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mnewman

Post subject: ArcFlash Hazard to IEEE1584  Assumptions Posted: Fri Mar 26, 2010 12:17 am 

Joined: Sun Jul 22, 2007 5:00 pm Posts: 33 Location: Stratham, SW Australia

Can anyone please confirm my following assumptions concerning arcflash hazard evaluation calculations to IEEE1584 :
1.) If the switchboard is certified as 'arc fault proof' or 'arc fault contained' then arcflash 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 arcflash 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 arcsource and specifies PPE to prevent a 2nd degree burn.
4.) No PPE will protect against the concusive airblast 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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JBD

Post subject: Posted: Fri Mar 26, 2010 6:23 am 

Joined: Mon Jan 18, 2010 11:35 am Posts: 523 Location: Wisconsin

mnewman wrote: Can anyone please confirm my following assumptions concerning arcflash hazard evaluation calculations to IEEE1584 :
1.) If the switchboard is certified as 'arc fault proof' or 'arc fault contained' then arcflash 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 arcflash 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 arcsource and specifies PPE to prevent a 2nd degree burn.
4.) No PPE will protect against the concusive airblast 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 arcresistant 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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mnewman

Post subject: Posted: Thu May 20, 2010 4:35 pm 

Joined: Sun Jul 22, 2007 5:00 pm Posts: 33 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 worstcase incident energy reported.
NFPA 70E Section D.6 also states that the calcualtions for incident energy be done at 100% bolted shortcircuit current and at 38% of bolted shortcircuit current. Does this mean the value of Ibf in the equation D.8.2(a) is done at 100% and 38% of bolted shortcircuit current to get the arcing current. Is 100% and then 85% of this arcing current then used to calculate incident energy and the worstcase reported ?
For voltages below 600V, NFPA 70E calcualtes a "second arc flash current" at 38% of bolted shortcircuit current.
For voltages bd


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JBD

Post subject: Posted: Thu May 20, 2010 7:49 pm 

Joined: Mon Jan 18, 2010 11:35 am Posts: 523 Location: Wisconsin

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 worstcase incident energy reported. NFPA 70E Section D.6 also states that the calcualtions for incident energy be done at 100% bolted shortcircuit current and at 38% of bolted shortcircuit current. Does this mean the value of Ibf in the equation D.8.2(a) is done at 100% and 38% of bolted shortcircuit current to get the arcing current. Is 100% and then 85% of this arcing current then used to calculate incident energy and the worstcase reported ?
Basically, yes.


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jghrist

Post subject: Posted: Fri May 21, 2010 7:16 am 

Joined: Wed Jun 04, 2008 9:17 am Posts: 428 Location: Spartanburg, South Carolina

mnewman wrote: NFPA 70E Section D.6 also states that the calcualtions for incident energy be done at 100% bolted shortcircuit current and at 38% of bolted shortcircuit current. Does this mean the value of Ibf in the equation D.8.2(a) is done at 100% and 38% of bolted shortcircuit current to get the arcing current. Is 100% and then 85% of this arcing current then used to calculate incident energy and the worstcase reported ?
I assume that your reference is D.5 and D7.2(a) in NFPA 70E2009. 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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Jim Phillips (brainfiller)

Post subject: Posted: Fri May 21, 2010 10:12 am 

Plasma Level 

Joined: Mon Sep 17, 2007 5:00 pm Posts: 1498 Location: Scottsdale, Arizona

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.
_________________ Jim Phillips, P.E. Brainfiller.com


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mnewman

Post subject: Posted: Sat May 22, 2010 4:58 pm 

Joined: Sun Jul 22, 2007 5:00 pm Posts: 33 Location: Stratham, SW Australia

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 ‘FlashProtection 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 shortcircuit 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 shortcircuit current when calculating the arcing current in systems where Un = < 480V ?
Which standard is mandatory in USA for arcflash evaluations ?
The Standards I refer to are :
NFPA 70E Standard for Electircal Safety in the Workplace – 2004 Edition.
IEEE Std 15842002 IEEE Guide for Performing ArcFlash Hazard Calculations.


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jghrist

Post subject: Posted: Sun May 23, 2010 6:34 pm 

Joined: Wed Jun 04, 2008 9:17 am Posts: 428 Location: Spartanburg, South Carolina

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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mnewman

Post subject: Posted: Sun May 23, 2010 8:33 pm 

Joined: Sun Jul 22, 2007 5:00 pm Posts: 33 Location: Stratham, SW Australia

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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Jim Phillips (brainfiller)

Post subject: Posted: Mon May 24, 2010 3:39 am 

Plasma Level 

Joined: Mon Sep 17, 2007 5:00 pm Posts: 1498 Location: Scottsdale, Arizona

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 Ampseconds 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.
_________________ Jim Phillips, P.E. Brainfiller.com


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mnewman

Post subject: Posted: Mon May 24, 2010 3:55 pm 

Joined: Sun Jul 22, 2007 5:00 pm Posts: 33 Location: Stratham, SW Australia

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 arcgap is outside the range of the IEEE 1584 empirical model, the theoretically derived Lee equations must be used to determine the ‘incident energy’ and ‘flashprotection boundary’.
Arcs are considered unenclosed ( inair ) 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 arcgap is outside the range of the IEEE 1584 model and the Lee method is used.
c) arcs can only be considered to be ‘inbox’ 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 ‘inair’.
The conclusion is that if switchgear is not tested and certified to withstand the arcblast, 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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jghrist

Post subject: Posted: Mon May 24, 2010 8:00 pm 

Joined: Wed Jun 04, 2008 9:17 am Posts: 428 Location: Spartanburg, South Carolina

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 arcgap is outside the range of the IEEE 1584 empirical model, the theoretically derived Lee equations must be used to determine the ‘incident energy’ and ‘flashprotection boundary’. This is correct. Quote: Arcs are considered unenclosed ( inair ) 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 arcgap is outside the range of the IEEE 1584 model and the Lee method is used.
c) arcs can only be considered to be ‘inbox’ 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 arcinabox. 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 ‘inair’.
The conclusion is that if switchgear is not tested and certified to withstand the arcblast, 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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stevenal

Post subject: Posted: Tue May 25, 2010 8:03 am 

Joined: Tue Jan 13, 2009 5:00 pm Posts: 532

I think the OPs point was that the Lee method doesn't allow for an inbox versus open air calculation. I believe this is true.


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jghrist

Post subject: Posted: Tue May 25, 2010 10:46 am 

Joined: Wed Jun 04, 2008 9:17 am Posts: 428 Location: Spartanburg, South Carolina

stevenal wrote: I think the OPs point was that the Lee method doesn't allow for an inbox 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 unenclosed 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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stevenal

Post subject: Posted: Tue May 25, 2010 1:19 pm 

Joined: Tue Jan 13, 2009 5:00 pm Posts: 532

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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