Arc Flash Calculations

By on

Arc Flash
Working in the vicinity of electrical equipment poses an hazard.  In addition to electric shock hazard, fault currents passing through air causes Arc Flash.

During arc flash, temperatures of 20,000 oC for several metres around equipment can exist, resulting in severe burn injuries.  Quickly expanding air and vaporization of metals causes high pressures, sound and shrapnel leading to multiple types of injuries.

To protect people a risk analysis on the arc flash hazard is carried out to better understand what may occur.  Appropriate procedures and personal protective equipment are then employed to minimise potential injuries.

When evaluating arc flash there are two main considerations, 1) the incident energy in the arc flash and 2) flash protection boundary.  I will be looking at the calculation of these two parameters in terms of IEE Std. 1584 'IEEE Guide for Performing Arc-Flash Hazard Calculations'.

IEE 1584 Calculation Method

Symbols and their meaning:

V  -    system voltage (kV)
Ibf    -    bolted three phase fault current (kA)
Ia   -    arcing current (kA)
E   -    incident energy (J/cm2)
EB  - incident energy at boundary distance (J/cm2)
En  - normalized incident energy (J/cm2)
t    -    arcing time (s)
D   - distance from arc to person (mm)
CB  - distance of boundary from arc point (mm)

Arcing Current

In order to determine both the flash protection boundary and incident energy, first the arcing current needs to be calculated.

For system voltages under 1000 V:

$log 10 ( I a )=K+0.662 log 10 ( I bf )+0.0966V+0.000526G+0.5588Vlo g 10 ( I bf )−0.00304 G log 10 ( I bf )$

Where:

K = -0.153 for open configurations and -0.097 for box configurations

For system voltages above 1000 V:

$log 10 ( I a )=0.00402+0.983 log 10 ( I bf )$

The arcing current is then given by:

$I a = 10 log 10 ( I a )$

Incident Energy

To calculated the incident energy, first normalized incident energy is calculated.  The result of this calculation is the incident energy for a arc duration to 0.2 s and at a distance from the arc point to the person of 610 mm.  Once the normalized energy has been calculated, this is then adjusted to reflect the actual arcing time and distance.

The normalized incident energy is given by:

$log 10 ( E n )= K 1 + K 2 +1.081 log 0 ( I a )+0.0011G$

$E n = 10 log 10 ( E n )$

Where:

K1 = -0.792 for open configurations and -0.555 for boxed configurations

K2 = 0 for ungrounded or high resistance ground systems and -0.113 for grounded systems

G = the conductor gap (see table below)

The incident energy is then given by:

$E=4.184 C f E n ( t 0.2 )( 610 x D x )$    - voltages ≤ 15 kV and gaps ≤ 152 mm

Where:

Cf    = 1.0 for voltages greater than 1000 V and 1.5 for voltages less than 1000 V

x    = is a distance exponent from the table below

$E=2.142× 10 6 V I bf ( t D 2 )$ - voltages > 15 kV or gaps > 152mm

Note: for voltages greater than 15 kV, it is not necessary to first calculate En

Note: to convert J/cm2 to cal/cm2 multiply by 0.239

Gap and distance factors

Voltage (kV)

Equipment Type
Typical gap between
conductors (mm)
Distance factor,
x
0.208 - 1

Open Air 10-40 2.000
Switchgear 32 1.473
MCC and Panels 25 1.641
Cable 13 2.000
>1 - 5 Open Air 102 2.000
Switchgear 13-102 0.973
Cables 13 2.000
>5 -15 Open Air 13-153 2.000
Switchgear 153 0.973
Cables 13 2.000

Flash Protection Boundary

The flash boundary is the distance from the arc point to a location when the incident energy is less than a given value.  Knowing the incident energy, the flash boundary distance is given by:

$D B = [ 4.184 C f E n ( t 0.2 )( 610 x E B ) ] 1 x$- voltages ≤ 15 kV and gaps ≤ 152 mm

$D B = 2.142× 10 6 V I bf ( t E B )$- voltages > 15 kV or gaps > 152 mm

Where:

Cf = 1.0 for voltages greater than 1000 V and 1.0 for voltages less than 1000 V

x = is a distance exponent from the table above

EB can be set at 5.0 J/cm2 for bare skin or at the rating of any proposed PPE.  The limit of 5.0 J/cm2 is that at which a person is likely to receive second degree burns.

Summary

To carry out arc flash calculations, first workout the arcing current, then the incident energy and finally the arc flash boundary.  Pretty straight forward.

To make life a little easier I have created a calculator to do the grunt work.  You can find it here:

Arc Flash Calculator

It anyone has an comments or spots any errors in the above, please chip in below.

More interesting Notes:

Steven has over twenty five years experience working on some of the largest construction projects. He has a deep technical understanding of electrical engineering and is keen to share this knowledge. About the author

myElectrical Engineering

Motor Insulation

Insulation on a motor prevents interconnection of windings and the winding to earth.  When looking at motors, it is important to understand how the insulation...

Aluminium Windings - Dry Type Transformers

The other day I was talking to a colleague who is a building services consultant.  Despite regularly specifying dry-type/cast resin transformers he was...

Getting Started with Patents

If you have a great idea or invent something the last thing you want is someone to steal the idea. One of the things you can do is protect the intellectual...

Cable Sheath and Armour Loss

When sizing cables, the heat generated  by losses within any sheath or armour need to be evaluated. When significant, it becomes a factor to be considered...

How to Calculate Motor Starting Time

Request to look at induction motor starting time have come up a few times on the site. Hopefully in this post, I give you guys some idea on how to calculate...

Capacitors - Energy Storage Application

Capacitors have numerous applications in electrical and electronic applications.  This note examines the use of capacitors to store electrical energy....

IEC 61439 Verification Methods

The (relatively new) switchgear and control gear standard, IEC 61439 'Low-voltage switchgear and controlgear assemblies' has three methods which can be...

The 27 km, Euro 6 billion  Collider lies on the border between France and Switzerland, took nearly 30 years to complete.  Some of the lofty goals for the...

New Mail Chimp

We've been sending out Newsletters on a regular basis for a few weeks now. To do this we have been using Google's Feedburner service. While Feedburner...

Induction Motor Equivalent Circuit

Induction motors are frequently used in both industrial and domestic applications.  Within the induction motor, an electrical current in the rotor is induced...

Have some knowledge to share

If you have some expert knowledge or experience, why not consider sharing this with our community.

By writing an electrical note, you will be educating our users and at the same time promoting your expertise within the engineering community.

To get started and understand our policy, you can read our How to Write an Electrical Note
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 9, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "https://myelectrical.com/notes/entryid/130/arc-flash-calculations-arc-flash-safety", "fetch_time": 1726272611000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2024-38/segments/1725700651540.77/warc/CC-MAIN-20240913233654-20240914023654-00184.warc.gz", "warc_record_offset": 376877017, "warc_record_length": 17600, "token_count": 1776, "char_count": 6958, "score": 3.890625, "int_score": 4, "crawl": "CC-MAIN-2024-38", "snapshot_type": "latest", "language": "en", "language_score": 0.8348734974861145, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00015-of-00064.parquet"}