This article begins a series in which I dive deep into the details of the newly published 2018 Edition of IEEE 1584—IEEE Guide for performing Arc-Flash Hazard Calculations. This standard is used globally for calculating the arcing current, incident energy and arc flash boundary, which are part of an arc flash study (risk assessment). The incident energy and arc flash boundary are also part of NFPA 70E’s list of information for equipment (arc flash warning) labels.
When an arc flash occurs, the size of the enclosure affects the arc flash hazard. The smaller the enclosure, the more concentrated the energy is—focusing it more toward the worker, resulting in greater incident energy exposure. The opposite is also true. Larger enclosures have less of a focusing effect, so the incident energy would be less, assuming all other parameters were the same.
The first edition of IEEE 1584 was published in 2002, and it accounted for the enclosure by considering three different sizes. Depending on whether the equipment was classified as medium voltage, low-voltage switchgear or low-voltage distribution (such as panels and motor control centers), the equations had a slight variation for each classification. At the time, this was considered a major advancement for arc flash studies.
Although there were three enclosure sizes, there was no method to consider any other size.
The 2018 edition provides greater flexibility by not only including more enclosure sizes for more types of equipment, but it also provides a new method to adjust the calculations for any enclosure size. This adjustment is known as the enclosure size correction factor (CF).
Without using a CF, the calculations are based on a default size that is normalized to 20 inches by 20 inches by 20 inches. If the actual enclosure is larger, the CF can be determined and used to calculate a more accurate (and lower) value of incident energy and arc flash boundary. Not including the CF would result in higher (more conservative) values—unless the enclosure is classified as shallow, as discussed below.
Enclosure size CF
Three steps make up the CF calculation: determine the equivalent height and width, determine the equivalent enclosure size, and calculate the CF.
Equivalent height and width
This calculation depends on the enclosure’s size as well as the electrode configuration (subject of the next article in this series). There are four different calculation methods depending on the enclosure size, which is divided into four categories that I like to refer to as small (20 inches), medium (20–26 inches), large (26–45 inches) and extra large (greater than 45 inches). Each size category has its own unique set of calculations.
Equivalent enclosure size
Once the equivalent height and width are determined, the equivalent enclosure size (EES) is calculated. This is the average of the equivalent height and width. For example, if the height is 36 inches and the width is 28 inches, the EES is (36 + 28) divided by 2, or 32 inches.
There are two methods for calculating the CF, depending on one additional parameter—whether the enclosure is “typical” or “shallow” as IEEE 1584 defines. A shallow enclosure has a height and width less than 20 inches, a voltage less than 600 volts, and a depth of 8 inches or less. For this case, the CF uses a variation of the CF equation that actually decreases the incident energy and arc flash boundary.
All other cases use the CF calculation for a “typical” enclosure, which will decrease the incident energy and arc flash boundary with increasing enclosure size.
The table above illustrates how the incident energy and arc flash boundary vary with enclosure size. This example is based on a medium size enclosure with a 30-kiloampere bolted fault current, 30-cycle arc duration, 32-millimeter conductor gap, 24-inch working distance, and an electrode configuration that is VCB—vertical conductor in a metal box.
The new 2018 edition of IEEE 1584 is a game changer for arc flash studies. Accounting for the enclosure size is just one of the many new additions.
Continue reading here for part 2.