I have not located a thread directly addressing what I have come across, but my "sanity check" calculations below have me dumbfounded.

Reputable companies have arc flash literature on the internet declaring that an arc flash can produce heat up to 35,000 degrees F: https://electrification.us.abb.com/sites/geis/files/gallery/The-Basics-of-Arc-Flash-Article_GE_Industrial_Solutions_0.pdf.

Here is my attempted conversion from 35,000 degrees F to cal:
F -> K: (35,000-32)*5/9 + 273.15 = 19,700 K
K -> J: 19,700*(Boltzmann constant(J/K)) = 2.7198E-19 J
J -> cal: 2.7198E-19/4.184 = 6.50E-20 cal

Clearly I'm making a fundamental error here - there's no way that an arc flash with that much heat has that little energy. Can someone explain how this analysis is wrong? Thanks.

I think that's for one particle of gas at equilibrium. Multiply by the # of particles of gas involved.

Thanks stevenal.
Here is my estimate based on that information:

Suppose the gas is oxygen, and the arc flash ionized gas ball consists of 10 mol of it.
1 mole of O2 gas occupies 22.4 L at normal pressure and temperature (1 atm and 77 degrees F) (based on this: https://socratic.org/questions/56afb21b11ef6b589ee3b3bf).
10*2/22.4 = 0.89 mol.
But an arc flash is high temperature, and volume is proportional to temperature.
V1/T1 = V2/T2 = (22.4L)/(77 degrees F) =V2/(35,000 degrees F).
V2 = (35,000 degrees F)*(22.4 L)/(77 degrees F) = 10,181 L.
So, (10*2 oxygen atoms)/10,181 = 1.964E-3 mol oxygen.
1.964E-3*6.02E23 (Avogadro's #) = 1.18E21 oxygen atoms.
1.18E21*2.72E-19 J (answer from before) = 322 J.
322 J/4.184 = 77 cal.

This answer at least seems physically possible, thank you for the guidance. Would 77 cal/cm^2 incident energy and 35,000 degrees F be reasonable for a severe arc flash?

No.

The 35,000 number is based on the immediate localized vaporization of metals and would quickly decrease as the resulting 'plasma cloud' begins to expand.
Your 77 number is based on being at least 18" away from the point of vaporization.

You are are taking a scare tactic number and trying to use it in a rational manner.

The 35,000F number has been removed from NFPA 70E since it was not helpful and no one could find a scientific citation to support this long quoted part of NFPA 70E. Plasma temperature and plasma density are both needed to predict so temperature alone isn't very useful even if it was accurate.

Most of the physicists we have consulted (Dr. Gordon, and Dr. Sweeting) indicate that arc in air is closer to 5000 or 10000F. But the calorie/cm2 or J/cm2 is the total heat energy but the temp of the gas ball is deceiving since up to 80% of the energy in our common arc test is infrared until it hits a surface so there is no "temperature" per se.

Stick with the IEEE numbers. ArcPro uses a physics finite elements model based on plasma density, temperature and column size, and length to predict total energy in points in the space but temperature is not a good analogy in my estimation. Comparisons to the "surface of the sun" is even less useful.

Hi All

Greetings from North of the border. As your calculations are above me as an industrial electrician, just wondering about Cal/cm2 as compared to temperature in degrees F? In many cases in my sessions up here in Ontario, the tradespeople will ask as a comparison form Cal/cm2 to degrees Farenheit. Basically using simple examples such as the temperature of their BBQ, oven, etc. Considering to my knowledge there is no direct calculation from Cal/cm2 to temperature in Farenheit. I hope that this makes sense. Thank you, Len