault Current

\[ I_{total} = I_{cable/bus} + I_{motor,total} \]

The total available fault current is the sum of the fault current through the distribution system plus all motor contributions.

IEC 60909 Network Studies (MCP tool, mode='network')

The fe_calc_short_circuit tool can also solve an inline network to IEC 60909-0 (equivalent voltage source at the fault). The worked examples below include these studies.

\[ Z_Q = \frac{c_Q\,U_{nQ}^2}{S''_{kQ}}, \qquad K_T = \frac{0.95\,c_{max}}{1 + 0.6\,x_T}, \qquad I''_k = \frac{c\,U_n}{\sqrt{3}\,|Z_k|} \] \[ I''_{k2} = \frac{c\,U_n}{|2Z_{(1)}|}, \qquad I''_{k1} = \frac{\sqrt{3}\,c\,U_n}{|Z_{(1)} + Z_{(2)} + Z_{(0)} + 3R_F|}, \qquad I''_{kE2E} = \frac{\sqrt{3}\,c\,U_n}{|Z_{(1)} + 2Z_{(0)}|} \] \[ \kappa = 1.02 + 0.98\,e^{-3R/X}, \qquad i_p = \kappa\sqrt{2}\,I''_k, \qquad I_{th} = I''_k\sqrt{m + n} \]

Voltage factor \(c_{max} = 1.10\) (low voltage with 10 % tolerance, and high voltage) and \(c_{min} = 0.95\) (low voltage) / \(1.00\) (high voltage). Impedances are referred to the fault-bus voltage by \((U_{fault}/U_{element})^2\). Behind a Dyn transformer the grid's zero sequence is blocked, so \(Z_{(0)} = K_T Z_T\) plus the LV line zero-sequence impedances (3 × positive sequence unless given).

Key Reference Tables

Table 1 — Transformer Impedance (Z%) — Sample Values

kVA Primary Secondary Z%
25480V208/120V1.50
50480V208/120V2.07
75480V208/120V1.73
100480V208/120V1.63
150600V277/480V3.83
300480V208/120V2.56
500480V208/120V3.35
750480V208/120V4.60
1000480V208/120V5.22
150013800V277/480V5.50
250013800V277/480V5.75
500013800V277/480V6.50
1000013800V277/480V7.50

The lookup takes the nearest tabulated kVA, the primary voltage class (480 V, 600 V, 2400 V, 4160 V, 4800 V, 13.8 kV) and the secondary system (208 V → 208/120 V, 240 V → 240 V, 480 V → 277/480 V, 2400 V, 4160 V). When that combination is not tabulated the calculator uses 5.75 %. These are typical values carried by the calculator, not a published standard table. Always enter the nameplate %Z when it is known; for the Bussmann worst case enter the nameplate value reduced by the −10 % manufacturing tolerance (e.g. 3.5 % → 3.15 %).

Table 2 — Conductor C Values (three single conductors in conduit)

Size Copper Aluminum
Steel conduitNonmagnetic conduit Steel conduitNonmagnetic conduit
600 V5 kV15 kV600 V5 kV15 kV 600 V5 kV15 kV600 V5 kV15 kV
14 AWG389——389——237——237——
12 AWG617——617——376——376——
10 AWG981——982——599——599——
8 AWG1,5571,551—1,5591,555—951950—952951—
6 AWG2,4252,4062,3892,4302,4182,4071,4811,4761,4721,4821,4791,476
4 AWG3,8063,7513,6963,8263,7893,7532,3462,3332,3192,3502,3422,333
3 AWG4,7744,6744,5774,8114,7454,6792,9522,9282,9042,9612,9452,929
2 AWG5,9075,7365,5746,0445,9265,8093,7133,6703,6263,7303,7023,673
1 AWG7,2937,0296,7597,4937,3077,1094,6454,5754,4984,6784,6324,580
1/08,9258,5447,9739,3179,0348,5905,7775,6705,4935,8385,7665,646
2/010,75510,0629,39011,42410,87810,3197,1876,9686,7337,3017,1536,986
3/012,84411,80411,02213,92313,04812,3608,8268,4678,1639,1108,8518,627
4/015,08213,60612,54316,67315,35114,34710,74110,1679,70011,17410,74910,387
250 kcmil16,48314,92513,64418,59417,12115,86612,12211,46010,84912,86212,34311,847
300 kcmil18,17716,29314,76920,86818,97517,40913,91013,00912,19314,92314,18313,492
350 kcmil19,70417,38515,67822,73720,52618,67215,48414,28013,28816,81315,85814,955
400 kcmil20,56618,23516,36624,29721,78619,73116,67115,35514,18818,50617,32116,234
500 kcmil22,18519,17217,49226,70623,27721,33018,75616,82815,65721,39119,50318,315
600 kcmil22,96520,56717,96228,03325,20422,09720,09318,42816,48423,45121,71819,635
750 kcmil24,13721,38718,88929,73526,45323,40821,76619,68517,68625,97623,70221,437
1000 kcmil25,27822,53919,92331,49128,08324,88723,47821,23519,00628,77926,10923,482

Source: Cooper Bussmann SPD “Selecting Protective Devices” (2005), Point-to-Point Method, “C” Values for Conductors (the same table appears in the Bussmann Electrical Plan Review, 2002). C = 1 / impedance per foot, from IEEE Std 241-1990 and IEEE Std 242-1986 R and X. The column is chosen from the system voltage: up to 600 V → 600 V, up to 5 kV → 5 kV, above → 15 kV. Bussmann publishes no values for 14–10 AWG at 5 kV or 14–8 AWG at 15 kV; the calculator rejects those combinations.

Table 3 — Busway C Values

Ampacity Plug-in Feeder High impedance
CopperAluminumCopperAluminumCopper
225 A28,70023,00018,70012,000—
400 A38,90034,70023,90021,300—
600 A41,00038,30036,50031,300—
800 A46,10057,50049,30044,100—
1000 A69,40089,30062,90056,20015,600
1200 A94,30097,10076,90069,90016,100
1350 A119,000104,20090,10084,00017,500
1600 A129,900120,500101,00090,90019,200
2000 A142,900135,100134,200125,00020,400
2500 A143,800156,300180,500166,70021,700
3000 A144,900175,400204,100188,70023,800
4000 A——277,800256,400—

Source: Cooper Bussmann SPD (2005), Point-to-Point Method, “C” Values for Busway (“one over the impedance per foot … in a survey of industry”). Low-voltage busway only: the calculator rejects busway on systems above 600 V and the blank combinations.

Table 4 — Motor Full-Load Current (A), NEC Table 430.250

HP 200 V 208 V 230 V 460 V 575 V 2300 V
0.52.52.42.21.10.9—
0.753.73.53.21.61.3—
14.84.64.22.11.7—
1.56.96.6632.4—
27.87.56.83.42.7—
31110.69.64.83.9—
517.516.715.27.66.1—
7.525.324.222119—
1032.230.8281411—
1548.346.2422117—
2062.159.4542722—
2578.274.8683427—
309288804032—
401201141045241—
501501431306552—
60177169154776216
75221211192967720
1002852732481249926
12535934331215612531
15041439636018014437
20055252848024019249

Three-phase squirrel-cage induction motors, NFPA 70 Table 430.250. The column follows the system voltage: 195–204 V → 200 V, 205–216 V → 208 V, 220–240 V → 230 V, 440–480 V → 460 V, 550–600 V → 575 V, 2300–2400 V → 2300 V. Other system voltages have no column and motors are rejected there. The contribution is 4 × FLA (Bussmann Note 1).

Worked Examples

The 25 examples below are the validation cases for this calculation. Each one was produced by the calculation engine from the inputs shown, every figure was checked against the independent reference quoted at the end of the example, and the same cases are run against every engine build (see validation scorecard, tool fe_calc_short_circuit; worked examples for every calculation).

#CaseMethodFault pointFault (kA)ip (kA)Rating (kA)GoverningResult
11500 kVA, 480 V transformer terminalsBussmann point-to-pointtransformer terminals31.378–35interrupting ratingPASS
2Bussmann SPD example — fault X1 with −10 % Z toleranceBussmann point-to-pointtransformer terminals57.277–65interrupting ratingPASS
3Bussmann SPD example — fault X2 after 25 ft of 6 × 500 kcmilBussmann point-to-pointend of cable 155.137–65interrupting ratingPASS
440 kA source, 50 ft of 1/0 CuBussmann point-to-pointend of cable 122.117–––
5500 kVA, 208 V — feeder then branch circuitBussmann point-to-pointend of cable 23.555–10interrupting ratingPASS
665 kA source, 3 × 4/0 per phase — 42 kA panel failsBussmann point-to-pointend of cable 143.040–42interrupting ratingFAIL
72500 kVA — 2000 A busway then 1/0 tapBussmann point-to-pointend of cable 129.736–65interrupting ratingPASS
8750 kVA, 13.8 kV primary, %Z not givenBussmann point-to-pointtransformer terminals15.689–22interrupting ratingPASS
910 kA at 208 V through 30 m of 12 AWGBussmann point-to-pointend of cable 10.700–––
101000 kVA, 480 V — 100 ft of 500 kcmil CuBussmann point-to-pointend of cable 115.608–––
1142 kA source — 200 ft of 3 × 500 kcmil, two 50 HP motorsBussmann point-to-pointend of cable 129.378–––
12500 kVA, 480 → 208 V (table %Z) — 25 ft of 600 A busway, 50 ft of 2/0Bussmann point-to-pointend of cable 114.722–––
1350 kA source — 1600 A Al feeder busway, 2 × 500 kcmil Al in PVC, four 25 HP motorsBussmann point-to-pointend of cable 130.241–35interrupting ratingPASS
144160 V — 20 kA source, 300 m of 4/0 Cu (5 kV column)Bussmann point-to-pointend of cable 112.482–––
15500 kVA, 4160 → 480 V fed from 10 kA — then 50 ft of 4/0Bussmann point-to-pointend of cable 18.396–10interrupting ratingPASS
16Bussmann example — 300 kVA, 208 V, faults #1 and #2Bussmann point-to-pointend of cable 218.670–––
17IEC 60909 — 1.6 MVA, 415 V board, three-phase maximumIEC 60909 3ph maxLV (0.415 kV)36.23487.05–mva method cross checkPASS
18IEC 60909 — line-to-line fault on the 415 V boardIEC 60909 LL maxLV (0.415 kV)31.379–––
19IEC 60909 — line-to-earth fault behind a Dyn11 transformerIEC 60909 LG maxLV (0.415 kV)37.537–––
20IEC 60909 — double line-to-earth fault on the 415 V boardIEC 60909 LLG maxLV (0.415 kV)36.956–––
21IEC 60909 — line-to-earth fault through 5 mΩIEC 60909 LG maxLV (0.415 kV)28.950–––
22IEC 60909 — minimum case on the 415 V boardIEC 60909 3ph minLV (0.415 kV)29.69871.13––
23IEC 60909 — fault at the 11 kV infeedIEC 60909 3ph maxGRID (11 kV)13.12232.40––
24IEC 60909 — 60 Hz, 13.8 kV cable, 2.5 MVA, 480 V MCC rated 30 kAIEC 60909 3ph maxMCC (0.48 kV)33.67962.8530fault ratingFAIL
25IEC 60909 — two 1 MVA transformers in parallel, MVA cross-checkIEC 60909 3ph maxLV (0.415 kV)44.143106.13–mva method cross checkPASS

