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 Post subject: Arc Flash Boundary at 40 cal/cm2
PostPosted: Tue Apr 12, 2016 11:38 am 
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We have some labels (which I hope to change out sooner, rather than later) that just tell you the voltage and table category of the equipment. For something labeled a category 4, I am going to assume the equipment is at 40 cal/cm2, and we need to use that value for our arc flash boundary. But what I'm wondering is what is the arc flash boundary given an incident energy of 40 cal/cm2?


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 Post subject: Re: Arc Flash Boundary at 40 cal/cm2
PostPosted: Tue Apr 12, 2016 4:45 pm 
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Voltrael wrote:
We have some labels (which I hope to change out sooner, rather than later) that just tell you the voltage and table category of the equipment. For something labeled a category 4, I am going to assume the equipment is at 40 cal/cm2, and we need to use that value for our arc flash boundary. But what I'm wondering is what is the arc flash boundary given an incident energy of 40 cal/cm2?


The hands down easiest approach is to use the assumptions section of the tables in 70E which at least in the 2015 edition give you the arc flash boundaries. That is the method that I recommend. If you don't have a copy it is a little time consuming but you can page through the entire thing or jump chapter-by-chapter to get the values from the nfpa.org web site.
.
Keep in mind that there is no one-size-fits all result. In 2015, we have 3 subtables: a task/risk table, an equipment table, and a PPE table. You get the PPE level from the equipment table which also includes all the assumptions, arc flash boundaries, etc. The PPE table is very similar to 2012 and earlier editions. The equipment won't change so you can just jump to that table to get the information you seek.

You can't just convert to a single result. The issue is what is known as "radiative view factors". A long enclosed box such switchgear tends to "focus" the thermal radiation compared to a relatively flat shallow box such as a panelboard. Thus IEEE 1584 includes different exponents to deal with differences in enclosure shapes. So the arc flash boundary for a row of medium voltage starters is going to be a lot farther away than a panelboard even if both are rated for 40 cal/cm^2, even if the working distance was set to be the same for both of them (which it isn't).

As per IEEE 1584 (and Annex D in NFPA 70E) the first step in doing an empirical arc flash calculation is to determine the arcing current followed by the normalized incident energy (normalized to 610 mm, 0.2 seconds arcing time). The next step (and what we need for this discussion) is to run the denormalizing formula backwards. E = Cf * En * (t / 0.2) * (610^x)/(D^x) where:
Cf is a calculation factor (1.0 for >= 1 kV, 1.5 for < 1 kV). I'm assuming you have this.
En is the normalized energy (what you need).
t is arcing time. This is given in the 70E tables.
x is the calculation exponent and D is the working distance.. This information is given in IEEE 1584 and restated again in the table in Annex D.

Once you have recovered the normalized distance, rerun the formula "forwards" while adjusting the distance D until you get to 1.2 cal/cm^2 or do the math to solve for D in the above calculation to determine the arc flash boundary directly.

Note that while doing all this work though you will have to look for the arcing time on the tables in the exact same place where the arc flash boundary is already given.


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 Post subject: Re: Arc Flash Boundary at 40 cal/cm2
PostPosted: Tue Apr 12, 2016 5:13 pm 
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Voltrael wrote:
We have some labels (which I hope to change out sooner, rather than later) that just tell you the voltage and table category of the equipment. For something labeled a category 4, I am going to assume the equipment is at 40 cal/cm2, and we need to use that value for our arc flash boundary. But what I'm wondering is what is the arc flash boundary given an incident energy of 40 cal/cm2?


If incident energy at working distance value is known, system voltage is less than 15kV and gap between conductors is less than 6 inches, the arc flash boundary can be calculated using equation below derived from IEEE 1584:

Db = (E/Eb)^(1/x) * D,

where
Db - arc flash boundary in mm
E - incident energy exposure in J/cm2
Eb - incident energy in J/cm2 at the boundary distance
D - distance from possible arc point to person in mm
x - distance exponent from IEEE 1584 Table 4

If incident energy at working distance is equal to 40 cal/cm2, than arc flash boundary is roughly 10 times the working distance.

