It is currently Thu Jan 27, 2022 11:07 pm

Author Message

 Post subject: Arc Flash Calculation on Delta High LegPosted: Wed May 02, 2012 5:03 pm
 Sparks Level

Joined: Tue Apr 03, 2012 1:28 pm
Posts: 53
Location: Louisville KY
Okay...I am a bit confused. I was overlooking a study at a facilty that paid for a full-blown arc flash analysis. It appears that the company may have used EasyPower software to perform the analysis. To me the numbers do not add up. Can someone confirm to me the following:
1. In calculating short circuit current with a Delta high-leg system, the following data was provided:

300 KVA transformer with 6.37% impedence and an X/R ratio of 3, 480 v primary/ 240 v secondary 3-phase delta high leg.
The listed s.c.c is 9.127 kAmps. The secondary feeds a panelboard that feeds a bus duct system. The panelboard is 10 foot away from the transformer in steel conduit and is fed via 4 per phase 350 mcm cables.

I have calculated this several times and get higher s.c.c. values than what is given by the software.

Also, this panelboard is listed at an arc flash incident energy of 73 cal/cm2. I am assuming the high level is because the 480/240V transformer has no secondary protection until it terminates in the panelboard. I am assuming that the software calculated a high danger level depending on primary breaker protection. The breaker feeding the transformer primary is a Square D Type MH 125-1200 A with an Int rating of 65 kA and a 600 a trip. The breaker lists a thermal curve (fixed) with a 5-10x trip instantaneous and the data sheet says 6(3600 a). (???).

Even at a 1 second trip time for this breaker, I do not come up with 73 cal/cm2 or any where near that value. i am more along a 48 cal level at 18 inches.

Any guidance here? Are the calculations different for a Delta High-leg secondary? I wouldn't think so since we are still dealing with overall impedence to calculate short circuit current...right?

-Ken

Top

 Post subject: Posted: Thu May 03, 2012 5:57 am
 Sparks Level

Joined: Wed Jan 28, 2009 2:19 pm
Posts: 73
Location: Georgia
The original company may have looked at several scenarios (back-up generator, increased utility contribution, decreased utility contribution, motors on/off, etc.) in their evaluation and only posted the results for the worst case. Just a thought.

Top

 Post subject: Posted: Thu May 03, 2012 7:32 am
 Sparks Level

Joined: Tue Apr 03, 2012 1:28 pm
Posts: 53
Location: Louisville KY
Good point but isn't the account calculation part pretty straightforward? Transformer impedance and KVA and voltage of the secondary determines this...unless the high leg delta affects the calculation differently (not sure on this one...)

Top

 Post subject: Posted: Thu May 03, 2012 9:53 am
 Arc Level

Joined: Mon Jan 18, 2010 11:35 am
Posts: 551
Location: Wisconsin
For discussion, if you ignore the 120V single phase connections, you can usually treat the 240V high leg as a standard delta.
What SCA are you getting? The 9.12kA value sounds pretty reasonable.

When evaluating clearing time, remember IEEE recommends performing calculations at 85% and 100% Arcing fault values, and then using the high result.

Top

 Post subject: Posted: Thu May 03, 2012 10:52 am
 Sparks Level

Joined: Tue Feb 24, 2009 7:05 am
Posts: 252
Kenneth Sellars wrote:
1. In calculating short circuit current with a Delta high-leg system, the following data was provided:

300 KVA transformer with 6.37% impedence and an X/R ratio of 3, 480 v primary/ 240 v secondary 3-phase delta high leg.
The listed s.c.c is 9.127 kAmps. The secondary feeds a panelboard that feeds a bus duct system. The panelboard is 10 foot away from the transformer in steel conduit and is fed via 4 per phase 350 mcm cables.

I have calculated this several times and get higher s.c.c. values than what is given by the software.

If the bolted short-circuit current upstream of the transformer is about 23.9 kA at 480V, and you don't count motor contribution from downstream, then the listed scc looks good to me at the panelboard.

As to the Ie, at 9.1kA Ibf, 240V, 18 in working distance, arc-in-a-box, solidly grounded system, 25 mm arc gap and a clearing time of 5 second on the CB on the 480V, you can get around 74 cal/cm^2. Why they haven't used the 2 second rule? Don't know.

Top

 Post subject: Posted: Thu May 03, 2012 2:13 pm
 Arc Level

Joined: Tue Jan 13, 2009 5:00 pm
Posts: 579
A three phase to ground fault will be unbalanced so you won't have a single short circuit current. I would expect the high leg fed by two transformers to have the highest current.

Top

 Post subject: Posted: Thu May 03, 2012 4:59 pm
 Sparks Level

Joined: Tue Apr 03, 2012 1:28 pm
Posts: 53
Location: Louisville KY
Exactly what I was thinking...I cannot obtain a value of 74 cal unless I make the c/b trip time ridiculously high. I finally got the data from the breaker, and on the trip curve, this breaker should go at about 0.033 seconds...which yields a very low calorie level...which is what I would expect with such a low short circuit current. Not really sure why this company gave the results that they did.