Example 1 — 1500 kVA, 480 V transformer terminals

Step 1 of the point-to-point method alone: the let-through of a transformer on an infinite primary, against a 35 kA rating.

Given: 480 V three-phase, 1,500 kVA transformer, 13,800 V primary (infinite source), 5.75 % Z. Path from the source: none (fault at the source); equipment interrupting rating 35 kA. C values from the Bussmann SPD tables (cables: three single conductors).

Full-load currentIFL = S × 1000 / (√3 × EL-L) = 1,500 × 1000 / (√3 × 480) = 1,804.2 A
Transformer let-through%Z = 5.75 % (nameplate); ISC = IFL × 100 / %Z = 1,804.2 × 100 / 5.75 = 31,378 A (infinite primary)
Fault currentISC at the transformer terminals = 31.378 kA rms symmetrical
Interrupting ratingNEC 110.9: ISC ≤ equipment rating → 31.378 vs limit 35.000 kA PASS
Result1/1 checks pass → PASS; ISC = 31.378 kA at the transformer terminals; governing: interrupting rating
ReferenceHand calculation: I_FL = 1500 kVA / (√3 × 480 V) = 1804.2 A; Isc = I_FL × 100 / 5.75 = 31 378 A (infinite primary, Bussmann point-to-point step 1); 31.378 / 35 kA = 89.7 % (validation case c0009-bussmann-1500kva-480v-terminal)

Example 2 — Bussmann SPD example — fault X1 with −10 % Z tolerance

Reproduces the handbook's worst-case transformer let-through by entering the nameplate 3.5 % reduced by the −10 % tolerance (3.15 %).

Given: 480 V three-phase, 1,500 kVA transformer, 13,800 V primary (infinite source), 3.15 % Z. Path from the source: none (fault at the source); equipment interrupting rating 65 kA. C values from the Bussmann SPD tables (cables: three single conductors).

Full-load currentIFL = S × 1000 / (√3 × EL-L) = 1,500 × 1000 / (√3 × 480) = 1,804.2 A
Transformer let-through%Z = 3.15 % (nameplate); ISC = IFL × 100 / %Z = 1,804.2 × 100 / 3.15 = 57,277 A (infinite primary)
Fault currentISC at the transformer terminals = 57.277 kA rms symmetrical
Interrupting ratingNEC 110.9: ISC ≤ equipment rating → 57.277 vs limit 65.000 kA PASS
Result1/1 checks pass → PASS; ISC = 57.277 kA at the transformer terminals; governing: interrupting rating
ReferenceEaton Bussmann SPD (Selecting Protective Devices), Point-to-Point Method, Example 1, fault X1: I_FL = 1500 × 1000 / (480 × 1.732) = 1804 A; multiplier = 100 / (0.9 × 3.5) = 31.746; Isc = 1804.2 × 31.746 = 57 277 A (handbook: 57 270 A with I_FL rounded to 1804 A). Motor contribution (4 × FLA in the handbook) is not part of this case. (validation case c0009-bussmann-spd-example-x1-tolerance)

Example 3 — Bussmann SPD example — fault X2 after 25 ft of 6 × 500 kcmil

The handbook's second fault point: six parallel 500 kcmil copper conductors in steel conduit from the 57.3 kA X1 bus.

Given: 480 V three-phase, 1,500 kVA transformer, 13,800 V primary (infinite source), 3.15 % Z. Path from the source: 6 × 500 Cu in steel conduit, 7.62 m; equipment interrupting rating 65 kA. C values from the Bussmann SPD tables (cables: three single conductors).

Full-load currentIFL = S × 1000 / (√3 × EL-L) = 1,500 × 1000 / (√3 × 480) = 1,804.2 A
Transformer let-through%Z = 3.15 % (nameplate); ISC = IFL × 100 / %Z = 1,804.2 × 100 / 3.15 = 57,277 A (infinite primary)
Cable 16 × 500 Cu, 7.62 m = 25.00 ft, C = 22,185 (steel conduit, 600V): f = 1.732 × L × I / (C × n × E) = 1.732 × 25.00 × 57,277 / (22,185 × 6 × 480) = 0.0388; M = 1 / (1 + f) = 0.9626; I = 57,277 × 0.9626 = 55,137 A
Fault currentISC at the end of cable 1 = 55.137 kA rms symmetrical
Interrupting ratingNEC 110.9: ISC ≤ equipment rating → 55.137 vs limit 65.000 kA PASS
Result1/1 checks pass → PASS; ISC = 55.137 kA at the end of cable 1; governing: interrupting rating
ReferenceEaton Bussmann SPD (Selecting Protective Devices, 2005), Point-to-Point Method, Example 1, fault X2: Isc at X1 = 57 277 A (1500 kVA, 3.5 % × 0.9); 7.62 m = 25 ft of 6 × 500 kcmil Cu, three single conductors in steel conduit, C = 22 185 (SPD Table 2, 600 V): f = 1.732 × 25 × 57 277 / (6 × 22 185 × 480) = 0.0388; M = 1/(1 + f) = 0.9626; I = 57 277 × 0.9626 = 55 137 A; 55.137 / 65 kA = 84.8 %. Motor contribution (4 × FLA in the handbook) is not part of this case. (validation case c0009-bussmann-spd-example-x2-6x500kcmil)

Example 4 — 40 kA source, 50 ft of 1/0 Cu

One cable from a known fault level: the f factor and multiplier the method is built on.

Given: 480 V three-phase, known fault level 40 kA rms. Path from the source: 1/0 Cu in steel conduit, 15.24 m. C values from the Bussmann SPD tables (cables: three single conductors).

SourceKnown fault level ISC = 40 kA = 40,000 A
Cable 11 × 1/0 Cu, 15.24 m = 50.00 ft, C = 8,925 (steel conduit, 600V): f = 1.732 × L × I / (C × n × E) = 1.732 × 50.00 × 40,000 / (8,925 × 1 × 480) = 0.8086; M = 1 / (1 + f) = 0.5529; I = 40,000 × 0.5529 = 22,117 A
Fault currentISC at the end of cable 1 = 22.117 kA rms symmetrical
ResultISC = 22.117 kA at the end of cable 1
ReferenceHand calculation, Bussmann point-to-point: f = 1.732 × 50 ft × 40 000 A / (C 8925 × 1 × 480 V) = 0.8086; M = 1/(1 + f) = 0.5529 (the tool reports the divisor 1 + f = 1.8086); I = 40 kA × 0.5529 = 22.117 kA (C for 1/0 Cu, three single conductors in steel conduit, Bussmann SPD table) (validation case c0009-bussmann-known-fault-1-0-cable)

Example 5 — 500 kVA, 208 V — feeder then branch circuit

Two cables in series: each f factor uses the fault current at the start of that cable.

Given: 208 V three-phase, 500 kVA transformer, 480 V primary (infinite source), 4 % Z. Path from the source: 4/0 Cu in steel conduit, 15 m; 6 Cu in steel conduit, 20 m; equipment interrupting rating 10 kA. C values from the Bussmann SPD tables (cables: three single conductors).

