Arc Flash Incident Energy (Lee Method)
Computes arc flash incident energy by the Ralph Lee method, E = 5.12×10⁵ × V × I_bf × t ÷ d², with voltage in kV, bolted fault current in kA, fault clearing time in seconds and working distance in millimetres. Lee's method models an open-air arc as an ideal radiant heat source, ignoring the energy an enclosure reflects back; IEEE 1584 therefore keeps it only as the legacy model, recommended for open-air arcs and for voltages above 15 kV where the empirical equations do not apply. The result in cal/cm² sets the PPE category: 1.2 cal/cm² is the second-degree burn threshold and the value that bounds the arc flash boundary. The cal/cm² form is adopted (the J/cm² variant uses 2.142×10⁶ and is 4.184 times larger). Enter the voltage, the fault current, the clearing time and the working distance.
Result
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Lee method: incident energy and the 1.2 cal/cm² line
An arc flash study comes last, after the short-circuit calculation and protection coordination. Holding the fault current and the clearing time, the engineer has to turn both into calories per square centimetre at the face of whoever opens the panel. That figure becomes the label on the door, sets the clothing and marks the boundary nobody crosses unprotected. Underestimate it and an electrician in an 8 cal/cm² shirt ends up in front of a 25 cal enclosure. Overestimate it badly and maintenance gets buried in 40 cal gear nobody can wear through a six-hour shift.
Ralph Lee's equation reads E = 5.12×10⁵ × V × I_bf × t ÷ d², with voltage in kV, bolted fault current in kA, clearing time in seconds and working distance in millimetres. It treats the arc as an ideal radiant point source in open air, at the maximum power transfer condition where the arc drops half the system voltage. The cal/cm² form appears here; the joule variant carries 2.142×10⁶ and returns a value 4.184 times larger, a mix-up that has produced wrong labels. Defaults of 480 V, 20 kA, 200 ms and 455 mm yield 4.748 cal/cm². The reference threshold sits at 1.2 cal/cm², the energy causing a second-degree burn on bare skin; NFPA 70E clothing classes land at 4, 8, 25 and 40 cal/cm².
Lee models no enclosure. An arc inside a cubicle throws part of its energy back out through the opening, which is why IEEE 1584 replaced this calculation with test-derived equations across 208 V to 15 kV, keeping Lee for open-air arcs and for voltages beyond the tested range. It further assumes every amp of bolted fault current turns into arcing current, and at low voltage that never holds. Inverse-square falloff runs optimistic: testing puts the distance exponent between 1.47 and 2, so the method underestimates at longer reach. Watch units — 480 V goes in as 0.48, and typing 480 multiplies the answer by a thousand.
Frequently asked questions
The screen defaults give 4.748 cal/cm². What PPE does that call for?
Why does voltage go in as kV and distance as millimetres?
Which moves the result more, distance or clearing time?
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