Little Experiment

[wwhitney]

118.3 volts / 1.7 amps = 69.6 ohms for GES ?

No. That computation would only apply if you disconnected the neutral-ground bond at the service, shorted the line conductor to your GES, and measured the resulting line-neutral voltage at 118.3 (since the neutral is still connected to the utility GES) and the current at 1.7A.

Did you measure the current on the GES with no load? Is your current measurement accurate in the 1-2A range? Are you on city water with a metallic water service? Because with a current division of 1.7A / 32.8A, the GES resistance seems very low, as per my earlier computation.

Cheers, Wayne

 

[ptonsparky]

Can I use a series parallel calc?
I know I have the 34.5 amps going into the neutral buss. It then splits into my parallel neutral return and the GES. Guessing I need to calculate the VD on the 1/0 and the 2 separately. Then apply...still thinking..

As accurate as my Fluke 337 can be. Current does drop to 0 at no load and does indicate .1 to .2 amps on GEC as lighting and/or a small fan is cycled.

 

[wwhitney]

Can I use a series parallel calc?

That's basically what I did in post #16. I can draw out the circuit if you'd like.

I don't see any problem with the approach in post #16, other than that the answer of 2.1 ohms seems too low.

If you take the 120 mOhms wire resistance you calculated in post 20 as correct, then the GES path resistance is in parallel with it, and the current divides between the neutral and the GES path as 1.7/32.8. The voltage drop along either path will be the same, I.e. 1.7*R = 32.8*0.120. So the GES path resistance R = 2.3 ohms.

Cheers, Wayne

 

[ptonsparky]

Post 20 is the combined resistance of the 1/0 and 2.

I would not expect my Ufer is that effective, but I've never attempted to calculate or measured one before.

 

[wwhitney]

Post 20 is the combined resistance of the 1/0 and 2.

You put in the diameter of #2, but the calculator uses twice the one way distance? So that would give you the calculated resistance and voltage drop if your ungrounded were #2 as well, instead of 1/0.

The observed one-way resistance on the #2 was higher (110 mOhms per post #16, vs 60 mOhms per the calculator), perhaps that includes the transformer impedance? Introducing an unknown transformer impedance into the analysis of post #16 gives another variable without an additional constraint, so then I think the math problem is undetermined.

Cheers, Wayne

 

[ptonsparky]

I put in the average of 1/0 & 2. That gives me the combined resistance both ways when using a reduced neutral.

Maybe I'll pull the GEC and see what happens, just for the S&G. I suspect the VD to go up.

 

[wwhitney]

I put in the average of 1/0 & 2. That gives me the combined resistance both ways when using a reduced neutral.

In which field? Your screenshot shows a wire diameter of 0.29125", and NEC Chapter 9 Table 8 says a #2 conductor has a diameter of 0.292", and a #1/0 conductor has a diameter of 0.372".

But that alerts me to an error in post #16, I thought #1/0 was 3 sizes different from #2 (as in 2 vs -1), but it's just 2 sizes different. So the area ratio is less than 2. Maybe I'll draw up a diagram and redo the calculations there.

Cheers, Wayne

 

[oldsparky52]

/Okay, from a guy that quit thinking years ago, and dropped out of college when I hit the brick wall in calculus (circa 1971).

VD=5.5 volts. 1/2 on hot and 1/2 on neutral so the neutral VD is 2.75 volts with a 123.8 starting voltage and a 34.5-amp load.

VD=IR
2.75=34.5*R; 2.75/34.5=R; 0.0797=RT

1/RT = 1/R1 + 1/R2 (neutral is a parallel circuit)

1/.0797=1/R1+1/R2; 12.55=1/R1+1/R2= RT

12.55=1/R1+1/R2

The current divides between the two paths on a 95%/5% split (approximately).

So R1=95% of 12.55 and R2 = 5% of 12.55

12.55*95%=11.92 ohms
12.55*5%=0.6275 ohms

This is the 3rd time I've done this and came up with different answers every time. so I don't know what I'm doing wrong, but I'm not working it again. If I'm correct or incorrect I would love to hear where and why. Thanks.

 

[wwhitney]

VD=5.5 volts. 1/2 on hot and 1/2 on neutral so the neutral VD is 2.75 volts with a 123.8 starting voltage and a 34.5-amp load.