Full-load currentIFL = S × 1000 / (√3 × EL-L) = 500 × 1000 / (√3 × 208) = 1,387.9 A
Transformer let-through%Z = 4 % (nameplate); ISC = IFL × 100 / %Z = 1,387.9 × 100 / 4 = 34,697 A (infinite primary)
Cable 11 × 4/0 Cu, 15 m = 49.21 ft, C = 15,082 (steel conduit, 600V): f = 1.732 × L × I / (C × n × E) = 1.732 × 49.21 × 34,697 / (15,082 × 1 × 208) = 0.9427; M = 1 / (1 + f) = 0.5147; I = 34,697 × 0.5147 = 17,860 A
Cable 21 × 6 Cu, 20 m = 65.62 ft, C = 2,425 (steel conduit, 600V): f = 1.732 × L × I / (C × n × E) = 1.732 × 65.62 × 17,860 / (2,425 × 1 × 208) = 4.0240; M = 1 / (1 + f) = 0.1990; I = 17,860 × 0.1990 = 3,555 A
Fault currentISC at the end of cable 2 = 3.555 kA rms symmetrical
Interrupting ratingNEC 110.9: ISC ≤ equipment rating → 3.555 vs limit 10.000 kA PASS
Result1/1 checks pass → PASS; ISC = 3.555 kA at the end of cable 2; governing: interrupting rating
ReferenceHand calculation, Bussmann point-to-point: I_FL = 500 kVA / (√3 × 208 V) = 1387.9 A; Isc = 1387.9 × 100/4 = 34 697 A; 4/0 Cu (C 15 082), 15 m = 49.21 ft: f = 1.732 × 49.21 × 34 697 / (15 082 × 208) = 0.9427, I = 34 697 / 1.9427 = 17 860 A; 6 AWG Cu (C 2425), 20 m = 65.62 ft: f = 1.732 × 65.62 × 17 860 / (2425 × 208) = 4.024, I = 17 860 / 5.024 = 3555 A; 3.555 / 10 kA = 35.5 % (validation case c0009-bussmann-208v-two-cables-in-series)

Example 6 — 65 kA source, 3 × 4/0 per phase — 42 kA panel fails

Parallel sets divide the f factor by n; the remaining fault exceeds a 42 kA rating, so the interrupting-rating check fails.

Given: 480 V three-phase, known fault level 65 kA rms. Path from the source: 3 × 4/0 Cu in steel conduit, 30 m; equipment interrupting rating 42 kA. C values from the Bussmann SPD tables (cables: three single conductors).

SourceKnown fault level ISC = 65 kA = 65,000 A
Cable 13 × 4/0 Cu, 30 m = 98.43 ft, C = 15,082 (steel conduit, 600V): f = 1.732 × L × I / (C × n × E) = 1.732 × 98.43 × 65,000 / (15,082 × 3 × 480) = 0.5102; M = 1 / (1 + f) = 0.6622; I = 65,000 × 0.6622 = 43,040 A
Fault currentISC at the end of cable 1 = 43.040 kA rms symmetrical
Interrupting ratingNEC 110.9: ISC ≤ equipment rating → 43.040 vs limit 42.000 kA FAIL
Result0/1 checks pass → FAIL; ISC = 43.040 kA at the end of cable 1; governing: interrupting rating
ReferenceHand calculation, Bussmann point-to-point: 30 m = 98.43 ft; f = 1.732 × 98.43 × 65 000 / (15 082 × 3 × 480) = 0.5102; I = 65 000 / 1.5102 = 43 040 A > 42 kA (102.5 %) — fails NEC 110.9 (validation case c0009-bussmann-parallel-4-0-rating-fail)

Example 7 — 2500 kVA — 2000 A busway then 1/0 tap

A busway segment (f without the n term) followed by a cable, from a large transformer.

Given: 480 V three-phase, 2,500 kVA transformer, 13,800 V primary (infinite source), 5.75 % Z. Path from the source: 2,000 A plug-in Cu busway, 15 m; 1/0 Cu in steel conduit, 10 m; equipment interrupting rating 65 kA. C values from the Bussmann SPD tables (cables: three single conductors).

Full-load currentIFL = S × 1000 / (√3 × EL-L) = 2,500 × 1000 / (√3 × 480) = 3,007.0 A
Transformer let-through%Z = 5.75 % (nameplate); ISC = IFL × 100 / %Z = 3,007.0 × 100 / 5.75 = 52,296 A (infinite primary)
Busbar 12,000 A plug-in Cu busway, 15 m = 49.21 ft, C = 142,900: f = 1.732 × L × I / (C × E) = 1.732 × 49.21 × 52,296 / (142,900 × 480) = 0.0650; M = 1 / (1 + f) = 0.9390; I = 52,296 × 0.9390 = 49,105 A
Cable 11 × 1/0 Cu, 10 m = 32.81 ft, C = 8,925 (steel conduit, 600V): f = 1.732 × L × I / (C × n × E) = 1.732 × 32.81 × 49,105 / (8,925 × 1 × 480) = 0.6513; M = 1 / (1 + f) = 0.6056; I = 49,105 × 0.6056 = 29,736 A
Fault currentISC at the end of cable 1 = 29.736 kA rms symmetrical
Interrupting ratingNEC 110.9: ISC ≤ equipment rating → 29.736 vs limit 65.000 kA PASS
Result1/1 checks pass → PASS; ISC = 29.736 kA at the end of cable 1; governing: interrupting rating
ReferenceHand calculation, Bussmann point-to-point: I_FL = 2500 kVA / (√3 × 480) = 3007.0 A; Isc = 3007.0 × 100/5.75 = 52 296 A; 2000 A plug-in Cu busway C = 142 900 (Bussmann SPD busway table), 15 m = 49.21 ft: f = 1.732 × 49.21 × 52 296 / (142 900 × 480) = 0.0650, I = 49 105 A; 1/0 Cu C = 8925, 10 m = 32.81 ft: f = 1.732 × 32.81 × 49 105 / (8925 × 480) = 0.6513, I = 49 105 / 1.6513 = 29 736 A (validation case c0009-bussmann-2000a-busway-then-cable)

Example 8 — 750 kVA, 13.8 kV primary, %Z not given

With no nameplate %Z and no table entry for this kVA/primary voltage, the method's 5.75 % default applies (and is reported as an assumption).

Given: 480 V three-phase, 750 kVA transformer, 13,800 V primary (infinite source), %Z not given. Path from the source: none (fault at the source); equipment interrupting rating 22 kA. C values from the Bussmann SPD tables (cables: three single conductors).

Full-load currentIFL = S × 1000 / (√3 × EL-L) = 750 × 1000 / (√3 × 480) = 902.1 A
Transformer let-through%Z = 5.75 % (not given and not tabulated for this kVA and primary voltage → method default); ISC = IFL × 100 / %Z = 902.1 × 100 / 5.75 = 15,689 A (infinite primary)
Fault currentISC at the transformer terminals = 15.689 kA rms symmetrical
Interrupting ratingNEC 110.9: ISC ≤ equipment rating → 15.689 vs limit 22.000 kA PASS
Result1/1 checks pass → PASS; ISC = 15.689 kA at the transformer terminals; governing: interrupting rating
ReferenceHand calculation: no %Z table entry for 750 kVA with a 13.8 kV primary → 5.75 % (method default); I_FL = 750 kVA / (√3 × 480) = 902.1 A; Isc = 902.1 × 100/5.75 = 15 689 A; 15.689 / 22 kA = 71.3 % (validation case c0009-bussmann-default-impedance-750kva)

Example 9 — 10 kA at 208 V through 30 m of 12 AWG

A small, long branch circuit: f ≫ 1, the fault drops below 1 kA and the method warns about coordination.

Given: 208 V three-phase, known fault level 10 kA rms. Path from the source: 12 Cu in steel conduit, 30 m. C values from the Bussmann SPD tables (cables: three single conductors).

SourceKnown fault level ISC = 10 kA = 10,000 A
Cable 11 × 12 Cu, 30 m = 98.43 ft, C = 617 (steel conduit, 600V): f = 1.732 × L × I / (C × n × E) = 1.732 × 98.43 × 10,000 / (617 × 1 × 208) = 13.2833; M = 1 / (1 + f) = 0.0700; I = 10,000 × 0.0700 = 700 A
Fault currentISC at the end of cable 1 = 0.700 kA rms symmetrical
ResultISC = 0.700 kA at the end of cable 1. Low fault current may affect protective device coordination
ReferenceHand calculation, Bussmann point-to-point: 30 m = 98.43 ft; f = 1.732 × 98.43 × 10 000 / (617 × 1 × 208) = 13.283; M = 1/(1 + 13.283) = 0.0700; I = 10 000 × 0.0700 = 700 A (C for 12 AWG Cu = 617, Bussmann SPD) (validation case c0009-bussmann-12awg-branch-low-fault)

Example 10 — 1000 kVA, 480 V — 100 ft of 500 kcmil Cu

Transformer let-through reduced by one large feeder (help page example 1).

Given: 480 V three-phase, 1,000 kVA transformer, 4,160 V primary (infinite source), 5.75 % Z. Path from the source: 500 Cu in steel conduit, 30.48 m. C values from the Bussmann SPD tables (cables: three single conductors).

Full-load currentIFL = S × 1000 / (√3 × EL-L) = 1,000 × 1000 / (√3 × 480) = 1,202.8 A
Transformer let-through%Z = 5.75 % (nameplate); ISC = IFL × 100 / %Z = 1,202.8 × 100 / 5.75 = 20,918 A (infinite primary)
Cable 11 × 500 Cu, 30.48 m = 100.00 ft, C = 22,185 (steel conduit, 600V): f = 1.732 × L × I / (C × n × E) = 1.732 × 100.00 × 20,918 / (22,185 × 1 × 480) = 0.3402; M = 1 / (1 + f) = 0.7461; I = 20,918 × 0.7461 = 15,608 A
Fault currentISC at the end of cable 1 = 15.608 kA rms symmetrical
ResultISC = 15.608 kA at the end of cable 1
ReferenceHand calculation, Bussmann point-to-point: I_FL = 1000 kVA / (√3 × 480) = 1202.8 A; Isc = 1202.8 × 100/5.75 = 20 919 A; 500 kcmil Cu, three single conductors in steel conduit, C = 22 185 (Bussmann SPD Table 2, 600 V); 30.48 m = 100 ft: f = 1.732 × 100 × 20 919 / (22 185 × 1 × 480) = 0.3402; I = 20 919 / 1.3402 = 15 608 A (validation case c0009-bussmann-1000kva-500kcmil-100ft)

Example 11 — 42 kA source — 200 ft of 3 × 500 kcmil, two 50 HP motors

Parallel feeder from a known fault level, then motor contribution as 4 × NEC Table 430.250 full-load current (help page example 2).

