IEEE 1584 for Dummies...I'm trying guys, but need your help.

[WarEagle]

IEEE 1584 for Dummies...I'm trying guys, but need your help.

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
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 = lg(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.
Please help guys. Thanks.​

 

[charlie b]

Sorry, but I can't help with this. I don't have a copy of the IEEE standard, and I wouldn't use it anyway. I have access to the SKM Power Tools program, including the Arc Flash module, and that is what I would use.

I will question one step in your math, however. You went from,

lg[Ia] = (bunches of numbers) = 0.2197

to

Ia = log (.2197) = -.6581

If the log of Ia is .2197, you don't calculate Ia by taking the log of .2197. You need the inverse function of the log function. You need to raise 10 to the power of .2197.

 

[nhee2]

charlie b

Sorry, but I can't help with this. I don't have a copy of the IEEE standard, and I wouldn't use it anyway. I have access to the SKM Power Tools program, including the Arc Flash module, and that is what I would use.​

I may be hijacking the thread - but am curious - at what point/size job do you consider SKM or ETAP or whatever a necessity? For instance - I'm an EE/controls engineer in a non-EE firm - our designs are fairly simple on the power distribution side - single radial feed, maybe an MCC. Short circuit calcs can be performed by hand/excel. Most work is in the instrument/controls side. We don't have SKM. But I have noted to folks that it would be a worthwhile investment if a couple of larger projects come in. We have had a request for an arc-flash study, and have ended up referring the customer to a partner firm since we don't have (a) the software and (b) the experience. Having the software would solve one of those limitations...

 

[charlie b]

I would say that getting the software becomes a necessity when you can no longer be confident in the manual calculations. For example, if you have a MCC, then how are you accounting for the motor contribution to the fault currents? The situation will also become difficult to handle manually when you start adding step-down transformers in multiple locations. Finally, if you have to deal with selective coordination or arc flash calculations, I would never even try to do it manually.

 

[pfalcon]

First clean up the page. Don't put math information into a paragraph. Provide line breaks:

(one set of information per line)
Estimate catagory rating of a 3 phase 200A disconnect?
...

(Do not use the evil hyphen for pauses)
X-former is 12.47kVA @ 0.480V (V)

(Provide symbols in the given section when known)
Short circuit/bolt fault current (Ibf) == 2000A (an assumption for example sake)

(Use brackets, braces, and spaces to aid grouping depth)
lg(Ia) = -0.097 + 0.662 * (lg[2] + 0.0966 * [.480] + 0.000526 * [25] + 0.5588 * [.480] * [lg{2} - 0.00304 * {25} * {lg(2)} ?! ]) !?
lg(Ia) = -0.097 + 0.662 * lg(2) + 0.0966 * 0.480 + 0.000525 * 25 + 0.5588 * 0.480 * lg(2) - 0.00304 * 25 * lg(2)

(and as pointed out by Charlie)
lg(Ia) = 0.2197
Ia = 10^0.2197 = 1.658

 

[nhee2]

I would say that getting the software becomes a necessity when you can no longer be confident in the manual calculations. For example, if you have a MCC, then how are you accounting for the motor contribution to the fault currents? The situation will also become difficult to handle manually when you start adding step-down transformers in multiple locations. Finally, if you have to deal with selective coordination or arc flash calculations, I would never even try to do it manually.

Charlie B thanks for feedback. I would typically assume 4 x FLA for motor contribution.