Hot phase and voltage drop

[Twoskinsoneman]

Was talking to an electrician the other day. On a 480 sub panel he had a b-phase wire of the feeder way over heated. Could barely touch the wire for more than 2 seconds. He checked the voltage across the main breaker. The drop across the b phase of the breaker was twice the "a" and "c" phases but it was only 5 millivolts. He ordered a new breaker. (very expensive).

I asked what the current differences were and he said he didn't check. He said the voltage drop confirms the breaker is bad... Anyway can a 5 millivolt drop really confirm this to be a bad breaker? I am dubious but this guy has much more experience than me and I thought I would ask here.

BTW he is still waiting for the breaker to come in so I don't know what came of it yet.

 

[G._S._Ohm]

a b-phase wire of the feeder way over heated.

So the current in this wire must be higher than in the others.

The drop across the b phase of the breaker was twice the "a" and "c" phases but it was only 5 millivolts.

Assuming all three of the breaker contacts were in the same [good] condition this means the current in b is roughly twice that of the others, which means that the temperature rise above ambient for the b phase wire is roughly 4x that of the other phases.

BTW, for relay contacts at rated currents less than 30 mV means good contacts.
Apparently for these breakers at rated current, the contact resistance plus the voltage drop across the internal resistance that does the thermal trip amounts to only about 5 mV.
Therefore I think it's possible that the force pressing contacts together in breakers is much higher than the force pressing contacts together in relays.

I don't suppose you could get the current flow in the phases and the breaker current ratings?
With these numbers the breaker internal resistance values can be figured for breakers used at these current levels.
Since a lot of money is riding on a good outcome it'd be helpful to have some breaker specifications that help confirm future diagnoses for these types of problems.

 

[Rick Christopherson]

I think your friend may have bought a very expensive breaker without doing all of his homework. You shouldn't be getting much drop across the breaker, so his measurements are still indeterminate. The fact that they are so close together doesn't prove the breaker is bad. That small of a difference could mean higher current flow, or it could mean nothing. However, the small drop for all of the phases indicates that there is not a high resistance connection inside the breaker. If the currents had been measured and found to be similar, he should have checked the lug connections for a high resistance connection.

 

[hurk27]

Rarely does the contact Resistance in the breaker result into the wire heating up, my bet is the terminal the wire is landed in. not saying it can't just most designs of breakers today, contact resistance result in a hot breaker that prematurely trips.

 

[G._S._Ohm]

I only hinted at this in my post and it is the main point of this whole exercise. My bad.

The wire is hot because the current is higher in that wire than in the others because the voltage on that phase is high or the load impedance in that phase is low, or both.
The higher breaker drop is a symptom of the higher current or of high contact resistance, or both.

Comparative measurements of wire temperature have their place as a beginning in collecting clues but measurements in volts, amps and ohms are more credible.

 

[hurk27]

I only hinted at this in my post and it is the main point of this whole exercise. My bad.

The wire is hot because the current is higher in that wire than in the others because the voltage on that phase is high or the load impedance in that phase is low, or both.
The higher breaker drop is a symptom of the higher current or of high contact resistance, or both.

Comparative measurements of wire temperature have their place as a beginning in collecting clues but measurements in volts, amps and ohms are more credible.

What you posted is good info for any electrician to use in trouble shooting breakers, I found one other thing to look for that I stumbled upon that about drove me nuts.

I was called to a Long John Silvers restaurant that had a breaker for their roof top AC unit that kept tripping, I took the current readings (RMS Mode) and did some I.R. scans the breaker was running about 140 degrees C, the connections were tight as was the buss screw, (QO 60 amp 3-pole bolt on) the current on the breaker was about the FLA of the unit (36 amps) had about 6ma drop across the breaker but very little across the terminals, the only thing I could figure was the contacts were heating the thermo over load, I replaced the breaker and after a few hours it did the same thing it heated up to about 140 degrees and tripped, so ok maybe a bad (NEW) breaker, I changed it again, same results, this got me wondering what it could be, messing around I set my current meter to peek reading and was astounded at the reading, almost 92 amps, so it was time for a recorder, I borrowed a recorder and there was the problem, it showed a 3450 hz pulse of about 125 amps peek, call an AC man out and he confirmed that the compressor had a bad bearing, after he changed out the compressor the average current was 23 amps.

I was still not sure why the average current of 36 amps cause this much heating, but I guess 125 amps at a frequency of 3450hz anything can happen.

Or just that my meter couldn't read the true average current at a frequency of 3450hz?

 

[G._S._Ohm]

What you posted is good info for any electrician to use in trouble shooting breakers, I found one other thing to look for that I stumbled upon that about drove me nuts.

