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 Post subject: I need help badly for calculation. IEEE 1584 for DummiesPosted: Tue Aug 05, 2014 6:55 am

Joined: Tue Aug 05, 2014 6:48 am
Posts: 3
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: I need help badly for calculation. IEEE 1584 for DummiePosted: Tue Aug 05, 2014 7:57 am
 Plasma Level

Joined: Tue Oct 26, 2010 9:08 am
Posts: 2174
Location: North Carolina
WarEagle wrote:
Equation Factors
Utility is 12.47kV ACSR

ACSR comes in a variety of sizes just as with any other wire. Southwire puts out a very good book that brings all the various formulas together in one place. Calculating impedance on an overhead line is complicated by the fact that it depends on both the cable and geometry but it is pretty straightforward empirical equations that work pretty well. So you only have half the data...still don't know the utility available fault current, which is often a challenge to get. If you never look at the primary side you can often assume infinite available current, which gets us to the next step.

Quote:
%Z impedance – 2.3%
X-former is 12.47kVA – 480V
Short circuit/bolt fault current – 2000A (an assumption for example sake)

I have no idea why you would assume this. Calculating short circuit current is the first step in the process and trivially easy to do, especially if you don't have any inductive loads or you are not considering cable impedance. It also seems pretty darned low from experience.

Quote:
3/0 copper feeding line side of 200A disconnect
.01 clearing time

First off, what about the housing of the wiring? If it's rigid metallic conduit, it affects the eddy currents of the cable and thus the inductive impedance. Second, what overcurrent device? If it's sized where it trips almost "instantaneously", the assumption with most fuses is that they trip in about 4-8 milliseconds. With small (100 A or less) breakers, they often trip in 1-2 cycles (16-35 ms). With larger more recent vintage vacuum breakers for larger sizes or even a lot of air break types, 3 cycles plus one cycle for the relay (4 cycles = 65 ms) is the norm. The inherently problem here though from what you are describing is that you have no overcurrent device on the secondary side of the transformer, which is an altogether common design practice. It results in relying on the upstream protection of the transformer which out of necessity (avoiding the transformer's inrush) is painfully slow, resulting in trip times more like 0.250-0.500 seconds. So your example is highly unlikely to actually occur in the real world.

Quote:
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

Quote:

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…..

Simple. You moved the logarithm from the left to right side. That's not what you should have done. The inverse of a logarithm is the exponent. Since we are in natural log space, you use e^(xxx) to invert it. This is why your math didn't work. Regardless at the extremely low end of current, the empirical formula often produces even more silly nonsense such as Iarc > Ibf. However this doesn't matter as you don't need to compute this last result at all. Just carry lg(Ia) through to the next equation intact (0.2197).

Quote:
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..

Propagation of errors.

Quote:
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.

No. H/RC or "categories" have nothing to do with an engineering analysis. Category ratings come from the task tables in 70E. Incident energy calculations come from an engineering analysis, simple as that.

Finally, keep in mind that out of the dozen or so power system analysis studies you can do, you are starting with a high level one. In order, you should be doing:
1. Short circuit analysis. There are lots of methods here. For equipment sizing generally the accepted practices are ANSI or IEC methods which are quick (can do them with calculator and paper). Second method is comprehensive which results in the most accurate results but really requires software to do it. For arc flash, the comprehensive method is what is used in software programs.
2. Coordination analysis. The basic method is looking at time-current curves. But there are many protective devices which confound this technique, including something as simple as a current limiting fuse. So again...not as easy as it looks initially. You need this data though to calculate how long it takes to interrupt an arcing fault.
3. Arc flash hazard analysis. This is what you are trying to do without the benefit of #1 or #2. So among other things, you can't apply the calculation twice, once at the Ibf and once again at the arcing current using two different opening times since you didn't do step #2. This is further complicated by the fact that in many cases, you need to also perform an analysis of the underlying equations. IEEE 1584 will produce a result but even without your math error, garbage in = garbage out. So you need to determine whether or not you can even APPLY the equation in the first place to a particular system. It is purely empirical. It is based on a limited set of test data encompassing only 3 enclosure sizes plus "open air". Thus it definitely has limits. It works pretty good on switchgear, MCC's, most panelboards, and true open air conditions. It is of questionable value when evaluating for instance industrial control panels and junction boxes which don't fit nicely into one of those categories.
4. Arc flash RISK analysis. Fortunately this has nothing at all to do with steps 1-3. It has everything to do with looking at equipment condition and likelihood of a failure as a general rule. As of 70E-2012, this step is alluded to but not explicitly called for. As of 70E-2015, it will be required. My recommendation is to use the CCPS (LOPA) method published by AIChE. Why this one and not others? Three reasons. First, many risk standards are based on more or less continuous risk such as when exposures or incidents happen frequently such as several times a shift as opposed to the AIChE method which looks at rare, "bad events" with severe consequences in large chemical plants. Second, most analysis methods have absolutely terribly vague definitions. Just look at the one in Appendix F of 70E and try to figure out how to rank something or what to do with the rankings. Third, most risk analysis methods are designed for other applications such as robotics (ANSI-RIA, one of the best), mechanical equipment (ISO), or packaging machinery (PMMA) and would be basically impossible to apply to electrical equipment without nonsensical results. In this respect the underyling LOPA method is actually pretty bad since it assumes we are dealing with pipes and pumps with toxic chemicals but the appendices of the CCPS in particular fill in gaps and produce a generalized model that can be directly applied to arc flash hazards.

