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 Post subject: Meaningful Arc Flash Calculations Possible for 208/240 8KVA?
PostPosted: Wed May 28, 2014 1:38 pm 
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We have someone that wants us to put NFPA 7E / ieee 1584 arc flash info on some control cabinets we made / are making.

The Enclosure is fed from a single phase 2P+G 208/240VAC, 6400W/8000VA UPS.

The Enclosure has a 40A 208/240VAC Disconnect Breaker that feeds several 480W 24VDC 20A Power Supplies.

The enclosure is labeled with 10K SCCR (in some cases 5K),
based on the branch circuit side device specifications.

The UPS manufacturer doesn't have SCCR, Bolted MVA, Transformer Impedance,
... information available for their UPS smaller than about 15KA.

A 50A Single Phase Breaker feeds the UPS input disconnect (2P+G).

The 480 to 240/208 transformer that feeds the UPS input disconnect information is not available to us,
and is considered by the end user to be out of our scope, so we don't need to know;
{... and the transformer hasn't been selected yet in most cases, by the time they want the ARC Flash info from us}.


The Shock Protection Boundary from Table 130.3C: 50 V–300 V Limited Approach Boundary 3.5ft,
Restricted Approach Boundary Avoid contact, Prohibited Approach Boundary Avoid contact seems fine.

However, as far as the information I have available, and using formula D.2(e),
with 8KVA and 6 cycle clearing time, e.g.: POWER((53*(0.008*0.125)*0.1),0.5)
I get less than 0.9in for the ARC Flash Boundary?

Providing the above information, can anyone direct us to any reference information (or even software products)
that would allow us to make any kind of meaningful e.g. NFPA 70E 130.5 / Annex D Arc Flash Hazard Analysis Calculations?


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 Post subject: Re: Meaningful Arc Flash Calculations Possible for 208/240 8
PostPosted: Wed May 28, 2014 2:50 pm 
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Location: Rutland, VT
A couple of items:

1. Arc flash is current under existing NFPA 70E and IEEE 1584 is only calculated as a 3 phase fault and resulting incident energy. That being said, using the 3 phase calculation for a single phase system will produce a conservative result.

2. Arc flash calculations done by IEEE 1584 will depend on the available bolted fault current at the location under study.

3. The equation you used from NFPA 70E is D.2e which if you refer to Table D.1 is only valid to calculate the arc flash boundary in open air. I believe from your description you have an arc in a box, so should be using D.7 equations, which require the available bolted fault current (See #2 above)

In my opinion and I think most people who do this type of analysis, will tell you that it needs to done using the available fault current at the location the equipment is installed in. This fault current could change depending on the electrical system parameters where the equipment is installed. What I mean by this is there maybe an emergency generator that could supply this eqpt and that fault current and resulting arc flash will be different than normal configuration


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 Post subject: Re: Meaningful Arc Flash Calculations Possible for 208/240 8
PostPosted: Fri May 30, 2014 7:41 am 
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Location: North Carolina
First, sounds like an equipment manfuacturer. The end user has to do arc flash, sorry to say.

Anitech Systems wrote:
Providing the above information, can anyone direct us to any reference information (or even software products)
that would allow us to make any kind of meaningful e.g. NFPA 70E 130.5 / Annex D Arc Flash Hazard Analysis Calculations?


IEEE 1584 equation is based on exactly ONE test where they managed to achieve a stable arc at 208 V and that is used as the "assumption" for the entire range of 50-300 V data. In other words, IEEE 1584 is invalid. The reason that there is so little data is simply because arcs are very unstable at low voltages (usually need around 150 V or so to even initiate an arc so there is a good chance it will not restrike and may do so for several cycles). In addition at this point there is no single phase test data used within the IEEE 1584 test data set. So using that methodology is at best questionable and the standard itself says so.

At this point I recommend using NESC-2012 table which is in turn based on actual test data. Incident energy = 4 cal/cm^2.

As no 480 V data...all information downstream of the transformer is invalid because you cannot determine the available fault current. Worse still if you assume a very high fault current this corresponds to fast trip times which also results in incident energy since trip time is short. However if the available fault current is very large then trip times are very long which results in a very high incident energy. Only the actual field results are valid. You just have to stonewall the customers and trust me, this comes up over and over again. Even in my case where I'm the plant electrical engineer and the project engineers think that it should just be a specification on a piece of equipment when I have no idea what they are actually buying ahead of time.

As to figuring out what to do with UPS's or VFD's, on the front end where everything is still copper, you can safely use the available fault current as it goes through the disconnect, terminals, cable, etc. If it's a line interactive or "offline" type where essentially the UPS passes line current through directly to the load then the fault current also passes straight through so this is relatively easy to model. If on the other hand it is a double-conversion type or a drive then there is no direct connection. You can usually safely assume that the available fault current is about twice the drive's maximum current capability since the semiconductors are rated for RMS current before they rupture in very short order (usually 1 or 2 cycles at most), if the drive even lets it get that far. Drive protection will kick in well before then. The only other factor to consider are the capacitors (DC link) and/or batteries. Batteries can be treated separately as a DC system and you need to know series resistance among other factors. Capacitors can be looked at as the amount of stored energy. Convert joules to calories. Then take the working distance and calculate the area of a sphere with the working distance as a radius in square cenitmeters. Divide by 2 to assume a hemisphere (assumes capacitor is in an enclosure). Then your final answer = calories available / area of hemisphere which will be in cal/cm^2. This is ultraconservative again because only a portion of the heat energy actually gets radiated (most theoretical guesses estimate about 25%) but the best you will be able to do.


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