AN ARC FLASH PROTECTION
AN ARC FLASH PROTECTION
(OP)
Can someone help with the formulas to calculate the following:
1. arc current
2. incident energy and
3. flash protection boundary
the system fault level is 16KA @ 11kv
Regars,
Basonia
1. arc current
2. incident energy and
3. flash protection boundary
the system fault level is 16KA @ 11kv
Regars,
Basonia






RE: AN ARC FLASH PROTECTION
My $0.02
RE: AN ARC FLASH PROTECTION
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RE: AN ARC FLASH PROTECTION
If equipment is a switchgear with 25 mm gap, working distance is at 18 inches, and 4 cycles (0.066 sec) total interrupting time of breaker or fuse:
Predicted arcing current = 15.41 kA
Incident energy exposure = 1.77 cal/cm2
Flash protection boundary = 28 inches
PPE level cat = 1
Please verify calcs as I made it quick.
RE: AN ARC FLASH PROTECTION
RE: AN ARC FLASH PROTECTION
IgIa= 0.00402+0.983*IgIbf
1.What is Ig in this formula and what is the simplified formula to calculate Ia ( arcing current)?
I have been trying to calculate arcing current using this formula above and comparing the answer with the one in www.arcadvisor.com calculator but, could not find the same answers.I think, perhaps am missing something in the formula, Can some clearify the formula?
Regards, BASONIA
RE: AN ARC FLASH PROTECTION
RE: AN ARC FLASH PROTECTION
RE: AN ARC FLASH PROTECTION