sharing a neutral

[duhhuh]

there are 3 ckts feeding three supervisor offices (one office is on ckt 2, other two are on ckt 6) and another office with 4 computer work stations (on ckt 4), which share the same neutral.

there are computers in each office, and paper shredders/printers in each supervisors office. the office on ckt 2 also has a coffee pot (900w) which "burned up". this is the second coffee pot to do so.

Question: I put a circuit analyzer on it, all checked out. 121.8V, .1 impedence. When I introduced a load of 15 amps, voltage dropped to 112v.
Only the two coffee pots (in the same office) have 'burned up''. there has been no effect on the computers/printers/shredders/phones.
What could be the cause?

People have coffee pots or other items in their offices all the time as well as code permits sharing a neutral on 3 phase circuits...

thanks,
mjr

 

[ultramegabob]

I suspect a loose neutral connection somewhere.

 

[Mr. Bill]

Here's a PowerPoint slide that explains this. Just click in the page to move to the next slide. The example is for a 120/240V single phase system. A floating neutral on a 3-phase system would be similar, just slightly different math.
http://code-elec.com/userimages/Lost Neutral.ppt

 

[Mr. Bill]

With a loose common neutral, the higher resistance items (coffee pot) will draw a higher voltage. When the voltage gets over the operating tolerance it's only a matter of time before it fails. Under voltage can sometimes be a problem too.

 

[ches2443]

After you had taken MR. BILL advice, I would take a amp reading on the neutral with every running. If this checks out, then you might look into nonelectrical cause to the problem. Because if it was lack of voltage then it seems to me that the coffee pots would under heat not over heat.

 

[don_resqcapt19]

With a loose common neutral, the higher resistance items (coffee pot) will draw a higher voltage. When the voltage gets over the operating tolerance it's only a matter of time before it fails. Under voltage can sometimes be a problem too.

Given the high wattage of coffee pots compared to the other items listed, I would not expect that the coffee pot to be the higher resistance item.

 

[76nemo]

Are you assuming the VD with a tester like the Ideal 61-165?

I have to start with this question, which I see the most. Coffee pots and many others appliances with heating elements have TCO's built into them. If they run dry/empty for a long enough period, the TCO's will open up, and do their intended duty. These TCO's are about 30 cents a piece. I have to go back and reread your post. If I read it correctly, the coffee pots are of the main concern.

 

[76nemo]

I suspect a loose neutral connection somewhere.

An impedance checker like the Ideal 61-165 checks the neutral and grounding impedance as well as the hot.

 

[76nemo]

I reread your post, and my initial opinion is the branch circuit is not the problem.


Reitterate: JMO

 

[Mr. Bill]

Given the high wattage of coffee pots compared to the other items listed, I would not expect that the coffee pot to be the higher resistance item.

Yeah, I still get wattage and resistance mixed up in my head. My brain hears them as the same. Even after I went thru that powerpoint link I posted.

Is the coffee pot on at night or any other time when the other loads are off? And computer loads cycle a lot. They don't often get near the nameplate amps. When the other 2 circuits are light loads I think that would cause the voltage at the coffee pot to measure about 208V and the current would also increase proportionally. This would create a very high wattage at the coffee pot. And it gets fried.

 

[ohm]

I too vote for a loose neutral. But, be very careful about assigning a resistance value to a COLD element or incandescent lamp. Until the element heats up a lamp draws 80 times it's running current. That's why the tungsten rating is used on light switches.

The same applies using a battery powered tester to assign an impedance to a cold heating element/lamp.

Have you ever been in a dark room and turn on a large lamp? Switch it off & on ten times and depending on the time in the cycle it may seem like a large abnormal spark half the time, but it's normal.

Measure the resistance of a cold light bulb and see if you can apply Ohm's Law to it.

 

[Mr. Bill]

I too vote for a loose neutral. But, be very careful about assigning a resistance value to a COLD element or incandescent lamp. Until the element heats up a lamp draws 80 times it's running current. That's why the tungsten rating is used on light switches.

Where do you get this value of 80 times it's running current from? And how many cycles does this last? I remember my dad hearing some guy tell him to leave his house lights on all the time because it draws less energy then turning them on/off during the day due to the start up current. I told my dad to turn off his lights when he's not using them and that will save him energy.