Since the hot and neutral are different sizes, the voltage drops will not be equal.

Cheers, Wayne

 

[oldsparky52]

Since the hot and neutral are different sizes, the voltage drops will not be equal.

Cheers, Wayne

LOL, well that blows the rest of it (and I already reviewing see I did something else wrong).

 

[wwhitney]

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

 

[oldsparky52]

/Okay, from a guy that quit thinking years ago, and dropped out of college when I hit the brick wall in calculus (circa 1971).

VD=5.5 volts. 1/2 on hot and 1/2 on neutral so the neutral VD is 2.75 volts with a 123.8 starting voltage and a 34.5-amp load.

VD=IR
2.75=34.5*R; 2.75/34.5=R; 0.0797=RT

1/RT = 1/R1 + 1/R2 (neutral is a parallel circuit)

1/.0797=1/R1+1/R2; 12.55=1/R1+1/R2= RT

12.55=1/R1+1/R2

The current divides between the two paths on a 95%/5% split (approximately).

So R1=95% of 12.55 and R2 = 5% of 12.55

12.55*95%=11.92 ohms
12.55*5%=0.6275 ohms

This is the 3rd time I've done this and came up with different answers every time. so I don't know what I'm doing wrong, but I'm not working it again. If I'm correct or incorrect I would love to hear where and why. Thanks.

I should have said 1/R1=95% of 12.55 and 1/R2 = 5% of 12.55.

1/R1 = 11.92; 1/11.92 = R1; = 0.084 ohms
1/R2 = 0.6275; 1/0.625 =R2 = 1.6 ohms

 

[kwired]

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

This here. There are several series and parallel components. Fill in the known values and calculate the unknown values. Hint your feeder neutral and the earth are in parallel with one another but are not 123.8 across them nor is it the 118.3 volts across them, some of that drop is occurring across the ungrounded conductor but since the neutral is not same size as ungrounded, if you were to disconnect the grounding electrodes that drop would not be same on remaining two conductors either.

If you can measure the voltage between the neutral conductor at the building distribution and the source neutral terminal you then have a voltage reference and a current reference for each parellel path and just need to calculate the resistance of the two paths.

 

[LarryFine]

View attachment 2559537

Why is the load neutral grounded (upper-right corner)? That's why I asked why there was any GEC current.

 

[wwhitney]

Why is the load neutral grounded (upper-right corner)? That's why I asked why there was any GEC current.

Because a service requires a GES connected to the service neutral? The measured GEC current was 1.7A. Schematically, the GEC is the upper right most vertical conductor in my drawing.

The GES and earthing conductor on the upper left are the utility's, at the pole, and possibly all the poles if the distribution is MGN and the primary and secondary neutrals are both earthed and thus connected at the pole.

Cheers, Wayne

 

[LarryFine]

I must apologize. I don't know why, but for some reason, I thought we were talking about feeder conductors between a house and a detached building, not service conductors between a utility and a house.

 

[wwhitney]

I thought we were talking about feeder conductors between a house and a detached building, not service conductors between a utility and a house.

You know, looking back, it's not clear which this is--the comment about "POCO meter 300' away" could mean a meter on the house, or it could mean a meter at the pole, with the service entrance 300' away.

But if it's a feeder with an EGC, then the OP has a bigger problem. If it's an old feeder without EGC, then it's the same as a service for this discussion.

Cheers, Wayne

 

[kwired]

I must apologize. I don't know why, but for some reason, I thought we were talking about feeder conductors between a house and a detached building, not service conductors between a utility and a house.

Those usually were grounded again at the separate structure until more recently (2011 NEC maybe?) where code no longer allows it except, existing installs can remain, with conditions

 

[ptonsparky]

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

 

[kwired]

You know, looking back, it's not clear which this is--the comment about "POCO meter 300' away" could mean a meter on the house, or it could mean a meter at the pole, with the service entrance 300' away.

But if it's a feeder with an EGC, then the OP has a bigger problem. If it's an old feeder without EGC, then it's the same as a service for this discussion.

Cheers, Wayne

Or it can be like I just mentioned, old three wire feeder that was allowed to be grounded again at a separate structure.