Loose neutral damage

[FionaZuppa]

Not sure where your going here. You lose a N at say the point of delivery, it is entirely possible to see 240V on one leg if there is nothing running on the other leg.
Go over Jaggedbens post again.

It happens more than you can imagine. Tree falls, tears neutral in half, the hot legs still together, can’t run equipment long like that.

No MWBC involved

Can you show a diagram of this? If you rip out the N of a 1ph CT xrmer, anything that was wired to the N for half voltage will not suddenly run line-line. MWBC would be the only thing that would run wonky, where turning 'on' just one device (line-CT) does nothing, but turning both 'on' runs just fine (line1-device1-CT-device2-line2).

 

[jaggedben]

Can you show a diagram of this? If you rip out the N of a 1ph CT xrmer, anything that was wired to the N for half voltage will not suddenly run line-line. MWBC would be the only thing that would run wonky, where turning 'on' just one device (line-CT) does nothing, but turning both 'on' runs just fine (line1-device1-CT-device2-line2).

The path is:

L1 <-> load on L1 <-> neutral bar in panel <-> load on L2 <-> L2

Since L1 <-> L2 is 240V and N is no longer referenced to anything, the two loads are in series at 240V with the neutral bar simply completing the circuit.

 

[leepalmer]

Happens probably more often than one thinks,

Actually had it happen to one of our jobs. Then within a couple years of that I witnessed 2 other incidents. The common problem with all these incidents was the cable vision/internet coax shielding taking on the current returning to transformer. Phone lines with shielding also. Especially homes not connected to a community metallic water line EGC.
So if anyone here ever runs across some telling them that the coax on the side of the house looks melted, this is a tell-tale sign that the neutral is compromised.

So between connections oxidizing to tree limbs wearing thru an overhead triplex neutral, one can see how often this occurs.

So, if the N is moved to that location, no xfrmr "pole" will be able to return to the CT, and the 120v ckts will not function.
drift hunters

 

[ptonsparky]

There has to be a load, two actually. One on each L-N. When the neutral connection is removed, they become a series circuit.

I'm slow typist..

 

[Fred B]

Right, so if the N is lifted there then no xfrmr "pole" has a path back to the CT, hence the 120v ckts won't run.

Not properly anyway. Unless you have every circuit completly balaced (unlikely) and no multiwire circuits, you will have "attempted operation" and voltage fluctuations high/low voltage on either hot leg depending on loads attached. Everything is somewhat tied together in the service panel.

So why would any device on a 120v ckt see 240v and fry? 120v BC's are two CCC's, one hot (black) and one neutral (white), and you lift the neutral, put a volt meter across that receptacle and you get zero.

This only relevant on a single branch circuit not if loss of primary or service neutral or a MWBC. Every circuit that is 120V is semi interconnected via the neutral and grounding bus and via the bonded N/G point. You would see this effect if you have on the branch circuits with multiple neutrals tied together in a second point of reference such as a Jbox. Lift one neutral at the device end and you will get some variations of voltage on that neutral in relation to the EGC depending on loads on the other circuits linked in the Jbox. Seen it and with a shocking effect of 30V to 90V referenced N/G (apprentice got a shock on N/G contact.)

The N does not magically connect to the other end of the CT xfrmr. What am I missing?

It doesn't magically connect as you describe but there us still a reference to ground on the system as a whole. Imbalaces from single phase 120 loads will still try to return to source through any means possible, even if it is a high impedance means through the earth (NEV). So not only would you need to lift the xfer neutral but also all references to ground by removing all N/G bonds everywhere in the system. The higher the impedance you could be limiting equipment damage as not enough getting thru to cause dammage and essentially is "off" but will get some measurable current thru.
Electricity is as desperate as an addict searching for a fix to get back to the source if there is any means at all for it to get there.

 

[Hv&Lv]

Can you show a diagram of this? If you rip out the N of a 1ph CT xrmer, anything that was wired to the N for half voltage will not suddenly run line-line. MWBC would be the only thing that would run wonky, where turning 'on' just one device (line-CT) does nothing, but turning both 'on' runs just fine (line1-device1-CT-device2-line2).

