amperage on a neutural

8/8/2019 Amperage on a Neutural

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amperage on a neutural

I measured the amperage on a neutural that is part of a 3 wire 100 amp panel, 120/240 volt.One hot carried 21 amps, second leg carried 38amps and the neutural carried 26 amps. Doesthis sound normal for the nuetural to carry so much current?

SPARKYNSJ 12-04-2009 08:15 PM

Of the current measured on the ungrounded conductors, how much is flowing in L-L loads andhow much is L-N?

Roger

roger 12-04-2009 08:50 PM

Quote:

Something else is wrong. 38-21=17.

What type tester was used here?

chris kennedy 12-04-2009 09:07 PM

Originally Posted by roger (Post 1134610)Of the current measured on the ungrounded conductors, how much is flowing in L-Lloads and how much is L-N?

Roger

Maybe some L-N loads got turned on in the middle of the metering.

480sparky 12-04-2009 09:28 PM

Actually fed from two legs of 120/208 3-phase?

zbang 12-04-2009 09:35 PM

Page 1 of 13Mike Holt's Forum - amperage on a neutural

1/14/2010http://forums.mikeholt.com/printthread.php?t=120522&pp=40


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Quote:

Sounds logical and mathematical to me!

brian john 12-04-2009 09:52 PM

Originally Posted by zbang (Post 1134626)

Actually fed from two legs of 120/208 3-phase?

Quote:

As well as myself but, the OP does say "3 wire 100 amp panel, 120/240 volt" .

Roger

roger 12-04-2009 09:57 PM

Originally Posted by brian john (Post 1134635)Sounds logical and mathematical to me!

Could this be some of a neighbors neutral current ?

Either:

Turn off the main and see if it all goes away.

Or

Measure current on the GEC in the panel.

SG-1 12-04-2009 11:34 PM

If the panel is fed with 2 phases of a 3-phase wye, then the neutral should be about 33, not 26.So the amps don't add up for either services.

Did you check the GEC for amperage?

hardworkingstiff 12-04-2009 11:40 PM

Quote:

erickench 12-04-2009 11:44 PM




3/26

This makes alot of sense.

Originally Posted by roger (Post 1134610)Of the current measured on the ungrounded conductors, how much is flowing in L-Lloads and how much is L-N?

Roger

Quote:

Why Eric, does that make so much sense?

hardworkingstiff 12-05-2009 12:05 AM

Originally Posted by erickench (Post 1134675)Quote:

This makes alot of sense.

Originally Posted by roger (Post 1134610)Of the current measured on the ungrounded conductors, how much is flowing inL-L loads and how much is L-N?

Roger

Quote:

And since it does please explain why.

Roger

roger 12-05-2009 09:35 AM

Originally Posted by erickench (Post 1134675)This makes alot of sense.

Quote:

roger 12-05-2009 09:43 AM




4/26

But knowing how much of the ungrounded current is L-N and how it is divided between the legs

will help in knowing what the neutral load should be and then start looking from there.

Roger

Originally Posted by chris kennedy (Post 1134614)Something else is wrong. 38-21=17.

091205-0851 EST

SPARKYNSJ:

Additional information is needed.

We really need to know if source is single or three phase. Measure each line to neutral voltagewith a DVM that resolves 0.1 V. Then measure the line to line voltage with the same meter. If the sum of the two voltages from line to neutral are close to equal to the line to line voltage,then it is a single phase center tapped transformer supply. If the sum of the two line to neutralvoltages is closer to the line to line voltage divided by 0.866, then it is a 3 phase Y source.

Next remove all loads from the main panel. Measure the current from the ground bus of themain panel to all grounding paths. Most particularly the water pipe. If there is negligible current,then the neutral current is a result of loads on the main panel.

If the supply is three phase, both line to neutral voltages are very close in value (like within 1V), and all loads are resistive, then assuming 120 deg phase difference the neutral current isabout 38 - 0.5*21 = 38 - 10.5 = 27.5 A. Not too far from the 26 A measured.

Separately assume it is really single phase, then an adequate reactive load can cause the sameresult.

.

gar 12-05-2009 10:17 AM

Quote:

On single-phase distribution, shouldn't the neutral always be A-B=N (use absolute value)? I


Originally Posted by roger (Post 1134778)But knowing how much of the ungrounded current is L-N and how it is divided betweenthe legs will help in knowing what the neutral load should be and then start looking fromthere.

Roger




5/26

don't see how L-N loads affect this since when they(branch circuit neutrals) all hit the busbar,the service entry N amps would be the difference in the two ungrounded conductors.

Quote:

Exactly, and when we find out how much of the current is L-L in the ungrounded conductors wecan then figure out how much current "should" be L-N, then we would need to figure out howthese L-N loads are divided between the legs and this would let us know how much current we

"should" be reading on the service neutral.

If the OP hadn't said 120/240 v I would think the three phase scenario would be the answer asalready stated.

Roger

roger 12-05-2009 10:42 AM

Originally Posted by hardworkingstiff (Post 1134809)On single-phase distribution, shouldn't the neutral always be A-B=N (use absolutevalue)? I don't see how L-N loads affect this since when they(branch circuit neutrals) allhit the busbar, the service entry N amps would be the difference in the two ungroundedconductors.

