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 Post subject: Arc-Flash Hazard to IEEE1584 - Assumptions
PostPosted: Fri Mar 26, 2010 12:17 am 

Joined: Sun Jul 22, 2007 5:00 pm
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Location: Stratham, SW Australia
Can anyone please confirm my following assumptions concerning arc-flash hazard evaluation calculations to IEEE1584 :

1.) If the switchboard is certified as 'arc fault proof' or 'arc fault contained' then arc-flash studies are not required if personnel do not open any switchgear panels.

2.) If the switchboard is not certified as 'arc fault proof' or 'arc fault contained' then arc-flash studies to IEEE1584 assume that there is no barrier between the arc source ( gap between busbars ) and personnel.

3.) IEEE1584 quantifies the arc flash 'incident energy' at a defined 'working distance' from the arc-source and specifies PPE to prevent a 2nd degree burn.

4.) No PPE will protect against the concusive air-blast for incident energy above 167 J / cm^2 ( 40 cal / cm^2 ).

What standard quantifies the hazard and specifies PPE to protect against the burning oil expelled from exploding oil filled switchgear ?


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PostPosted: Fri Mar 26, 2010 6:23 am 
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mnewman wrote:
Can anyone please confirm my following assumptions concerning arc-flash hazard evaluation calculations to IEEE1584 :

1.) If the switchboard is certified as 'arc fault proof' or 'arc fault contained' then arc-flash studies are not required if personnel do not open any switchgear panels.

2.) If the switchboard is not certified as 'arc fault proof' or 'arc fault contained' then arc-flash studies to IEEE1584 assume that there is no barrier between the arc source ( gap between busbars ) and personnel.

3.) IEEE1584 quantifies the arc flash 'incident energy' at a defined 'working distance' from the arc-source and specifies PPE to prevent a 2nd degree burn.

4.) No PPE will protect against the concusive air-blast for incident energy above 167 J / cm^2 ( 40 cal / cm^2 ).

What standard quantifies the hazard and specifies PPE to protect against the burning oil expelled from exploding oil filled switchgear ?


IEEE1584 is a group of formulas for calculating arc fault incident energy. NFPA70E is about using the values obtained from IEEE1584, or other methods.

You need to look into NFPA 70E for your answers.

1) Maybe, maybe not. Some arc-resistant gear simply redirects the flash up towards the ceiling, where it may be reflected back down towards the operator. Other gear is only rated for a specific amount of SCA for a specific time (i.e. 20kA for 1 sec).

2) Yes.

3) The standards for PPE requirements are in NFPA70E.

4) IEEE1584 provides arc flash values, it does not calculate other factors such as blast force. NFPA 70E is also concerned with flash values, it warns there may be additional hazards but does not specifically address them.


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PostPosted: Thu May 20, 2010 4:35 pm 

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Location: Stratham, SW Australia
Thanks JBD; I have another question :
Both IEEE1584 and NFPA 70E specify that the incident energy be calulated at a "second arcing current" of 85% of the arcing current and the worst-case incident energy reported.
NFPA 70E Section D.6 also states that the calcualtions for incident energy be done at 100% bolted short-circuit current and at 38% of bolted short-circuit current. Does this mean the value of Ibf in the equation D.8.2(a) is done at 100% and 38% of bolted short-circuit current to get the arcing current. Is 100% and then 85% of this arcing current then used to calculate incident energy and the worst-case reported ?




For voltages below 600V, NFPA 70E calcualtes a "second arc flash current" at 38% of bolted short-circuit current.
For voltages bd


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PostPosted: Thu May 20, 2010 7:49 pm 
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mnewman wrote:
Thanks JBD; I have another question :
Both IEEE1584 and NFPA 70E specify that the incident energy be calulated at a "second arcing current" of 85% of the arcing current and the worst-case incident energy reported.
NFPA 70E Section D.6 also states that the calcualtions for incident energy be done at 100% bolted short-circuit current and at 38% of bolted short-circuit current. Does this mean the value of Ibf in the equation D.8.2(a) is done at 100% and 38% of bolted short-circuit current to get the arcing current. Is 100% and then 85% of this arcing current then used to calculate incident energy and the worst-case reported ?


