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 Post subject: I. M. Onderdonk’s equation
PostPosted: Wed Jan 28, 2015 3:56 pm 
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I have a small 4.2 KV/120V, 250 VA instrument transformer that is connected directly to the generator buss by 18 AWG wire, approximately 24 inches long. No high voltage fuses. Transformer is located in a SWBD attached to the generator. Transformer had an internal arc fault on the primary terminal board, between primary windings and studs on the terminal board. Transformer exploded and 18 AWG wires were vaporized.

I know the short circuit current for the generator, 6000 amps. I’m trying to determine the incident energy and arc flash boundary for the transformer with the SWBD covers removed.

Because I know the short circuit current of the generator, and the gauge of the wire used to hookup to the transformer. Do you think I can use I. M. Onderdonk’s equation for fusing time current for copper conductors, solved for time, to determine the Arc Exposure time?


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 Post subject: Re: I. M. Onderdonk’s equation
PostPosted: Fri Jan 30, 2015 9:04 am 
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Two problems with this.

First, the failure is clearly shorted turns in the PT. This is a bolted fault, not an arcing fault. As such the energy stays inside the wire and/or transformer until everything melts at which point magnetic propulsion kicks in and blows it apart. Because there is no air gap involved, you don't get a power arc and no arc flash.

Second problem is that generally speaking melting time for wire is usually a long time and will almost certainly exceed 2 seconds which is the "cutoff" recommended in IEEE 1584, so it should not be used in an arc flash calculation even if an arcing fault happened in the transformer. Further, the impedance of the wire, transformer, and the arc itself should be taken into account, not just the short circuit of the generator. At least as far as the arcing current goes, IEEE 1584 has equations specifically to deal with this but at or above 1 kV, it is assumed that bolted fault and arcing fault currents are the same value.

Using wires as a "fuse" is not uncommon in generation/transmission/distribution work. But geneally the place where they are used is when the protection circuit has already operated or will never operate, and the goal is just to remove the defective equipment from the circuit. Speed is relatively unimportant (to a point), and arc flash is not a consideration since this is not an "arcing fualt" scenario.


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 Post subject: Re: I. M. Onderdonk’s equation
PostPosted: Sat Jan 31, 2015 3:52 pm 
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Thanks for the reply. A failure analysis was done on the transformer, the PT windings were not shorted and are in cased in a epoxy/ sand mix. The failure actually occurred internally to the transformer, right below the HV terminal board assembly, before copper foil wires were connected to the primary windings. The HV terminal board assembly was actually propelled from the transformer. My guess it from the copper foiled wires vaporizing in the arc.


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 Post subject: Re: I. M. Onderdonk’s equation
PostPosted: Mon Feb 02, 2015 11:32 am 
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Based on the description that the transformer was propelled, this is definitely matching the description of a bolted fault. Arcs can propel themselves (or more specifically, the plasma surrounding the arc), leaving an "arc track" behind but the only propulsion is generated by heated air surrounding the arc which is contained such as within an enclosure. When the interrupting ratings are exceeded, and considering heating is a function of I^2*R, this is more and more sounding like a bolted fault. In this type of fault, the conductors are heated to very high temperatures (they melt) and magnetic forces (also proportional to I^2) stress equipment to it's mechanical limits. You tend to see things ripped apart, bus bars bent, cables bent and/or ripped apart, etc., from a bolted fault. Transformers often fail right at the line side of the termination onto the transformer or some place in the connection from the bus to the coil.