Given: 480 V three-phase, known fault level 42 kA rms. Path from the source: 3 × 500 Cu in steel conduit, 60.96 m, motors 2 × 50 HP. C values from the Bussmann SPD tables (cables: three single conductors).

SourceKnown fault level ISC = 42 kA = 42,000 A
Cable 13 × 500 Cu, 60.96 m = 200.00 ft, C = 22,185 (steel conduit, 600V): f = 1.732 × L × I / (C × n × E) = 1.732 × 200.00 × 42,000 / (22,185 × 3 × 480) = 0.4554; M = 1 / (1 + f) = 0.6871; I = 42,000 × 0.6871 = 28,858 A
Motor full-load currentNEC Table 430.250, 460 V column: Σ quantity × FLA = 2 × 50 HP × 65 A = 130.0 A
Motor contributionBussmann Note 1: 4 × ΣFLA = 4 × 130.0 = 520 A; I = 28,858 A + 520 A = 29,378 A
Fault currentISC at the end of cable 1 = 29.378 kA rms symmetrical
ResultISC = 29.378 kA at the end of cable 1
ReferenceHand calculation, Bussmann point-to-point: 60.96 m = 200 ft; 500 kcmil Cu steel conduit C = 22 185 (SPD Table 2): f = 1.732 × 200 × 42 000 / (22 185 × 3 × 480) = 0.4554; I = 42 000 / 1.4554 = 28 858 A. Motors (Bussmann SPD Note 1: 4 × total motor FLA): 50 HP at 460 V = 65 A (NEC Table 430.250); 4 × 2 × 65 = 520 A; total 28 858 + 520 = 29 378 A (validation case c0009-bussmann-42ka-3x500-two-50hp-motors)

Example 12 — 500 kVA, 480 → 208 V (table %Z) — 25 ft of 600 A busway, 50 ft of 2/0

%Z omitted so the component's typical-impedance table is used for a 208Y/120 V secondary, then plug-in busway and a branch cable (help page example 3).

Given: 208 V three-phase, 500 kVA transformer, 480 V primary (infinite source), %Z not given. Path from the source: 600 A plug-in Cu busway, 7.62 m; 2/0 Cu in steel conduit, 15.24 m. C values from the Bussmann SPD tables (cables: three single conductors).

Full-load currentIFL = S × 1000 / (√3 × EL-L) = 500 × 1000 / (√3 × 208) = 1,387.9 A
Transformer let-through%Z = 3.35 % (not given → the component's typical-%Z table, nearest kVA and voltage class); ISC = IFL × 100 / %Z = 1,387.9 × 100 / 3.35 = 41,429 A (infinite primary)
Busbar 1600 A plug-in Cu busway, 7.62 m = 25.00 ft, C = 41,000: f = 1.732 × L × I / (C × E) = 1.732 × 25.00 × 41,429 / (41,000 × 208) = 0.2103; M = 1 / (1 + f) = 0.8262; I = 41,429 × 0.8262 = 34,229 A
Cable 11 × 2/0 Cu, 15.24 m = 50.00 ft, C = 10,755 (steel conduit, 600V): f = 1.732 × L × I / (C × n × E) = 1.732 × 50.00 × 34,229 / (10,755 × 1 × 208) = 1.3251; M = 1 / (1 + f) = 0.4301; I = 34,229 × 0.4301 = 14,722 A
Fault currentISC at the end of cable 1 = 14.722 kA rms symmetrical
ResultISC = 14.722 kA at the end of cable 1
ReferenceHand calculation, Bussmann point-to-point: %Z = 3.35 % (c0009 Table 1 typical value, 500 kVA, 480 V primary, 208Y/120 V secondary — the value the original help-page example used); I_FL = 500 kVA / (√3 × 208) = 1387.9 A; Isc = 1387.9 × 100/3.35 = 41 429 A; 600 A plug-in Cu busway C = 41 000 (Bussmann SPD Table 1), 7.62 m = 25 ft: f = 1.732 × 25 × 41 429 / (41 000 × 208) = 0.2104, I = 41 429 / 1.2104 = 34 229 A; 2/0 Cu steel conduit C = 10 755 (SPD Table 2), 15.24 m = 50 ft: f = 1.732 × 50 × 34 229 / (10 755 × 208) = 1.3251, I = 34 229 / 2.3251 = 14 722 A (validation case c0009-bussmann-500kva-208v-table-z-600a-busway)

Example 13 — 50 kA source — 1600 A Al feeder busway, 2 × 500 kcmil Al in PVC, four 25 HP motors

Aluminium busway and aluminium conductors in nonmagnetic conduit, plus motor contribution, against a 35 kA rating.

Given: 480 V three-phase, known fault level 50 kA rms. Path from the source: 1,600 A feeder Al busway, 20 m; 2 × 500 Al in nonmagnetic conduit, 40 m, motors 4 × 25 HP; equipment interrupting rating 35 kA. C values from the Bussmann SPD tables (cables: three single conductors).

SourceKnown fault level ISC = 50 kA = 50,000 A
Busbar 11,600 A feeder Al busway, 20 m = 65.62 ft, C = 90,900: f = 1.732 × L × I / (C × E) = 1.732 × 65.62 × 50,000 / (90,900 × 480) = 0.1302; M = 1 / (1 + f) = 0.8848; I = 50,000 × 0.8848 = 44,239 A
Cable 12 × 500 Al, 40 m = 131.23 ft, C = 21,391 (nonmagnetic conduit, 600V): f = 1.732 × L × I / (C × n × E) = 1.732 × 131.23 × 44,239 / (21,391 × 2 × 480) = 0.4897; M = 1 / (1 + f) = 0.6713; I = 44,239 × 0.6713 = 29,697 A
Motor full-load currentNEC Table 430.250, 460 V column: Σ quantity × FLA = 4 × 25 HP × 34 A = 136.0 A
Motor contributionBussmann Note 1: 4 × ΣFLA = 4 × 136.0 = 544 A; I = 29,697 A + 544 A = 30,241 A
Fault currentISC at the end of cable 1 = 30.241 kA rms symmetrical
Interrupting ratingNEC 110.9: ISC ≤ equipment rating → 30.241 vs limit 35.000 kA PASS
Result1/1 checks pass → PASS; ISC = 30.241 kA at the end of cable 1; governing: interrupting rating
ReferenceHand calculation, Bussmann point-to-point: 1600 A Al feeder busway C = 90 900 (Bussmann SPD Table 1), 20 m = 65.62 ft: f = 1.732 × 65.62 × 50 000 / (90 900 × 480) = 0.1302, I = 50 000 / 1.1302 = 44 239 A; 500 kcmil Al, three single conductors in nonmagnetic conduit, C = 21 391 (SPD Table 2, 600 V), 40 m = 131.23 ft, n = 2: f = 1.732 × 131.23 × 44 239 / (21 391 × 2 × 480) = 0.4897, I = 44 239 / 1.4897 = 29 697 A; motors 4 × 25 HP at 460 V = 34 A each (NEC Table 430.250), 4 × 136 = 544 A (SPD Note 1); total 30 241 A ≤ 35 kA (86.4 %) (validation case c0009-bussmann-aluminium-feeder-busway-nonmagnetic)

Example 14 — 4160 V — 20 kA source, 300 m of 4/0 Cu (5 kV column)

A medium-voltage system takes its C value from the 5 kV column of the SPD conductor table.

Given: 4,160 V three-phase, known fault level 20 kA rms. Path from the source: 4/0 Cu in steel conduit, 300 m. C values from the Bussmann SPD tables (cables: three single conductors).

SourceKnown fault level ISC = 20 kA = 20,000 A
Cable 11 × 4/0 Cu, 300 m = 984.25 ft, C = 13,606 (steel conduit, 5kV): f = 1.732 × L × I / (C × n × E) = 1.732 × 984.25 × 20,000 / (13,606 × 1 × 4,160) = 0.6024; M = 1 / (1 + f) = 0.6241; I = 20,000 × 0.6241 = 12,482 A
Fault currentISC at the end of cable 1 = 12.482 kA rms symmetrical
ResultISC = 12.482 kA at the end of cable 1
ReferenceHand calculation, Bussmann point-to-point: 4/0 Cu, three single conductors in steel conduit, 5 kV column C = 13 606 (Bussmann SPD Table 2); 300 m = 984.25 ft: f = 1.732 × 984.25 × 20 000 / (13 606 × 1 × 4160) = 0.6024; I = 20 000 / 1.6024 = 12 482 A (validation case c0009-bussmann-4160v-5kv-cable-column)

Example 15 — 500 kVA, 4160 → 480 V fed from 10 kA — then 50 ft of 4/0

A transformer with a known primary fault level (Bussmann second-transformer procedure): source and transformer impedances in series, then a feeder, against a 10 kA rating.

Given: 480 V three-phase, 500 kVA transformer, 4,160 V primary (10 kA available at the primary), 5.75 % Z. Path from the source: 4/0 Cu in steel conduit, 15.24 m; equipment interrupting rating 10 kA. C values from the Bussmann SPD tables (cables: three single conductors).

