Cable Fault Damage and NEC 110.10

Status
Not open for further replies.
R

ricomichigan

Member
Location
South Carolina
I have a 480V panel with 40kA fault current available. I have several 20A circuit breakers in this panel with #12 wire going to my loads. An electrical inspector has informed me that according to NEC 110.10, my wire is not protected against a fault. His reasoning is that a fault could occur on the load side of the 20A breaker and the 20A breaker (open time of 1 cycle) will let through enough fault current to damage the insulation on the wire. I have seen the fault current ratings of insulated wire and they are low (#12 is rated for 2768A for 1 cycle). I have had a circuit breaker manufacturer's representative tell me that since the circuit breaker is UL 489 listed that the wire will not be damaged. Has anyone dealt with this issue before?
 
ricomichigan said:
I have had a circuit breaker manufacturer's representative tell me that since the circuit breaker is UL 489 listed that the wire will not be damaged. Has anyone dealt with this issue before?
Click to expand...
I don't think that UL 489 looks at short circuit let through currents and conductor damage. I don't even think that there is a specified maximum clearing time for short circuits. The pick up current for the magnetic or instantaneous trip part of small breakers of various brands varies from 8 times the breaker rating to 40 times the breaker rating.
Don
 
If the Inspector is postulating a short circuit within the conductor itself, then the conductor must necessarily already be damaged, and the overcurrent protection device cannot prevent damage to something that is already damaged. On the other hand, if the Inspector is postulating a short circuit somewhere downstream, such as at a load, then the fault path will include the conductors, the fault current will be reduced by the resistance of the conductors, and the available fault current will therefore be far less than the 40kA that is available at the panel.

This is a non-issue.
 
Thanks for the reply, Charlie B, but it continues to be an issue. I have indeed pointed out that a fault in the cable damages the cable and adding some insulation damage to an already damaged cable is a moot point. But he says there could be a fault at a splice or junction box 10 feet away from the source panel and at that point the fault current would still be above the maximum allowable cable fault current.
 
Thank you for your reply, don_resqcapt19. The circuit breaker catalog shows the time-current curve for a 20A breaker to clear in one cycle (maximum) in a fault condition . The rep states that the UL testing includes connecting a short piece of wire to the circuit breaker and verifies that the wire insulation is not damaged.
 
The circuit breakers are rated for 40kA. The issue is that even though the breaker is rated for the available fault current, it will only open in one cycle therby letting through an amount of I2T energy. That energy will heat the wire and cause insulation damage before the breaker trips in one cycle.
 
Who said we are trying to protect the wire from any damage? 110.10 says it has to protect the components from "excessive damage". The 20A breaker should do its job of preventing a fire or explosion and deenegrizing a faulted circuit.

Even the greenest inspector should know that a 20A breaker is allowed on #12 wire.

Steve
 
The AIC rating of a breaker has nothing to do with the amount of current that it "lets through" during a fault. The 40K rating says that it will operate succesfully when there is 40K available on the line side lugs.

All breakers have internal resistance (primarily their contacts) that increases as soon as the contacts begin to seperate and continues to increase until the arc is quenched (resistance of the arc), this resistance reduces the amount of let through current.

One of the big differences between UL testing of circuit breakers and fuses, is that the breakers are tested with 4' of conductor on the load side. To pass the UL489 test the conductor can not be damaged nor can it pull loose from the terminations.
 
Last edited:
Jim,
It is my understanding that there is no maximum current required for the magnetic short circuit trip point in the UL standard. Is that correct?
I know that the magnetic trip currents for some of the more common breakers vary from 8 times the device rating to 40 times the device rating. Any fault currents less than the magnetic trip point will cause the breaker to trip based on its time/current curve.
Don
 
In the 1999 handbook is says that 110.10 was changed to substitute the word current for withstand. This change was because equipment was marked
showing short circuit amps. It also says that the basic purpose of the OC device is "to open the circuit before conductors or conductor insulation is damaged".
It goes on to say that the overcurrent condition must be eliminated before the conductor insulation damage point is reached. System components consist of wire, bus structures, switching, protection and disconnect devices and distribution equipment. When the available short circuit current exceeds the rating of an electrical component, the OC device must limit the let through energy to within the rating of that electrical component.

What happened to the spell check?
 
It seems that inspector doesn't like you, but beyond that I think he is full of you know what. Where I live (SLC) the inspector has the burden to prove to you why he is enforcing something, not just throwing out some code refferrence.
 
Isn't this the section that says installing things within the listing requirements satisfies the code requirement. If so, what part of the listing requirement would be violated? There was another thread where the inspector required a certain length of cable to connect to some device, and I never understood where that came from.

It DOES seem like a short enough section of wire can draw enough current in a cycle to cause damage, and this code section is supposed to prevent that, but I don't really see how to comply. :confused:
 
Don,

I agree that an exact magnetic trip point is not defined. But in this case we are talking about a #12 wire being protected by a 20A breaker. There is no doubt a 20A breaker will be operating in its magnetic mode when subjected to a 40K fault.

Ul has tested the wire and the breaker as an assembly when establishing the SCCR. There is no violation of 110.10.
 
Jim,
There is no doubt a 20A breaker will be operating in its magnetic mode when subjected to a 40K fault.
Click to expand...
Does UL 489 even require that the breaker have a magnetic mode? Even if the breaker has a magnetic mode, what is the let throgh current. Is is possible that it could be greater than the ~2700 amp 1 cycle withstand for #12?

Don
 
Don,

To my knowledge, magnetic tripping is not a specific requirement of UL489. I believe that there are very few specific time vs current requirements other than "not damaging the conductor".
 
ricomichigan said:
..An electrical inspector has informed me that according to NEC 110.10, my wire is not protected against a fault. (if) the 20A breaker (open time of 1 cycle) will let through enough fault current ..(#12 is rated for 2768A for 1 cycle).
Click to expand...
Using E/Z (120/0.044)* for ASCC, #12cu can not generate 2768A ASCC after 21.6 feet, or after 13.5 feet for #12 aluminum. *(Ze=Tbl.9 pf=1, steel conduit XL insignificant)

Wouldn't this inspector risk getting fired if they did not prove either a defective device, or nicked wire was exposed or faulting, within 22 conductor feet, in violation of 110.10?
 
Last edited:
Status
Not open for further replies.
Top