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.