Arcing Short-Circuit Current

Published On
Mar 15, 2016

This article is the second in a series that provides a step-by-step approach for performing arc flash calculations. The first part, “A Calculated Introduction,” appeared in the January 2016 issue of ELECTRICAL CONTRACTOR.


Arcing short-circuit currents


The first step in performing arc flash calculations is to determine the magnitude of arcing short-circuit current that could flow during an arc flash. This is the current that flows across an air gap between the conductors. It is typically created from either the conducting objects being blown apart or by the conductor melting. During an arcing fault, the arc has resistance, which causes the arcing short-circuit current to be smaller than the bolted short-circuit current.


IEEE 1584—IEEE Guide for Performing Arc-Flash Hazard Calculations provides two sets of equations for calculating the arcing short-circuit current. One is for systems operating from 208 volts (V) to 1,000V, and the other is used for systems operating between 1,000 and 15,000V.


The calculation is based on using the bolted, three-phase, short-circuit current obtained from a conventional short-
circuit study. This value—along with other variables, such as the arc gap distance, and whether the arc flash occurs in open air or in an enclosure—is entered into the equation. According to the equations, if the arc occurs in an enclosure, the conducting plasma is more concentrated, so the resulting arcing current will be less than if the arc occurs in open air. For systems operating up through 1,000V, a “K” factor is used in the equations to account for this difference. For an arc flash occurring in open air, a K value of –0.153 is used. If the calculation is for the arc occurring in a box, which represents an equipment enclosure, then the K value is –0.097.


The equations for calculating the arcing short-circuit current for systems through 1 kilovolt (kV) can seem complex. However, it is really a series of small steps that, taken individually, are much simpler than they appear. I developed an arcing short-circuit calculation worksheet (see above) based on IEEE 1584 equations that breaks the calculation into individual steps. Each step can be solved separately to determine the final answer.


IEEE 1584 2002 equation for estimating the arcing short-circuit current


log Ia = K + 0.662 log Ibf + 0.0966V + 0.000526G + 0.5588V(log Ibf) – 0.00304G (log Ibf)


Ia = 10 log Ia


log = log10

Ia = arcing current in kiloamperes (kA)


K = –0.153 for open air and –0.097 for arcs in a box


Ibf = bolted, three-phase available short-circuit current in kA


V = system voltage in kV


G = conductor gap in millimeters (mm)


Example arcing short-circuit current calculation


A 480V panel has an available “bolted” three-phase fault current of 30,000 amperes (A) based on a conventional short-circuit study. The arcing short-circuit current will be calculated.


Step 1: Enter the bolted, three-phase short-circuit current Ibf. Multiply the logarithm of Ibf by the constant 0.662.


Step 2: Enter the system voltage in kilovolts, and multiply it by the constant 0.0966.


Step 3: Enter the conductor gap distance based on the equipment type. For a panel, IEEE 1584 suggests using 25 mm. Multiply it by the constant 0.000526.


Step 4: Enter the three-phase, bolted short-circuit current Ibf in kiloamperes and the system voltage in kilovolts. Take the logarithm of Ibf and multiply it by the voltage and by the constant 0.5588. The logarithm defines an order of magnitude, and most scientific calculators have a “LOG” button that makes this an easy calculation.


Step 5: Enter the gap distance G in millimeters and Ibf. Multiply the logarithm of Ibf by the gap distance G, and multiply it by the constant –0.00304.


Step 6: Since this is an enclosure, select the K value of 0.097 for a “box.”


Step 7: Add steps one through six.


Step 8: Raise number 10 to the value that you found in Step 7.


The final result is the arcing current in kiloamperes.


For arcing short-circuit current calculations on systems operating at voltages between 1 kV and 15 kV, IEEE 1584 has a much simpler equation that requires only the bolted short-circuit current: log Ia = 0.00402 + 0.983 X log Ibf, and the total arcing short-circuit current is Iarcing = 10 log10(Ia).


The next column in this series will address incident-energy calculations.

Stay Informed Join our Newsletter

Having trouble finding time to sit down with the latest issue of
ELECTRICAL CONTRACTOR? Don't worry, we'll come to you.