The resistance load line of a typical DC circuit can be analytically presented as:
where Vs stands for an open source voltage and Rs is system resistance including source and feeders. Introducing voltage drop across an arc Varc, arcing current can be expressed as:
where Isc is the available fault current at the point of arc and is equal to Vs/Rs. It was demonstrated that there is a minimum voltage needed to maintain an arc. That minimum depends on the current magnitude, gap width, and orientation of the electrodes. This transitional point can be expressed as:
G - conductor gap distance in millimeters
It - the transitional point current measured in amperes
Above that minimum, the arc V-I characteristic can be expressed as:
Varc - arcing voltage in volts
Iarc - dc arcing current in amperes
To find the point where the arc V-I characteristic crosses the circuit load line, solve equations (2) and (4) using the iterative method. As the first approximation, assume Varc is equal to half of the system voltage Vs. Then, follow the steps below:
Cycle through the steps 1 and 2 above until the answers for Varc converge.
The energy released during an arc flash event is roughly proportional to the arc duration. The upstream protective device operation controls the arc flash duration. A fuse or properly maintained overcurrent protective device has a predictable time to open the circuit with a specific arc current value. Hence, arcing current impacts the released energy directly through the current itself and through interaction with the overcurrent protective device.
Additionally, circuit time constant affects current rise and protection device performance characteristics, thus impacting the arc duration. In this case, time current characteristic of the upstream protective device clearing the fault may have to be adjusted for the time constant. If this occurs, the process of determining the protective device operating time is cumbersome. First, the time-current characteristic of the protective device has to be analytically presented as a function of the available fault current. The relationship between the effective RMS current, the available fault current, and the number of time constants can be expressed as:
where the K factor is expressed in numbers of time constants n = tarc/tconst:
This creates a dilemma due to the fact that one cannot determine the arcing time without the RMS value of the arcing current, and one cannot solve for the RMS current without the arcing time represented by the n term in equation (6). This requires an iterative solution. As a first approximation, begin by assuming that Irms equals Iarc, determining tarc from the analytical expression for the protection device time-current characteristics tarc=f(Irms), determining the number of time constants n and calculating K from equation (6), substituting its value into equation (5) to calculate the new RMS current, and then solving for the arc duration again. Once the first approximation of the arc duration has been made, determine the new number of time-constants n, re-calculate the K term and substitute its value into equation (5). This produces a new Irms. Re-calculate for a new tarc by using the new Irms and continue until the answers converge.
The power in the arc can be calculated by:
where the Parc is measured in watts.
The energy in the arc is a function of power and time:
where the Earc is measured in Joules.
Incident energy exposure for an open-air arc where the heat transfer depends on the spherical energy density is then expressed as:
Eiair - incident energy from an open air arc at distance D in Joules/cm2
D - distance from the arc in centimeters
This formula assumes the radiant heat transfer. Not all of the arc energy will be transferred as radiant heat especially within the short time interval after the arc was ignited. Therefore, the equation (9) will produce a conservative but safe estimate of incident energy exposure.
For the arc in a box, the enclosure has a focusing effect on the incident energy. For the selected enclosure type and test distance, the incident energy from an arc flash in a box can be calculated by:
Eibox - the incident energy from an arc flash in a box in Joules/cm2
D - distance from the arc in centimeters
A and K are obtained from optimal values listed in table below:
Equation (9) written in terms of arc flash boundary, becomes:
AFB - arc flash boundary is measured in centimeters
Et - threshold incident energy to second degree burn in J/cm2 calculated by:
Likewise, equation (10) written in terms of arc flash boundary, becomes:
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