Resistive neutral on a 240/120 single phase system

[PetrosA]

120729-1411 EDT

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Find the power company transformer. Run a long test lead from the ground rod at the transformer. You might measure the current in the transformer grounding wire. It should not be very much. This wire will be a moderately heavy wire relative to expected normal current, and whatever current is in the wire won't produce much voltage drop from the transformer center tap to the ground rod. If there is high current, then you know there is some kind of problem.

One thing to keep in mind is that, if you are dealing with a broken or weak neutral, it can be broken or weak before or after the utility transformer. Either way you can measure current on the transformer ground wire and there can be shocks felt on the property. Gar's next step will help determine where that break or weakness is.

This test lead is taken to the main panel. Note: there could be a large voltage between the test lead wire and the main panel neutral. Be careful and use suitable equipment.

Measure the voltage difference between the main panel neutral bus and the transformer ground rod. This is the voltage drop across the transformer neutral to main panel neutral, assuming not much voltage drop on the transformer grounding conductor. If the first experiments with the light bulbs or meters did not indicate a large voltage change with normal load changes, then while monitoring the neutral path voltage connect a 100 W load to one phase. Note: the change in neutral voltage difference. At 8 A at my home I get about 0.2 V change. At 200 A this would extrapolate to about 5 V. If with the 100 W load you get a very small change, then try a 1500 heater, about 12 A.

Note: my calculated neutral path resistance is about 0.2/8 = 0.025 ohms.

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At this point you will have enough information to determine whether this is a bad neutral, and whether it's before or after the utility transformer since you've effectively checked the neutral from the service back to the transformer. If the current and voltage readings are similar when taken to the neutral or the test lead and a neutral symptom still exists, then the break is on the primary side of the transformer (In those areas of the country where single phase 120/240 is derived from a primary phase and neutral...).

Reading the OP, it seems to me that calculating all this would really only help if the broken or weak neutral is between the service and transformer. Otherwise, with neutral current returning from the transformer back through the utility grounds the variations in soil conditions (mineral content, moisture level, breakage point in the network etc.) could hugely affect any calculations by introducing an enormous number of unknowns.

 

[steve66]

From a real world perspective, current can and does flow through the grounding system... even on properly connected wiring systems. Current takes all paths, not just the one least impeding. Current on those paths is inversely proportional to each path's impedance.

Smart:

Where would this current go? Where would it come from?

I will concede there could be some small leakage currents coming from the primary side of the transformer, but I doubt they are enough to consider.

And there isn't any way to calculate actual currents and voltages unless you know the leakage current.

I think some of the calculations and numbers thrown out on this thread have the cart before the horse. All the loop equations are pointless unless someone drawings the circuit and labels the currents to make sure everyone is on the same page.

 

[steve66]

GAR & Mivey have wrote loop equations for a neutral that is open, and then shorted to ground.

That's not correct for the circuit the OP talked about.

 

[Smart $]

... Where would this current go? Where would it come from?

The neutral service conductor is connected to grounding systems on both ends. Common parallel paths are metal water lines, communications' grounds, and earth.

I think some of the calculations and numbers thrown out on this thread have the cart before the horse. All the loop equations are pointless unless someone drawings the circuit and labels the currents to make sure everyone is on the same page.

Not totally pointless. Understanding typically starts with the basic and graduates to the more complex. Not much difference here than Electrical 101 texts giving simple neutral current formulas for which accuracy is highly dependent on unity power factor.

 

[Smart $]

GAR & Mivey have wrote loop equations for a neutral that is open, and then shorted to ground.

That's not correct for the circuit the OP talked about.

OP topic provided resistance values for 3-wire system with resistive neutral (actually an open neutral with resistive parallel earth return path). Gar's calculation is correct for the values provided... but an assumption was made by the OP'er that the 25 ohm resistance of the ground rod was also 25 ohms from the neutral bar of the consumer's panel to the neutral connection on the transformer, it is the only parallel neutral current path, and the L1 and L2 currents have a unity power factor.

 

[gar]

120730-1224 EDT

steve66:

The analysis that I previously posted was using the circuit values provided in the original post. The purpose of the particular analysis was to show NSTech one circuit analysis method that could solve the question as presented, and that he might be able to follow the steps.

Although presented as an open neutral it was really a neutral of 25 ohms resistance, just thru the ground. And the 25 ohms was obviously an arbitrary value, but derived from a grounding electrode spec limit.

In the real world you would never want the neutral resistance to get that high. Under normal load conditions I think you should like the neutral voltage drop to be under one volt or so. Under motor startup or someone with an instant water heater the neutral voltage drop can be expected to be higher.

 

[steve66]

120730-1224 EDT

steve66:

The analysis that I previously posted was using the circuit values provided in the original post. The purpose of the particular analysis was to show NSTech one circuit analysis method that could solve the question as presented, and that he might be able to follow the steps.

Although presented as an open neutral it was really a neutral of 25 ohms resistance, just thru the ground. And the 25 ohms was obviously an arbitrary value, but derived from a grounding electrode spec limit.