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Michael Furtak, C.E.T.
http://arcadvisor.com


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 Post subject: Re: Arc Flash Boundary at 40 cal/cm2
PostPosted: Mon Apr 18, 2016 4:52 am 

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The tables in 70E would be a good place to start. But the problem I have with the tables is you need the available fault current and the clearing time of your protective device to be correctly following them. To go through all that trouble you should have a study done. If you have labels than I would assume a report was already done and you should have that information somewhere, if not in the report than the engineer that performed the study should be able to get that for you. I don't like to assume anything when it comes to keeping my techs safe though. When I train our techs I show them where they can find any information they need in their arc flash reports and instruct them to call me if they have any questions.


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 Post subject: Re: Arc Flash Boundary at 40 cal/cm2
PostPosted: Mon Apr 18, 2016 10:44 am 
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lovetacycle wrote:
The tables in 70E would be a good place to start. But the problem I have with the tables is you need the available fault current and the clearing time of your protective device to be correctly following them. To go through all that trouble you should have a study done. If you have labels than I would assume a report was already done and you should have that information somewhere, if not in the report than the engineer that performed the study should be able to get that for you. I don't like to assume anything when it comes to keeping my techs safe though. When I train our techs I show them where they can find any information they need in their arc flash reports and instruct them to call me if they have any questions.


First off the OP's question is how to go from a presumably table result back to an acceptable arc flash boundary. For all I know the incident energy analysis was done and converted into H/RC values which are then applied and that all of the calculation data is unavailable, leaving the OP with the option to spend a large amount of money building a new model or else doing something to recover an arc flash boundary which was not on the original labels.

Second a tabular approach is one of the best approaches for the "low voltage" case where empirical and theoretical modelling have failed. In this case the best approach is to look at boundary conditions (how bad can it get for a given set of inputs) and establish a table of PPE based on that information. IEEE 1584 is useless below 250 V, below a certain fault current, for single phase cases, and for anything above 15 kV. When I say this by the way I'm not meaning tabular == NFPA 70E. This is not always the best one to use.

Third a tabular approach is the best to use when it is not practical to be able to do an incident energy analysis in the first place. In all of these cases, a tabular approach specifically such as NFPA 70E is better than nothing at all and that may be the only practical choice. Four examples come immediately to mind. The first is when a service vendor goes into plants where an incident energy analysis has not been done so the contractor has to determine a reasonable incident energy on the spot with almost no information to go on. The second case is during startups or during a preliminary incident energy analysis where previously none existed so some sort of baseline needs to be established without benefit of having an analysis. Frequently due to performance-based contracts it may not be possible to determine the exact equipment design specifications until much later in the process and with delivery times sometimes under 4-5 weeks or even shorter in rush/emergency conditions, there's just not enough time to do an engineering study ahead of time. The third case is when a particular employer is simply not willing to spend money on an incident energy analysis. There are a variety of reasons why they may be unwilling to pay for a "good" study so either a combination of the tabular approach and/or an engineering study or the tabular approach alone may be the only option. The fourth and final case is when the distribution system is either in a condition where it is being rapidly changed under emergency conditions (such as loss of a distribution substation) or when constant and routine changes occur to the system to the point where it exceeds the ability to analyze either all scenarios or even just to handle the situation at hand. A case in point is a "typical" gold mine in the Western part of the U.S. These operations start up and go into service with 18-24 month production cycles before they close down. In that environment you'd be lucky to get a completed arc flash study in the first year and by the time everyone is trained and up to speed in year 2, they're thinking about the next job and not the current one.

In each of the four previously mentioned cases the common thread is that there is not enough time or money to do an engineering study. In that case the table method is a fall back strategy. I think that there is plenty of interest in developing better tables than those in 70E but so far that goal seems to be elusive. NESC comes a long way in that the only required information for most of the entries is knowledge of maximum short circuit current and system voltage, both of which are easily had quickly. But for industrial enclosed panels which are of interest (such as 480/600 V class equipment), NESC falls short as well.