In contrast in an arc flash there is a lot of heat and smoke. There are shiny melted and resolidified spots on the conductors where the arcing occured. There may be arc tracks leading from the initial point of failure up to the source of the arc flash where the arc plasma (not the equipment) was magnetically propelled along parallel bus bars. Doors are blown off and/or sheet metal is warped from the pressure of the arc blast but components aren't launched from this process because it's purely a "pressure vessel" type of an issue. There is usually lots of heat-related distortion and damage, and lots of soot on everything. In contrast although a bolted fault will certainly have a lot of melting, twisted bus bars and wires, etc., the general level of heat-related damage is not on everything but isolated to the conductors themselves and equipment near the conductors. Arcing faults generate huge amounts of radiant heat that hits every surface and all equipment in the affected area more or less "evenly" and the damage occurs over a much larger area instead of concentrating near the conductors. Arcing faults don't have the huge magnetic forces (in general) so very little buckling of conductors occurs and the conductors look more or less "intact", just burned. Because of the pressure that occurs and the much higher heating, arcing faults don't tend to stay "confined" to an enclosure and will vent themselves (knock the doors off). Bolted faults tend to stay confined to an enclosure except when the components are launched by magnetic forces. I would even go so far as to suggest that you may want to check your AIC ratings on components because based on the description while I have no doubt that the transformer board and/or the cables may have indeed been affected by the fault, the transformer may not have in fact been the source of the fault and it may have started elsewhere, but without looking at the situation myself, I can't judge the source. Many people tend to look for the center of the destruction but this is no help in accident investigation. Even if the fault was at the transformer for instance if a transient (lightning) happened elsewhere, it is very common for an arc to at least start at the bus or on the first couple turns of a transformer where the insulation is weakest (conductors are closest together). At 4 kV if it formed a corona discharge, it would instantly turn the epoxy into a semiconductive carbon material which would then make it a bolted line-to-line fault. Need to check your BIL's and insulation coordination.

See the photos in this article for the differences between the two fault types:
http://ecmweb.com/content/flash-burn

If you want to treat the wire as the overcurrent protective device, then the equation is the right one but be very careful with opening time. If it exceeds 2 seconds, then use 2 seconds for the IEEE 1584 empirical equation.


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 Post subject: Re: I. M. Onderdonk’s equation
PostPosted: Mon Feb 02, 2015 12:04 pm 
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Did some more checking. There are two equations, Oderdonk and Preece. They both look at wires in free air. However if it is enclosed in a plastic tube (insulated), this tends to insulate the wire and will change the results.

http://home.earthlink.net/~jimlux/hv/fuses.htm
http://www.cooperindustries.com/content ... tion_1.pdf

Seems like Oderdonk results in the longest fusing time overall compared to any other method. If the goal is to predict an arcing fault then without actual test data the goal would be to have the longest fusing time.

However there is also a huge problem with this approach. All of these methods predict the fusing or melting time of the wire. None of them predict the opening time of the circuit. In a "real" fuse the wire is surrounded by a mix of sand and/or boric acid and these materials provide a source of gases and/or cooling media to reduce the air temperature which quenches the arc more quickly. Also the phyiscal shape of the bus inside the fuse is designed in such a way that multiple arcs form simultaneously all along the length of the fuse body, simultaneously quenching the arc. In a wire in free air, none of this exists. The concern over using these equations to predict "fusing" or melting time is strictly ensuring that the wire does NOT melt. However even once the wire melts, not much changes. Liquid copper and even copper vapor are excellent conductors. Just because it underwent a phase change, the wire is still a wire. As the wire changes shape either by convection or gravitational forces after melting or vaporizing, it will stretch out the conductive path, and may arc, and will eventually fail. This may seem like a nearly instantaneous effect but with arc flash, milliseconds matter. So even if the melting time of the wire is predictable, the opening time of the circuit is what is important. Preece and Oderdonk's equations do not predict this. I might feel comfortable with their use for predicting opening times on the order of several seconds but with arc flash timing down in the millisecond range, I don't believe it can yield valid results by knowing the melting time and guessing at the opening time based on it.

It would be a lot simpler to just use some small short circuit-only fuses for the PT which is used in a lot of other substation gear instead of relying on the melting of a wire. Then opening times will be very predictable.


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 Post subject: Re: I. M. Onderdonk’s equation
PostPosted: Mon Feb 02, 2015 4:49 pm 
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For an example of the unsuitability of thin wire used as a fuse, consider the method generally used to initiate arcs for test purposes. A thin wire is connected to provide an initial short circuit. The wire vaporizes immediately upon energization, but the arc plasma continues to conduct until an upstream device clears the fault.


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