Sub panel question

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Re: Sub panel question

For just 5' of 4 AWG (R1) from 120V meter to Load (R2) and 5' of 4 AWG (R3) from load back to meter, I get:

5' 4 AWG = 0.00154 ohms (R1)
5' 4 AWG = 0.00154 ohms (R3)

Voltage Drop (vd) across each of R1 and R3 is:

E = I * R

Evd = 30A * 0.00154 ohms = 0.0462 vd @ R1
Evd = 30A * 0.00154 ohms = 0.0462 vd @ R3
add the R1 and R3 together 0.0924 vd.

120V @ meter minus 0.0924 vd = 119.9076 vd @ R2

R = E / I

R = 119.9076 vd / 30A = 3.99692 ohms @ R2.

R1 @ .00154 + R2 @ 3.99692 + R3 @ 00154 = 4 ohms

E = I * R

120V = 30A * 4 Ohms
 
Re: Sub panel question

Glenn,
I don't know what I did with my numbers. They are not correct. My point is that if the coupling is a high resistance or open connection the only voltage to drive current through the stickman is the voltage drop on the grounded conductor, so the amount of current will be very small. With the correct voltage drop of 0.0462 and a stickman resistance of 500 ohms, the current through him would only be 0.0462/500 or 0.0000924 amps.
don
 
Re: Sub panel question

Would someone with the "AC savvy" make comments of the actual impedances etc. as opposed to the typical DC ohms Law of this 5 feet of 4 AWG and the RMC senerio ?

Glenn
 
Re: Sub panel question

Originally posted by charlie b:
The path you described will not carry current in the manner you describe. Let me return to my original statement about the fundamental principal at work here. Once current has left the source (i.e., via the hot leg), it seeks all available paths back to the source (i.e., normally via the neutral or cold leg). But once it has made it there, it does not go any further. I know it will continue to go around the same loop over and over; I don?t mean that. But if current goes from the sub-panel neutral bus to the main panel neutral bus, it?s next move is to the source. That's the end of the circuit. It will not proceed from the main panel neutral to the sub-panel (via the bonding jumper and the feeder EGC). The motive force for current flow (i.e., voltage) will drive current the other way (i.e., back to the source).
Click to expand...
Thanks for the reply charlie.

I agree, the next move is back to the source which is the utility transformer. Yes, the service neutral is the best route back and most of the current will go that way, but some current will follow other paths. As has been mentioned before on this forum, if one of the grounding electrodes is a metal water line connected to the neighbor's metal water line for example, some of the current will follow the GEC, the metal water lines, the neighbor's GEC and the neighbor's service neutral. Another path would be through the GEC to any grounding electrode, then through the earth to the utility ground and back to the transformer that way. This would be a high resistance path but a few misguided electrons will take that path because it is there. And another path to the transformer is the one I am describing, through a feeder EGC, then a branch circuit EGC, through a metal appliance case, through an unfortunate person, through the earth, through the utility ground, and back to the transformer. The current would not follow the service ground back to the service panel in this case, because as you say that wouldn't get it any closer to home, but a small amount of current would use the path I describe through the utility ground because it is an available path from the service panel back to the utility transformer.

Now, the conclusion I am coming to is that this current I describe would be insignificantly small because of the resistance involved in that path, but I'm wondering if that wouldn't also be true of current in the path you describe in your original post through an unfortunate person if the path includes the earth.

I am thinking that the real danger created by the illegal bond in a sub-panel is the possibility it creates for a path back to the service panel as you describe via an unfortunate person that does not use the earth but some more conductive path such as a metal water line, or some other conductive element that is in contact with something that is bonded to the service panel.
 
Re: Sub panel question

eprice:

I still think that you are visualizing paths that cannot carry current in the direction that you are describing. So let?s pin down a couple of basic principles, and see where we can go from there.

First, no current EVER flows in the GEC (i.e., the wire that connects the main panel?s ground bus to the grounding electrode ? be that electrode a ground rod or a water pipe or whatever), and no current EVER flows through the grounding electrode itself, UNLESS some piece of equipment has a fault to ground (e.g., hot wire to case), or unless there is a lightning strike. Reason: There is not a complete path. The resistance between all current-carrying components (e.g., the transformers, panels, wires, and loads) and planet Earth is essentially infinite.