Basically, yes.


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PostPosted: Fri May 21, 2010 7:16 am 
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mnewman wrote:
NFPA 70E Section D.6 also states that the calcualtions for incident energy be done at 100% bolted short-circuit current and at 38% of bolted short-circuit current. Does this mean the value of Ibf in the equation D.8.2(a) is done at 100% and 38% of bolted short-circuit current to get the arcing current. Is 100% and then 85% of this arcing current then used to calculate incident energy and the worst-case reported ?


I assume that your reference is D.5 and D7.2(a) in NFPA 70E-2009. D.5 is an example calculation using the Doughty/Neal paper. D.7 is an example calculation using the IEEE Std. 1584 method. IEEE Std. 1584 does not use the 38% of bolted fault current in its calculations. It uses a second arc current equal to 85% of Ia.


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PostPosted: Fri May 21, 2010 10:12 am 
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Here is a bit more info.

The Doughty/Neal paper from 1997 gave us a method to calculate incident energy using the bolted short circuit current. It was a landmark paper for arc flash theory back then.

It was a great approach at the time but the shortcoming was when you look at a time current curve to determine the duration of the arc flash, using the "bolted" short circuit current can give the false impression that a device could trip instantaneously when in fact, it could be timing out due to the lower arcing current (which was not calculated back then)

The 38% came from a review of data that indicated the lowest 480 V arcing current could be as low as 38% of bolted current so that number was used for evaluating the time current curve. It was not used for the acutal calculation.

As an example, if you had 10 kA for a bolted short circuit, that number is used in the actual Doughty/Neal calculation. However, when looking at the TCC, you would use 38% of 10kA i.e. 3800 Amps to see if the device still trips at the same speed. 38% was more of a band aid since there were no arcing current calculations back then.

IEEE 1584 built upon the Daughty/Neal equations and included a method to calculate the arcing current. The IEEE equations now use arcing current for the incident energy calculations and the arcing current is also used when evaluating the TCC to determine clearing time.

It was determined that although the arcing equations are a good step forward, they are not perfect so an 85% was added. This is simply a sensitivity analysis to see if a 15% reduction in the calculated arcing current would cause the device to go from instantaneous to time delay mode. It is considered that if the actual instantaneous trip is within 15% of the calculated arcing current, it is too close for comfort.

Most software calculates the incident energy using 100% arcing current and the time associated with 100% and also 85% of arcing current and the time associated with 85%. Whichever one is greater gets used in the study.

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PostPosted: Sat May 22, 2010 4:58 pm 

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Thank you Jghrist and Jim Phillips for your most helpful posts.

I am confused by the two standards IEEE 1584 and NFPA 70E.
NFPA 70E calculates ‘Flash Protection Boundary’ only for an open arc as
Dc = ( 2.65 x MVAbf x t ) ^ 0.5 for Un=<600V while IEEE 1584 calculates the ‘Flash-Protection Boundary’ based the incident energy which is in turn based on arcing fault current. The IEEE 1584 formulas include Un =<600V and Un > 600V, and also open arcs and arcs ‘in a box’.

The ‘incident energy’ formulas are also different in the two standards.
NFPA 70E recommends using both 100% and 38% of prospective short-circuit current when calculating incident energy using formulas which are not in IEEE 1584.

Section D.8 of NFPA 70E – 2004, also includes what appears to be the identical methodology and equations as IEEE 1584 for calculating ‘arcing fault current’, ‘incident energy’, and the resulting ‘flash protection boundary’ while stating that it “is not part of the requirements of this NFPA document”.

The NFPA 70E equations mentioned in my above first two paragraphs appear to be what IEEE 1582 Section 7.2 refers to as ‘Theory Based Model’ of Ralph Lee, while the IEEE 1584 ( and MFPA 70E Section D.8 ) refers to IEEE 1582 Section 7.3 ‘Empirically derived models based on a curve fitting program’ from extensive laboratory testing.

