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Incident energy at the arc flash boundary in arc flash analysis 1
- Thread starter Skythian
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as an example, someone stands on a sunny day for roughly 20 minutes ( I can stay much longer on the beach
) making himself exposed to approx 1000W/m^2 x 20min x 60 sec = 0.1W/cm^2 x 1200sec = 120J/cm^2, or 29cal/cm^2. In case when severity of burn was measured in terms of incident energy only, how could one survive like this without wearing "PPE 4" suit?
In your beach episode, your skin absorbed a total of 29 cal/cm^2 over a 20-minute period. That's only 0.02 cal/cm^2 per second. You're convecting and evaporating the heat away the whole time, and other than the rosy pink (or tan) result, you don't cook yourself.
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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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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- #5
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?
The incident energy in cal/cm² takes time of exposure into account, and it is the total energy, not the energy per unit of time, that determines the arc flash hazard.
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.
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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.
I'd say assume linearity for real arc flash calculations. Usually you limit the exposure time to 2 seconds anyway on the assumption that if there is an arc flash the worker won't stand there in front of the arc longer than 2 seconds unless his movements are restricted.
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- #11
Hi Zogzog,
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?
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?
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- #13
thank you for submitting the link and thank you for your patience responding to my question. You seem to really an expert in arc flash studies.
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.
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.
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- #15
Thank you, I'll think about it. The main reason I started this discussion was to see what do you guys think about using 1.2 cal/cm2 as a minimum incident energy for a second degree burn, and if anybody else can see the problem with it as well ( looks like me only is having the problem with it so far ).
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.
I don't see that 1.2 cal/cm2 value is terribly critical to overall concept of arc-flash safety. If you accept the need for a boundary inside of which PPE is required, then a heat energy value must be selected. Any value will be a compromise. I would be surprised if the calculation for the incident energy and the arc-flash boundary is accurate to more than plus/minus 20% anyway. When tests were duplicated using identical current and identical test setups, there was a range of results.
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.
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.
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- #17
ronpaulrocks
Electrical
Skythian, follow the [URL unfurl="true"]http://arcflashforum.com/threads/2221/[/url] 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.
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- #18
LionelHutz
Electrical
There is an article in the March/April Electical Line Magazine that discusses this. It basically discusses how using the fixed 1.2cal/cm^2 value is flawed.
ronpaulrocks
Electrical
attached is a copy of the article from the March/April 2012 issue of Electrical Line Magazine (Vol.18, No.2) for those interested reading it.
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