IEEE 1584-2018 Arc-Flash calculation Steps [1]

 

According to IEEE 1584-2018 Guide follow 11 Steps to perform calculation of Arc-Flash Boundary (AFB) and Incident Energy (IE), as below.

Step 1：Collect the system and installation data

While the data required for this study is similar to data collected for typical short-circuit and protective-device coordination studies, it goes further in that all low-voltage distribution and control equipment plus its feeders and large branch circuits must be included.

Step 2：Determine the system modes of operation

In a site with a simple radial distribution system there is only one mode of operation.

Step 3：Determine the bolted fault currents

Input  all  data  from  the  single-line  diagrams  and  the  data  collection  effort  into  a  short-circuit  program. Run and get the faulted currents.

Step 4：Determine typical gap and enclosure size based upon system voltages and classes of equipment

Typical gaps between conductors (or bus gaps) as shown in Table 1 which also provides information on the enclosure sizes used for each voltage class.

Table 1 Classes of equipment and typical bus gaps

Equipment class	Typical bus Gaps (mm)	Enclosure Size (H × W × D)
		SI units (metric)
15 kV switchgear	152	1143 × 762 × 762 mm
15 kV MCC	152	914.4  × 914.4  × 914.4 mm
5 kV switchgear	104	914.4  × 914.4  × 914.4 mm
5 kV switchgear	104	1143  × 762  × 762 mm
5 kV MCC	104	660.4  × 660.4  × 660.4 mm
Low-voltage switchgear	32	508  × 508  × 508 mm
Shallow low-voltage MCCs and panel boards	25	355.6  × 304.8mm  × ≤203.2 mm
Deep low-voltage MCCs and panel boards	25	355.6  × 304.8mm  × >203.2 mm
Cable junction box	13	355.6  × 304.8 mm × ≤203.2 mm  or 355.6  × 304.8 mm × >203.2 mm

 

Step 5： Determine the equipment electrode configuration

The following electrode configurations are defined and listed according to their order of use within the incident energy model：

— VCB： Vertical conductors/electrodes inside a metal box/enclosure

— VCBB： Vertical conductors/electrodes terminated in an insulating barrier inside a metal box/enclosure

— HCB： Horizontal conductors/electrodes inside a metal box/enclosure

— VOA： Vertical conductors/electrodes in open air

— HOA： Horizontal conductors/electrodes in open air

Step 6： Determine the working distances

Typical working distances can be found in Table 2 based on the class of equipment.

Table 2 Classes of equipment and typical working distances

Equipment class	Working distance (mm)
15 kV switchgear	914.4
15 kV MCC	914.4
5 kV switchgear	914.4
5 kV MCC	914.4
Low-voltage switchgear	609.6
Shallow low-voltage MCCs and panelboards	457.2
Deep low-voltage MCCs and panelboards	457.2
Cable junction box	457.2

 

Step 7： Calculation of arcing current

To determine the arcing current：

1.           Determine the applicable equipment electrode configuration.

2.           If the system voltage is 600 V < V_oc ≤ 15 000 V, use Equation (1) to find intermediate values at 600 V, 2700 V, and 14300 V. Use Equation (2), Equation (3), Equation (4) to find the final value of the arcing current.

3.           If the system voltage is 208 V ≤ V_oc ≤ 600 V, use Equation (1) to find the intermediate value (600 V only) and Equation (5) to find the final value.

Formula as below：

Find intermediate valu(1)

where

I _bf： the bolted fault current for three-phase faults (symmetrical rms) (kA)

I _{arc_600}： the average rms arcing current at V_oc = 600 V (kA)

I _{arc_2700}： the average rms arcing current at V_oc = 2700 V (kA)

I _{arc_14300}： the average rms arcing current at V_oc =14 300 V (kA)

G： the gap distance between electrodes (mm)

k1 to k10： are the coefficients provided in Table 3

lg： log₁₀

Table 3—Coefficients for Equation (1)

Find final value (600 V < V_oc ≤ 15000 V)       (2)

 (3)                     (4)

 where

I _{arc_1}： the first I _arc interpolation term between 600 V and 2700 V (kA)

I _{arc_2}： the second I _arc interpolation term used when V_oc > 2700 V (kA)

I _{arc_3}： the third I _arc interpolation term used when V_oc is < 2700 V (kA)

V_oc： the open-circuit voltage (system voltage) (kV)

When 0.600 < V_oc ≤ 2.7, the final value of arcing current is given as follows：

I _arc = I _{arc_3}

When V_oc > 2.7, the final value of arcing current is given as follows：

I _arc = I _{arc_2}

Find final value (V_oc ≤ 600 V)

                    (5)

where

V_oc：the open-circuit voltage (kV)

I_bf： the bolted fault current for three-phase faults (symmetrical rms) (kA)

I_arc： the final rms arcing current at the specified V_oc (kA)

I_{arc_600}： the rms arcing current at V_oc=600 V found using Equation (1) (kA)

Step 8： Determine the arc duration

The arc duration is most commonly dependent on the operating time of a time-overcurrent protective device.