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Arc flash on a 15 kV line: the Lee equation against the OSHA look-up table (OSHA final rule preamble, Table 11)

Tool	Arc Flash Screening
Source	Occupational Safety and Health Administration (2014). "Electric Power Generation, Transmission, and Distribution; Electrical Protective Equipment; Final Rule." Federal Register, Vol. 79, No. 70, 11 April 2014, pp. 20316–20743. Table 11, p. 20465; 29 CFR 1910.269, Appendix E, Table 6.
Level	Professional

Problem

A lineworker in rubber insulating gloves works 15 in (380 mm) from a conductor of a 15,000 V distribution line. The available fault current is 5,000 A and the protective device clears it in 34.5 cycles. In the rulemaking a commenter showed that four methods gave 2.9, 5, 40 and 153 cal/cm² for this one case; OSHA repeated the calculation. What incident energy does the Lee (NFPA 70E Annex D) equation give, what does OSHA's own table for single-phase arcs in open air give, and how far away is the 1.2 cal/cm² boundary by the equation?

Inputs to Enter

Exposure: overhead line, single-phase arc in open air; line work with rubber insulating gloves (Table 6)
System voltage 15,000 V; bolted fault current 5 kA (entered); clearing time 34.5 cycles at 60 Hz (0.575 s)
Working distance 15 in (381 mm); boundary energy 1.2 cal/cm²; arc ratings 4, 8, 25, 40 cal/cm²
Reduced-current case: 50 % of the fault current (2.5 kA) cleared in 0.75 s (the tool's illustrative second case)

Open the Arc Flash Screening tool (loaded with this example)

Answers

Lee equation: E = 793 · 5 · 15 · 0.575 / 15² = 152.0 cal/cm² (636 J/cm²) (book: 152, Table 11; the commenter's figure 153)
OSHA Appendix E Table 6, 4.0 to 15.0 kV, 5 kA row: 46 cycles for 4 cal/cm², so 34.5 cycles gives 4 cal/cm² or less (book: 5.0 from the proposed table; the final table lists 4)
OSHA rates the Lee equation acceptable but conservative (Y-C) for a single-phase arc from 601 V to 15 kV; the table level is carried forward and a 4 cal/cm² arc rating covers it
Arc flash boundary at 1.2 cal/cm² by the Lee equation (the table carries no distance): 168.8 in = 14.07 ft (4.288 m)
Reduced case, 2.5 kA for 45 cycles: the 5 kA row still gives 4 cal/cm² or less; Lee 99.1 cal/cm², so the full current governs

Notes

The tool's default inputs are this case; press Submit. The preamble (p. 20466) explains the gap: the equation is for a three-phase arc (dividing by the 2.2 conversion factor leaves 70 cal/cm²) and becomes more conservative as the voltage rises. The page is a screening estimate with the public methods and is not an IEEE 1584 study.