Assuming the IEEE 1584 method is used, is it accepted practice to use 100% and 38% of bolted short-circuit current when calculating the arcing current in systems where Un = < 480V ?

Which standard is mandatory in USA for arc-flash evaluations ?

The Standards I refer to are :
NFPA 70E Standard for Electircal Safety in the Workplace – 2004 Edition.
IEEE Std 1584-2002 IEEE Guide for Performing Arc-Flash Hazard Calculations.


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PostPosted: Sun May 23, 2010 6:34 pm 
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NFPA 70E does not specify how to calculate IE or arc flash boundaries. It give some examples in Appendix D of calculations using different methods, but Appendix D is not part of the mandatory standard.


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PostPosted: Sun May 23, 2010 8:33 pm 

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NFPA 70E - 2004, Article 130.3 (A) defines 'Flash Protection Boundary' = Dc = ( 2.65 * MVAbf * arc duration )^0.5.

IEEE 1584 - 2002 defines 'Flash Protection Boundary' as either :
Db = ( 4.184 Cf En ( t / 0.2 ) ( 610^x / Eb ))^( 1/x ) for LV, and...
Db = ( 2.142 * 10^6*V*Ibf ( t / Eb ) )^0.5 for HV or outside gap range ( Lee Method )

Which is the mandatory formula ?


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PostPosted: Mon May 24, 2010 3:39 am 
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mnewman wrote:
NFPA 70E - 2004, Article 130.3 (A) defines 'Flash Protection Boundary' = Dc = ( 2.65 * MVAbf * arc duration )^0.5.

IEEE 1584 - 2002 defines 'Flash Protection Boundary' as either :
Db = ( 4.184 Cf En ( t / 0.2 ) ( 610^x / Eb ))^( 1/x ) for LV, and...
Db = ( 2.142 * 10^6*V*Ibf ( t / Eb ) )^0.5 for HV or outside gap range ( Lee Method )

Which is the mandatory formula ?


Actually none of the equations are mandatory like JGhrist mentioned. NFPA 70E requires that an Arc Flash Protection (AFPB) be established and used but does not mandate a specific method.

I had a pretty lengthy conversation with an NFPA staff member many years ago (2004/2005 time frame)about using the IEEE equations. The person mentioned the only equations that are part of NFPA 70E (back during the 2004 Edition) were the 2 "Lee Equations" for the AFPB. I quickly pointed out they are from the 1980's and if they are going to include equations, why don't they use the most current ones from IEEE. The very next edition of NFPA 70E - 2009 saw the Lee equations moved to the the appendix so they are no longer "officially" part of the standard.

The "4 foot" rule in 70E is the only method that remains within the text of 70E however it's applicability is quite limited i.e. no more than 1,667 Amp-seconds total current and clearing time.

The "latest and greatest" in the industry still remains the IEEE 1584 equation but it requires a bit of effort.

Keep in mind, the only purpose of the AFPB is to keep unprotected people out of the danger zone if an arc flash hazard is present. With that in mind, when in doubt, keep people further back than you think necessary. It does not affect the person performing the work, it just means the unprotected person that is not suppose to be in the area is really not in the area.

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PostPosted: Mon May 24, 2010 3:55 pm 

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Thank you Jghrist and Jim Phillips for your posts.

Can you please confirm my understanding of IEEE 1584 as defined below :

Where system voltage is above 15 kV, or the arc-gap is outside the range of the IEEE 1584 empirical model, the theoretically derived Lee equations must be used to determine the ‘incident energy’ and ‘flash-protection boundary’.

Arcs are considered unenclosed ( in-air ) when :

a) system voltage is between 1000 V and 15 kV, when using the IEEE 1584 model.

b) system voltage is above 15 kV or the arc-gap is outside the range of the IEEE 1584 model and the Lee method is used.

c) arcs can only be considered to be ‘in-box’ where system voltage does not exceed 15 kV.

