This weeks question was begun by "Matt780"

[url='http://arcflashforum.brainfiller.com/threads/poll-metal-enclosed-switchgear-above-15-kv.3234/#post-15002'][See Matt's original post][/url]

I thought I would move it to the "Question of the Week" to attract more attention to it.

IEEE 1584 is only for voltages up to 15 kV and the utility programs tend to be used more for single phase transmission and distribution lines and not three phase equipment.

This week's question focuses on how to model switchgear above 15 kV. Whether you have this type of equipment or not (most don't) what do you think would be the best method to you use for equipment above 15 kV?

Matt's question:


What method should be used for incident energy calculations for metal clad switchgear above 15-kV?

Using the 1584 method
Using the Ralph Lee Method
Using the NFPA 70E Task Based Tables
Classifying equipment as "Extreme Danger" (above 40 cal/cm^2)
Other Method (please explain)

Lee is 300% at the 15 kV boundary of 1584 data and the error increases with increasing voltage...overly conservative.
1584 does not apply (suggests Lee).
70E task table currently give values up to 38 kV but are going away in 2015 outside of 1584 limits (15 kV).
Recommend that since equipment design is same as used in generation/distribution equipment, use tables in NESC (IEEE C2) or ArcPro, which is what NESC values come from. Note that ArcPro recommends tripling values in enclosed gear but things get tricky with medium voltage gear because enclosure sizes are so large that it begins to be a mix of open air and enclosed conditions. Even 1584 gives no guidance above 1 kV for enclosed gear. C2 also has equipment specific tables based on research done by EPRI or PG&E.

Generally speaking medium voltage gear tends to have lower incident energies than 480/600 volt gear and it gets better as voltage increases. Marking anything, even >40 cal gear as "extreme danger" shows a total lack of understanding of the actual danger and discredits the engineer. There is a totally unfounded rumor that equipment over 40 cals kills on contact or even looking at it cross eyed due to a hazard called arc blast even though the physics do not support a 40 cal boundary.

PaulEngr wrote:

..... Note that ArcPro recommends tripling values in enclosed gear but things get tricky with medium voltage gear because enclosure sizes are so large that it begins to be a mix of open air and enclosed conditions. .....

This doesn't really have anything to do with switchgear, but the concept of enclosed versus open air. When you say the the size of the container is so large that it begins to be a mix of open air and enclosed conditions, where is the "so large" boundary?

I have some 1 and 2 MW inverters that are in enclosures/containers that are air-tight. The inverter is installed in an ~500 cu ft enclosure. There is a lot of other equipment inside the enclosure (filters, contactors, busbars, fuses, etc) with ample clearances, but of course it was designed to be as compact as feasible. There are several access panels and doors. There is a mix of voltages inside the enclosure (24 VDC, 120 VAC, 220 VAC, 315 - 480 VAC, and 500 - 1000 VDC).

Then I have battery containers that are the size of a 40' container trailer.

I have said that if you open one panel or door, it is enclosed conditions. If you open ALL the access panels and doors it is more like open air conditions.

Back to the original question, I believe most (all?) of the major arc flash programs that use IEEE 1584 automatically revert to the Ralph Lee equations at 15 kV. Not that it is the best solution above 15 kV (which is the subject of this poll) but that is what most people will end up with unless they intentionally do something different.

ArcPro.

Give me a few tens of thousands of dollars and I'll run the tests and find out for you. In fact you can just donate it to the joint IEEE/NFPA arc flash research program and save a lot of double dipping. Seriously, that group has been testing different size enclosures and finding that enclosure size plays a role in terms of incident energy. However so far nothing has been released publicly yet so there is not yet any information to go on.

The closest thing that we have is some equipment-specific data published by EPRI and PG&E, which is incorporated into the tables in IEEE C2-2012 (NESC).



I would count this all as enclosed. If we accept the interpretation that arc flash is being modeled as a thermal radiation problem (and it is), then as the source of the arc is more tightly packed into an enclosure, the extra/close surfaces act as both absorbers to some degree and reflectors to another degree. This tends to act more or less like a reflector like you see in a light fixture and concentrates the heat in a more directional way as opposed to a cable in open air that offers nearly perfectly spherical radiation. So the enclosure consideration will most likely have a lot more to deal with the amount of open area in the enclosure and this would have to be considered from an optical (infrared light) perspective. Changing from enclosed to open air would be for instance in a large 480 V CT cabinet with bus bars spaced 4" apart. This is reflected in the very low incident energy rating given in NESC for this type of cabinet. This is all speculation on my part from the open literature on the subject though so I might be way off track in understanding why "open" vs. "enclosed" conditions are so different.

How much working distance do we need to consider while calculating energy based on Lee method?