Incident energy at the arc flash boundary in arc flash analysis
Incident energy at the arc flash boundary in arc flash analysis
(OP)
Can anybody explain why 1.2 cal/cm^2 incident energy was selected in solving the equation for the arc flash boundary in IEEE P1584 and NFPA 70E? Does it imply that an exposure to less than the 1.2 cal/cm^2 is not enough to cause a second degree burn?






RE: Incident energy at the arc flash boundary in arc flash analysis
RE: Incident energy at the arc flash boundary in arc flash analysis
RE: Incident energy at the arc flash boundary in arc flash analysis
In an arc flash incident, you get it all at once, in much less than one second. If the flash exposure did last one second and you received 1.2 cal/cm^2, that's 60 times higher than what you got on the beach per unit of time. It's similar to the difference between sitting on a surface that is 100°F for 20 minutes, or sitting on a surface that is 1000°F for just long enough to jump back up and run for water.
It isn't just the amount of energy you absorb, it's the amount per unit of time that relates to the burn.
Good on ya,
Goober Dave
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RE: Incident energy at the arc flash boundary in arc flash analysis
thanks, DRWeig
I agree with what you are saying but you are talking now about heat flux intensity having an impact on time to 2nd degree burn. Indeed, the incident energy intensity is measured in cal/cm^2/sec ( or cal/cm^2 per second however you like). My problem with the IEEE and NFPA guides is that none of them really considers the intensity of the heat flux. Both of them use 1.2 cal/cm^2 incident energy as a threshold for a second degree burn IRRELEVANT to time duration over which that energy was absorbed.
Don't you think that less than 1.2 cal/cm^2, in fact MUCH less than 1.2 cal/cm^2, can cause second degree burn on bare skin if delivered in less than one (1) second, similar to what you are saying that indeed more than 1.2 cal/cm^2 is required to cause second degree burn when delivered over extended time interval longer than one second?
RE: Incident energy at the arc flash boundary in arc flash analysis
The beach analogy doesn't work because injury is not linearly related with time for low radiant intensity levels. You probably won't get a second degree burn by sitting 4 feet away from a 100 W lamp for 2.6 hours.
RE: Incident energy at the arc flash boundary in arc flash analysis
I understand you are saying that it takes one second exposure time into account. Would it be same 1.2 cal/cm^2 for other than 1 second exposure time, in particular less then one second?
according to this statement, my beach example would seem to make sense.
Do you mean at some point of a time injury would then become linearly related with time? What would be the threshold time for such transition? By the way, the beach analylogu doesn't work for me neither, I've used it only to highlight the glitch (as far as I am concerned) in IEEE 1584 and NFPA 70E accepting 1.2 cal/cm^2 as a minimum amount of incident energy for a second degree burn on bare skin.
RE: Incident energy at the arc flash boundary in arc flash analysis
RE: Incident energy at the arc flash boundary in arc flash analysis
could you please show analytical expression / formula describing the linearity between injury and time you would recommend for "real arc flash calculations"?
RE: Incident energy at the arc flash boundary in arc flash analysis
RE: Incident energy at the arc flash boundary in arc flash analysis
are you sure 1.2 cal/cm2 in IEEE 1584 and NFPA 70E is based on 1/10th of a second and not on 1 second?
IEEE 1584 states "if a butane lighter is held 1 cm away from a person's finger for one second and the finger is in the blue flame, a square centimeter area of the finger will be exposed to about 5.0 J/cm^2 or 1.2 cal/cm^2"..
Also, Stoll curve ("Heat Transfer Through Fabrics", Alice Stoll, September 1970, Figure 11 Human skin tolerance time to absorbed thermal energy delivered in a rectangular heat pulse.) indicates 1.2 cal/cm2 per second as a threshold energy for a second degree burn.
Let me know please what Ralph Lee paper you are referring to?
RE: Incident energy at the arc flash boundary in arc flash analysis
RE: Incident energy at the arc flash boundary in arc flash analysis
Now, if you don't mind of course, lets work out an example from the Ralph Lee article to make sure we stay on the same line.
1) Table II reads 1.99 in. arc sphere diameter for 7.5MW arc power
2) Table III reads 43C temperature rise in skin in 0.1sec at 91.4cm distance from center for 2 in. arc sphere diameter.
3) Figure 6 shows that 43C temperature rise would result in curable burn (37C + 43C = 80C). Above it is "grey area" so to speak up until not curable burn line.
4) Using the figures above, I've come up with 17 cal/cm2/sec arc power, or 1.7 cal/cm2 arc energy @ 0.1 sec. To be specific:
ArcPower = 7.5 x 10^6 / (4pi x 91.4^2) = 71 W/cm2 = 71 x 0.24 = 17 cal/cm2/sec
ArcEnergy = ArcPower x 0.1 sec = 7.1 J/cm2 = 1.7 cal/cm2 exposure for 0.1 sec.
please confirm this is what you are coming up as well.
RE: Incident energy at the arc flash boundary in arc flash analysis
RE: Incident energy at the arc flash boundary in arc flash analysis
After reading the Ralph Lees article, I noticed that the author assumes all arcing power is being converted into arc flash power, with the arc flash boundary equation expressed as function of arcing power. I have a problem with such an assumption ( arc flash power equals to arcing power ) especially for short time durations in the order of 100msec. That would be the same as assume that, when turning incandescent light bulb on, all the energy supplied to the light bulb would instantly be converted into ligh energy emitted by the bulb filament ( indeed it looks instantly to human eye but not that instanly through oscilloscope connected to light sensor viewing the bulb). My calculations show that only at most 1/3 of the arcing energy is being converted into arc flash energy ( incident energy ) within 100msec after the arc was initiated. The calculations include light energy emitted during the 100msec "heating" cycle, and "cooling" cycle thereafter. Taking that 1/3 correction factor into account, 1.7 cal/cm2 arcing power for 1/10th of a second reported in the Ralph Lee paper translates into approx 0.6 cal/cm2 incident energy exposure for 1/10th of a second, which is roughly half of the 1.2cal/cm2 threshold used by IEEE 1584 and NFPA 70E.
RE: Incident energy at the arc flash boundary in arc flash analysis
I think one could pick any element of the IEEE 1584 calculation procedure and be able to make a fairly strong case that it could be improved. Someday perhaps the updated version will be released.
The IEEE 1584 equations are based on testing that was done after the publication of the Ralph Lee paper. His calculations are basically theoretical. Also, the IEEE 1584 equations were developed for 15 kV and below. Above 15 kV, they basically revert to the Lee equations, IIRC.
For someone doing arc-flash hazard analysis, I'd recommend following the IEEE 1584 approach unless there exists a solid, test-based rationale for modifications.
The 1.2 cal/cm2 value, I believe is based on prior testing done in medical research.
RE: Incident energy at the arc flash boundary in arc flash analysis
Skythian, follow the http://arcflashforum.com/threads/2221/ link to find the answer to your question. Indeed you are right, much less than 1.2 cal/cm2 suffice to cause second degree burn during very short time very intense exposure to arc flash. For example, only 0.3 cal/cm2 for a 1/100th of a second are required to cause same degree of burn on bare skin as 1.2 cal/cm2 incident energy exposure for a one second.
RE: Incident energy at the arc flash boundary in arc flash analysis
BINGO!
RE: Incident energy at the arc flash boundary in arc flash analysis
RE: Incident energy at the arc flash boundary in arc flash analysis