Since you've provided a value of 80 times it's running current I had to bring out my calculator. 100W A-lamp at 120V draws 0.833A. This times 80 = 66.7A.

The time-current curve for a type FA circuit breaker from Square-D shows the instantaneous trip area at about 300-600A for a 15A or 20A breaker. I've seen (5) to (10) 100W lamps on a single circuit and never had the circuit breaker trip before. So I don't think the 80 times value is accurate even for a fraction of a cycle. I do think that the resistance of the lamp is different when it's cold or hot. But I've never tested to be sure.

If you could reference somewhere that you heard the 80 times value from I'm still curious to see it.

 

[Dennis Alwon]

I don't see where a loose neutral should burn up a coffee pot and have no effect on the computers, etc in the room. I understand the VD etc but how does it burn up a element and not pop a computer?

 

[Former Jeanyus]

How about a loose neutral connection at the outlet for the coffee maker?

 

[ohm]

Where do you get this value of 80 times it's running current from? And how many cycles does this last? I remember my dad hearing some guy tell him to leave his house lights on all the time because it draws less energy then turning them on/off during the day due to the start up current. I told my dad to turn off his lights when he's not using them and that will save him energy.

Since you've provided a value of 80 times it's running current I had to bring out my calculator. 100W A-lamp at 120V draws 0.833A. This times 80 = 66.7A.

The time-current curve for a type FA circuit breaker from Square-D shows the instantaneous trip area at about 300-600A for a 15A or 20A breaker. I've seen (5) to (10) 100W lamps on a single circuit and never had the circuit breaker trip before. So I don't think the 80 times value is accurate even for a fraction of a cycle. I do think that the resistance of the lamp is different when it's cold or hot. But I've never tested to be sure.

If you could reference somewhere that you heard the 80 times value from I'm still curious to see it.

Mr. Bill I can't recall where I heard the 80 X number but I found something in the 14th edition of "Electrical Wiring..Residential by Ray C. Mullin pg 134:
"A tungsten filament lamp draws very high momentary inrush current at the instant the circuit is energized. This is because the cold resistance of tungsten is very low. For instance the cold resistance of a typical 100W lamp is aprox. 9.5 ohms. This same lamp has a a hot resistance of 144 ohms" It says the normal operating current is 0.83A but the max. inrush could be 17.9A.

It drops to normal current in 6 cycles (0.10 sec). Therefore "T" rated switches are a good idea on loaded incandescent circuits and Ohms Law (hope my namesake doesn't hear this) could mislead you on impedance testing.

Sorry about the 80 X, I'll keep looking.

 

[ohm]

I don't see where a loose neutral should burn up a coffee pot and have no effect on the computers, etc in the room. I understand the VD etc but how does it burn up a element and not pop a computer?

Dennis, if the computer load is more than the coffee pot more voltage will be imposed across the coffee pot than the computer, if you have a loose neutral.

 

[steelersman]

Here's a PowerPoint slide that explains this. Just click in the page to move to the next slide. The example is for a 120/240V single phase system. A floating neutral on a 3-phase system would be similar, just slightly different math.
http://code-elec.com/userimages/Lost Neutral.ppt

Interesting except for the fact that it used a "W" in place of the ohm symbol so for all of the values for R it had a "W" after it. Typo.

 

[Dennis Alwon]

Dennis, if the computer load is more than the coffee pot more voltage will be imposed across the coffee pot than the computer, if you have a loose neutral.

I am not sure I understand why esp. if they are the same phase.

BTW, I agree there is probably a loose neutral I just don't see the effect on the pot.

 

[steelersman]

Dennis, if the computer load is more than the coffee pot more voltage will be imposed across the coffee pot than the computer, if you have a loose neutral.

actually you are backwards. the higher load (watts) item will have higher voltage with loose neutral. Look at the powerpoint thin that Bill provided and you'll see.

 

[ohm]

I am not sure I understand why esp. if they are the same phase.

BTW, I agree there is probably a loose neutral I just don't see the effect on the pot.

If the're the same phase it's not a loose neutral. They would both feel the same high voltage ...if they were both turned on.