Full-load currentIFL = S × 1000 / (√3 × EL-L) = 500 × 1000 / (√3 × 480) = 601.4 A
Transformer f%Z = 5.75 % (nameplate); known primary fault IP = 10,000 A: f = IP × VP × 1.732 × %Z / (100 000 × kVA) = 10,000 × 4,160 × 1.732 × 5.75 / (100 000 × 500) = 8.2859
Transformer let-throughM = 1 / (1 + f) = 0.1077; ISC = (VP / VS) × M × IP = (4,160 / 480) × 0.1077 × 10,000 = 9,333 A
Cable 11 × 4/0 Cu, 15.24 m = 50.00 ft, C = 15,082 (steel conduit, 600V): f = 1.732 × L × I / (C × n × E) = 1.732 × 50.00 × 9,333 / (15,082 × 1 × 480) = 0.1116; M = 1 / (1 + f) = 0.8996; I = 9,333 × 0.8996 = 8,396 A
Fault currentISC at the end of cable 1 = 8.396 kA rms symmetrical
Interrupting ratingNEC 110.9: ISC ≤ equipment rating → 8.396 vs limit 10.000 kA PASS
Result1/1 checks pass → PASS; ISC = 8.396 kA at the end of cable 1; governing: interrupting rating
ReferenceBussmann SPD (Selecting Protective Devices, 2005), Point-to-Point Method, Procedure for second transformer in system: Step A f = I_sca(P) × V_P × 1.732 × %Z / (100 000 × kVA) = 10 000 × 4160 × 1.732 × 5.75 / (100 000 × 500) = 8.2859; Step B M = 1/(1 + f) = 0.10769; Step C I_sca(S) = (V_P/V_S) × M × I_sca(P) = (4160/480) × 0.10769 × 10 000 = 9333 A (ohmic check at 480 V: Z_source = 0.0031975 Ω + Z_T = 0.0575 × 480²/500 000 = 0.026496 Ω → 9333 A). 4/0 Cu steel conduit C = 15 082 (SPD Table 2), 15.24 m = 50 ft: f = 1.732 × 50 × 9333 / (15 082 × 480) = 0.1117, I = 9333 / 1.1117 = 8396 A ≤ 10 kA (84.0 %) (validation case c0009-bussmann-known-primary-fault-500kva)

Example 16 — Bussmann example — 300 kVA, 208 V, faults #1 and #2

The published Bussmann point-to-point example: 300 kVA 2 % (× 0.9 tolerance) transformer, 20 ft of 2 × 500 kcmil to the main panel, 20 ft of 2 AWG to the branch panel.

Given: 208 V three-phase, 300 kVA transformer, 480 V primary (infinite source), 1.8 % Z. Path from the source: 2 × 500 Cu in steel conduit, 6.096 m; 2 Cu in steel conduit, 6.096 m. C values from the Bussmann SPD tables (cables: three single conductors).

Full-load currentIFL = S × 1000 / (√3 × EL-L) = 300 × 1000 / (√3 × 208) = 832.7 A
Transformer let-through%Z = 1.8 % (nameplate); ISC = IFL × 100 / %Z = 832.7 × 100 / 1.8 = 46,262 A (infinite primary)
Cable 12 × 500 Cu, 6.096 m = 20.00 ft, C = 22,185 (steel conduit, 600V): f = 1.732 × L × I / (C × n × E) = 1.732 × 20.00 × 46,262 / (22,185 × 2 × 208) = 0.1736; M = 1 / (1 + f) = 0.8520; I = 46,262 × 0.8520 = 39,418 A
Cable 21 × 2 Cu, 6.096 m = 20.00 ft, C = 5,907 (steel conduit, 600V): f = 1.732 × L × I / (C × n × E) = 1.732 × 20.00 × 39,418 / (5,907 × 1 × 208) = 1.1113; M = 1 / (1 + f) = 0.4736; I = 39,418 × 0.4736 = 18,670 A
Fault currentISC at the end of cable 2 = 18.670 kA rms symmetrical
ResultISC = 18.670 kA at the end of cable 2
ReferenceCooper Bussmann, Point-to-Point Method example of 3-phase short-circuit calculation (Electrical Plan Review 2002; same example in SPD): I_FLA = 300 × 1000 / (208 × 1.732) = 833 A; multiplier = 100 / (0.9 × 2) = 55.55; Isca = 46 273 A (handbook, I_FL rounded) / 46 262 A unrounded; fault #1: f = 1.732 × 20 × 46 262 / (22 185 × 2 × 208) = 0.174, M = 0.852, I = 39 418 A (handbook 39 425 A); fault #2: f = 1.732 × 20 × 39 418 / (5907 × 1 × 208) = 1.11, M = 0.474, I = 18 670 A (handbook 18 687 A). (validation case c0009-bussmann-epr-example-300kva-208v)

Example 17 — IEC 60909 — 1.6 MVA, 415 V board, three-phase maximum

The full IEC 60909 maximum case on a radial network: c_max, network transformer correction K_T, κ, ip and Ith, cross-checked against the MVA method.

Given: IEC 60909 maximum case, three-phase fault at LV; 50 Hz. Buses: GRID (11 kV, infeed S''kQ = 250 MVA, X/R 10); LV (0.415 kV). Branches: TX1 GRID→LV transformer 1.6 MVA, 6 % Z, X/R 8, Dyn11. MVA-method cross-check requested.

Network feeder GRIDZQ = cQ UnQ² / S''kQ = 1.1 × 11² / 250 = 0.5324 Ω, split at X/R 10 → 0.05298 + j0.5298 Ω; referred to 0.415 kV × (0.415/11)² → 0.00007540 + j0.0007540 Ω
Transformer TX1ZT = 6 % × 11² / 1.6 = 4.537 Ω at X/R 8 → 0.5628 + j4.502 Ω; xT = 0.05954; KT = 0.95 cmax / (1 + 0.6 xT) = 0.95 × 1.1 / (1 + 0.6 × 0.05954) = 1.00896; KT ZT referred to 0.415 kV = 0.0008082 + j0.006466 Ω
Short-circuit impedanceZk = Z(1) = Σ = 0.0008836 + j0.007220 Ω (engine Thévenin: 0.0008840 + j0.007220 Ω); |Zk| = 0.007274 Ω, X/R = 8.171
Initial currentI''k = c Un / (√3 |Zk|) = 1.1 × 415 / (√3 × 0.007274) = 36.234 kA (c = cmax = 1.1)
Peak currentR/X = 0.1224; κ = 1.02 + 0.98 e−3R/X = 1.6988; ip = κ √2 I''k = 1.6988 × √2 × 36.234 = 87.05 kA
Thermal currentf = 50 Hz, Tk = 1 s, n = 1: m = [e4·50·Tk·ln(κ−1) − 1] / (2·50·Tk·ln(κ−1)) = 0.0279; Ith = I''k √(m + n) = 36.234 × √1.0279 = 36.736 kA
MVA-method checkNative MVA method (c = 1.0, no KT): 33.524 kA; (36.234 − 33.524) / 33.524 = +8.08 %, accepted band -5 … +15 % → 8.08 vs limit 15.00 % PASS
Result1/1 checks pass → PASS; I''k = 36.234 kA (three-phase, max case) at LV (0.415 kV); governing: mva method cross check
ReferenceHand calculation to IEC 60909-0:2016 (equivalent voltage source, c_max = 1.10 LV): Z_Q = 1.1 × 0.415² / 250 split at X/R 10 = 0.0000754 + j0.000754 Ω; Z_T = 6 % × 0.415² / 1.6 split at X/R 8 = 0.000801 + j0.006409 Ω; K_T = 0.95 × 1.1 / (1 + 0.6 × 0.05954) = 1.00896; Z_1 = Z_Q + K_T·Z_T = 0.000884 + j0.007220 Ω; Ik'' = 1.1 × 415 / (√3 × 0.007274) = 36.234 kA; R/X = 0.1224, κ = 1.02 + 0.98 e^(−3 × 0.1224) = 1.6988; ip = κ √2 Ik'' = 87.05 kA; m = [e^(4·50·1·ln 0.6988) − 1]/(2·50·1·ln 0.6988) = 0.0279, Ith = Ik'' √(m + 1) = 36.736 kA. MVA method (c = 1, no K_T): Z = 0.415²/250 (X/R 10) + Z_T → 33.524 kA; (36.234 − 33.524)/33.524 = +8.08 % (validation case psas-iec60909-lv-3ph-max)

Example 18 — IEC 60909 — line-to-line fault on the 415 V board

Line-to-line fault with Z_2 = Z_1: √3/2 of the three-phase current.

Given: IEC 60909 maximum case, line-to-line fault at LV; 50 Hz. Buses: GRID (11 kV, infeed S''kQ = 250 MVA, X/R 10); LV (0.415 kV). Branches: TX1 GRID→LV transformer 1.6 MVA, 6 % Z, X/R 8, Dyn11.

Network feeder GRIDZQ = cQ UnQ² / S''kQ = 1.1 × 11² / 250 = 0.5324 Ω, split at X/R 10 → 0.05298 + j0.5298 Ω; referred to 0.415 kV × (0.415/11)² → 0.00007540 + j0.0007540 Ω
Transformer TX1ZT = 6 % × 11² / 1.6 = 4.537 Ω at X/R 8 → 0.5628 + j4.502 Ω; xT = 0.05954; KT = 0.95 cmax / (1 + 0.6 xT) = 0.95 × 1.1 / (1 + 0.6 × 0.05954) = 1.00896; KT ZT referred to 0.415 kV = 0.0008082 + j0.006466 Ω
Short-circuit impedanceZk = Z(1) = Σ = 0.0008836 + j0.007220 Ω (engine Thévenin: 0.0008840 + j0.007220 Ω); |Zk| = 0.007274 Ω, X/R = 8.171
Initial currentZ(2) = Z(1): I''k2 = c Un / |2 Z(1)| = 1.1 × 415 / (2 × 0.007274) = 31.379 kA (= √3/2 × I''k3)
ResultI''k = 31.379 kA (line-to-line, max case) at LV (0.415 kV)
ReferenceHand calculation to IEC 60909-0: line-to-line Ik2'' = c Un / |2 Z1| = 1.1 × 415 / (2 × 0.007274) = 31.379 kA = (√3/2) × 36.234 kA (Z2 = Z1) (validation case psas-iec60909-lv-ll-max)

Example 19 — IEC 60909 — line-to-earth fault behind a Dyn11 transformer

Line-to-earth fault: the delta winding blocks the grid's zero sequence, so Z_0 is the corrected transformer impedance alone and Ik1'' exceeds Ik3''.

