***My initial post may have been posted in the wrong part of the forum, so I'm posting here***
Hi guys. I am a long time reader and this is my very first post. I'm an EE out of Auburn University, graduating in 2006. I know this post is long but I need help and I hope you all don't think this is a stupid question/post. Hopefully this will help someone years down the road. I am trying to understand the concept of doing arc flash calculations using IEEE 1584 equations. I saw NFPAâ€™s equation and it was very simple once you had the data to input, but itâ€™s only limited from 1650kA. Before I attempt to use an actual arc flash program, I want to understand the engineering behind the calculator. Pleases keep in mind that I will not do any arc flash analysis without shadowing an experience mentor and going to a few training classes.
PLEASE help me and Iâ€™m sorry if this post is redundant. I did a lot of searching and reading last night and I could never find an answer, so my head is hurting this morning lol. I would like to attempt to calculate an area at one of our facilities in the field. This is an example run, Iâ€™m not inputting anything into any model. I want to calculate what the estimated category rating is at the 3 phase 200A disconnect at the pole downstream of a 325kVA, 1 Phase transformer bank (75kV xformer bank), with DeltaWye grounded. This disconnect is located on the same pole as the xformers bank.
Equation Factors Utility is 12.47kV ACSR %Z impedance â€“ 2.3% Xformer is 12.47kVA â€“ 480V Short circuit/bolt fault current â€“ 2000A (an assumption for example sake) 3/0 copper feeding line side of 200A disconnect .01 clearing time
IEEE 1584 Equations For systems between 0.208 and 1 kV: lg Ia = K + 0.662(lg Ibf) + 0.0966(V) + 0.000526(G) + 0.5588(V)(lg Ibf)  0.00304(G)(lg Ibf)
Based on factors above: Determine arc fault currents
lg[Ia] = 0.097 + 0.662 (lg(2kA) + 0.0966(.480) + .000526(25) + .5588(.480)(lg(2kA)  .00304(25)(lg(2kA)) = 0.097 + 0.662 (0.3010) + 0.0464 + .0132 + .0807  .0229 = 0.2197 Ia = log (.2197) = .6581
This is my first issue guys. I highly doubt this number should be negative once completed. Maybe I messed up a decimal point somewhere, maybe miscalculated something? Can someone please confirm my math? If the top part is incorrect, of course everything else will the wrong going forward. Thatâ€™s ok, as long as I can get assistance with the part thatâ€™s incorrect, itâ€™s simple to fix my math ïŠ. Letâ€™s keep moving forward shall weâ€¦..
Based on factors above: Determine incident energy
(2) lg En = K1 + K2 + 1.081(lg Ia) + 0.0011(G)
Lg[En] = 0.555 + (.113) + 1.081 (lg (.6581) + .0011(25) = .668 + 1.081(ERROR) + .0275 En = unknown ïŒ
This is my 2nd problem. Since I have the incorrect arc fault current, itâ€™s giving me an â€œerrorâ€ once I try to compute it in the equation. The log of the negative Ia is giving me an error. I will still continue to the follow through on the following equations just to see if my â€œtheoryâ€. Please forgive me guys, Iâ€™m learning, I hate to ask stupid questions. Your feedback is greatly appreciated.
(1) E = 4.184(Cf)(En)(t/0.2)(610x/Dx)
E = 4.184(1.5)(unknown)(0.1/0.2)(610^1.641/455^1.641) = 6.276(unknown)(.5)(37215/23003) = 6.276(unknown)(.8089) = and this is where Iâ€™m stuck again.
Based on what I read, this Incident energy is in joule/cm^2, so if I had everything correct above, I would take the E answer and then multiply by 0.239 cal/cm^2 to get a estimated category rating. Is this correct? Once again please help me. I now report to a boss who isnâ€™t an EE and he thinks that you can simply plug in an equation to get the answer. Iâ€™ve been trying to tell him that there are SO MANY variables to determine the true value, thus the need for the arc flash program, but he still wants me to calculate an â€œexampleâ€ by hand.
Please help guys. Thanks.