For switchgear up to 15 kV which is not tested to withstand arcing faults, what is the difference between a closed and open door ?
My understanding of IEEE 1584 and the Lee method is the calculations for incident energy assume only air between the arc and the worker when the door is closed before the arcing fault occurs. If the arc blast pressure blows the initially closed door off, then the same pressure may also blow the switchgear enclosure walls, roof, and floor off, then the arc would be ‘in-air’.

The conclusion is that if switchgear is not tested and certified to withstand the arc-blast, and the switchgear enclosure door is initially closed, it is not sure that only the switchgear door will be blown off and most of the arc blast energy focused in one direction.


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PostPosted: Mon May 24, 2010 8:00 pm 
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mnewman wrote:
Thank you Jghrist and Jim Phillips for your posts.

Can you please confirm my understanding of IEEE 1584 as defined below :

Where system voltage is above 15 kV, or the arc-gap is outside the range of the IEEE 1584 empirical model, the theoretically derived Lee equations must be used to determine the ‘incident energy’ and ‘flash-protection boundary’.

This is correct.
Quote:
Arcs are considered unenclosed ( in-air ) when :

a) system voltage is between 1000 V and 15 kV, when using the IEEE 1584 model.

b) system voltage is above 15 kV or the arc-gap is outside the range of the IEEE 1584 model and the Lee method is used.

c) arcs can only be considered to be ‘in-box’ where system voltage does not exceed 15 kV.

This is not correct. Whether or not the arc is considered open air or in a box depends on the equipment configuration, not the voltage. If the equipment is enclosed, then use arc-in-a-box. If the equipment is not enclosed, like a pole mounted switch, then use open air.

Quote:
For switchgear up to 15 kV which is not tested to withstand arcing faults, what is the difference between a closed and open door ?
My understanding of IEEE 1584 and the Lee method is the calculations for incident energy assume only air between the arc and the worker when the door is closed before the arcing fault occurs. If the arc blast pressure blows the initially closed door off, then the same pressure may also blow the switchgear enclosure walls, roof, and floor off, then the arc would be ‘in-air’.

The conclusion is that if switchgear is not tested and certified to withstand the arc-blast, and the switchgear enclosure door is initially closed, it is not sure that only the switchgear door will be blown off and most of the arc blast energy focused in one direction.

The calculations do not consider the shielding effect of doors or panels. The calculations assume that there is nothing between the worker and the arc. If there is a possibility of an arc because of interaction with the equipment, then you have to have PPE for the IE, with or without doors being closed. Since there is no generally accepted method of calculating IE with the doors closed, and a blast may blow off the doors, the consensus, I think, is to use the IE as calculated without doors.

There is a difference between arc pressure and arc incident energy. A low current arc with a long duration can have the same IE as a high current arc with a short duration. The high current arc will have a much higher arc pressure and may blow off doors where a low current, long duration arc of the same IE may not blow off the doors.


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PostPosted: Tue May 25, 2010 8:03 am 
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I think the OPs point was that the Lee method doesn't allow for an in-box versus open air calculation. I believe this is true.


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PostPosted: Tue May 25, 2010 10:46 am 
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stevenal wrote:
I think the OPs point was that the Lee method doesn't allow for an in-box versus open air calculation. I believe this is true.

That's true, but the way the statement is worded, you might assume that if you had a 35 kV enclosed switch, you should consider it un-enclosed because it is over 15 kV. This is not correct. The IE would be higher because of the concentrating effect of the enclosure. The voltage is outside the limits of the empirically derived IEEE 1584 equations and the Lee equation isn't applicable because it is enclosed. Somewhat of a quandary.


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PostPosted: Tue May 25, 2010 1:19 pm 
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The 1584 spreadsheet will accept any input for equipment class at this voltage level. Changing is from 1 (open air) to 3 (switchgear) simply has no affect on the results. Perhaps they consider Lee to be conservative enough to cover all classes.


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