Smoke in the Machine

Status
Not open for further replies.

Karl H

Senior Member
Location
San Diego,CA
I've been a service commercial/industrial electrician for 21 years now. I've had a problem that has confused me since I was a 2nd year apprentice. As everyone knows, there are 5 types of electricians; those that don't know, those that don't care, those that don't know and don't care but only want a paycheck, those that don't know but can make-up logical fairy tales on the spot and finally, those that know but who we never seem to encounter when it comes to the NEC or electrical theory. Of all of these types...I've come across the entire lot, except of course, the ones who are truly educated. Ponder the following situation; you have a single phase 3 wire ckt, A and B phase are under load, the grounded conductor is carrying the un balanced current and someone before you didn't follow ART 300-13 "Device Removal"... you open a J-box, the solderless pressure connector "pops-off" (in search of freedom no doubt), the customers new 120v 16I machine gets force fed 208v, followed by the facility manager frantically asking, "What did you do??????" You answer, "Open a J- Box!!!!!!" The above actually happened to me in 1986 and ever since then, I've been wondering how one of the phases can double its voltage in an open grounded conductor scenario? I can recreate it "safely" with two keyless fixtures; one with a 60watt bulb and one with a 150 watt bulb, two separate ckts sharing one common grounded conductor, with a switch to open the grounded conductor. Can someone please explain to me, how does this elevation in voltage occur? I hope someone out there can explain this to me without telling me "in an open neutral...electron fairies get their wings!!!!!!" lol
 

480sparky

Senior Member
Location
Iowegia
Draw the entire circuit out with the two loads and a neutral. Then calculate all four variables on each portion of each circuit... voltage (E), amperage (I), power (W) and resistance (R). Some will be a given, some you will need to calculate.

Then erase the neutral and recalulate all the E, I, W and R varaibles again

Ol' Georg will be proud of you when you figure it out.
 

haskindm

Senior Member
Location
Maryland
Sparky has put you on the right track. Remember that when you lost the neutral, instead of having two parallel 120 volt loads you had two loads hooked in series on a 208-volt circuit.
 

al hildenbrand

Senior Member
Location
Minnesota
Occupation
Electrical Contractor, Electrical Consultant, Electrical Engineer
The choice of the bulbs in your bench experiment could have a greater difference from each other.

That is, try a 25 Watt and a 200 Watt (or even bigger) bulb.

To get a high voltage swing on one of the two series loads, that load has to be a much higher total resistance than the other series load.

Saying this another way, thinking of your experience back in the Eighties, the "brand new 16I machine" was probably in a low load state, that is, the machine, at its line terminals looked like a high resistance.

And.

Whatever was running on the other side of the multiwire branch circuit was at full load, that is, the line terminals saw a low resistance.

Break the grounded conductor and the bulk (but not every last volt) of the voltage will appear on the line terminals of the 16I.
 

charlie b

Moderator
Staff member
Location
Lockport, IL
Occupation
Retired Electrical Engineer
Once the grounded conductor is no longer in the circuit, what is left is the following circuit path: From the source, through load #1, through load #2, and back to the source. This is a series circuit, with the two loads (i.e., two resistors) being in series.

The voltage provided by the source will be shared by the two loads. If the two loads are identical, then the half the voltage will be dropped across each load. If one load is much higher than the other (such as Al suggested with a 25 watt and 200 watt bulb), then the voltage will be divided between the two in accordance with their resistance values. The formula is that the voltage across each load is equal to its resistance divided by the total resistance, and that result multiplied by the available voltage. In this instance, the voltage across the 25 watt bulb would be 27 volts, and the voltage across the 200 watt bulb would be 213 volts.
 

al hildenbrand

Senior Member
Location
Minnesota
Occupation
Electrical Contractor, Electrical Consultant, Electrical Engineer
:D The fun part about your bench experiment is the 25 Watt lamp will FLASH and blow. . . :D
 

don_resqcapt19

Moderator
Staff member
Location
Illinois
Occupation
retired electrician
Charlie,
I think you have the drops across the lamps backwards. The 25 watt lamp will have the higher resistance and the greatest voltage across it.
 

LarryFine

Master Electrician Electric Contractor Richmond VA
Location
Henrico County, VA
Occupation
Electrical Contractor
charlie b said:
In this instance, the voltage across the 25 watt bulb would be 27 volts, and the voltage across the 200 watt bulb would be 213 volts.
Sorry, Charlie. The higher-wattage load has the lower resistance, and would have the lower voltage drop across it. The 25-watt bulb is the one that dies a premature death.

