Little Experiment

[ptonsparky]

Technically this is a farm building.

 

[kwired]

Sorry, you guys have been working at this while I took a picnic lunch to a neighboring city park. Had a nap in the sunshine, followed by a chocolate chip malt.

It is a three wire service from the utility pole to the garage. There is a house on the property but the only common is the utility meter. No piping, no water, not communication cable. Milbank makes some nice lugs that allow a double tap so no disconnect at the pole. . Not required yet and being a cheapskate, I didn't put one in.

I have a CEE at the house as well and have never noticed any current flow. Then again I have never purposely loaded up one leg with every thing else off. Full sized neutral. 4/0

FWIW, I am the last transformer on the last pole of the POCO. Nearest household on the POCO primary feed is about a mile away.

I didn't think the applied voltage of 118 was quite right

I spent the afternoon outdoors as well, short sleeves. Not going to be anywhere near that nice the next three or four days the way it is sounding.

 

[synchro]

OK, here's a redo of my post #16, with a crude diagram below to show the circuit and my terminology (same as post #16).

The area ratio of #1/0 to #2 is 70.41/43.23 = 1.63. So I'm going to assume that R2 = 1.63 R1. That should be true for the service conductor portion, but the other various conductors and connections in series may make that not true. But we need a constraint of that form to be able to solve the problem with the information supplied.

The voltage difference between the neutral side of the load and the neutral side of the transformer coil is fixed, which means that the current divides between the neutral and the earth path inversely with resistance. As the current ratio is 32.8 / 1.7 = 19.3, we have R3 = 19.3 R2. Again I'm ignoring short conductor sections at either end that are actually in series with the parallel resistances.

The total voltage drop under load is 5.5V, attributable to R1 in series with (R2 parallel to R3). The latter is (1/(1/1+1/19.3)) * R2 = 0.95 R2. So the total non-load resistance in the series circuit is R1 + 0.95(1.63 R1) = 2.55 R1. V = I R gives us 5.5V = 34.5A * 2.55 R1, or R1 = 63 mOhms. Whence R2 = 102 mOhms, and R3 = 2.0 Ohms.

Cheers, Wayne

View attachment 2559537

I believe that Tom should be able to measure the resistance of the neutral conductor in parallel with GES more directly by using the line voltage on the other ungrounded conductor (i.e., phase) as a reference. For example, let the load shown in Wayne's diagram above be across L1 and N. Then with no load current being drawn from L2, measure the difference in the L2-N voltage when the 34.5A L1 load current is turned on vs. when it's off. Call this voltage difference Vd_{L2_N}. Then the resistance of the neutral and GES in parallel would be Vd_{L2_N} / 34.5A.
And then using the current divider formula similar to what Wayne has done, the neutral conductor resistance would be (Vd_{L2_N} / 34.5A) x R3/(R2 + R3) and the GES resistance would be (Vd_{L2_N} / 34.5A) x R2/(R2 + R3).

The impedance of the transformer is assumed to be small enough that it can be ignored in this test.

 

[ptonsparky]

I believe that Tom should be able to measure the resistance of the neutral conductor in parallel with GES more directly by using the line voltage on the other ungrounded conductor (i.e., phase) as a reference. For example, let the load shown in Wayne's diagram above be across L1 and N. Then with no load current being drawn from L2, measure the difference in the L2-N voltage when the 34.5A L1 load current is turned on vs. when it's off. Call this voltage difference Vd_{L2_N}. Then the resistance of the neutral and GES in parallel would be Vd_{L2_N} / 34.5A.
And then using the current divider formula similar to what Wayne has done, the neutral conductor resistance would be (Vd_{L2_N} / 34.5A) x R3/(R2 + R3) and the GES resistance would be (Vd_{L2_N} / 34.5A) x R2/(R2 + R3).

The impedance of the transformer is assumed to be small enough that it can be ignored in this test.

I probably took that measurement, just didn't record it. When I repeat this I'll use more of my meters for simultaneous readings. I have enough of them.

Thanks to all.

 

[LarryFine]

It is a three wire service from the utility pole to the garage.

Okay, now that I know, my opinion is that the GEC current must be caused by a voltage difference between the garage neutral and (either or both) the house neutral and GEC, and the pole neutral and electrode.

