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 Post subject: Bolted Fault Current Question
PostPosted: Mon Aug 04, 2008 6:49 pm 
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I have taken a short circiut calculation study by a consultant done via ETAP. I would like to use the bolted fault current to do the arc flash hazard analysis in IEEE1584. Do I take the ini. rms symm current(3-phase) or the breaking rms current value as the bolted fault current? Please help as I am not familiar with ETAP.
Thanks!


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PostPosted: Tue Aug 05, 2008 10:45 am 
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I am not familiar with ETAP specifically but the formula / proceedure requires the bolted three phase RMS symmetrical short circuit current.


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PostPosted: Sun Aug 16, 2009 6:29 am 
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can i use the bolted three phase RMS symmetrical short circuit current which calculated throught the IEC 60909 method? i found that is using different method will produce different result, around 4kA apart. Which method shall be used to calculate the bolted three phase RMS symmetrical short circuit current? Thanks.... :)


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PostPosted: Tue Aug 18, 2009 4:50 pm 
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Welcome to Fault Current calculations. There are several methods available for fault current. The easiest one for Arc Flash calcs is called the MVA method. Its limited because it only gives you 3 phase bolted RMS Sym, but that's all you need for Arc Flash. What you see with the different methods is normal, and there is no specific 'right' answer. You can calculate twice with both the low and the high values. But in my experience, your worse case will probably be with the lowest fault current as that increases your breaker clearing times and raises the AF value.


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PostPosted: Mon Sep 14, 2009 10:42 am 
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Read http://www.arcadvisor.com/faq/about_mva_method.html about MVA method for short circuit analysis. The method can also be used to resolve unbalanced faults ( line to line, line to ground etc. ). Another free and handy tool for short circuit analysis can be accessed from http://www.circuitprotection.ca/resources.html

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


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PostPosted: Tue Sep 15, 2009 6:03 am 
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No one has answered the question yet, so I will make an attempt. Actual IE exposure will occur from the time of the fault to the time of interruption, and the decaying current will have an effect. Current at the time of interruption will lead to the least conservative IE value (less than actual), therefore use the initial current value to get a more conservative result.


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PostPosted: Tue Sep 15, 2009 2:38 pm 
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stevenal wrote:
No one has answered the question yet, so I will make an attempt. Actual IE exposure will occur from the time of the fault to the time of interruption, and the decaying current will have an effect. Current at the time of interruption will lead to the least conservative IE value (less than actual), therefore use the initial current value to get a more conservative result.


Don't forget that if the current gets low enough for the TCC to quit the instaneous (or similar) region at the lower current, the real IE can go up. If it stays in that region, then you're correct.


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