Given: IEC 60909 maximum case, line-to-earth fault at LV; 50 Hz. Buses: GRID (11 kV, infeed S''kQ = 250 MVA, X/R 10); LV (0.415 kV). Branches: TX1 GRID→LV transformer 1.6 MVA, 6 % Z, X/R 8, Dyn11.

Network feeder GRIDZQ = cQ UnQ² / S''kQ = 1.1 × 11² / 250 = 0.5324 Ω, split at X/R 10 → 0.05298 + j0.5298 Ω; referred to 0.415 kV × (0.415/11)² → 0.00007540 + j0.0007540 Ω
Transformer TX1ZT = 6 % × 11² / 1.6 = 4.537 Ω at X/R 8 → 0.5628 + j4.502 Ω; xT = 0.05954; KT = 0.95 cmax / (1 + 0.6 xT) = 0.95 × 1.1 / (1 + 0.6 × 0.05954) = 1.00896; KT ZT referred to 0.415 kV = 0.0008082 + j0.006466 Ω
Short-circuit impedanceZk = Z(1) = Σ = 0.0008836 + j0.007220 Ω (engine Thévenin: 0.0008840 + j0.007220 Ω); |Zk| = 0.007274 Ω, X/R = 8.171
Zero sequenceDelta primary winding: the grid's zero sequence is blocked, Z(0) = KT Z0T (Z0T = ZT) + 3 × line impedances on the LV side = 0.0008082 + j0.006466 Ω
Initial currentI''k1 = √3 c Un / |Z(1) + Z(2) + Z(0)|; |2Z(1) + Z(0)| = |0.002576 + j0.02091| = 0.02106 Ω = √3 × 1.1 × 415 / 0.02106 = 37.537 kA; earth current 3I0 = 37.537 kA
Fault powerS''k = √3 Un I''k = √3 × 0.415 × 37.537 = 26.98 MVA
ResultI''k = 37.537 kA (line-to-earth, max case) at LV (0.415 kV)
ReferenceHand calculation to IEC 60909-0:2016 clause 8.2.3 (Z_2 = Z_1; Dyn: Z_0 = K_T Z_0T, Z_0T = Z_T): Z_Q = 1.1 × 0.415² / 250 split at X/R 10 = 0.0000754 + j0.000754 Ω; Z_T = 6 % × 0.415² / 1.6 split at X/R 8 = 0.000801 + j0.006409 Ω; K_T = 0.95 × 1.1 / (1 + 0.6 × 0.05954) = 1.00896; Z_1 = Z_Q + K_T·Z_T = 0.000884 + j0.007220 Ω; Z_0 = 1.00896 × (0.000801 + j0.006409) = 0.000808 + j0.006466 Ω; Ik1'' = √3 c Un / |2Z_1 + Z_0| = √3 × 1.1 × 415 / |0.002576 + j0.020906| = 37.537 kA (1.036 × Ik3''); S = √3 × 0.415 × 37.537 = 26.98 MVA (validation case psas-iec60909-lv-lg-dyn11)

Example 20 — IEC 60909 — double line-to-earth fault on the 415 V board

Two-phase-to-earth fault by symmetrical components: the larger faulted-phase current and the earth current 3I0.

Given: IEC 60909 maximum case, double line-to-earth fault at LV; 50 Hz. Buses: GRID (11 kV, infeed S''kQ = 250 MVA, X/R 10); LV (0.415 kV). Branches: TX1 GRID→LV transformer 1.6 MVA, 6 % Z, X/R 8, Dyn11.

Network feeder GRIDZQ = cQ UnQ² / S''kQ = 1.1 × 11² / 250 = 0.5324 Ω, split at X/R 10 → 0.05298 + j0.5298 Ω; referred to 0.415 kV × (0.415/11)² → 0.00007540 + j0.0007540 Ω
Transformer TX1ZT = 6 % × 11² / 1.6 = 4.537 Ω at X/R 8 → 0.5628 + j4.502 Ω; xT = 0.05954; KT = 0.95 cmax / (1 + 0.6 xT) = 0.95 × 1.1 / (1 + 0.6 × 0.05954) = 1.00896; KT ZT referred to 0.415 kV = 0.0008082 + j0.006466 Ω
Short-circuit impedanceZk = Z(1) = Σ = 0.0008836 + j0.007220 Ω (engine Thévenin: 0.0008840 + j0.007220 Ω); |Zk| = 0.007274 Ω, X/R = 8.171
Zero sequenceDelta primary winding: the grid's zero sequence is blocked, Z(0) = KT Z0T (Z0T = ZT) + 3 × line impedances on the LV side = 0.0008082 + j0.006466 Ω
Initial currentI''k2EL2 = c Un |Z(0) − aZ(2)| / |Z(1)Z(2) + Z(1)Z(0) + Z(2)Z(0)|; |Z(1)Z(2) + Z(1)Z(0) + Z(2)Z(0)| = 0.0001477 Ω², |Z(0) − aZ(2)| = 0.01196 Ω, |Z(1) + 2Z(0)| = 0.02031 Ω → larger faulted-phase current 36.956 kA; I''kE2E = √3 c Un / |Z(1) + 2Z(0)| = 38.937 kA earth current
Fault powerS''k = √3 Un I''k = √3 × 0.415 × 36.956 = 26.56 MVA
ResultI''k = 36.956 kA (double line-to-earth, max case) at LV (0.415 kV)
ReferenceHand calculation to IEC 60909-0:2016 clause 8.2.2 (Z_2 = Z_1): Z_Q = 1.1 × 0.415² / 250 split at X/R 10 = 0.0000754 + j0.000754 Ω; Z_T = 6 % × 0.415² / 1.6 split at X/R 8 = 0.000801 + j0.006409 Ω; K_T = 0.95 × 1.1 / (1 + 0.6 × 0.05954) = 1.00896; Z_1 = Z_Q + K_T·Z_T = 0.000884 + j0.007220 Ω, Z_0 = 0.000808 + j0.006466 Ω; I''k2EL2 = c Un |Z_0 − a Z_2| / |Z_1Z_2 + Z_1Z_0 + Z_2Z_0| = 36.956 kA (L3: 36.901 kA); I''kE2E = √3 c Un / |Z_1 + 2Z_0| = √3 × 1.1 × 415 / |0.002500 + j0.020152| = 38.938 kA (validation case psas-iec60909-lv-llg-dyn11)

Example 21 — IEC 60909 — line-to-earth fault through 5 mΩ

A fault resistance adds 3R_F to the line-to-earth loop and cuts the current by almost a quarter.

Given: IEC 60909 maximum case, line-to-earth fault through RF = 0.005 Ω at LV; 50 Hz. Buses: GRID (11 kV, infeed S''kQ = 250 MVA, X/R 10); LV (0.415 kV). Branches: TX1 GRID→LV transformer 1.6 MVA, 6 % Z, X/R 8, Dyn11.

Network feeder GRIDZQ = cQ UnQ² / S''kQ = 1.1 × 11² / 250 = 0.5324 Ω, split at X/R 10 → 0.05298 + j0.5298 Ω; referred to 0.415 kV × (0.415/11)² → 0.00007540 + j0.0007540 Ω
Transformer TX1ZT = 6 % × 11² / 1.6 = 4.537 Ω at X/R 8 → 0.5628 + j4.502 Ω; xT = 0.05954; KT = 0.95 cmax / (1 + 0.6 xT) = 0.95 × 1.1 / (1 + 0.6 × 0.05954) = 1.00896; KT ZT referred to 0.415 kV = 0.0008082 + j0.006466 Ω
Short-circuit impedanceZk = Z(1) = Σ = 0.0008836 + j0.007220 Ω (engine Thévenin: 0.005884 + j0.007220 Ω including RF); |Zk| = 0.007274 Ω, X/R = 8.171
Zero sequenceDelta primary winding: the grid's zero sequence is blocked, Z(0) = KT Z0T (Z0T = ZT) + 3 × line impedances on the LV side = 0.0008082 + j0.006466 Ω
Initial currentI''k1 = √3 c Un / |Z(1) + Z(2) + Z(0) + 3RF|; |2Z(1) + Z(0) + 3RF| = |0.01758 + j0.02091| = 0.02731 Ω = √3 × 1.1 × 415 / 0.02731 = 28.950 kA; earth current 3I0 = 28.950 kA
Fault powerS''k = √3 Un I''k = √3 × 0.415 × 28.950 = 20.81 MVA
ResultI''k = 28.950 kA (line-to-earth, max case) at LV (0.415 kV)
ReferenceHand calculation, symmetrical components: Z_Q = 1.1 × 0.415² / 250 split at X/R 10 = 0.0000754 + j0.000754 Ω; Z_T = 6 % × 0.415² / 1.6 split at X/R 8 = 0.000801 + j0.006409 Ω; K_T = 0.95 × 1.1 / (1 + 0.6 × 0.05954) = 1.00896; Z_1 = Z_Q + K_T·Z_T = 0.000884 + j0.007220 Ω, Z_0 = 0.000808 + j0.006466 Ω; Ik1'' = √3 c Un / |2Z_1 + Z_0 + 3R_F| = √3 × 1.1 × 415 / |0.017576 + j0.020906| = 28.950 kA (R_F = 0.005 Ω) (validation case psas-iec60909-lv-lg-fault-resistance)

Example 22 — IEC 60909 — minimum case on the 415 V board

Minimum case: the weaker grid (150 MVA, X/R 8) and c_min = 0.95 at LV give the current protection must still detect.