In the real world you would never want the neutral resistance to get that high. Under normal load conditions I think you should like the neutral voltage drop to be under one volt or so. Under motor startup or someone with an instant water heater the neutral voltage drop can be expected to be higher.

The equations still don't fit the circuit the OP talked about. The resistance to ground of the grounding electrode system has nothing to do with the neutral point. Its an open neutral - there is no connection at the neutral point.

The circuit is a simple single loop.

No current flows to ground or through the 25 ohm resistance, and the 25 ohm resistance has no effect on the voltages or currents in the circuit.

 

[Smart $]

... The resistance to ground of the grounding electrode system has nothing to do with the neutral point.

Its an open neutral - there is no connection at the neutral point.

This is where you are in error. The neutral connection at both ends of the service neutral are bonded to the respective grounding systems. Neutral current flows on all available paths. When the service neutral has full integrity, the majority of neutral current flows on the service neutral, as it is the path of least resistance. Some neutral current flows through the grounding system, as it presents higher resistance parallel paths. When the service neutral loses integrity (i.e. "open" for example), neutral current still flows on the parallel paths and likely increases to make up for the reduced or non-existent service neutral current.

The circuit is a simple single loop.

No current flows to ground or through the 25 ohm resistance, and the 25 ohm resistance has no effect on the voltages or currents in the circuit.

You need to re-assess your position on this.

 

[steve66]

This is where you are in error. The neutral connection at both ends of the service neutral are bonded to the respective grounding systems. Neutral current flows on all available paths. When the service neutral has full integrity, the majority of neutral current flows on the service neutral, as it is the path of least resistance. Some neutral current flows through the grounding system, as it presents higher resistance parallel paths. When the service neutral loses integrity (i.e. "open" for example), neutral current still flows on the parallel paths and likely increases to make up for the reduced or non-existent service neutral current.


You need to re-assess your position on this.

Smart: Thanks for the detailed description. Now I'm following you all.

For some reason, I had an open neutral on the load side of the panel stuck in my mind. Completely different situation.

That's why I asked where would the current come from, or where would it go. I was stuck on a neutral open at the load, and I couldn't figure out where you were all seeing a current flow through the 25 ohms.

 

[steve66]

Greetings. Often I end up at a customer site where people and/or animals are experiencing shocks. I will sometimes find an "open" neutral that is the culprit. But even though the neutral might be effectively disconnected at the meter enclosure or transfer switch there is still almost always a connection between the center tap of the utility transformer and the neutral bar of the electrical service panel through ground rods. So here is my question: Lets say there are 120 V loads on Line 1 with a total resistance of 10 ohms and Line 2 is at 30 ohms. With an open neutral the voltage and amperage on this voltage divider would be easy to calculate. BUT here is where I am confused: The center tap of the transformer is connected to the utility ground and their neutral is connected to multiple grounds effectively bringing their neutral on the supply side to near zero ohms. Then lets say the building that is fed with the "open" neutral has a single ground rod of 25 ohms. Now the simple voltage divider becomes more complex because there is a 25 ohm resistance from the neutral bar on the service panel to the center tap of the transformer. Can someone explain to me the formula that will take into account all 3 resistances? Thanks!

Millman's theorem makes short work out of this calculation. Its a lot easier than using loop analysis and working simultaneous equations.

Here is a link to the formula;

http://www.allaboutcircuits.com/vol_1/chpt_10/6.html

One voltage would be entered as 120 volts, while the other would be -120 volts.

So for this problem, its: (120/10 + 0/25 + -120/30) / (1/10 + 1/25 + 1/30) = 46.15 volts across the 25 ohm resistor. Its that simple.

Once you have the voltage across the 25 ohm resistance, its easy to solve the voltages across the other two resistors by subtracting from the 120 volts sources. That gives the 166 V and 73 V GAR got.

The nice thing about Millman's is that it can easily be used for more than 3 branches - just extend the equation. It's also easy to substitute in AC voltages and impedences. So it could easily take PF into account. An equation for a 3 phase circuit with different power factors in the legs would be:

((120@0/Z1) + (120@120/Z2) + (120@240)/Z3) ) / (1/Z1 + 1/Z2 + 1/Z3 + 1/ZN)


And its very easy to write a program that uses Millimans formula on anything programmable.

 

[mivey]

I agree that a non-unity power factor caused by reactive loads will make the calculation posed inaccurate.

The formulas I posted work fine with non-unity pf loads. The only assumptions are that the voltages are from a single-phase center-tapped type source and that they are approximately balanced. The assumption of a balanced source is a reasonable assumption under most circumstances and certainly for the scenario posed. If you felt froggy, you could add some impedance to Z1 & Z2 to account for unbalanced source impedance.

Millman's theorem makes short work out of this calculation. Its a lot easier than using loop analysis and working simultaneous equations.

No doubt. But the next step with the loop analysis is to also consider L-L load impedance.