Arc flash calculation for 208V System fed from 125KVA transformer 2

[HamidEle]

I am doing an arc flash study for a 480/208V system. There are some small transformer, 15KVA, 75KVA. But ETAP gave me results that indicates Incident energy levels are above Cat.2, within which I would have to do some adjustment to reduce the available arc flash levels.

But this aslo is confusing me. Becasue as per IEEE 1584, Equipment under 240V need not be considered unless it involves at least one 125KVA or larger low-impedance transformer in its immediate power supply.

The client engineer indicated the above point either, but as an engineer, we have to take any possible injuries into the consideration. Even though it amy be over conservative.

What would you recommend in this situation?

 

[dpc]

I'd recommend following the recommendations in IEEE 1584 unless you feel you have greater expertise.

NFPA 70E has a task-based table that can be applied to the small 208 V systems.

 

[majesus]

You didn't state where the arc flash is being calculated? But I'm assuming you mean the secondary.

Etap will calculate an Arc Flash on a 3 phase bus where ever it can determine the 3phase bolt short circuit regardless of the XFMR's size. However, as there is exceptions to everthing, this is pasted directly from ETAP's help file:

Range of Operation

These calculations follow the methodology described in IEEE 1584-2002. The same limitations of this method apply to the quick incident energy calculator.


1. If any of the following: the Bus Nominal kV, Bolted Fault Current or Fault Clearing Time are set to zero the calculation is not triggered, and there are no displayed results. This applies to either set of parameters (User-Defined and Calculated).

2. If the bus nominal kV is less than 0.208 kV, this message is displayed: "The Bus kV is outside the range allowed by the Empirical IEEE Method" and there are no displayed results. This applies to either set of parameters (User-Defined and Calculated).

3. If the bolted fault current is outside the range of 0.7 kA to 106 kA and the bus nominal kV is between 0.208 and 15 kV the following message is displayed: "The fault current is outside the range allowed by the Empirical IEEE Method" and there are no displayed results. This applies to either set of parameters (User-Defined and Calculated).




Now that being said, in your model, have you included all the upstream overcurrent protections before these XFMRs? What is protecting the XFMR from short circuit?

 

[HamidEle]

There is a breaker at the upstream protecting the Transfomer from short circuit. There is no breaker at the secondary main. The arc flash was calculated on the 208V Panel bus.

 

[majesus]

What is the breaker's size? What is the CB's clearing time on that bus? (YOu'll find that info in the bus' properties, under the arcflash tab.)

If you link a snap shot of the SLD photo, I'll simulate it myself and see what I'm getting.

Maj

 

[HamidEle]

Majesus,

The breaker size is 150A (Transformer primary side). The transformer is 75KVA(480V/208V).Ignore the cable impedance. Fault clearing time is 6.083 Second. Grounded. The incident energy level by ETAP is 45.401 Cal/cm2,which exceeds the Max PPE rating.

Let's see what you are getting.

Thanks

 

[Zogzog]

1584 makes that statement because they believe the arc will not be self sustaining, like the arc you get unplugging a vacuum cleaner while it is running.Your calculations are assuming the arc will be self sustaining and is relying on the protective device to clear the fault.

Now I have discussed this with some 70E members and some people that have done thousands of arc flash lab tests that discagree with the 1584 rule, so I am not sure what to belive.

If I were you, I would use the 70E tables for the system below the 1584 rule as dpc mentioned.

 

[HamidEle]

it is really hard to determine. But for the safety of personnel, it is better to be conservative than sorry.

 

[majesus]

I'll simulate it Monday Morning at work.
(now I just have to remember)
Maj

 

[Zogzog]

You work in a test lab majesus? What do you mean you will simulate it Monday?

 

[WDeanN]

"Fault clearing time is 6.083 Second."

Try applying the 2 second rule, if you insist on calculating it.
Also "The client engineer indicated the above point"
I would say do what the client says. If you feel like you must go further, then do so with a footnote, and spell out all of your assumptions, and why you felt the need to present additional concerns.

"But for the safety of personnel, it is better to be conservative than sorry."
There are times when the additional PPE will create more of a hazard than necessary. You have to also consider this in you deliberations. If you continually provide overly conservative results, the people in the field will begin to disregard your calculations, and may end up under dressed. I try to find out if any events have happened in the past, and compare those events against what I am finding in the course of my study.

 

[Zogzog]

Good points! I always get a good laugh out of clearing times like this, who is going to stand there for 6 seconds saying "Ohhhh, look at the pretty lights"?

 

[majesus]

HamidEle,

I simulated it at work on my copy of Etap, and it seems that the Arcing Current is in the overlap of the CB's Magnetic and Thermal tripping region.

I did do several runs, with various XFMR's impedance. If I use say Z=2.6%, then the CB trips in its instantaneous region. If XFMR settin is at z=5%, then it can be a long trip time. Here are the results:

This is for Z=5%

http://i48.photobucket.com/albums/f223/majesus/simulation2.jpg

And this is for Z=2.6%

http://i48.photobucket.com/albums/f223/majesus/Simulation.jpg

As DPC mentioned, I'd go with what IEEE recommends.

A good read if you haven't seen it yet is this post:

http://www.eng-tips.com/viewthread.cfm?qid=171682&page=1

 

[cirtcele]

Hi Majesus

How do you calculate the short circuit current with ETAP where you have to select either 1/2 cycle or 5 cycle option. Namely, this should depend on the protection device.
I also appreciated that you read the help lines of the software provider to see who does what.
thanks

 

[GMIEE]

What I don't understand about the 2 second rule is aren't these questions all about arc flash? Arc flash occurs in an almost instantaneous manner does it not? How then does the 2 second rule apply to anything more than the arc itself?

 

[Rags004]

The reason the IEEE 1584 has the 125 kVA exception is because it is hard to sustain an arc-flash at 240V or less without a certain amout of available fault current. The question is how much. IEEE uses 10,000 A bolted fault. However, we have a documents published that claim they have been able to sustain an arc with as little as 4200A bolted fault. We have been using a cut-off of 4000A bolted fault.
Most of the transformers 75 kVA or less that I have seen usually have an impedance of around 5-6 % thus with the source impedance your not likely to have a very high bolted fault. However, if you have a low impedance transformers 75 or 45 kVA transformer you will get a enough fault current that you should go with the calculated values.

Bob

 

[dpc]

I'd just like to clarify that the "2 second rule" is not a rule at all. It is actually a suggestion contained in an "annex" to IEEE 1584 and is not part of the actual IEEE 1584 standard.

Arcing can continue until the fault clears (think welding). It is not an instantaneous event.

 

[Zogzog]

Sure it can but you are not going to stand there and watch it, you should be moving away from the arc after 2 seconds.

 

[burnt2x]

Zog,
Gave you a star! Sharp and quick thinking does it! I could be the fastest man in an event like that!

 

[Homerjs78]

The 1584 also lists a caveat to the 2s statement regarding how quickly a person exposed to the event is able to “move”. This maybe should not be applied across the board but only in special cases where it is assumed to apply.