GFCI and Branch Circuit Protection

[fifty60]

There is equipment that uses open a nichrome wire heating heating elements. I am being told by an NRTL that the wire on the heater failing is a "normal part of the heater life cycle" and that I should use GFCI breakers.

I've never really used GFCI's on anything other than outlets. I am not able to replace the entire branch circuit over current protective breaker with a GFCI am I?

The equipment can be considered as a branch circuit, but I am looking at the branches inside the equipment as individual branch circuits. One of which is the heater circuit. Will I need to use 1 normal breaker and 1 GFCI breaker? Can I size my wire of a GFCI alone?

 

[charlie b]

I am being told by an NRTL that . . . I should use GFCI breakers.

Whatever for? If a wire that serves as a heating element were to fail, it would create an open circuit, and a GFCI device will not trip under such conditions.

 

[GoldDigger]

Whatever for? If a wire that serves as a heating element were to fail, it would create an open circuit, and a GFCI device will not trip under such conditions.

Unless it is an inadequately supported stretched coil of wire that can go sproing when it melts and flail around until it touches the heater case. And that should be a marginal design for UL too.

 

[don_resqcapt19]

Whatever for? If a wire that serves as a heating element were to fail, it would create an open circuit, and a GFCI device will not trip under such conditions.

One or both ends of the heating element are still energized when the element fails and if an energized end touches grounded metal, there will be current flow. Probably too little current to trip a normal breaker, but enough so that a GFCI would clear the circuit.

 

[fifty60]

Would periodical insulation resistance or voltage withstand testing prevent this type of failure?

 

[GoldDigger]

Would periodical insulation resistance or voltage withstand testing prevent this type of failure?

No. The wire is not typically insulated and the ceramic or mica standoff insulators are not what fails.
The thinning cross section of the resistance wire will not cause any leakage, nor will it change the overall resistance enough to be reliably detected.
But if you see a visible hot spot on the heater wire it is probably a thin spot on its way to burning out.
Do not confuse a hot spot due to thinning with one which results from bending the wire sharply back on itself, putting more heat into a small volume.
If the heating element is not visible, I know of no way to predict an upcoming failure.
A semi-destructive test of overvolting the heater to produce more current may cause a failure waiting to happen to happen immediately instead (sooner rather than later, anyway).