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

Statistics: Posted by Leonard — Mon Jun 15, 2020 1:17 pm

]]>

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.

Statistics: Posted by elihuiv — Mon Jun 15, 2020 11:46 am

]]>

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.

Statistics: Posted by JBD — Fri Jun 12, 2020 6:56 am

]]>

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?

Statistics: Posted by bpjmal1 — Thu Jun 11, 2020 11:52 am

]]>

]]>

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.

Statistics: Posted by bpjmal1 — Thu Jun 11, 2020 6:05 am

]]>

Annex I - Development of Special Model for Circuit Breakers

Equation I.4, solving for Ibf.....should there not be a term after "log"? All I see is an exponent that I can't really read.

Thanks!

yes indeed the only way the equation makes sense is if it reads Ibf = I1 = 10^(log * I1) not Ibf = I1 = 10*log^ I1

Statistics: Posted by arcad — Wed May 20, 2020 7:34 am

]]>

Thank you Jim.....I'm glad I was able to make a contribution

One other question while you're on topic.....the equation itself, to me, doesn't make sense......the term I1 is being solved for as a function of itself.....it's like having an equation such as X = 4X instead of X = 4Y......in the first equation, if X = 2, then it can't also be equal to 8.

I'll look forward to your response, and thanks!

Ernie Bowles

EB Consulting

After considerable thought my conclusion is - you might be the first to have actually read Annex I

This material is a carry over from the 2002 standard and the information was from a specific manufacturer and from what I understand was based on that manufacturer's devices. I believe it was an attempt to simplify what was known in 2002 and of course we know orders of magnitude more now.

Specific to your question: I agree equation I.4 seems like something is not correct. It is the same equation from 2002. I don't know if the "dot" should be a different operator such as + or what happened. I believe the people behind the original work are no longer involved.

The simple (?) solution is ignore it and focus on the 2018 equations.

Great question Ernie, it certainly got me scratching my head! (which now hurts)

btw, this is my personal view and can't be taken as an official position of IEEE or any other standards organization.

Thanks

- Jim

Statistics: Posted by Jim Phillips (brainfiller) — Tue May 19, 2020 2:27 pm

]]>

One other question while you're on topic.....the equation itself, to me, doesn't make sense......the term I1 is being solved for as a function of itself.....it's like having an equation such as X = 4X instead of X = 4Y......in the first equation, if X = 2, then it can't also be equal to 8.

I'll look forward to your response, and thanks!

Ernie Bowles

EB Consulting

Statistics: Posted by ebowles — Tue May 19, 2020 10:51 am

]]>

Annex I - Development of Special Model for Circuit Breakers

Equation I.4, solving for Ibf.....should there not be a term after "log"? All I see is an exponent that I can't really read.

Thanks!

I believe it may be a typo (unofficially - I have not confirmed with the committee yet)

This is verbatim (almost) from the 2002 Edition of IEEE 1584 page 62. In that edition the equation was:

Ibf = I1 = 10^(log * I1)

It looks like in the 2018 edition a superscript was lost. "log" should be part of the superscript. log * I1 not log^I1

Good catch - thanks. I will report back once I confirm.

Statistics: Posted by Jim Phillips (brainfiller) — Tue May 19, 2020 10:00 am

]]>

The thermal energy in a fault is proportional to I^2 t. But the IE result for any arc flash study always shows7 that the IE is proportional to I. I am very curious why is there such an anomaly.

There are many factors that come into play regarding incident energy. First let’s define the term incident energy. According to 2018 IEEE 1584, Incident Energy is defined as:

“The amount of

This means that the incident energy is not the total energy at the arc flash based sole on the current and time as in I^2. It is the energy that travels from the event towards a surface (worker) which means a few things such as.

-Not all of the thermal energy from the arc flash reaches the worker

-The energy dramatically decreases with distance from the arc flash

-The rate of decrease depends on the enclosure, opening size, voltage, electrode orientation and many other factors.

I hope this helps your understanding. Feel free to check back in if you have any other questions. Lots’ of people here ready to help.

Statistics: Posted by Jim Phillips (brainfiller) — Tue May 19, 2020 9:40 am

]]>

]]>

Equation I.4, solving for Ibf.....should there not be a term after "log"? All I see is an exponent that I can't really read.

Thanks!

Statistics: Posted by ebowles — Mon May 18, 2020 8:51 am

]]>

I apologize in advance if this has been asked, but can someone make heads or tails out of the very last equation in the Standard (also on the last page)? From what I can tell, it doesn't appear mathematically correct as it seems to be missing a term. Perhaps it's just my copy, but I don't think so. Any clarification would be greatly appreciated!

Thanks in advance.

Ernie Bowles

EB Consulting

could you please let me know what equation you are questioning exactly?

Statistics: Posted by arcad — Mon May 11, 2020 5:13 pm

]]>

Thanks in advance.

Ernie Bowles

EB Consulting

Statistics: Posted by ebowles — Thu May 07, 2020 2:28 pm

]]>