Wound rotor induction motors have an external circuit coming from the motor rotor, through slip rings, to a variable resistor bank. The slip ring enclosure and the resistor bank cabinet have an arc flash risk. Is there an established method/standard for calculating the incident energy along the motor secondary circuit? Have you ever run into the situation before?

I do not have any experience with wound rotor induction motors.

Any one cares to enlighten me? Thanks.

Statistics: Posted by RECS — Thu Aug 13, 2020 11:35 am

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Hi Jim: Does the IEEE 1584 committee have any plans for higher voltage overhead lines? Like Arc Pro?

Thanks for all you do!

There has been some discussion about extending to the 35-38 kV range for equipment but it is just an informal discussion. So nothing anytime soon. There is nothing about higher voltage overhead lines. Too much left to do at lower voltages first.

Statistics: Posted by Jim Phillips (brainfiller) — Tue Aug 11, 2020 9:26 am

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Statistics: Posted by K. Engholm — Mon Aug 10, 2020 11:18 am

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My understanding is that the first priority will probably be some DC content.

My personal desire would be acquiring data at higher fault current values, hopefully all the way to 200kA.

Both of these type of projects require testing few laboratories can provide and have significant cost associated. As Jim stated, this is expensive and funds are limited.

BUT keep your cards and letters coming to whoever you know in the IEEE working group. It is important that whatever expansion is undertaken it addresses real needs in the industry.

Yes DC is high on the list. I have been pushing DC for years and a while ago it up on the radar screen. Single phase is also on the list as mentioned. Going to 200 kA will be quite difficult due to lab constraints. We were limited this time around. As I understand, one of the labs had some damage at higher currents and throttled the upper current capabilities back. It is also expensive as we all know. Stay tuned, it all moves slow but it is moving in the right direction!

Statistics: Posted by Jim Phillips (brainfiller) — Mon Aug 10, 2020 7:55 am

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My personal desire would be acquiring data at higher fault current values, hopefully all the way to 200kA.

Both of these type of projects require testing few laboratories can provide and have significant cost associated. As Jim stated, this is expensive and funds are limited.

BUT keep your cards and letters coming to whoever you know in the IEEE working group. It is important that whatever expansion is undertaken it addresses real needs in the industry.

Statistics: Posted by Electricidad — Mon Aug 10, 2020 7:47 am

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Are there any plans for the next (3rd) edition of IEEE 1584? What would it include?

We are still catching our breath from the 2018 Edition. Many items were left out of the 2018 Edition due to budget reasons. The IEEE/NFPA Collaboration for this project had an ambitions test plan with a proposed budget of around $6.5 Million. Fundraising was cut short during the 2007-2008 economic situation so the final budget was about half of what was planned. This meant many areas were left for a future date.

IEEE has something called "dot" standards. The hope is rather than starting the whole process over again, we can address outstanding areas one at a time with it's own unique "dot" standard. We already have IEEE 1584.1 "dot-one" which I co-authored the first draft with a colleague years ago. It is undergoing revision right now.

We hope to someday add additional dot standards for areas such as single phase, DC, higher voltages and other areas although nothing has officially begun yet. The research is a continuous process and we will all keep moving the knowledge base forward.

**Usual disclaimer: Although I'm Vice Chair of IEEE 1584, this is my personal view and may or may not represent the view of IEEE or any other standards organization.

Statistics: Posted by Jim Phillips (brainfiller) — Fri Aug 07, 2020 7:39 am

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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

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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

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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

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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

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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

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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

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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

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