Added: Don caught it first, but I didn't see his post until after I responded; he gets the cookie.
 

Karl H

Senior Member
Location
San Diego,CA
Smoke in the Machine

I have read the posts for my question and am very grateful to all of you for offering your advice and trying to educate me. Thank you. I've put my code book on the shelf to cool down. It is so refresting to finally be amongst educated people in the field and I've got another question to pose. I currently have the biggest calculator I could buy and am now focused on Ohms Law.

I see how a parallel ckt losing the grounded conductor now becomes a series 208v ckt. So, am I to understand it that the elevated voltage is on the lowest resistive load? What I don't get is, if the two ckts that have lost the common grounded conductor are on A and B phase and A phase has the elevated voltage, how does the voltage of B transfer to A? And does that mean that all the circuits on A phase in the supply panel also experience the elevated voltage? Or is it only limited to the circuit(s) with the 16I machine now in series? If the latter is true, this makes no sense to me, then again...this is why I'm here! :grin:
 

al hildenbrand

Senior Member
Location
Minnesota
Occupation
Electrical Contractor, Electrical Consultant, Electrical Engineer
Karl H said:
So, am I to understand it that the elevated voltage is on the lowest resistive load?
No, the elevated voltage is on the largest resistive load.
Karl H said:
What I don't get is, if the two ckts that have lost the common grounded conductor are on A and B phase and A phase has the elevated voltage, how does the voltage of B transfer to A?
When the common grounded conductor is disconnected from the two loads, while the two loads remain connected to each other and the A and B phase, the two load's resistance become a simple "voltage divider".

That is, there is a single voltage between A and B. The two resistive loads add together, as far as the voltage between A and B is concerned. You have 208 Volts between A and B. If the two resistive loads total to, say, 104 Ohms, then every 1.0 Ohm will have 2.0 Volts across it.
Karl H said:
And does that mean that all the circuits on A phase in the supply panel also experience the elevated voltage?
No, the voltage swing on the two resistive loads occurs only on the other side of the break in the common grounded conductor.
Karl H said:
Or is it only limited to the circuit(s) with the 16I machine now in series?
No, it is not limited to the 16I machine. Any loads will do.
 

quogueelectric

Senior Member
Location
new york
I cant believe it

I cant believe it

don_resqcapt19 said:
Charlie,
I think you have the drops across the lamps backwards. The 25 watt lamp will have the higher resistance and the greatest voltage across it.
I agree twice in the same day. It must be a full moon . Itis too late for me to explain tonite But this is one of my favorites I will be back.
 

quogueelectric

Senior Member
Location
new york
I will be back tomorrow

I will be back tomorrow

Karl H said:
I've been a service commercial/industrial electrician for 21 years now. I've had a problem that has confused me since I was a 2nd year apprentice. As everyone knows, there are 5 types of electricians; those that don't know, those that don't care, those that don't know and don't care but only want a paycheck, those that don't know but can make-up logical fairy tales on the spot and finally, those that know but who we never seem to encounter when it comes to the NEC or electrical theory. Of all of these types...I've come across the entire lot, except of course, the ones who are truly educated. Ponder the following situation; you have a single phase 3 wire ckt, A and B phase are under load, the grounded conductor is carrying the un balanced current and someone before you didn't follow ART 300-13 "Device Removal"... you open a J-box, the solderless pressure connector "pops-off" (in search of freedom no doubt), the customers new 120v 16I machine gets force fed 208v, followed by the facility manager frantically asking, "What did you do??????" You answer, "Open a J- Box!!!!!!" The above actually happened to me in 1986 and ever since then, I've been wondering how one of the phases can double its voltage in an open grounded conductor scenario? I can recreate it "safely" with two keyless fixtures; one with a 60watt bulb and one with a 150 watt bulb, two separate ckts sharing one common grounded conductor, with a switch to open the grounded conductor. Can someone please explain to me, how does this elevation in voltage occur? I hope someone out there can explain this to me without telling me "in an open neutral...electron fairies get their wings!!!!!!" lol
But recreate this with one 100w lamp on 1 phase and 5 100w lamps on the other phase .This will also make this more realistic of reallife situations. I also made the math so simple you are gonna puke . I will be back tommorow. I made this exact presentation in 1984 for a jatc convention no electron fairies here. 208 was a lie at least until the equipment on the other phase melted into a direct short then the see saw goes the other way
 
Last edited:

480sparky

Senior Member
Location
Iowegia
For those with PowerPoint, I created a basic demonstration of this.