Top

 Post subject: Posted: Thu May 03, 2012 5:00 pm
 Sparks Level

Joined: Tue Apr 03, 2012 1:28 pm
Posts: 53
Location: Louisville KY
Any special way to calculate this?

stevenal wrote:
A three phase to ground fault will be unbalanced so you won't have a single short circuit current. I would expect the high leg fed by two transformers to have the highest current.

Top

 Post subject: Posted: Fri May 04, 2012 4:06 pm
 Arc Level

Joined: Tue Jan 13, 2009 5:00 pm
Posts: 579
Sorry. My books show equivalent sequence networks for lots of possible faults, but the three phase to mid-winding fault is not covered. Anyone else know how to calculate this?

Top

 Post subject: Posted: Sat May 05, 2012 4:20 am
 Plasma Level

Joined: Tue Oct 26, 2010 9:08 am
Posts: 2174
Location: North Carolina
Sequence networks assume that you can separate the system into three networks, two balanced and one unbalanced. High leg systems just change the modelling of the third ("0") network. The positive and negative sequence networks are identical. Just grind through the formulas (multiplying by "a") to calculate Z0 based on your starting formulas. The major difference is that unlike balanced systems (ungrounded deltas or grounded wye's), you will have zero sequence current flowing even under normal conditions.

However, the major advantage of symmetrical components theory is that for BALANCED systems, all 3 phases are effectively identical in how they are treated. Analysis of a single phase fault is applicable regardless of which phase it occurs on using symmetrical components theory.

With a high leg delta system however, symmetrical components theory falls apart. The high leg must be treated differently from the two lower phase legs for every fault. You have to do the math twice in every situation. Thus there is no inherent value in doing symmetrical component theory in high leg delta systems. You are not moving towards an inherently mathematically simpler system.

So...the reason you probably don't see anyone doing it is that there's no value in doing so.

Top

 Post subject: Posted: Sat May 05, 2012 7:28 am
 Sparks Level

Joined: Tue Apr 03, 2012 1:28 pm
Posts: 53
Location: Louisville KY
So, Paul, can programs like EasyPower be trusted in these scenarios? Do they take into acccount the dynamics involved in high-leg faults? I may have to send this question to EasyPower to find my answer...just curious if you might know.

...is there a good reference book for these type of transformer calculations?

Top

 Post subject: Posted: Mon May 07, 2012 4:28 pm
 Arc Level

Joined: Tue Jan 13, 2009 5:00 pm
Posts: 579
Paul,

I suggested sequence network analysis, since I understood it was applicable to unbalanced faults. If you have a better way, or any way at all, of analyzing a three phase to ground fault on these systems; please share it.

If the only change is in the zero sequence network and the networks connect conventionally, there would be no difference in the three phase fault which involves only the positive sequence network. All phases see the same magnitude of current regardless of grounding. Is this your position? Thanks.

Top

 Post subject: Posted: Mon May 07, 2012 6:04 pm
 Plasma Level

Joined: Tue Oct 26, 2010 9:08 am
Posts: 2174
Location: North Carolina
I am not 100% sure of what Easypower would and would not handle because I'm not familiar with it. Every modeling package out there has its quirks and it is best to know what those are.

The theory of symmetrical components supports any particular configuration, but the modeling software itself makes some assumptions and may make life difficult if it is not general enough. I can think of no better example than the one you are describing because I can't think of a way to set up a high leg distribution system in the two packages I'm more familiar with (SKM and Easytap).

All that being said if you are only concerned with L-L and L-L-L faults, using the appropriate transformer configurations, EasyPower would still produce valid results because the results are the same electrically (which is also true if performing the calculations by hand). The positive and negative sequence networks should be effectively the same. If however you attempted to calculate a L-G or L-G-L fault, you will get incorrect results unless you model your particular grounding system correctly (zero sequence network).

Fortunately if your goal is to model bolted faults (short circuit) and worst case arcing faults, you're in luck...everything works. It will only be if you attempt to model ground faults and your grounding is not modeled correctly that you will get invalid results.

As to a book...the best one I have is from Alstom...used to be Areva T&D's relay book. It is free to download. It goes into detail not only on doing the basic calculations but also gives detailed data on almost any electrical component in other chapters. There are better references for doing overhead distribution lines but for everything else it's invaluable. Link below for downloading.
[url='ftp://[email protected]/NPAG/']ftp://tdeftp_public:[/url][url='ftp://[email protected]/NPAG/'][URL='ftp://[email protected]/NPAG/']tdeftp_public[/url][/URL][url='ftp://[email protected]/NPAG/']@ftp.areva-td.com/NPAG/[/url]

Top

 Post subject: Posted: Mon May 14, 2012 3:30 pm

Joined: Tue Jul 19, 2011 2:38 pm
Posts: 8
Location: Brisbane, Australia
JBD wrote:
For discussion, if you ignore the 120V single phase connections, you can usually treat the 240V high leg as a standard delta.
What SCA are you getting? The 9.12kA value sounds pretty reasonable.

When evaluating clearing time, remember IEEE recommends performing calculations at 85% and 100% Arcing fault values, and then using the high result.

