Arc Flash Calculator

Methods and model range

IEEE 1584-2018 applies to three-phase AC equipment within the ranges below. Inputs outside them are still calculated, with a warning, since the equations are then extrapolated. For voltages above above 15 kV, the calculator uses the Lee method, which assumes the arcing current equals the bolted fault current and is generally very conservative.

Typical equipment values

IEEE 1584-2018 typical bus gaps, enclosure sizes and working distances. Shallow and deep enclosures are entered as 8" and 12" deep, since only whether the depth is 8" or less matters. Use the actual equipment data when available.

Arcing current

• V_oc is the open-circuit (system) voltage in kV, I_bf the bolted fault current in kA and G the gap in mm.
• Coefficients k₁ to k₁₀ depend on the electrode configuration and are given at 600 V, 2.7 kV and 14.3 kV (IEEE 1584-2018 Table 1).

Intermediate arcing current (Eq. 1)

I_arc,Voc = 10^{(k₁ + k₂ lg I_bf + k₃ lg G)} × (k₄I_bf⁶ + k₅I_bf⁵ + k₆I_bf⁴ + k₇I_bf³ + k₈I_bf² + k₉I_bf + k₁₀)
for V_oc = 600 V, 2.7 kV and 14.3 kV

Final arcing current, 600 V and below (Eq. 25)

I_arc = 1 / √[(0.6/V_oc)² × (1/I_arc,600² − (0.6² − V_oc²) / (0.6² × I_bf²))]

Final arcing current, above 600 V (Eq. 16 to 18)

I₁ = (I_arc,2700 − I_arc,600) / 2.1 × (V_oc − 2.7) + I_arc,2700
I₂ = (I_arc,14300 − I_arc,2700) / 11.6 × (V_oc − 14.3) + I_arc,14300
I₃ = I₁(2.7 − V_oc) / 2.1 + I₂(V_oc − 0.6) / 2.1
I_arc = I₃ up to 2.7 kV, I₂ above

Reduced arcing current (Eq. 2)

VarCf = k₁V_oc⁶ + k₂V_oc⁵ + k₃V_oc⁴ + k₄V_oc³ + k₅V_oc² + k₆V_oc + k₇ (Table 2)
I_arc,min = I_arc × (1 − 0.5 × VarCf)
Above 600 V, each intermediate arcing current is reduced before interpolation

Enclosure size correction factor

Applies to VCB, VCBB and HCB. Open-air configurations use CF = 1. Dimensions are in mm unless noted.

Enclosure type

Shallow when V_oc < 600 V, height and width < 508 mm (20 in) and depth ≤ 203.2 mm (8 in)
Typical otherwise

Equivalent width and height in inches (Table 6, Eq. 11 and 12)

Below 508 mm: 20 in (typical) or 0.03937 × d (shallow)
508 to 660.4 mm: 0.03937 × d
660.4 to 1244.6 mm: [660.4 + (d − 660.4)(V_oc + A) / B] / 25.4
Above 1244.6 mm: as at 1244.6 mm
VCB height above 660.4 mm: 0.03937 × d, at most 49 in

Equivalent enclosure size (Eq. 13)

EES = (H₁ + W₁) / 2
A, B = 4, 20 (VCB); 10, 24 (VCBB); 10, 22 (HCB)

Correction factor (Eq. 14 and 15)

Typical: CF = b₁EES² + b₂EES + b₃
Shallow: CF = 1 / (b₁EES² + b₂EES + b₃) (Table 7)

Incident energy and arc flash boundary

• T is the arcing duration in ms and D the distance in mm. Coefficients k₁ to k₁₃ are given at 600 V, 2.7 kV and 14.3 kV (Tables 3 to 5). k₁₂ is the distance exponent.
• At 600 V and below, the k₃ term uses I_arc,600 and the k₁₃ term the final I_arc, even for the reduced arcing current. For above 600 V, both use the intermediate arcing current at each voltage.

Incident energy (Eq. 3 to 6)

E = 12.552/50 × T × 10^{X + k₁₂ lg D} J/cm²
X = k₁ + k₂ lg G + k₃I_arc / (k₄I_bf⁷ + k₅I_bf⁶ + … + k₁₀I_bf) + k₁₁ lg I_bf + k₁₃ lg I_arc + lg(1/CF)
E (cal/cm²) = E (J/cm²) / 4.184 = 0.06 × T × 10^{X + k₁₂ lg D}

Arc flash boundary (Eq. 7 to 10)

AFB = 10^{[X − lg(20/T)] / (−k₁₂)} mm
20/T is the incident energy term at 1.2 cal/cm²

Distance for a target incident energy

D_target = 10^{[X − lg(E_target / (0.06 T))] / (−k₁₂)} mm
E_target in cal/cm²; at 1.2 cal/cm² this is the arc flash boundary

Above 600 V (Eq. 19 to 24)

E, AFB and D_target are calculated at 600 V, 2.7 kV and 14.3 kV, then interpolated to V_oc with the same equations as the arcing current

Each case uses its own arcing current and arcing duration. The case with the larger incident energy at the working distance governs, and its arc flash boundary is reported.

Lee method (above 15 kV)

V in kV, I_bf in kA, t in seconds and D in mm. The arcing current is taken as the bolted fault current, and the gap, electrode configuration and enclosure are not used.

Incident energy

E = 2.142 × 10⁶ × V × I_bf × t / D² J/cm²

Arc flash boundary and target distance

AFB = √(2.142 × 10⁶ × V × I_bf × t / E_B) mm
E_B = 1.2 cal/cm² (5.02 J/cm²), or the target incident energy

Worst-case boundary from a known incident energy

With the incident energy E_D known at working distance D, everything in X cancels at 600 V and below, so the boundary depends only on the distance exponent. A smaller exponent means energy falls off more slowly with distance, giving a larger boundary.

600 V and below

AFB = D × (E_D / 1.2)^{1 / −k₁₂}
VCB and VOA share the smallest exponent, 1.598 which leads to the governing worst case

Above 600 V

E is interpolated from three distance exponents, and supsequently also depends on bolted fault current. The calculator searches bolted fault currents from 200A to 65kA (the IEEE 1584-2018 model range) and arcing durations up to 2s, using both the full and reduced arcing currents, and reports the largest boundary. It uses the smallest typical enclosure, since that enclosure produces the known incident energy in the shortest arcing duration, which keeps the most cases within the 2s limit.

Lee method

AFB = D × (E_D / 1.2)^1/2
IEEE 1584-2018 gives no typical working distances above 15 kV. Therefore, the typical working distance of 36" for 15 kV equipment is utilized. Custom working distances can be applied as well.