I was called to a Long John Silvers restaurant that had a breaker for their roof top AC unit that kept tripping, I took the current readings (RMS Mode) and did some I.R. scans the breaker was running about 140 degrees C, the connections were tight as was the buss screw, (QO 60 amp 3-pole bolt on) the current on the breaker was about the FLA of the unit (36 amps) had about 6ma drop across the breaker but very little across the terminals, the only thing I could figure was the contacts were heating the thermo over load, I replaced the breaker and after a few hours it did the same thing it heated up to about 140 degrees and tripped, so ok maybe a bad (NEW) breaker, I changed it again, same results, this got me wondering what it could be, messing around I set my current meter to peek reading and was astounded at the reading, almost 92 amps, so it was time for a recorder, I borrowed a recorder and there was the problem, it showed a 3450 hz pulse of about 125 amps peek, call an AC man out and he confirmed that the compressor had a bad bearing, after he changed out the compressor the average current was 23 amps.

I was still not sure why the average current of 36 amps cause this much heating, but I guess 125 amps at a frequency of 3450hz anything can happen.

Or just that my meter couldn't read the true average current at a frequency of 3450hz?

Heating effect is due to I squared.
I guess if the "average" RMS current [which is not the same as "average current", which is 0 for AC] is 23 A but there are peaks of 125 A, the effect of 125 A squared can swamp out the effect of 23 A if it persists for a significant fraction of the 60 Hz period and so you end up reading 36 A. The 125 A is "hidden" in the 36 A reading.
I'd have to see the waveform to confirm. The RMS value can be calculated graphically with paper and pencil.

While I have your attention, what is normal breaker heating at the 23 A?
6 ma should be 6 mV? So the breaker resistance is 6 mV/36 A = 170 uohms?

But I can't imagine how some mechanical thing spinning and running on 60 Hz can generate 3 khz.

 

[jeremysterling]

.

.... a recorder ... showed a 3450 hz pulse of about 125 amps peek...

I was still not sure why the average current of 36 amps cause this much heating, but I guess 125 amps at a frequency of 3450hz anything can happen.

Or just that my meter couldn't read the true average current at a frequency of 3450hz?

Does the 3450 number actually represent the number of cycles that the current reached 125A RMS? That would be just over 5 seconds, right? That would trip the breaker.

I started a thread a couple weeks ago about measuring inrush and motor starting current that turned into an enlightening discussion.

Perhaps the mods could move hurk's post to its own thread.

As far as the OP, I agree with the other statements that the increased mV drop on B phase is likely directly proportional to the larger load on that phase and not to blame on the breaker contacts or terminations.

 

[hurk27]

Heating effect is due to I squared.
I guess if the "average" RMS current [which is not the same as "average current", which is 0 for AC] is 23 A but there are peaks of 125 A, the effect of 125 A squared can swamp out the effect of 23 A if it persists for a significant fraction of the 60 Hz period and so you end up reading 36 A. The 125 A is "hidden" in the 36 A reading.
I'd have to see the waveform to confirm. The RMS value can be calculated graphically with paper and pencil.

While I have your attention, what is normal breaker heating at the 23 A?
6 ma should be 6 mV? So the breaker resistance is 6 mV/36 A = 170 uohms?

But I can't imagine how some mechanical thing spinning and running on 60 Hz can generate 3 khz.

The 3450 current spikes was the RPM of the compressor motor hitting that bad spot in the bearing, I would assume?

 

[Twoskinsoneman]

Ever if there was more current on the b phase would you expect that to result in a significantly higher VD across the breaker?

I don't understand that. I understand you would have a little more voltage drop in that circuit in general because of the higher current but across the breaker? I wouldn't think the VD across the breaker should be any different based on the current.

 

[SG-1]

Suppose I have a 3000A circuit breaker. The contact resistance is 6uOhms.
V=IxR 1000A x .000006 Ohms = 6mV

At 1000A the volt drop is 6mV.
At 2000A the volt drop is 12mV.
At 3000A the volt drop is 18mV.

The breaker contact has a fixed resistance ( in a perfect world ). Just like a resistor, more current means more voltage dropped across the resistor.

 

[dbuckley]

And if you've got a voltage of 6mv and a current of 1000A you've got 6W of heat being generated across the contact resistance, which should be heat-sinked along the bars and tha attached cables.

 

[G._S._Ohm]

I wouldn't think the VD across the breaker should be any different based on the current.

The breaker has internal resistance and contact impedance, and so more current should equal more voltage drop. Ideally, 2x the I should equal 2x the voltage drop.

Try it, like I did.
Put a spare breaker in series with 120 V in series with a 10 A load, like a toaster.

Measure the voltage drop across the breaker, being careful that your voltage measurement leads only touch the breaker terminals and not the connections where the current flows. It's a four terminal Kelvin measurement and you don't want to measure breaker terminal contact impedance, only breaker internal resistance. That's what the Kelvin clips that come with micro-ohmmeters do.
Then internal resistance equals V/I.

Then add another 10 A load in parallel with the first and again check the voltage drop.

While you're at it, you can also measure the terminal contact impedances. Just put one voltmeter lead on one side of the terminal's mating surface and the other on the other side of the terminal's mating surface.
An ideal connection should read zero volts regardless of the current.

For a badly installed wirenut I measured this contact impedance at 0.5 vac with 10 A flowing, so 50 milliohms. This one wire didn't have score marks on it like the other two wires.
And this 5 W being dissipated in something with the surface area of a wirenut will result in a pretty high temp. rise above ambient.