Given: IEC 60909 minimum case, three-phase fault at LV; 50 Hz. Buses: GRID (11 kV, infeed S''kQ = 250 MVA, X/R 10 (min 150 MVA, X/R 8)); LV (0.415 kV). Branches: TX1 GRID→LV transformer 1.6 MVA, 6 % Z, X/R 8, Dyn11.

Network feeder GRIDZQ = cQ UnQ² / S''kQ = 1 × 11² / 150 = 0.8067 Ω, split at X/R 8 → 0.1001 + j0.8004 Ω; referred to 0.415 kV × (0.415/11)² → 0.0001424 + j0.001139 Ω
Transformer TX1ZT = 6 % × 11² / 1.6 = 4.537 Ω at X/R 8 → 0.5628 + j4.502 Ω; xT = 0.05954; KT = 0.95 cmax / (1 + 0.6 xT) = 0.95 × 1.1 / (1 + 0.6 × 0.05954) = 1.00896; KT ZT referred to 0.415 kV = 0.0008082 + j0.006466 Ω
Short-circuit impedanceZk = Z(1) = Σ = 0.0009507 + j0.007605 Ω (engine Thévenin: 0.0009510 + j0.007605 Ω); |Zk| = 0.007664 Ω, X/R = 8.000
Initial currentI''k = c Un / (√3 |Zk|) = 0.95 × 415 / (√3 × 0.007664) = 29.698 kA (c = cmin = 0.95)
Peak currentR/X = 0.1250; κ = 1.02 + 0.98 e−3R/X = 1.6935; ip = κ √2 I''k = 1.6935 × √2 × 29.698 = 71.13 kA
Thermal currentf = 50 Hz, Tk = 1 s, n = 1: m = [e4·50·Tk·ln(κ−1) − 1] / (2·50·Tk·ln(κ−1)) = 0.0273; Ith = I''k √(m + n) = 29.698 × √1.0273 = 30.101 kA
ResultI''k = 29.698 kA (three-phase, min case) at LV (0.415 kV)
ReferenceHand calculation to IEC 60909-0:2016: Z_Qmin = c_Q,min U² / S''kQ,min = 1.0 × 0.415² / 150 split at X/R 8 = 0.0001424 + j0.001139 Ω; K_T = 1.00896 (clause 6.3.3, c_max of the LV side); Z = Z_Qmin + K_T Z_T = 0.000951 + j0.007605 Ω; Ik''min = c_min Un / (√3 |Z|) = 0.95 × 415 / (√3 × 0.007664) = 29.698 kA; κ = 1.02 + 0.98 e^(−3 × 0.1250) = 1.6935, ip = 71.13 kA (validation case psas-iec60909-lv-3ph-min)

Example 23 — IEC 60909 — fault at the 11 kV infeed

A fault at the network feeder's own connection point returns its declared fault level.

Given: IEC 60909 maximum case, three-phase fault at GRID; 50 Hz. Buses: GRID (11 kV, infeed S''kQ = 250 MVA, X/R 10); LV (0.415 kV). Branches: TX1 GRID→LV transformer 1.6 MVA, 6 % Z, X/R 8, Dyn.

Network feeder GRIDZQ = cQ UnQ² / S''kQ = 1.1 × 11² / 250 = 0.5324 Ω, split at X/R 10 → 0.05298 + j0.5298 Ω
Short-circuit impedanceZk = Z(1) = Σ = 0.05298 + j0.5298 Ω (engine Thévenin: 0.05298 + j0.5298 Ω); |Zk| = 0.5324 Ω, X/R = 10.000
Initial currentI''k = c Un / (√3 |Zk|) = 1.1 × 11,000 / (√3 × 0.5324) = 13.122 kA (c = cmax = 1.1)
Peak currentR/X = 0.1000; κ = 1.02 + 0.98 e−3R/X = 1.7460; ip = κ √2 I''k = 1.7460 × √2 × 13.122 = 32.40 kA
Thermal currentf = 50 Hz, Tk = 1 s, n = 1: m = [e4·50·Tk·ln(κ−1) − 1] / (2·50·Tk·ln(κ−1)) = 0.0341; Ith = I''k √(m + n) = 13.122 × √1.0341 = 13.344 kA
ResultI''k = 13.122 kA (three-phase, max case) at GRID (11 kV)
ReferenceHand calculation: at the feeder connection point Ik''Q = S''kQ / (√3 Un) = 250 MVA / (√3 × 11 kV) = 13.122 kA (IEC 60909-0 network feeder definition, Z_Q = c U²/S''kQ cancels c); R/X of the feeder 0.1 → X/R 10 (validation case psas-iec60909-source-bus)

Example 24 — IEC 60909 — 60 Hz, 13.8 kV cable, 2.5 MVA, 480 V MCC rated 30 kA

A three-element radial path (HV cable, transformer, LV cable) referred to 480 V; the MCC's 30 kA rating fails while the MVA cross-check agrees.

Given: IEC 60909 maximum case, three-phase fault at MCC; 60 Hz. Buses: UTIL (13.8 kV, infeed S''kQ = 500 MVA, X/R 15); HV (13.8 kV); SWBD (0.48 kV); MCC (0.48 kV, rated 30 kA). Branches: C1 UTIL–HV cable 1.5 km × (0.15 + j0.12) Ω/km; TX1 HV→SWBD transformer 2.5 MVA, 5.75 % Z, X/R 7, Dyn1; C2 SWBD–MCC cable 0.03 km × (0.0754 + j0.0798) Ω/km. MVA-method cross-check requested.

Network feeder UTILZQ = cQ UnQ² / S''kQ = 1.1 × 13.8² / 500 = 0.4190 Ω, split at X/R 15 → 0.02787 + j0.4180 Ω; referred to 0.48 kV × (0.48/13.8)² → 0.00003372 + j0.0005058 Ω
Cable C1ZL = (0.15 + j0.12) × 1.5 km = 0.2250 + j0.1800 Ω; referred to 0.48 kV → 0.0002722 + j0.0002178 Ω (conductor resistance at 20 °C)
Transformer TX1ZT = 5.75 % × 13.8² / 2.5 = 4.380 Ω at X/R 7 → 0.6194 + j4.336 Ω; xT = 0.05692; KT = 0.95 cmax / (1 + 0.6 xT) = 0.95 × 1.1 / (1 + 0.6 × 0.05692) = 1.01049; KT ZT referred to 0.48 kV = 0.0007573 + j0.005301 Ω
Cable C2ZL = (0.0754 + j0.0798) × 0.03 km = 0.002262 + j0.002394 Ω (conductor resistance at 20 °C)
Short-circuit impedanceZk = Z(1) = Σ = 0.003325 + j0.008418 Ω (engine Thévenin: 0.003325 + j0.008418 Ω); |Zk| = 0.009051 Ω, X/R = 2.532
Initial currentI''k = c Un / (√3 |Zk|) = 1.1 × 480 / (√3 × 0.009051) = 33.679 kA (c = cmax = 1.1)
Peak currentR/X = 0.3950; κ = 1.02 + 0.98 e−3R/X = 1.3196; ip = κ √2 I''k = 1.3196 × √2 × 33.679 = 62.85 kA
Thermal currentf = 60 Hz, Tk = 1 s, n = 1: m = [e4·60·Tk·ln(κ−1) − 1] / (2·60·Tk·ln(κ−1)) = 0.0073; Ith = I''k √(m + n) = 33.679 × √1.0073 = 33.802 kA
Equipment ratingI''k ≤ short-circuit rating of MCC → 33.679 vs limit 30.000 kA FAIL
MVA-method checkNative MVA method (c = 1.0, no KT): 30.952 kA; (33.679 − 30.952) / 30.952 = +8.81 %, accepted band -5 … +15 % → 8.81 vs limit 15.00 % PASS
Result1/2 checks pass → FAIL; I''k = 33.679 kA (three-phase, max case) at MCC (0.48 kV); governing: fault rating
ReferenceHand calculation to IEC 60909-0:2016, referred to 0.48 kV with t² = (0.48/13.8)² = 0.0012098: Z_Q = 1.1 × 13.8²/500 (X/R 15) = 0.027869 + j0.41804 Ω → × t² = 0.0000337 + j0.0005058 Ω; HV cable 1.5 km × (0.15 + j0.12) = 0.225 + j0.180 Ω → × t² = 0.0002722 + j0.0002178 Ω; Z_T = 5.75 % × 0.48²/2.5 (X/R 7) = 0.0007494 + j0.0052459 Ω, K_T = 0.95 × 1.1 / (1 + 0.6 × 0.05692) = 1.01049 → K_T Z_T = 0.0007573 + j0.0053010 Ω; LV cable 0.03 km × (0.0754 + j0.0798) = 0.002262 + j0.002394 Ω; Z_k = 0.003325 + j0.008418 Ω; Ik'' = 1.1 × 480 / (√3 × 0.009051) = 33.679 kA > 30 kA (112.3 %); R/X = 0.3950, κ = 1.02 + 0.98 e^(−1.185) = 1.3196, ip = κ √2 Ik'' = 62.85 kA (κ by method C equals the radial R/X value); Ith at f = 60 Hz, T_k = 1 s: m = [e^(4·60·1·ln 0.31964) − 1]/(2·60·1·ln 0.31964) = 0.00731, Ith = Ik'' √(m + 1) = 33.802 kA. MVA method (c = 1, no K_T) = 30.952 kA; difference +8.81 % (validation case psas-iec60909-60hz-mcc-rating-fail)

Example 25 — IEC 60909 — two 1 MVA transformers in parallel, MVA cross-check

Two identical transformers between the same buses: both carry fault current in the IEC 60909 solve and in the MVA-method cross-check.