You can view it from this link:

http://www.code-elec.com/content/00/01/53/48/38/userimages/Lost Neutral.ppt

Left-click to advance the slides when you are ready.
Hopefully, you can understand what I did with it, and that it makes sense.
I may get ambitious and make this into an animated .gif, but this will have to do for now.
I did this late at night, so I may have made some errors. Please correct me if it is wrong.
 
Last edited:

stickboy1375

Senior Member
Location
Litchfield, CT
120/240v example...

120/240v example...

205ecm18fig1.gif



Step 1. Calculate the resistance of each appliance:
R=E2?P
a. Hair dryer rated 1,275W at 120V
R1=(120V)2?1,275W=11.3 ohms
b. Television rated 600W at 120V
R=(120V)2?600W=24 ohms

Step 2. Calculate circuit resistance:
RT=R1+R2
RT=11.3 ohms+24 ohms=RT=35.3 ohms

Step 3. Calculate circuit current:
IT=ESource?RT
IT=240V?35.3 ohms=6.8A

Step 4. Calculate the voltage for each appliance:
E=IT?Rx
a. Hair dryer: 6.8A?11.3 ohms=76.8V
b. Television: 6.8A?24 ohms=163.2V

Step 5. Calculate power consumed:
P=E2?R
a. Hair dryer: P=(76.8)2?11.3=522W
b. Television: P=(163.2)2?24=1,110W

The TV will temporarily operate at 163V and consume 1,110W!
 

quogueelectric

Senior Member
Location
new york
Baloon festival tied me up all day

Baloon festival tied me up all day

lets call the lamps 120 watt lamps for ease of mathematics.. So 120 volts across 120w lamps draw 1 amp every day of the week. Put 1 lamp on a phase and 5 lamps on b phase and you have 1 amp on a 1 amp on b circulating across the same 240 v center tapped winding, 4 amps is the imbalance running back on the neutral. 600 watts total consuption. breakthe neutral and you create a series paralell circuit with 1 120w lamp in series with 5 paralelled 120w lamps. =1ohm in series with .2 ohms so a simple voltage divider puts 200 volts across the 1 lamp and 40 volts across the 5 lamps in series. what this does is cause the 1 120w lamp to draw 200 watts because of the elevated voltage and the 5 paralelled lamps to draw 16 watts apeice because of the reduced vovtage X5 is 80 watts total 200 +80 =280 watts total.So the 1 lamp will run really bright and the other 5 will run really dim. The higher voltage can cause a fire without tripping the over current protection.
 

quogueelectric

Senior Member
Location
new york
It is almost 1 am

It is almost 1 am

kbsparky said:
Actually, its a new moon, and with that the stars will be exceptionally bright and easy to see tonight. Along with the Perseid Meteor shower, which should be at its peak sometime tonight, after midnight local time. :smile:
I will go look now come on hound dogs. I will finish open neutral later
 

len149

Member
Motor and Computer

Motor and Computer

Using an extreme analogy to explain:
Picture a large motor and a computer on a two pole breaker sharing the neutral. 240 Volt between phases. But, the neutral allows for each phase to have a return path to ground splitting the voltage in two with a potential difference of 120v. each phase to ground. Now If, you think about the motor it's just wire wound up just like a piece of 16ga. Wire if the neutral opens the motor acts as a conductor and returns the path to the other phase. (Take the path of least resistances ) with very low resistance of the motor the IR loss would be small and the voltage across the motor would be small say 5 Volts and the motor would not run but, act like a conductor completing the circuit. Thus the computer would have 235 Volts which is the remainder of the voltage and all the smoke will leak out causing the computer to fail.
 

e57

Senior Member
Years ago I was being shown the ropes by the brother of my employer, and as he was enphisizing the import of making sure to only disconnect the neutral of which you are working on, (Of which I knew already) "follow it right to the neutral bar - DOH!!! Go inside and see if I blew anything up..." I do and find the customer sitting horrorfied in front of a smoking near Salvador Dali like computer screen.

That took some splainin' to the boss I tell you...

Sometimes it is just better to shut it all off....
 
Status
Not open for further replies.
Top