Almost correct. Both won’t run just fine.
I’m on the utility side and sometimes forget when I talk about a MWBC on the inside or after the meter,
our service coming in before the meter is nothing more than a multi wire as ptonsparky pointed out in post #16.

 

[ggunn]

Right, so if the N is lifted there then no xfrmr "pole" has a path back to the CT, hence the 120v ckts won't run. So why would any device on a 120v ckt see 240v and fry? 120v BC's are two CCC's, one hot (black) and one neutral (white), and you lift the neutral, put a volt meter across that receptacle and you get zero. The N does not magically connect to the other end of the CT xfrmr. What am I missing?

You are missing that all the 120V loads are connected to each other through the neutral. When the neutral is disconnected at the source the circuit becomes a voltage divider network with the voltage on the neutral dependent on the ratio of aggregate L1 to N and L2 to N resistances. The L1 to N loads have a return path through the L2 to N loads and vice versa.

 

[GoldDigger]

No single load will see240V. But with two loads in series the voltage in the lower wattage load approaches arbitrarily close to 240V as the impedance ratio increases. If the only loads are two 100 W incandescent bulbs, neither would light if only one was turned on.
When you turn in the second, on the opposite "phase", they might both light fine if their thermal characteristics match closely enough. But if one heats faster it could well hog the voltage drop and burn out. Then both would be dark again.


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[ggunn]

No single load will see240V.

Yeah, well, but put a megohn load on one side and 1 ohm on the other and one of them will get pretty dang close!

 

[ptonsparky]

No single load will see240V. But with two loads in series the voltage in the lower wattage load approaches arbitrarily close to 240V as the impedance ratio increases. If the only loads are two 100 W incandescent bulbs, neither would light if only one was turned on.
When you turn in the second, on the opposite "phase", they might both light fine if their thermal characteristics match closely enough. But if one heats faster it could well hog the voltage drop and burn out. Then both would be dark again.


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Back when test lights were used it was pretty obvious when one wattage was less than the other.

 

[FionaZuppa]

The path is:

L1 <-> load on L1 <-> neutral bar in panel <-> load on L2 <-> L2

Since L1 <-> L2 is 240V and N is no longer referenced to anything, the two loads are in series at 240V with the neutral bar simply completing the circuit.

And, the two loads are 120v loads, so it's Line1--> 120vLoad-1 -->N1--> bar in panel --> N2--> 120vLoad2--> Line2

Looks like MWBC, so how would 240v be measured across Load-1 or Load-2 ?

Both loads need to be "on" to have amps. One load "off" and you get nothing across either load.

 

[Hv&Lv]

And, the two loads are 120v loads, so it's Line1--> 120vLoad-1 -->N1--> bar in panel --> N2--> 120vLoad2--> Line2

Looks like MWBC, so how would 240v be measured across Load-1 or Load-2 ?

Both loads need to be "on" to have amps. One load "off" and you get nothing across either load.

The panels not the source.
The transformer on the pole or pad mount is the source.
Lose the neutral there.

 

[FionaZuppa]

You are missing that all the 120V loads are connected to each other through the neutral. When the neutral is disconnected at the source the circuit becomes a voltage divider network with the voltage on the neutral dependent on the ratio of aggregate L1 to N and L2 to N resistances. The L1 to N loads have a return path through the L2 to N loads and vice versa.

It's a MWBC, nothing more.

In 1ph CT, the two loads (typical) in a MWBC will never see 240v across one load.

So how would any 120v device "fry"?

One MWBC scenario with two very low probability events, the shared N lifts and one loads shorts itself, leaving 240v on the second load.

 

[FionaZuppa]

The panels not the source.
The transformer on the pole or pad mount is the source.
Lose the neutral there.

Same thing.

BC's become pseudo MWBC's, except the actual ckt's do not share a N downstream (diversity), they share it in the panel on N bar.

 

[ggunn]

It's a MWBC, nothing more.

In 1ph CT, the two loads (typical) in a MWBC will never see 240v across one load.

So how would any 120v device "fry"?

One MWBC scenario with two very low probability events, the shared N lifts and one loads shorts itself, leaving 240v on the second load.