Given: IEC 60909 maximum case, three-phase fault at LV; 50 Hz. Buses: GRID (11 kV, infeed S''kQ = 250 MVA, X/R 10); LV (0.415 kV). Branches: TX1 GRID→LV transformer 1 MVA, 6 % Z, X/R 8, Dyn11; TX2 GRID→LV transformer 1 MVA, 6 % Z, X/R 8, Dyn11. MVA-method cross-check requested.

Short-circuit impedanceThévenin impedance at LV from the network solve: Zk = 0.0007220 + j0.005927 Ω, X/R = 8.209
Initial currentI''k = c Un / (√3 |Zk|) = 1.1 × 415 / (√3 × 0.005971) = 44.143 kA (c = cmax = 1.1)
Peak currentR/X = 0.1218; κ = 1.02 + 0.98 e−3R/X = 1.7000; ip = κ √2 I''k = 1.7000 × √2 × 44.143 = 106.13 kA
Thermal currentf = 50 Hz, Tk = 1 s, n = 1: m = [e4·50·Tk·ln(κ−1) − 1] / (2·50·Tk·ln(κ−1)) = 0.0280; Ith = I''k √(m + n) = 44.143 × √1.0280 = 44.757 kA
MVA-method checkNative MVA method (c = 1.0, no KT): 40.919 kA; (44.143 − 40.919) / 40.919 = +7.88 %, accepted band -5 … +15 % → 7.88 vs limit 15.00 % PASS
Result1/1 checks pass → PASS; I''k = 44.143 kA (three-phase, max case) at LV (0.415 kV); governing: mva method cross check
ReferenceHand calculation to IEC 60909-0:2016 at 0.415 kV: Z_Q = 1.1 × 0.415²/250 (X/R 10) = 0.0000754 + j0.000754 Ω; each Z_T = 6 % × 0.415²/1.0 (X/R 8) = 0.001282 + j0.010254 Ω, K_T = 0.95 × 1.1/(1 + 0.6 × 0.05954) = 1.00896; two in parallel: K_T·Z_T/2 = 0.000647 + j0.005173 Ω; Z_1 = 0.000722 + j0.005927 Ω; Ik'' = 1.1 × 415/(√3 × 0.005971) = 44.143 kA; R/X = 0.1218, κ = 1.700, ip = 106.13 kA. MVA method (c = 1, no K_T): Z = 0.415²/250 (X/R 10) + Z_T/2 → 40.919 kA; (44.143 − 40.919)/40.919 = +7.88 % (validation case psas-iec60909-parallel-transformers-verify)

Standards & References

  • NFPA 70 (NEC) — National Electrical Code, Article 110.9 (Interrupting Rating), Article 110.10 (Circuit Impedance, Short-Circuit Current Ratings, and Other Characteristics)
  • IEEE Std 141 (Red Book) — Recommended Practice for Electric Power Distribution for Industrial Plants — Chapter 4: Short-Circuit Current Calculations
  • IEEE Std 242 (Buff Book) — Recommended Practice for Protection and Coordination of Industrial and Commercial Power Systems
  • IEC 60909 — Short-circuit currents in three-phase a.c. systems — calculation of currents
  • Cooper Bussmann SPD — Selecting Protective Devices (2005), Short-Circuit Current Calculations — Point-to-Point Method: Table 1 (busway C values), Table 2 (conductor C values), Note 1 (motor contribution 4 × FLA)
  • NFPA 70 (NEC) Table 430.250 — Full-load current, three-phase alternating-current motors

Tips & Limitations

Tips
  • When transformer impedance is unknown the calculator uses a table value, or 5.75 % where the kVA / voltage combination is not tabulated. Nameplate Z% is always preferred for accuracy.
  • Parallel conductors significantly increase fault current — the C value is multiplied by the number of parallel sets (\(C \times n\) in the f-factor), reducing the effective impedance.
  • Motor contribution is 4 × FLA (Bussmann; 4–6 × is commonly accepted). For large motor loads it can be a significant share of the total available fault current; a warning is raised above 20 %.
  • The point-to-point method tends to produce slightly conservative (higher) fault current values compared to per-unit or impedance methods, which provides a safety margin.
  • Always verify that equipment interrupting ratings exceed the calculated available fault current.
Limitations
  • Three-phase bolted faults only — this calculator does not compute single-phase, line-to-line, or line-to-ground fault currents (the MCP tool's IEC 60909 mode='network' does).
  • Steady-state symmetrical RMS — asymmetric (DC offset) and peak values are not calculated. Multiply symmetrical RMS by 1.6–2.3 for estimated peak.
  • No arc resistance — arcing fault current is typically 50–85% of bolted fault current; use the arc flash calculator (IEEE 1584) for arcing values.
  • Voltage classes — the transformer %Z table is looked up by voltage class (e.g. 460 V → 480 V); cable C values use the 600 V, 5 kV or 15 kV column for the system voltage.
  • Three single conductors only — C values are for three single conductors in steel or nonmagnetic conduit; three-conductor cable and free-air runs are not modelled.
  • Table coverage — a cable, busway or motor that the Bussmann / NEC tables do not cover (e.g. 12 AWG at 5 kV, 4000 A plug-in busway, motors on a 400 V system) is rejected with a message rather than estimated.

Developer Guide

Architecture

File layout:

  • components/c0009_short_circuit/component.py — Component registration, PARAMETER_SCHEMA, ShortCircuitCalculator class
  • components/c0009_short_circuit/calculations.py — Pure math functions, lookup tables (4 tables), input checks, and main calculate_short_circuit() function (no Django dependencies)
  • components/c0009_short_circuit/equations.py — build_equations() for PDF report generation (MathJax articles)
  • components/c0009_short_circuit/excel_generator.py — Excel report output

The calculator class is ShortCircuitCalculator, which extends ElectricalCalculator. It delegates to calculate_short_circuit(params) in calculations.py.

Lookup tables in calculations.py:

  • TRANSFORMER_IMPEDANCE_TABLE / LARGE_TRANSFORMER_IMPEDANCE — kVA → primary → secondary → Z%
  • CONDUCTOR_C_VALUES — material → conduit → size → (600 V, 5 kV, 15 kV) C, Bussmann SPD Table 2
  • BUSWAY_C_VALUES — ampacity → type → material → C, Bussmann SPD Table 1
  • MOTOR_FLC_TABLE — HP → rated voltage → FLA (A), NEC Table 430.250; MOTOR_CONTRIBUTION_MULTIPLIER = 4

Code Paths

Key functions in calculations.py:

Function Purpose
calculate_short_circuit(params) Main entry point — orchestrates all steps and returns full results dict
calculate_transformer_contribution(kva, primary_v, secondary_v, ...) Transformer FLC and Isc (Method 1 or Method 2)
calculate_cable_contribution(voltage, wire_size, length_ft, ...) Cable f-factor, 1 + f divisor, downstream fault current
calculate_busbar_contribution(voltage, ampacity, length_ft, ...) Busway f-factor, 1 + f divisor, downstream fault current
calculate_motor_contribution(motors, voltage) 4 × the sum of motor FLAs (NEC Table 430.250)
normalize_voltage(voltage) Map numeric voltage to nearest standard table key (e.g., 460 → '480V')
get_transformer_impedance(kva, primary_v, secondary_v) Table lookup for Z%: closest kVA, primary voltage class, secondary system designation
get_m_multiplier(f_factor) Returns the divisor 1 + f (fault current is divided by it, i.e. multiplied by M = 1/(1 + f))
get_conductor_c_value(wire_size, material, conduit, voltage) Conductor C value lookup (None where Bussmann publishes no value)
get_busway_c_value(ampacity, busway_type, material) Busway C value lookup (exact rating, None where not published)
get_motor_fla(hp, voltage) Motor full-load current, NEC Table 430.250 column for the system voltage
validate_inputs(params) Method fields and table coverage; messages returned through validate_parameters()

Request flow:

  1. View passes validated parameters to ShortCircuitCalculator(params)
  2. validate_parameters() checks the method's required fields and that every cable, busway and motor is covered by the tables
  3. calculate() delegates to calculate_short_circuit(params)
  4. Step 1: Transformer or source fault establishes initial Isc
  5. Step 2: Each cable in sequence reduces Isc via f/M
  6. Step 3: Each busbar in sequence reduces Isc via f/M
  7. Step 4: 4 × the motor FLAs are added to the final Isc
  8. Results dict returned with all intermediate steps and warnings

Testing

Running tests:

pytest calcs/test_c0009_short_circuit.py -v

How to extend:

  1. Add new wire sizes by extending CONDUCTOR_C_VALUES and the wire_size enum in PARAMETER_SCHEMA.
  2. Add busway sizes by extending BUSWAY_C_VALUES and the ampacity enum (values from Bussmann Table 1 only).
  3. Add motor sizes by extending MOTOR_FLC_TABLE (from NEC Table 430.250) and the hp enum.
  4. Write tests covering new table entries and edge cases (table boundaries).