Also known as voltage drop on the service neutral. Since both structures share the neutral on the line side of the meter, I'd guess most of it is to the pole neutral.

 

[synchro]

I forgot to give the resistances in terms of the measured currents through R2 and R3.
From post #43, the resistance of R2 and R3 in parallel is R_P = Vd_{L2_N} / 34.5A
Using the reciprocals of R2, R3, and Rp (i.e., their conductances, because conductances add when they are in parallel):

1/R_P = 1/R2 + 1/R3
1/R3 = (32.8A/1.7A) x 1/R2
1/R_P = 1/R2 + (32.8/1.7) x 1/R2 = 20.29/R2
R2 = 20.29 R_P
R3= (1.7/32.8) x R2 = (1.7/32.8) x 20.29 R_P = 1.052 R_P

 

[ptonsparky]

OK, here's a redo of my post #16, with a crude diagram below to show the circuit and my terminology (same as post #16).

The area ratio of #1/0 to #2 is 70.41/43.23 = 1.63. So I'm going to assume that R2 = 1.63 R1. That should be true for the service conductor portion, but the other various conductors and connections in series may make that not true. But we need a constraint of that form to be able to solve the problem with the information supplied.

The voltage difference between the neutral side of the load and the neutral side of the transformer coil is fixed, which means that the current divides between the neutral and the earth path inversely with resistance. As the current ratio is 32.8 / 1.7 = 19.3, we have R3 = 19.3 R2. Again I'm ignoring short conductor sections at either end that are actually in series with the parallel resistances.

The total voltage drop under load is 5.5V, attributable to R1 in series with (R2 parallel to R3). The latter is (1/(1/1+1/19.3)) * R2 = 0.95 R2. So the total non-load resistance in the series circuit is R1 + 0.95(1.63 R1) = 2.55 R1. V = I R gives us 5.5V = 34.5A * 2.55 R1, or R1 = 63 mOhms. Whence R2 = 102 mOhms, and R3 = 2.0 Ohms.

Cheers, Wayne

View attachment 2559537

Got it.

I think.

 

[ptonsparky]

Last question. Maybe

Are Ufers commonly that low?


That almost has to be working in conjunction with the house Ufer and subsequently via its 4/0 back to the transformer. Current does take all paths.

 

[oldsparky52]

Am I correct in thinking that the resistance on the GEC is in the wire and connections to earth (at both ends) because once connected to earth there are so many parallel paths the the resistance of the actual earth is almost zero?

 

[ptonsparky]

Am I correct in thinking that the resistance on the GEC is in the wire and connections to earth (at both ends) because once connected to earth there are so many parallel paths the the resistance of the actual earth is almost zero?

That is what I have thought. The connection to earth is the point of resistance, not earth itself.

 

[kwired]

Am I correct in thinking that the resistance on the GEC is in the wire and connections to earth (at both ends) because once connected to earth there are so many parallel paths the the resistance of the actual earth is almost zero?

Sort of yes. The hard part is making a low resistance connection, especially from a single rod or pipe electrode.

Keep in mind that generally the utility side is a huge MGN network so your current path is not typically just to one electrode but many electrodes. Every customer side electrode is essentially part of that network as well. That alone is what makes the service neutral a good reference as a general rule as long as it has no compromised connections, and we assume it is typically as balanced across the phases as possible. Ultimately the more current there is on the neutral though the more voltage drop will be on it and then your load side will be at some potential above true ground potential, unless you would happen to have a very low impedance electrode connected to that conductor, but in that case the GEC could be carrying significant amount of current because it is essentially the "equalizer" here.

 

[ptonsparky]

Note to self: Put amp clamp on home GEC the next time I do Little Experiment at garage.

 

[kwired]

Note to self: Put amp clamp on home GEC the next time I do Little Experiment at garage.

Load changes at the house will effect what it reads. Even if nobody home refrigerators, HVAC, etc. that automatically cycle can impact reading.

 

[ptonsparky]

Load changes at the house will effect what it reads. Even if nobody home refrigerators, HVAC, etc. that automatically cycle can impact reading.

Very true but I can time stamp them and as I've noted before, I have not noticed any measurable current on the home GEC.