About Arc-Flash Calculator
The arc-flash calculator estimates the thermal hazard of an arcing fault using the IEEE 1584-2002 simplified empirical equations. It predicts the arcing current from the available bolted three-phase fault current, the normalised and actual incident energy at the worker working distance, the arc-flash boundary where the energy drops to 1.2 cal/cm^2, and the NFPA 70E personal protective equipment (PPE) category implied by that energy.
This tool implements the IEEE 1584-2002 model and is intended for estimation and learning only. It is valid for systems from 0.208 kV to 15 kV with bolted fault currents of roughly 0.7 to 106 kA and gaps of 13 to 152 mm. For actual equipment labelling, perform a full arc-flash study to the current standard (IEEE 1584-2018) together with NFPA 70E, using verified protective-device clearing times and an accurate short-circuit model.
How It Works
- Enter the system voltage (kV), the bolted three-phase fault current (kA), and select the equipment class, which sets the gap, distance exponent x, and the box/open-air constants K and K1.
- Choose grounded or ungrounded, which sets the K2 constant, and enter the arcing time t (the protective device clearing time plus breaker opening time).
- The model computes arcing current Ia, then the normalised incident energy En at 610 mm and 0.2 s, and scales it to the working distance and arc time: E = 4.184 * Cf * En * (t/0.2) * (610^x / D^x), with Cf = 1.5 at/below 1 kV and 1.0 above 1 kV.
- The arc-flash boundary is solved for E = 1.2 cal/cm^2, and the incident energy is mapped to an NFPA 70E PPE category.
Worked Example
A 600 V (0.6 kV) metal-clad switchgear lineup has a bolted fault current of 20 kA, a 32 mm conductor gap, a grounded system, a working distance of 455 mm, and a 0.2 s arcing time (x = 1.473, K = -0.097, K1 = -0.555, K2 = -0.113). The arcing current is Ia = 14.08 kA. The normalised energy En = 4.064 J/cm^2, and the incident energy E = 4.184 * 1.5 * 4.064 * (0.2/0.2) * (610^1.473 / 455^1.473) = 39.28 cal/cm^2. The arc-flash boundary is 4859 mm (4.86 m). At 39.3 cal/cm^2 the hazard is PPE category 4. These values match the published IEEE 1584-2002 worked example.
Formulas
- Arcing current (IEEE 1584-2002)
lg(Ia) = K + 0.662*lg(Ibf) + 0.0966*V + 0.000526*G + 0.5588*V*lg(Ibf) - 0.00304*G*lg(Ibf)- Normalised incident energy
lg(En) = K1 + K2 + 1.081*lg(Ia) + 0.0011*G- Incident energy and arc-flash boundary
E = 4.184*Cf*En*(t/0.2)*(610^x / D^x); DB = [4.184*Cf*En*(t/0.2)*610^x / 1.2]^(1/x)
Standards & References
- IEEE Std 1584-2002 (Guide for Performing Arc-Flash Hazard Calculations)
- IEEE Std 1584-2018 (current revision)
- NFPA 70E (Standard for Electrical Safety in the Workplace)
Frequently Asked Questions
Which standard does this calculator use?
It implements the IEEE 1584-2002 simplified empirical model. The current revision is IEEE 1584-2018, which uses a different and more accurate set of equations. Use this tool for estimation and study; use a full 2018 analysis for actual arc-flash labels.
Why does a shorter arcing time matter so much?
Incident energy is directly proportional to arcing time, E ~ t. Halving the protective-device clearing time halves the incident energy and shrinks the arc-flash boundary, so fast-acting protection is one of the most effective ways to reduce the hazard.
What is the arc-flash boundary?
It is the distance from the arc source at which the incident energy falls to 1.2 cal/cm^2, the threshold for a second-degree (just-curable) burn on bare skin. Anyone closer than this boundary must wear arc-rated PPE rated for the calculated incident energy.
How do I get the arcing time?
The arcing time is the total fault duration: the protective device time-current clearing time at the arcing current plus the breaker mechanism opening time. Read it from the device coordination study at the predicted arcing current, not the bolted fault current.