Arc Flash and LV Transformers

[Chapmeister]

I'm having some trouble!
Can someone tell me if this is right? or point out the flaw in my calculations?
If I have a bolted fault current value on the secondary of a transformer, the closest upsptream protective device will be called on to interrupt the arc. If the closest device is on the primary (upstream in a piece of switchgear). In order to determine the trip time of the device, the arc current has to be reflected across the transformer using the LV/HV ratio, and then read from the TCC curve for the fuse. Using the fuse melt time, the NFPA arc flash calculations can be done. Check this out:

Here are the numbers I'm having trouble with:
Transformer Specs:
112.5kVA
600/220V (delta-wye)
5.0 %Z

Available Fault Current (from custom spreadsheet):
24.5kA at the switch gear
16.4kA at the primary (due to feeder impedance)
4.34kA on the secondary (due to transformer impedance)

Fuse Data:
C200HR English Electric (bought out by GEC)

My arcing current on the secondary is then 2.45kA. (using equation D.8.2(a) from the NFPA70E)

Do I then take the 2.45 * 220/600 = 0.898kA to get the primary side current. Reading from the TCC I get ~24 second melt time.
This gives a normalized incident engery of 0.6J/cm^2 and incident energy of 652 J/cm^2 at 24 inches (460mm). This number is WAAAAYYY bigger than I would have expected.

Where is the flaw in my procedure????

 

[stevenal]

I think you want the clearing time, not the minimum melt. Of course this ups your incident energy calculation. The other mistake I see is the assumption that the exposed individual remains bolted in place for the entire 24 + seconds. IEEE 1584 suggests using a maximum of 2 seconds, but longer if if a bucket truck is used, or if the individual has crawled into the equipment.

 

[davidbeach]

Low fault currents can mean very nasty arc flash hazards due to extended tripping times. Your math could be entirely correct.

 

[Chapmeister]

So a good remedy would be in install a faster acting protective device closer to the load with the high arc flash hazard rating?

Am I right to say the high-side current is just the voltage-ratio times the low side arcing current?

 

[davidbeach]

You are correct about the high side current. Faster acting protection is a good solution, though in many cases it can be much easier said than done.

 

[stevenal]

You can install faster acting protection on the low side, but there will always be that bus section between the transformer and the low side protection to deal with.

 

[dpc]

This is a common problem. I haven't checked your numbers, but you can generally count on the transformer secondary/main breaker area as having the highest arc-flash levels. The reason is as you state - the primary device must clear a fault on the secondary side and the fault does not produce enough primary amps to clear quickly.

In these smaller transformers, the problem is even worse due to low fault current.

IEEE-1584 (in the Annex) does state that it is reasonable to use a maximum exposure time when the calculations show a very long arc duration. They suggest 2 seconds as a maximum fault duration.

From a practical standpoint, the probability of sustaining a 220V fault for 24 seconds is very low.

 

[davidbeach]

stevenal, absolutely correct. A transformer diff zone that overlaps the low side protection will provide the necessary coverage. But adding a diff relay to his installation would be a major undertaking as there is not a high side breaker to trip. Maybe a transformer diff to trip a ground switch on the high side. Once the ground switch closes, there is no more fault current on the low side and the fuses will clear much faster for a bolted fault on the high side than for a fault on the low side.

 

[Chapmeister]

I don't recall reading anything about this "2 second rule" in the NFPA70E. Besides the IEEE1584 (I don't have a copy) is there somewhere with info on this rule-of-thumb?

 

[Chapmeister]

How does clearing time relate to the melting time?

 

[davidbeach]

Minimum melt is the lower line on a fuse curve, total clearing time is the upper line. At minimum melt, a fuse at the fastest end of the manufacturing tolerance will begin to melt. Melting takes time. At total clearing time, a fuse at the slowest end of the manufacturing tolerance will have melted and the arc (in the fuse) will have extinguished.

 

[stevenal]

If you only have the melt curve:

"For fuses, the manufacturer’s time-current curves may include both melting and clearing time. If so, use the
clearing time. If they show only the average melt time, add to that time 15%, up to 0.03 seconds, and 10%
above 0.03 seconds to determine total clearing time. If the arcing fault current is above the total clearing time
at the bottom of the curve (0.01 seconds), use 0.01 seconds for the time.

 

[Chapmeister]

I'm not quite sure I understand. The fuse charts I have been using usually only have one line per fuse size on the log-log grid.

For example, the Ferraz Shawmut AJT family of fuses:

http://www.ferrazshawmut.com/products/pdf_107/AJTSMartspot.pdf

All is says on the TCC graph is "Melting Time - Current Data, 600V Fuses". I've been assuming the lines represent an average melting time. I've been using my "Arcing Current" values to read off the melt time. I just read a website that claimed the clearing time is 10-15% more than the melting time. Does this sound right to anyone?

By this token, my 24 second melting time stated above would be inflated to 27.6 seconds, which is an even worse arc-hazard result!

 

[dpc]

Re-read the info provided by David Beach and Stevenal. The Shawmut curve is an "average melt" curve. You need to adjust this per IEEE-1584 recommendations to get a conservative clearing time, and yes, this will cause longer arc-duration and higher calculated arc energy.

 

[davidbeach]

You can get minimum melt and total clear curves from the fuse manufacturers but it isn't easy. Once you get to medium voltage they don't try to pull that average melt #*^% any more.

It would not be of any surprise that the worst arc flash hazard on a system would occur on the lowest voltage level used, behind a transformer just big enough that you have to evaluate the secondary of the transformer. (I'm too lazy to look it up at the moment, but there is a transformer size below which you don't need to evaluate 240 or 208V systems.)

 

[Chapmeister]

I'm only evaluating fault current and arc flash hazard for systems in the plant that can consume a max rating of 100kVA. So you're right, I'm not dealing with the 208V systems in the factory.

We use so many types of fuses here that I was hoping I wouldn't have to twist the arm of the fuse manufacturers to get the curves.

Do you think the add 15% to the average melt time is a valid method? It seems a bit loose and open ended to me.

Thanks for all the help, guys. I appreciated it.

I welcome any and all comments!

 

[davidbeach]

Are you sure you worded your first paragraph correctly?

 

[Chapmeister]

No, I got it completely backwards!

We're only evaluating the systems with potential power usage greater than 100kVA.

 

[dpc]

IEEE 1584 actually distinguishes between 208 V and 240 V systems. It recommends that arc energy calculations be done for **all** 240 V systems and for 208 V systems 125 kVA and larger.

Apparently this is based on the test results where arcs were more likely to be sustained at 240 V. I think 208 V testing produced very few sustained arcs.

 

[DaveScott]

This may be a dumb question: have you considered using fuses with a lower rating?