Secondly, if there is a fault within a component (e.g., hot wire to case), the current flowing in the fault path (i.e., source to panel to fault point to case to EGC to panel and back to source) should be high enough to trip the breaker and terminate the event.

Third, although it is true that current will take all available paths, it does not travel in all available directions. The direction of the push (i.e., from source hot to source neutral) is the only direction that the current can take. No matter what fault conditions exist, no matter if an unfortunate person is touching the wrong component at the wrong time, current cannot flow from the main panel, via the feeder EGC, in a direction that takes it to the branch circuit EGC. The voltage source will push any and all current within the EGC in the opposite direction ? towards the main panel. This is not a matter of ?how much resistance does one path have, in comparison to that of another path.? It is a matter of current goes in the direction in which it is pushed.

As to the potential danger of a fault path that includes planet Earth, there are (sad to say) a vast number of documented cases in which a person holding a device receives a shock, with the only possible path through their body necessarily including the dirt beneath their feet, and with fatal consequences. It only takes 0.1 amps to cause a fatality. On a 110 volt system, that means the total path resistance would have to be less than 1100 ohms. The resistance of the body itself can easily be under 300 ohms, and the resistance of the ground rod to earth should be under 25 ohms. Unless the person is wearing rubber boots and is standing on dry desert sands, the resistance of the dirt can easily be under 700 ohms. As I say, it has happened many times.
 
Re: Sub panel question

Charlie B: Please explain why you state the ground electrode conductor never carries any current, except under fault events?
 
Re: Sub panel question

Is not the GEC and the grounding electrode a parallel path of the Grounded System Conductor of any Utilty supplied Service ie. Utilty transformer to the Service Equipment grounding system ?

I have withnessed the Water GEC path carry approximately 2/3rd's of what the grounded ( neutral ) conductor should be carrying.

Still no comments of my other question of the difference of an AC impedance law and a DC ohms law.
for the described 5 foot of 4 AWG , 30 Amps, 120 volt system.
 
Re: Sub panel question

Originally posted by bennie: Charlie B: Please explain why you state the ground electrode conductor never carries any current, except under fault events?
Click to expand...
I mean that (1) Current must have a complete path from the source back to the source, and that (2) Under ?normal conditions," the GEC is not part of a complete path.

If, throughout a distribution system, there is no ?connection? (meaning either a direct short circuit or a high-resistance leakage path, possibly from degrading insulation) from a ?current-carrying conductor? (meaning both the ungrounded and the grounded) to any ?external metal parts that are not intended to carry current,? then there would be no complete path from the source back to the source, with the GEC being part of the path.

But let me now repeat something from my initial post on this thread: ?I?m going to use conversational terms here, and make statements that are not precise.? So starting now, I?ll be a bit more precise.

(1) Any closed loop of conductive materials will experience an induced current, if the loop is in the presence of a AC magnetic field. Put your watch on the desk next to your computer, and current will flow around the band. In an electrical distribution system, there are many such loops. They consist of the ground rod, the GEC, the EGCs, water pipes, and other metal objects, with dirt being involved in some of the loops. The magnetic fields that surround the transformer and every current-carrying conductor will induce currents in each loop, with the result that some measurable currents may be present in the ground rod.

(2) Also, any two metal objects will have between them a capacitance, any single metal object will have a self-inductance, and any two metal objects will have between them a mutual inductance. In an electrical distribution system, the amount of capacitive and inductive reactances that exist among the various cables, conduits, and other metal objects can be high enough for there to be measurable currents present in the ground rod.

So I confess: I lied. But I told you at the beginning that I would.
 
Re: Sub panel question

I mean that (1) Current must have a complete path from the source back to the source, and that (2) Under “normal conditions," the GEC is not part of a complete path.
Click to expand...
I know it isn't "normal", but in an area that still has a metallic water main system, this scenario could easily exist for some time before being detected.

Neutral2.gif


Ed
 
Re: Sub panel question

Ed, that's a great graphic. Were you aware that EPRI built such a dummy subdivision in MA to test how currents circulate on water pipes and neutrals?
Karl
 
Re: Sub panel question

WOW!,

Ive been out of town. Thank you all so much for the responses!!! By the way I'm a residential home inspector.

Thanks again,
John
 
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