ECN Forum Forums General Topics of the Trade General Discussion Area short circuit question ?

To anyone;
This is going to sound like a stupid question and it has to do with short-circuits and ground-faults.
What I want to know is this, What is really going on in the cable during a short-circuit or ground-fault ?
I read my Soares book on grounding and old text books but it really doesnt nail it down.
Does anyone have a link or web-site ??

Any help would be much appreciated

Thanks -Mark-

Mark,
Ya gotta come up with a better question.
What's going on in the cable during a fault?
Movement, heat, release of energy, welding of joints, damage to the vinyl insulation (and being a memory fabric, that's a bad thing all unto itself), capacitive and inductive coupling inducing voltages and currents along the entire length and any metal objects nearby. Other than that.....eh, nothing

George:
Well for one, article 250.130c says a single conductor can be run from a device to the panels neutral buss, and they used to accept running a grounding conducctor to the water pipe, when replacing a 2-wire receptacle with a 3-wire.
In another part of the code it says because of proximity all conductors are suppose to be as close together as possible to reduce the impedance and to ensure that the breaker will trip or the fuse will blow.
My question is when there is a fault how is the reactance of the circuit affected and if the proximity thing is important why is it permissible to run a conductor separate from the the conductor delivering the fault.

George:
After mulling over this question over night I think I can phrase it better.
Here is the scenario, a 15 amp circuit, 120 volts running at 80%, ie 12 amps, been like that forever, in other words it has a state of equilibrium. Suddenly there is a ground-fault, a bolted fault to ground, How doesnt matter, but now it see a fault.
My question is this what happens to the impedance, the reactance of the circuit does it change, or does it stay constant, and the R of the impedance does its value change, The magnetic field generated by the fault how does that affect the performance of the circuit, and what does the breaker see ? I found one book covering part of it but when it got to double-integrals, and I havent done that in 33 years, I was at a decided disadvantage.
Business has slowed down so this is how I been spending some of my free time.
Any help would be appreciated
Thanks again
-Mark-

Mark,
Wow, that's probably one great way to spend a little time. I learned most of what I know after I thought I knew it all.

During a fault, amperage goes very high, very fast, that means your 'instantaneous' impedance changes dramatically. Soares illustrates this far better than I ever could here. That is the reason the code does not allow (except in some rare cases, as you quoted) conductors to be separated from each other. In DC it makes no difference, in AC it can raise impedances (with high amperages) to points where it will not follow the intended path back to XO, which is where we want a fault to travel.

The code recognized that adding a ground to an outlet, it is safer to have one, even with high impedance, than to not have one, at least we have established a reference point. remember, that's a contingency, you would not get away with it in a new installation.

Short answer, to repeat Soares, impedance goes high in a fault, always. If you do something to raise impedance such as separating conductors, you make it impossible to have a viable ground path, so never separate conductors.

Magnetic path, in a fault, nearly every metal object within reach of the conductors will have a voltage and amperage imposed upon it, detracting from the fault clearing cabability of the OC device. It's best if we eliminate those as much as is practical even though that is very difficult.

One of the reasons I really don't care for the practice of allowing a metal 90 in the underground portion of a PVC run, during a fault, we waste precious amps in turning this into an electro-magnet....literally. If it is embedded in concrete the problem can be multiplied by that factor as well.

Uh, geeez, did that answer your question at all?

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[This message has been edited by Bill Addiss (edited 03-10-2002).]