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 Post subject: IEEE 1584. I need help with calculationsPosted: Tue Aug 05, 2014 7:00 am

Joined: Tue Aug 05, 2014 6:48 am
Posts: 3
***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 NFPA’s equation and it was very simple once you had the data to input, but it’s only limited from 16-50kA. 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 I’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, I’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 3-25kVA, 1 Phase transformer bank (75kV x-former bank), with Delta-Wye grounded. This disconnect is located on the same pole as the x-formers bank.

Equation Factors
Utility is 12.47kV ACSR
%Z impedance – 2.3%
X-former is 12.47kVA – 480V
Short circuit/bolt fault current – 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. That’s ok, as long as I can get assistance with the part that’s incorrect, it’s simple to fix my math . Let’s keep moving forward shall we…..

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 

This is my 2nd problem. Since I have the incorrect arc fault current, it’s giving me an “error” 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 “theory”. Please forgive me guys, I’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 I’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 isn’t an EE and he thinks that you can simply plug in an equation to get the answer. I’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 “example” by hand.

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 Post subject: Re: IEEE 1584. I need help with calculationsPosted: Tue Aug 05, 2014 8:06 am
 Plasma Level

Joined: Tue Oct 26, 2010 9:08 am
Posts: 2173
Location: North Carolina
[quote="WarEagle]

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. [/quote]

Longer version posted in your first post.

You were good up to the calculation of 0.2197. Then you have lg(Iarc) = 0.2197. Now you should have taken the exponent of both sides to remove the logarithm but for some reason you just moved the logarithm to the other side. Correct would be Iarc = E^0.2197 = 1.26 J/cm^2. Note that the empirical equation is good to about 1-2 significant digits at best, and that you can just carry the term "lg(Iarc)" rather than the value (Iarc) through into the denormalizing equation without modification so you don't really need to calculate exponents here. Otherwise the first thing you do when calculating the denormalized value is to take the logarithm again. I thought it was clear that this is why the equations are written the way they are.

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