Do you know which section of IEEE that states that? I found something similiar in NFPA 70e under appendix D for an example calculation but that is just an example and not a requirement. I'll be nice to see the quoted text as well. Cheers.

Top

 Post subject: Posted: Mon May 14, 2012 5:01 pm
 Plasma Level

Joined: Tue Oct 26, 2010 9:08 am
Posts: 2174
Location: North Carolina
Strudel wrote:
Do you know which section of IEEE that states that? I found something similiar in NFPA 70e under appendix D for an example calculation but that is just an example and not a requirement. I'll be nice to see the quoted text as well. Cheers.

Appendix D quotes several different procedures (and so does IEEE 1584). The quote comes from IEEE 1584. The inherent difficulty in the results is in predicting opening time. IEEE 1584 has you calculate the opening time for the overcurrent device twice, at 100% of arcing fault current, and again at 85% of that value. The worst case is used because this increases the reliability of the calculated incident energy up to around 90-95% confidence level. This is very well described in the notes in IEEE 1584.

Top

 Post subject: Posted: Thu Aug 09, 2012 5:17 pm
 Arc Level

Joined: Tue Jan 13, 2009 5:00 pm
Posts: 579
So this question is starting to come a little closer to home. Many of these delta connections serve combined single and three phase loads, and are made up of single phase units. The tapped "lighter" will be of a larger kVA size, while the two connected to the wild leg will be smaller. Or one will be left out altogether for an open wye-open delta connection. Any more advice on how to calculate fault currents on these unbalanced sources? Thanks.

Top

 Post subject: Posted: Sat Aug 18, 2012 5:57 am
 Plasma Level

Joined: Tue Oct 26, 2010 9:08 am
Posts: 2174
Location: North Carolina
Since you don't even have balanced transformers it is sounding more and more like you'd have to simply treat this as 3 single phase systems and take the worst case out of all 3. IEEE 1584 can't directly be used because of course what you are proposing is not balanced in the slightest. But you can at least get conservative results by looking at the worst case fault current.

This is already being done by most practitioners for single phase cases. There is no method for single phase arc flash calculation that I'm aware of aside from making a huge leap and assuming that you can divide by three (hint: you can't). You would be doing the same in your case by using the 3 phase calculations and knowing that the actual incident energy will be less. How much less is not really known.

Top

 Post subject: Re: Arc Flash Calculation on Delta High LegPosted: Sun Jan 20, 2019 10:06 pm

Joined: Sun Jan 20, 2019 10:01 pm
Posts: 1
To find the fault current for high-leg delta, one must consider the unbalance to ground via the ground connection, and one must consider the unbalance to ground due to the fault. This setup is done via two perfect transformers with one transformer representing the primary, two representing the secondary. With a little complex mathematics one can derive the fault current.

Top

 Post subject: Re: Arc Flash Calculation on Delta High LegPosted: Mon Jan 21, 2019 5:15 am
 Plasma Level

Joined: Tue Oct 26, 2010 9:08 am
Posts: 2174
Location: North Carolina
bmarshall7688 wrote:
To find the fault current for high-leg delta, one must consider the unbalance to ground via the ground connection, and one must consider the unbalance to ground due to the fault. This setup is done via two perfect transformers with one transformer representing the primary, two representing the secondary. With a little complex mathematics one can derive the fault current.

1. This is a 2012 thread.

2. How would you construct such a thing in Easypower or any other power system analysis software? Sure we can do the calculation by hand but that doesn't appear to be the case.

Top

 Display posts from previous: All posts1 day7 days2 weeks1 month3 months6 months1 year Sort by AuthorPost timeSubject AscendingDescending
 Page 1 of 1 [ 19 posts ]

 All times are UTC - 7 hours

 You cannot post new topics in this forumYou cannot reply to topics in this forumYou cannot edit your posts in this forumYou cannot delete your posts in this forumYou cannot post attachments in this forum

 Jump to:  Select a forum ------------------ Forum Library / Articles The Lounge    Question of the Week - What Do You Think?    Arcflashforum.com Feedback and Announcements    Off Topic Discussions    News in Electrical Safety Arc Flash and Electrical Safety    General Discussion    Electrical Safety Practices    Equipment to Reduce Arc Flash Dangers    Personal Protective Equipment (PPE) Arc Flash Studies    General Discussion    Arc Flash Labels    Software for Arc Flash Studies    System Modeling and Calculations    NEW! Electrode Configuration Library – 2018 IEEE 1584 Codes and Standards    CSA Z462 Workplace Electrical Safety    EAWR Electricity at Work Regulations, HSE - Europe    OSHA CFR Title 29    IEEE 1584 - Arc-Flash Hazard Calculations    NFPA 70 - National Electrical Code - NEC (R)    NESC - ANSI C2 - National Electrical Safety Code    NFPA 70E - Electrical Safety in the Workplace    2015 NFPA 70E Share It Here    Arc Flash Photos    Your Stories    What's Wrong Here? by Joe Tedesco
© 2022 Arcflash Forum / Brainfiller, Inc. | P.O. Box 12024 | Scottsdale, AZ 85267 USA | 800-874-8883