IEEE 1584-2018 Arc-Flash calculation Steps [2]

 

Step 9： Calculate the incident energy

To determine the incident energy：

1.      Determine the enclosure size correction factor.

2.      If the system voltage is 600 V < V_oc ≤ 15000 V, use Equation (6), Equation (7), and Equation (8) to find intermediate values. Use Equation (10), Equation (11), Equation (12) to find the final value of the incident energy.

3.      If the system voltage is 208 V ≤ V_oc ≤ 600 V, use Equation (9) to determine the final incident energy.

Formula as below：

Find intermediate values        (6)

(7)

         (8)

(9)

 where

E₆₀₀ is the incident energy at V_oc = 600 V (J/cm²)

E₂₇₀₀ is the incident energy at V_oc = 2700 V (J/cm²)

E₁₄₃₀₀ is the incident energy at V_oc = 14 300 V (J/cm²)

E_≤600 is the incident energy for V_oc ≤ 600 V (J/cm²)

T： the arc duration (ms)

G：the gap distance between conductors (electrodes) (mm)

I_{arc_600}： the rms arcing current for 600 V (kA)

I_{arc_2700}： the rms arcing current for 2700 V (kA)

I_{arc_14300}： the rms arcing current for 14 300 V (kA)

I_arc：rms arcing current for V_oc ≤ 600 V [using Equation (25)] (kA)

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

D： the distance between electrodes and calorimeters (working distance) (mm)

CF： correction factor for enclosure size (CF = 1 for VOA and HOA configurations)

lg： log₁₀

k1to k13： the coefficients provided in Table 4, Table 5, and Table 6.

Table 4 Coefficients for Equation (6), Equation (9), Equation (13), and Equation (16)

Table 5 Coefficients for Equation (7), Equation (14)

Table 6 Coefficients for Equation (8), Equation (15)

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

                  (10)                (11)                  (12)

 where

E₁： the first E interpolation term between 600 V and 2700 V (J/cm²)

E₂： the second E interpolation term used when V_oc is > 2700 V (J/cm²)

E₃： the third E interpolation term used when V_oc is < 2700 V (J/cm²)

When 0.600 < V_oc ≤ 2.7, the final values of incident energy is given as follows：

E = E₃

When V_oc > 2.7, the final values of incident energy is given as follows：

E = E₂

Find final value (V_oc ≤ 600 V)

The incident energy is given as follows：

E = E_≤600

where

E_≤600： the incident energy for V_oc ≤ 600 V determined using Equation (9) solved using the arc current determined from Equation (1) and Equation (5) (J/cm²)

E： the final incident energy at specified V_oc (J/cm²)

Step 10： Determine the arc-flash boundary for all equipment

To determine the arc-flash boundary：

1.    Determine the enclosure size correction factor

2.    If the system voltage is 600 V < V_oc ≤ 15000 V, use Equation (13), Equation (14), and Equation (15) to find intermediate values. Use Equation (17), Equation (18), Equation (19) to find the final value of the arc-flash boundary.

3.    If the system voltage is 208 V ≤ V_oc ≤ 600 V, use Equation (16) to determine of the final arc-flash boundary.

Formula as below：

Find intermediate values

(13)

(14)

 

(15)

(16)

where

AFB₆₀₀： the arc-flash boundary for V_oc = 600 V (mm)

AFB₂₇₀₀： the arc-flash boundary for V_oc = 2700 V (mm)

AFB₁₄₃₀₀： the arc-flash boundary for V_oc = 14 300 V (mm)

AFB_≤600： the arc-flash boundary for V_oc ≤ 600 V (mm)

G： the gap between electrodes (mm)

I_{arc_600}： the rms arcing current for 600 V (kA)

I_{arc_2700}： the rms arcing current for 2700 V (kA)

I_{arc_14300}： the rms arcing current for 14 300 V (kA)

I_arc： the rms arcing current for V_oc ≤ 600 V [obtained using Equation (5)] (kA)

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

CF：the correction factor for enclosure size (CF=1 for VOA and HOA configurations)

T：the arc duration (ms)

lg： log₁₀

k1 to k13： are the coefficients provided in Table 4, Table 5, and Table 6.

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

      (17)                                

       (18)

       (19)

 where

AFB₁： the first AFB interpolation term between 600 V and 2700 V (mm)

AFB₂： the second AFB interpolation term used when V_oc is greater than 2700 V (mm)

AFB₃： the third AFB interpolation term used when V_oc is less than 2700 V (mm)

When 0.600 < V_oc ≤ 2.7, the final values of arc-flash boundary are given as follows：

AFB = AFB₃

When V_oc > 2.7, the final values of arc-flash boundary are given as follows：

AFB = AFB₂

Find final value (V_oc ≤ 600 V)

The arc-flash boundary is given as follows：

AFB = AFB_≤600

where

AFB_≤600：arc-flash boundary for V_oc ≤ 600V determined using Equation(16) solved using the arc current determined from Equation (1) and Equation (5) (mm)

AFB：the final arc-flash boundary at specified V_oc (mm)

Step 11： Determine second arcing current

It must account for the arcing current variation. Repeat step 8, step 9, and step 10 using the reduced arcing current. It is possible that the incident energy and arc-flash boundary results obtained using the reduced arcing current are different. The final incident energy or arc-flash boundary is the higher of the two calculated values.

(20)

 where

VarC_f： the arcing current variation correction factor

I_arc：the final or intermediate rms arcing current(s) (kA) (see note)

I_{arc_min}： a second rms arcing current reduced based on the variation correction factor (kA)

V_oc： the open-circuit voltage between 0.208 kV and 15.0 kV

k1 to  k7： the coefficients provided in Table 7

Table 7 Coefficients for Equation (20)

NOTE—The correction factor (1 – (0.5 × VarC_f)) is applied as follows：

— 208 V ≤ V_oc ≤ 600 V： To I_arc (final current only)

— 600 V < V_oc ≤ 15000 V： To I_{arc_600}, I_{arc_2700}, and I_{arc_14300} (intermediate average arcing currents). The final I_arc value inherits the correction factor.

The “0.5” coefficient indicates that variation is applied to the average arcing current to obtain a lower-bound value arcing current.

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.