So far, so good. Now, a final consideration. Are your loads purely resistive or do they have an inductive component to them (motors and transformers)? If they are purely resistive then so far, so good. BUT if they are inductive then you need to consider the amount of short circuit current contributed by the loads as their magnetic fields discharge into the fault. This is often almost as large as the fault current from the power source at least during the first few transient cycles. And this affects opening time as well. Further, some components may be current limiting. This also affects the analysis, so this is a nontrivial problem in general.

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 Post subject: Re: I need help badly for calculation. IEEE 1584 for DummiePosted: Tue Aug 05, 2014 8:55 am

Joined: Tue Aug 05, 2014 6:48 am
Posts: 3
Paul,

Thanks for your response bud! Hopefully I can try to clarify some of my original post. The values that I put below are based on some assumptions that I created to see if my math was correct. Let me see if I can answer each question that you have. This will help me learn in the long term also.

1) The ACSR size for our facilities are typically 4/0. Yes you are right, the utility fault current is typically very hard to get, but if I use the "infinite system", the fault current is calculated as

75000/(480 x 1.73) = 90A
90/.023 = 3913A short circuit bolt fault

I knew this part, but like I said earlier, I just made up a number just for the sake of the equation to understand my math. Our onshore facilities may have a sum of 40-60HP max.

2) the line fuses are kearney 3 KS upstream of the x-former bank. They have been sized to clear almost instantaneously. Depending on voltage distance our typical gauge is 3/0 THHN 90c feeding the line side of the disconnect. So are you saying that I need the time curve data for the Kearney 3 ks fuses?

Thank you and thank you again for detailed response!! This is why I frequent here. I will print off your response and post it in my office for knowledge.

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 Post subject: Re: I need help badly for calculation. IEEE 1584 for DummiePosted: Tue Aug 05, 2014 10:38 pm
 Plasma Level

Joined: Tue Oct 26, 2010 9:08 am
Posts: 2174
Location: North Carolina
We use 4/0 ACSR too but as static lines. It is midrange in strength and weight these days. We use AAC 477 Cosmos as the current conductors. A little weak but good impedance. Need to know the stranding too such as Penguin. It matters a little.

Kearny fuses like most MV fuses are not that fast. And yes, you need that. "Instantaneous" in the fuse world is a 1-2 ms semiconductor DC fuse or 1/4 cycle AC fuse. You need to convert your impedances through the transformer to the line side to determine short circuit current on the line side or do the current conversion then check the fuse curve. Going from about 4 kA at 480 to 12.5 kv is lowering it to about 150 A, and Iarc is even less. Even for boric acid expulsion fuses this will be slow. Then you have to take 85% of Iarc and check the curve again with the lower Iarc and slower trip time. Typically this lower value drives the result. And none of this includes cable to the assumed fault location. If cable impedance approaches transformer impedance even slightly, current drops dramatically. This is the reason that a rookie mistake is to assume cable impedance does not matter. It often drives clearing time which has a disproportionate effect on incident energy relative to the lower arcing current. It is unfortunate that ANSI and IEC standard inverse curves do not match arc flash curves or impedance would be less important to arcing.

I once thought fuses were "really fast", too. Fast is some semiconductor fuses I have in a drive that melt in under 1 ms and so typically multiple fuses open at once but arc flash in the drive is under 0.01 cal/cm2 in a traction drive with a peak rating of about 3000 kW. The MG sets it replaced were around 13 cal/cm2, mostly driven by 3500 hp sync motor inductance.

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 Post subject: Re: I need help badly for calculation. IEEE 1584 for DummiePosted: Wed Oct 01, 2014 12:44 pm

Joined: Fri Sep 19, 2014 12:57 pm
Posts: 4
have to read this a couple times to digest but please WarEagle keep asking questions. I am in a similar boat. Graduated from a good school and found my way into a SC and coordination job of which i had zero background. Feel like i am being drug up the learning curve face first. There are so many tid bits of info out there and it is difficult to put them all together.

i often need to ask obvious questions just for a sanity check and will likely be posting some here myself soon

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