ECN Forum Forums General Topics of the Trade General Discussion Area How long can a circcuit be?

Don,


It may have to do with the dynamic impedance of the protective device.

Regardless, during a test UL requires 10" (I was mistaken in my previous statment of 4') of rated conductor to be subjected to the same fault current as the breaker. This is to prove that the breaker protects the conductor.

This is a quote from a George Gregory article:

JBD,
The AIC testing is just to prove that the breaker itself can withstand the fault current. In many cases the conductor will be damaged if the fault current is any where near the AIC rating of the breaker. Now if you are talking about the overcurrent trip points of the breaker, then the conductor, if properly sized, would be protected. As I recall UL Standard 489 for breakers only calls for trip testing at 135% and 600% of the breaker rating. There is no test or requirement that the breaker have a magnetic or instantaneous trip function.
Don
Don

I beg to differ with some sentiments expressed so far.

Just because the code sets no rule does not mean something is a good idea. Npw, if you are saying circuit length is a design issue- well, I couldn't agree more.

As for protecting the conductros:

We base our voltage drop calculations on the load imposed. A small load will result in little voltage drop.
Probably the biggest weakness in voltage drop calculations is that they are based upon the RESISTANCE of the conductor. Resistance is fine- when you're using DC. We're not- we use AC. That means IMPEDANCE is an issue.

A real world example I once saw: A 350 ft. circuit, #12 wire in 1/2" EMT, had a short between the wire and the pipe. Rather than trip the breaker, the fault continued.... in the process, it cut an extermely neat slot in the side of the EMT. After 'plasma cutting' for about 1/2 inch, it appears the wire bent too far from the pipe for the arc to continue, and the fault ended. Now, this exposed wire end was still "hot"; but, as the wire had been severed by the fault, parts of the circuit downstream from it no longer had power.

Using the GEMI program, one can see that, at the length involved, one could no longer rely upon the breaker to function under a 40 amp fault.

Now, for those of you who know how to weld, it is no surprise the 120 volts and 40 amps is plenty of electricity to work with.

I submit that an arc that will burn through steel is certainly a safety hazard. After a certain length of run, we need to consider whether the breaker will operate when needed.

Don,
It is a fact that to pass a UL test for an AIC rating the breaker must protect 10" of rated conductor as well as itself.

The quote I provide is:
from
IEEE TRANSACTIONS ON INDUSTRY APPLICATIONS, VOL. 35, NO. 1, JANUARY/FEBRUARY 1999 page 135

titled
Short-Circuit Ratings and Application Guidelines for Molded-Case Circuit Breakers

written by:
William M. Hall, Senior Member, IEEE, and George D. Gregory, Senior Member, IEEE

As far as New Zealand goes (and Ray might remember these) there were tables in the ECP (Electrical Code of Practice) 28, for the installation of cables, with a maximum length given for 5% voltage drop in a given run.
We have a test here in NZ, called an Earth-Loop Impedance test, that tests not only the connections to your panel, but back to the transformer.
It gives you a reading of the total impedance of the total loop, from that, you can measure your supply voltage and find your fault current.
Fault currents as mentioned above, are in the order of kilo-amps.
I did a test the other day at work and found that a simple fault will cause 250,000A, flow through cords and the like, if a breaker does not trip.
The Sub-station(66kV) is just over the fence, with a tranny just out the door (11kV/400V).

{Message edited for 2 typo's}


[This message has been edited by Trumpy (edited 11-02-2006).]

Mike, are you sure about the 250000 Amps??
That is looks more like the 250MVA level of the 11 kV bus!!
If your supply transformer is 100 or 200 kVA you get in the 10's of kA's if the supply cable is heavy gauge and short.

You can't even protect that faultlevel with a 120 kA HRC fuse.

What size is that 400V transformer? If it's 25kVA-50kVA, it probably isn't capable of more than 20kAIC or thereabouts. The impedances of the wires and cable and extension cords will drastically reduce the peak fault current from the maximum the transformer is theoretically capable of.

- Start with the transformer impedance. Then, add up all the line resistances between the transformer and your fault point using NEC tables. You'll certainly need to take the complex impedance into account, especially if your fault current is approaching the limit of your breakers.

JBD,
I just don't understand how the conductor is protected by a breaker with a high AIC unless the breaker is of the current limiting type. As I said before the lowest breaker AIC rating of 5000 amps will cause damage to a #14 conductor unless the breaker opens in less than 1/2 cycle (this is beased on the Insulated Cable Engineers
Association (ICEA) damage current level). Do standard breakers open that quickly? Not trying to be a pain, but just trying to apply what I think that I know to this case.
Don

resqcapt19, I don't want to say that it's impossible for there to be conditions that could possibly put 5000A into a #14 cable, but even if you hooked that #14 cable to an ideal 120V 0-impedance source, if the romex exceeded 4', resistance in that cable is too high to allow current levels to reach 5000'. In reality, transformer and feeder impedance drop current levels WELL below that point.



[This message has been edited by SteveFehr (edited 11-02-2006).]

Don,
It primarily has to do with system impedance. All OCPDs exhibit some current limiting effect as soon as they begin to open effectively changing the bolted fault into an arcing fault. And, yes small breakers can and do operate in less than 1/2 cycle.

Here is more information from the article:


[This message has been edited by JBD (edited 11-02-2006).]