USING IEEE std 1584-2002 (Guide for Performing Arc-Flash Hazard Calculations)

TO CALCULATE INCIDENT ENERGY FOR APPLICATIONS UNDER 1kV

Disclaimer: The NESC does not indicate how a utility is required to perform an arc-flash hazard analysis other than considering available fault current, clearing time, and worker distance. There are many variables associated with determining available fault current, clearing time and worker distance, and yet more variables which are considered in determining how to use the requirements of the NESC. Because MLGW’s immediate need to address compliance with the requirements of the NESC, I offer the following information, but do not claim significant expertise in this area.

Constants and definitions for voltage < 1000 Volts

Ibf = 3-phase short-circuit current in kA

V = line-to-line system voltage in kV

G = conductor gap in millimeters

K = -0.153 (open air arcs)

K = -0.097 (arcs in a box)

K1 = -0.792 (open air arcs)

K1 = -0.555 (arcs in a box)

K2 = 0 (ungrounded or high – Impedance)

K2 = -0.113 for grounded

Cf = 1.5 (for calculations under 1kV)

IEEE 1584 Factors for Equipment and Voltage Classes

System Type of Typical Voltage Equipment Conductor Distance (kV) Gap (mm) X-factor

0.208 to 1 Open-Air 10 - 40 2.000Switchgear 32 1.473MCC's and Panels 25 1.641Cables 13 2.000Open-Air 102 2.000

>1 to 5 Switchgear 13 - 102 0.973Cables 13 2.000Open-air 13 - 153 2.000

> 5 to 15 Switchgear 153 0.973Cables 13 2.000

IEEE 1584 Factors for Equipment and Voltage Classes

System Type of Typical Voltage Equipment Conductor Distance (kV) Gap (mm) X-factor

0.208 to 1 Open-Air 10 - 40 2.000Switchgear 32 1.473MCC's and Panels 25 1.641Cables 13 2.000Open Air 102 2.000

>1 to 5 Switchgear 13 - 102 0.973Cables 13 2.000Open-Air 13 - 153 2.000

> 5 to 15 Switchgear 153 0.973Cables 13 2.000

Example Application

IEEE std 1584-2002 (Guide for Performing Arc-Flash Hazard Calculations) Calculating Three-Phase Arcing Current (Voltage <= 1kV)

Ia = 10 log (Ia)

Where: Log (Ia)= K + [0.622 * log (Ibf)] + [0.0966*(V)] + [0.000526 * G] + [0.5588 * (V) * (log(bf))] – [0.00304 * G * log (Ibf)]

.662 x Log (Ibf ) = .662 x log (50) = 1.124720.0966 x V = .0966 x .48 = 0.046370.000526 x G = .000526 x 32 = 0.016830.5588 x V x log (Ibf ) = 0.5588 x .48 x log(50) = 0.45570-0.00304 x G x Log(Ibf ) = -0.00304 x 32 Log(50) = -0.16528+ K = -0.097 (arc in a box) = -0.097

Log(Ia) = sum of eqt components = 1.38134

Ia= 10 raised to 1.38134 power = 24.06

Log (Ia)= 0.662 Log(Ibf) + 0.0966V + 0.000526G + 0.5588V (log (Ibf))- .00304G (log (Ibf)) + K

IEEE std 1584-2002 (Guide for Performing Arc-Flash Hazard Calculations) Calculating Three-Phase Incident Energy for Voltage <= 15kV

Ea = 10log Ea

= Incident Energy

Where: Log Ea= K1 + K2 + [1.081 x (log Ia)] + [(0.0011)*G]

Normalized for 610mm (24”) and .2s

Ia = 24.06 kA K1 = -0.555 (arc in a box)G = 32 mm K2 = -0.113 (for grounded)

1.0811 x Log (Ia) = 1.0811 x log(24.06) = 1.381290.0011 x G = 0.0011 x 32 = 0.03520K1 = -0.55500K2 = -0.0113

Log (Ea) = sum of components = 0.85019Ea = J per cm squared 10 raised to the Log Ea = 7.2

multiply by .24 to get cal per cm squared = 1.728

Ea = 10log Ea

= Incident Energy Where: Log Ea= K1 + K2 + [1.081 x (log Ia)] + [(0.0011)*G]

(Result is normalized for clearance time of .2s and distance of 610 mm (24 inches))

Equation if working distance and clearance time are known

E= (4.184)*Cf * Ea*[(t/0.2)*(610x / Dx)]

D = 18” = 457.2mm

t = arcing time in seconds = .167 (11cycles)

X = 1.473 from table

Cf = 1.5 (for V <= 1kV)

Ea = 7.2 (from previous slide)

E = 57.9 J/cm2

E = 13.8 cal/cm2 (mult by .24)

Was 1.728 at 24 inches and .2s