As long as the lumped resistance from L1 to N and L2 to N are the same, everything is hunky if the neutral is disconnected; if the neutral were there there would be no current on it, anyway. If the resistances are not balanced, though, the voltage to neutral from the legs is entirely dependent on the difference in the lumped load resistances L1 to N and L2 to N, and the line with the higher resistance to neutral will get the higher voltage, although V(L1 to N) and V(L2 to N) will always sum to 240V. This is EE101 stuff.

 

[kwired]

And, the two loads are 120v loads, so it's Line1--> 120vLoad-1 -->N1--> bar in panel --> N2--> 120vLoad2--> Line2

Looks like MWBC, so how would 240v be measured across Load-1 or Load-2 ?

Both loads need to be "on" to have amps. One load "off" and you get nothing across either load.

It won't, but if resistance of one is low enough it can come close to being full 240 across the other load.

Remember you are putting two different loads in series with a 240 volt supply. If both loads are equal resistance then you get equal volts across each of them.

For simplicity lets assume true resistance loads. If one is 120 watt load @ 120 volts then it normally draws 1 amp and has 120 ohms resistance
lets say the other load is 1200 watts @ 120 - it draws 10 amps normally and has 12 ohm resistance.

Now put them in series and you have total resistance of 132 ohms. with 240 volts being the supply volts that means 1.8 amps flows through both resistors - they are in series. That 120 ohm resistance with 1.8 amps has 216 volts across it, and power in it will be 389 watts, getting close to four times the watts it was designed for. If it was marginally designed heat dissipation wise @ 120v it likely burning out in short time.

At same time the other resistor only has the remaining 24 volts across it, still 1.8 amps = power in that resistor is only 43 watts when it normally sees 1200 watts in normal voltage application, it will survive this the first one is what will fail.

Now put more loads in there like you will most likely have on a multiple outlet circuit or a feeder or service and it gets more complex as you have both series and parallel loads to deal with, but bottom line is lower resistance loads will have less voltage drop across them.

 

[FionaZuppa]

@kwired
Yes, imbalance. But I mentioned that, if one load shorts itself then 240v will (might) be on the other load, and poof.

Since what, about 1980, resi panels are fairly ok being "balanced" across both lines, as the practice became better at doing things?
There's likely two scenarios to be had.
1) the resi is an electrons saver so they keep everything off, in this case using one load means it won't turn on, then they try other siwtches and stuff, nothing works, they call someone.
2) the resi has all sorts of stuff on all the time, the more stuff 'on' the lower the imbalance will be across loads (bc's) that are looping amps from one pole to the other.

And then you have that most N bars in panels are stabbed to Earth, so unless the example of issue is a fully floating N bar, there's likely still amps flowing pole to Earth. You can create a test model of this using a small low-v CT xfrmr, just tap the CT at shared load point of connection to a ground rod, then lift off the CT at xfrmr, what do you get?

 

[wwhitney]

Simple example: a 120/240V panel is supplying (20) 120W incandescent bulbs (1A each) L1-N and (10) 120W incandescent bulbs L2-N. So the line currents are L1 = 20A, N = 10A, L2 = 10A. The voltages L1-N and L2-N are stable at 120V (ignoring supply impedance).

Now cut the neutral supply conductor to the panel. The current on N goes to 0, and the L1 loads and L2 loads are in series, so the L1 and L2 currents must match. The impedance of the L1 loads is 6 ohms (120V / 20A), and of the L2 loads is 12 ohms (120V / 10A). So the total series impedance is 18 ohms, and the current will be 240 / 18 = 13.3A. The voltage L1-N will be 6 ohms * 13.3 A = 80 volts, and the voltage L2-N will be 12 ohms * 13.3A = 160 volts. At least until the 160V damages some of the L2 loads.

Cheers, Wayne

 

[ptonsparky]

A couple lamp holder pigtails, two incandescent lamps of widely different wattages and a MWBC is the homework assignment today. Take measurements and report back.

I have to mow the South 40.

 

[ptonsparky]

Todays power supplies that have a range of 100 to 240, or even wider, have seemed to reduce the the damage.