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????

 

[Chapmeister]

I'll look in to hazard mitigation later. At this point I just wanted to make sure my analysis technique was correct. It seemed like I was finding a lot of "hot spots" in our factory, and I thought maybe I was doing it wrong.

 

[stevenal]

Likely a lower rated primary fuse would not survive a fully offset inrush current, or short term overloading. Usually there are not too many choices for a particular transformer. We are now trying to use system protection devices as PPE, something they were never intended or designed for.

 

[DaveScott]

Interesting: for information.
(Inrush current)

http://www.sandc.com/webzine/012003_1.asp

Typically inrush is 12x FLC for 0.1s, and 25x FLC for 0.01s.
The 112KVA Tr with 200A fuses is easily too big for 12xFLC, but about right for 25xFLC.
150A fuses would be cutting it fine for inrush.

Are there any fuses with a slightly better curve?

 

[Chapmeister]

Switching from a 200A fuse to a 150A fuse of the same type would reduce the melt time from ~24 seconds to about ~5.5 seconds.

This will reduce incident energy from 5.61 to 2.74 J/cm^2. This brings the PPE requirement down from >4 down to level 1.

The 112.5kVA transformer powers a 75hp grinder motor. Does this seem like a viable solution?

 

[RonShap]

Since the upstream fuse will clear the fault current, I believe that you need to calculate the arc flash fault current from the bolted fault current found at that fuse location in the distribution (24.5kA). An arc fault at the secondary of the transformer will draw its calculated value, but the upstream fuse that will attempt to clear the fault during that event downstream, will carry the calculated value at the fuse location.
That is my understanding. I generally use SKM software, so I haven't done it by hand in a while.

 

[DaveScott]

Roughly speaking, you could stall the grinder for about 5 minutes, before the 150Amp fuses would blow.

 

[Chapmeister]

"the upstream fuse that will attempt to clear the fault during that event downstream, will carry the calculated value at the fuse location."
--------------------------------
This idea is new to me. The calculated current value (bolted 3ph fault) at the fuse is much larger than the arcing current at the fault location on the transformer secondary.
I don't see how the fuse could possibly see the full bolted fault current based on a downstream fault. Or perhaps I'm misunderstanding?

I'm not sure how well the production line would respond to a stalled grinder... probably not a good thing.

 

[stevenal]

If the 150A fuse in question can carry the inrush, and short term overloading is not a factor; solution looks viable.

 

[dpc]

Your energy values don't seem to match your PPE levels.

5.61 Joules = 1.34 cal - This is a low arc-flash level.

Are you missing some zeros somewhere?

 

[Chapmeister]

Sorry, dpc, I got my numbers mixed up. The 5.61 and 2.74 values are actually the Flash Protection Boundry distances (in meters). This is, for those who don't know, the distance at which a curable burn will be recieved in the even of an arc flash.

Here are the proper numbers:

@ 24 seconds, E = 652 J/cm^2 --> PPE level 4
@ 5.5 seconds, E = 155 J/cm^2 --> PPE level 3

These are clearly huge amounts of energy. Something I wouldn't expect to see on the low side of a 112.5kVA transformer at 220V.

 

[davidbeach]

The fuse on the upstream side of the transformer will see the secondary fault current multiplied by the current ratio of the transformer. This will be significantly less than the bolted fault current available on the primary of the transformer. Kirchhoff wouldn't have it any other way. Convert to per unit; the transformer becomes an impedance and amps in (in per unit) equals amps out (in per unit). Fault on the secondary causes x per unit amps to flow, same x per unit amps are flowing on the primary side. You'll find that the SKM software can be used to find the maximum fault current at each location, or you can use it to find the fault currents at any portion of the system for a specific fault. Try that and you will find that the primary current for a secondary fault is considerably less than the current at the same location for a primary fault.

 

[RonShap]

Chapmeister,
Just to confirm your calculated results, in simulating your situation in SKM, I get approx 1050 J/cm^2 because it also calculates the worst case considering the arcing current low tolerance.

David,
Thanks for straightening me out. I agree.

 

[rbulsara]

I am not sure about the calculaitons, but per NFPA 70E and/or IEEE 1584 transformers rated 125kVA and below are assigned Category 0 for PPE and there is not need to do arc flash calcs.

 

[dpc]

rbulsara,

The 125 kVA limitation in IEEE 1584 applies to 208V and lower. Anything higher than 208 V is a different story.

 

[RonShap]

rbulsara,
The IEEE 1584 is that 208/120V systems should not be a concern. Well, they shouldn't be, but what do I tell the judge when he says it was. The text in the IEEE document is not shall or must, so I do the calc anyway.

 

[stevenal]

To clarify the voltage level: "Equipment below 240 V need not be considered unless it involves at least one 125 kVA or larger low impedance
transformer in its immediate power supply."

How does "need not" stack up against "should not"?

 

[rbulsara]

ronshop/dpc:

I think stevenal answered it. As for the judge goes all bets are off. On the otherhand I am sure if you show them recommendations of IEEE any san jury will equitt you of negligence. After all, the person who decides to work live an equipment bears some responsibilty.

If the judge or a jury is insane, it would not matter any way. Do you seriously think that just because your calculation turns out to be techncally correct and you are involved in a lawsuit it will be of any more help than showing a published standard??

 

[rbulsara]

The fact is arc flash calculation method is still evolving, in fact it is in embryonic stage, you do the best you can with available tehcniques.

The IEEE formulas are all imperical formulas or are curve fits. Meaning curves were plotted based on several test results and then mathematician were called in to find the equation that fits the curves. So there are infinite number of scenarios that are not yet covered. At the same time there is no sense in going your merry way to find arc flash energy based on some unfoudned formulas.

IEEE recommends ignoring below 125kVA and 240V because the tests proved so, I believe.

 

[Chapmeister]

RonShap,
The values I gave were for 100% arcing current/100% bolted fault current. I have been performing my calculations at 100% AND 38% bolted fault current. 38% is the lowest fault level at which an arc can sustain itself on a 480V system (note: "industry accepted level", see: Annex D.6(1) NFPA70E).

Also, from these two BFC values I'm performing my calculations at 100% and 85% arcing current (as stated in Annex D.8.2(c)), to "account for variations in the arcing current and the time for the overcurrent device to open."

Does this 85% idea cover the addition of the 15% to the melting time as discussed above??? Do you think double counting this 15% might make the results too conservative?

Addressing the other comments,
If you do what the standards say then you will be deemed as having performed your due diligence to the best of your ability and by industry accepted ideals. This will essentially clear you of liability, barring strange circumstances.

rbulsara,
It doesn't matter if I NEED to do the calc, I'm doing it anyway while I'm at it. I'm working my way though about 200 loads in our system on 600V feeders. I am not generally dealing with the 208V loads in the plant. The example I gave just happened to step down to 220V. We have lots of step downs to 415V, 460V, 480V too.

 

[stevenal]

The 85% is to account for fault impedance. You want to look at clearing time (not melting time) at each current level.