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 Post subject: Re: Arc Flash Boundary at 40 cal/cm2
PostPosted: Tue Apr 19, 2016 8:36 am 
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Okay, more information about my facility. I started working here last fall. I was finally able to obtain both the SKM model performed by the engineering firm associated with the mill and a copy of Power*Tools within the last month. Yesterday I received a label printer so I am able to print updated labels myself.

My plan is to extend the model (which is pretty much at the MV level) down to the 480 level and relabel everything. I'm also concerned they didn't include enough scenarios in the arc flash model given we have four turbine generators and a utility feed for the mill. I will be studying that to see what the potential impact on the arc flash modeling is.

Anyway, what I have found in use in the field are two labels.

One of them is fairly typical of what you would find in the recent past. It includes
"Arc Flash hazard Clothing Category" (which directly follows the tables)
Incident Energy
Flash Approach Boundary
Limited Approach Boundary
Restricted Approach Boundary
Prohibited Approach Boundary
Shock Hazard (voltage present)
PPE Required
Equipment ID

So I'm not too worried about that label. The label I update with will remove the clothing category, prohibited approach boundary and PPE required.

The other label I found has the following:
Shock Hazard (in VAC)
Arc Flash Hazard (As a Level, corresponding to the tables)

And that is it, just those two data points. I'm just trying to come up with a general rule to give to people to deal with these labels until I can replace them with proper labels. The guidance I have decided on is to tell people to assume a boundary of ten feet for Level 2 and forty feet for Level 4.

And I will try to work in updating these labels as quickly as possible.


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 Post subject: Re: Arc Flash Boundary at 40 cal/cm2
PostPosted: Tue Apr 19, 2016 12:08 pm 
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As far as number of scenarios go there are only a minority of cases to consider and we can start actually with the TCC's before even looking at scenarios. You only need to look for two critical points on the curves, assuming that short time settings are high enough not to create any additional "inflection pints".

1. What is the minimum fault current at which the clearing time is 2 seconds or greater?
2. What is the minimum fault current which causes an instantaneous trip (if any)?

This breaks time up into 3 regions delineated by specific fault currents. Going from lowest to highest:

A. As fault current is decreased and the clearing time is 2 seconds or greater, decreasing fault current decreases incident energy. Need to check the case for only the utility and only for a single turbine since typically one or the other may reach this condition. Note that this lower point is usually (but not always) the greatest incident energy.
B. All points between the two critical points. As the fault current increases (usually by adding generators), the net increase in arcing power is offset by a corresponding and larger decrease in fault time resulting in a net DECREASE in incident energy throughout this region. Thus counter-intuitively, the largest incident energy is at the lowest fault current. This occurs for all IEC and ANSI standard curves.
C. Points above an "instantaneous trip" value. At this point clearing time is fixed and increases in fault current follow the intuitive expectation that incident energy increases. This point may exceed the value at the zone between region A and B in some cases so it pays to look at "all turbines and the utility online" for this purpose.

Depending on your particular system, region A or C quite often does not exist, and in some cases there may be only a single region and no region B either. Thus only a very small handful of scenarios (say only utility, 1 turbine, and all turbines) may be necessary for analysis. Initially of course you need to survey "everything" to determine what the above system limits are and then after that for reporting purposes the extra scenarios are simply not necessary.

Furthermore typically the "generator bus" is very stiff and makes almost no difference downstream of feeder breakers fed from this bus so you may only need to run various scenarios for the generator bus. Downstream busses can then be analyzed with only a single scenario.

Finally in terms of your description my experience is that incident energy is relatively mild in medium voltage systems and becomes a much bigger problem as voltage decreases. By way of example try a very simple experiment in SKM. Create a simple fuse protected distribution transformer with a 4160 V secondary side and size it for 2500 kVA with a simple load such as a disconnect. Use say 5.75% Z for impedance. Check the incident energy at the disconnect on the secondary side which is only protected by the fusing on the primary side. Now with the exact same scenario and transformer, adjust the secondary voltage to 480 V, still with 5.75% Z. Recheck incident energy...it will probably be above 40 cal/cm^2 compared to the medium voltage scenario.


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