Arc Flash – Incident Energy at Working Distance

 
With the given values:
Cf = 1.5 for voltages at or below 1 kV.
En = 5.26 J/cm2
En = 4.4126 J/cm2 (reduced incident energy)
t = 0.025 s (from Table 1, See note 1)
x = 1.641 (from Table D.4.2)
D = 455 mm See Table 3.

E~=~4.184{C_f}{E_n}(t/0.2)(610^x/D^x)
E~=~4.184*1.5*5.26*(0.025/0.2)(610^1.641/455^1.641)
E~=~6.676J/cm2
E~=~1.5956cal/cm2

For the reduce incident energy

E~=~4.184*1.5*4.4126*(0.025/0.2)(610^1.641/455^1.641)
E~=~5.6J/cm2
E~=~1.3385cal/cm2

Now that the incident energy has been determined, the Arc Flash Boundary can be calculated. Arc Flash Boundary is the distance at which a person is likely to receive a second degree burn. The onset of a second degree burn is assumed to be when the skin receives 5.0 J/cm2 of incident energy.

For the empirically derived equation,

D_B=({4.18{4C_f}{E_n}(t/0.2)(610^x/E_B)})^{1/x}…..[3]

For the theoretically derived equation,

D_B=sqrt{2.142*10^6*V*{I_bf}*(t/E_B)}…..[4]

where:
DB = distance (mm) of the arc flash boundary from the arcing point
Cf = calculation factor
Cf = 1.0 for voltages above 1 kV
Cf = 1.5 for voltages at or below 1 kV
En = incident energy normalized
t = time, sec
x = distance exponent from Table D.4.2
EB = incident energy in J/cm2 at the distance of the arc flash boundary
V = system voltage, kV
Ibf = bolted three-phase available short-circuit current

The above equations could be used to select personal protective equipment (PPE), to ensure that it is adequate to prevent thermal injury at a specified distance in the event of an arc flash.

For our example, using the reduced arcing current and equation [3], the calculated arc flash boundary (AFB) is 506.4 mm requiring Category 1 PPE (see Table 130.7(C)(15)(A)(b)).

Notes:
1. This need to taken from the time-current curve of the protective device. In the absence of such information, Table 1 can be used. Latest models of MCCB can clear faults in 0.5 – 1 cycle. As per NFPA 70E, 2 seconds is a reasonable maximum time for calculations.