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 Post subject: Stand off Distance
PostPosted: Sat Sep 19, 2009 6:16 am 
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My company has completed ARC flash studies and has calculated the standoff distances (for unprotected workers) when work is being done that requires flash protection. My company's standard is not explicit in that it states a distance but does not elaborate. Example: one breaker on our 4kV switchgear requires a stand off distance of 141 Feet. I would believe that is for line of sight but people are interpreting it differently and in fact some have created a sphere with a radius of 141 feet when the breaker is being disconnected or grounds attached. It would seem to me if you are behind the switchgear, around a corner or behind a cinder block wall, the 141' does not apply. Does anyone have an interpretation on this, source, citation or any kind of standard? I am trying to clear up this 'gray' area so my people are protected yet not have to shutdown work in an entire building.


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PostPosted: Sat Sep 19, 2009 7:53 am 
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paulsalgad2 wrote:
Example: one breaker on our 4kV switchgear requires a stand off distance of 141 Feet.


Welcome to the Forum. I would like to see the calc that puts a 4kv flash extending that far. As far as your 'sphere', lets use some common sense. Lets apply some basic physics here. :rolleyes:


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PostPosted: Sat Sep 19, 2009 10:25 am 
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I would like to know what type of work is being done on equipment with an 141ft FPB, and if it is truly "energized work" how that is being justified.

What is the Ei at 18" or whatever the assumed working distance is?

To answer your question, it is a sphere, and has been shown that way in previous 70E versions so you won't find anything that says otherwise (Like you said, grey area), however as Chris said you should use some common sense.


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PostPosted: Sun Sep 20, 2009 11:41 am 
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You could set off 10 sticks of dynamite at 150ft and not get burned. The IEEE equation just doesn't make sense when it starts to spit out numbers like this. For 5KV switchgear I would think that a boundary of 30ft would give adequate protection.


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PostPosted: Mon Sep 21, 2009 5:13 am 
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We have similar issue with excessive boundaries for equipment directly connected to a generator or equipment where the theoretical fault clearing time hits the 2 second cut off prior to clearing the fault. The calculation times out at the 2 second maximum resulting in high calculated incident energy and a large boundary. Adjustments and mitigation are usually possible but not always. These locations have caused us numerous problems with regards to labeling. The plant questions a flash boundary of 100├óÔéČÔäó, as do we. This is more evidence a maximum boundary is needed in depended of what the equations say and the fact more testing is needed to refine the standard.


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PostPosted: Tue Sep 22, 2009 12:27 am 
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I cannot find it at the moment, but there has been an article to this effect. (It may have been Lee's "Pressure due to Arcs"?)
If I remember correctly, it stated that "substantial" walls would be sufficient to withstand the arc blast and could serve as a boundary limit. I think it said that a "substantial" wall was brick, block, or even some drywall walls.

I'll keep looking for the article and post more when I find it.


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PostPosted: Tue Sep 22, 2009 2:44 am 
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WDeanN wrote:
I cannot find it at the moment, but there has been an article to this effect. (It may have been Lee's "Pressure due to Arcs"?)
If I remember correctly, it stated that "substantial" walls would be sufficient to withstand the arc blast and could serve as a boundary limit. I think it said that a "substantial" wall was brick, block, or even some drywall walls.

I'll keep looking for the article and post more when I find it.


But the boundary is determined by incident energy level, not arc pressure. A long duration arc can have a high IE with relatively low pressure.

I don't think that the equations for incident energy or flash hazard boundary are really valid for large distances. The energy for an arc in a box is concentrated by the sides of the box, but this effect would not be much after a few multiples of the box depth. In a recent study, we calculated an IE of 68 cal/cm² for a 2 second 480 volt 13 kA arc, with a boundary of 17.6 feet. I doubt seriously that you would get second degree burns 17.6 feet away from this arc in two seconds.

I can't imagine getting burned 141 feet away from any arc, regardless of the current or duration. Particularly if there is anything between you and the arc or if it is a different room or building.


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PostPosted: Tue Sep 22, 2009 6:36 am 
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When students question the validity of these long distances in class, I ask them a few questions
"How far away is the sun ?"
[color="Red"]ANSWER-93 million miles[/color]

" How long would you have to sit in the sun to get a 2nd degree burn ?"
[color="red"]ANSWER - Not long depending on the time of day and your skin condition[/color]

" How much of the suns energy does it take to burn you ? "
[color="Red"]ANSWER - A very very very small amount of the suns energy is needed[/color]

So, to get a 2nd degree sunburn you need three things
*Distance
*Time
*Available energy

Sound familiar ?


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PostPosted: Tue Sep 22, 2009 8:25 pm 
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jghrist wrote:
But the boundary is determined by incident energy level, not arc pressure. A long duration arc can have a high IE with relatively low pressure.


The point is that a wall will block most of the energy released from the arc fault.
To use Capt. Jim's analogy, a wall will block you from getting sunburned.
Capt. Jim, I have red hair, and even I don't get a sunburn after just 2 seconds in the sun on the worst of days.

jghrist wrote:
I don't think that the equations for incident energy or flash hazard boundary are really valid for large distances. The energy for an arc in a box is concentrated by the sides of the box, but this effect would not be much after a few multiples of the box depth. In a recent study, we calculated an IE of 68 cal/cm² for a 2 second 480 volt 13 kA arc, with a boundary of 17.6 feet. I doubt seriously that you would get second degree burns 17.6 feet away from this arc in two seconds.


I think there are several flaws in the hazard boundary.
The first is that they didn't place the probes out far enough to get good data on the boundary distance. Looking at the data from the initial tests, you will see that in most of the tests, the farthest calorimeter was placed less than 5 ft. from the test arc.
The second is the assumption that you hit on. They took the measurements from 3 feet away, and simply extrapolated that out to the distances that we are seeing in the real world, with neglects the fact that once the arc gets out of the box, much of the "focusing" effect will be dissipated.


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PostPosted: Tue Sep 22, 2009 10:33 pm 
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[quote="WDeanN"]The point is that a wall will block most of the energy released from the arc fault.
To use Capt. Jim's analogy, a wall will block you from getting sunburned.
Capt. Jim, I have red hair, and even I don't get a sunburn after just 2 seconds in the sun on the worst of days.

True, but if you were closer to the sun, 2 seconds would be plenty of time.
My point is that the 3 variables make up the degree of damage
Too close
Too much energy
Too much time


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PostPosted: Wed Sep 23, 2009 8:55 am 
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The sun analogy is not a good one. A better analogy would be,
Does and electric arc welder create an arc?
Can you feel its heat from 150 feet away?

The lack of being practical is what makes this code appear inadequate.


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PostPosted: Thu Sep 24, 2009 1:46 am 
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You don't have the same power, by a large factor!

A welder consumes maybe 30 kW (or 30 kVA) of power, while an arc in a 600 V panel can release 1000 times more power if the ustream transfo is a 3000 kVA.

Not the same league...


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