# Arc Flash Calculations

**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 ^{o}C 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)I_{bf }- bolted three phase fault current (kA)I- arcing current (kA)_{a}E- incident energy (J/cm^{2})E- incident energy at boundary distance (J/cm_{B}^{2})E- normalized incident energy (J/cm_{n}^{2})t- arcing time (s)D- distance from arc to person (mm)

C_{B}- 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:

$${\mathrm{log}}_{10}({I}_{a})=K+0.662{\mathrm{log}}_{10}({I}_{bf})+0.0966V+0.000526G+0.5588Vlo{g}_{10}({I}_{bf})-0.00304\text{}G{\mathrm{log}}_{10}({I}_{bf})$$

Where:

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

For system voltages above 1000 V:

$${\mathrm{log}}_{10}({I}_{a})=0.00402+0.983{\mathrm{log}}_{10}({I}_{bf})$$

The arcing current is then given by:

$${I}_{a}={10}^{{\mathrm{log}}_{10}({I}_{a})}$$

### 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:

$${\mathrm{log}}_{10}({E}_{n})={K}_{1}+{K}_{2}+1.081{\mathrm{log}}_{0}({I}_{a})+0.0011G$$

$${E}_{n}={10}^{{\mathrm{log}}_{10}({E}_{n})}$$

Where:

K= -0.792 for open configurations and -0.555 for boxed configurations_{1}

K= 0 for ungrounded or high resistance ground systems and -0.113 for grounded systems_{2}

G= the conductor gap (see table below)

The incident energy is then given by:

$$E=4.184{C}_{f}{E}_{n}\left(\frac{t}{0.2}\right)\left(\frac{{610}^{x}}{{D}^{x}}\right)$$ - voltages ≤ 15 kV and gaps ≤ 152 mm

Where:

C

_{f }= 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$$E=2.142\times {10}^{6}V\text{}{I}_{bf}\left(\frac{t}{{D}^{2}}\right)$$ - voltages > 15 kV or gaps > 152mm

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

E_{n}Note: to convert J/cm

^{2}to cal/cm^{2}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:

$${D}_{B}={\left[4.184{C}_{f}{E}_{n}\left(\frac{t}{0.2}\right)\left(\frac{{610}^{x}}{{E}_{B}}\right)\right]}^{\frac{1}{x}}$$- voltages ≤ 15 kV and gaps ≤ 152 mm

$${D}_{B}=\sqrt{2.142\times {10}^{6}V\text{}{I}_{bf}\left(\frac{t}{{E}_{B}}\right)}$$- voltages > 15 kV or gaps > 152 mm

Where:

C_{f }= 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

*E _{B}* can be set at 5.0 J/cm

^{2}for bare skin or at the rating of any proposed PPE. The limit of 5.0 J/cm

^{2}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:

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

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