091205-0953 EST

hardworkingstiff:

Your statement is correct for resistive loads in the center tapped supply circuit you are assuming.

When you make an ordinary current measurement there is no phase information. Thus, if onecurrent is shifted in phase relative to the other there will be a different result than that of yourequation.

In fact your equation with the minus sign and using the absolute values of the currents impliesthat the two currents flowing into the neutral are 180 deg out of phase.

.

gar 12-05-2009 11:03 AM

Quote:

dkarst 12-05-2009 11:12 AM

Originally Posted by gar (Post 1134794)091205-0851 EST




6/26

Maybe I'm slow this morning but for 3 phase assumption it seems like part of phasor math ismissing... shouldn't there be a 21 cos30 ~ 18 component as well for a total of about 32Amps?

If the supply is three phase, both line to neutral voltages are very close in value (likewithin 1 V), and all loads are resistive, then assuming 120 deg phase difference theneutral current is about 38 - 0.5*21 = 38 - 10.5 = 27.5 A. Not too far from the 26 Ameasured.

.

Quote:

That (IMO) is the determing factor in a "single-phase" distribution. I've read the many pages onthe discussion of whether 2 legs and a neutral of a WYE is a single-phase distribution, and I stillmaintain this view.

How much phase shifting would you expect to see in a residence? I thought that's what the OPsuggested, maybe I missed that.

EDIT: I went to the OP, and it did not mention a residence, so my simplistic review is.....worthless, :-)


Originally Posted by gar (Post 1134829)091205-0953 EST

hardworkingstiff:

Your statement is correct for resistive loads in the center tapped supply circuit you areassuming.

When you make an ordinary current measurement there is no phase information. Thus, if one current is shifted in phase relative to the other there will be a different result thanthat of your equation.

In fact your equation with the minus sign and using the absolute values of the currentsimplies that the two currents flowing into the neutral are 180 deg out of phase.

.

091205-1026 EST

dkarst:

gar 12-05-2009 11:41 AM




7/26


8/26

The sum of the currents entering a node is zero. It doesn't matter if they are just passingthrough (so to speak) or getting returned on the neutral. (The circuit is a "black box".)

The imbalance in the currents is the result of having multiple powerfactors per load, morespecifically, per phase. The simplest solution is a 47 degree phase shift on the higher leg, butthere could be many combinations of both leading and lagging powerfactors that give the sameresult.

Quote:

Rick, we are not trying to calculate anything, the OP is questioning the neutral current he isreading, take L-L loads out of the equation and we will have a better starting point.

It would be nice to know if the OP did verify zero sequencing though.

Roger

roger 12-05-2009 02:39 PM

Originally Posted by Rick Christopherson (Post 1134925)Mathematically it makes no difference which loads are L-L or L-N unless you arecalculating the neutral current based on the individual powerfactors for each load.

Quote:

Roger -I'm must be short on coffee today. What does that mean?

cf

Cold Fusion 12-05-2009 02:51 PM

Originally Posted by roger (Post 1134943)...verify zero sequencing ...

Quote:

roger 12-05-2009 03:14 PM

Originally Posted by Cold Fusion (Post 1134951)Roger -I'm must be short on coffee today. What does that mean?




9/26

Make sure the sum off all conductors equal zero by putting a clamp around all of them. If aclamp on meter reads current when placed around all the conductors at once I would be lookinginto SG-1's (post number 8) suspicion.

Roger

cf

Quote:

Of course, thank you

cf

Cold Fusion 12-05-2009 03:16 PM

Originally Posted by roger (Post 1134969)Make sure the sum off all conductors equal zero ---

Quote:

Whether you realize it or not, you are trying to calculate something. By subtracting out the L-Lloads, all you are doing is, well, subtraction. The core problem remains the same, but you simplyreduce the magnitude of the line currents. It actually makes the problem more complicated thanit needs to be, given the information provided. As I already said, it is applicable only if you knowthe powerfactor of individual loads.

Quote:

That situation (a super-node) is applicable only if there is ground-neutral bonding at the panel inquestion.

Rick Christopherson 12-05-2009 04:04 PM

Originally Posted by roger (Post 1134943)Rick, we are not trying to calculate anything, the OP is questioning the neutral current heis reading, take L-L loads out of the equation and we will have a better starting point.

Originally Posted by roger (Post 1134943)It would be nice to know if the OP did verify zero sequencing though.....If a clamp on meter reads current when placed around all the conductors at once I wouldbe looking into SG-1's (post number 8) suspicion.




10/26

If that is in fact the case, then all you have done is forgotten one path into the node. Howeverthe existence of a super-node shared among the neighbors is far less common than electriciansmake it out to be. The reason why it gets presented so often is because it is the easy answer toa situation that those individuals do not fully understand. The easiest way to identify a super-node is to measure the current in the main ground before and after cutting the mainbreaker/disconnect. If there remains current in the ground wire, then you have a super-node.However, in nearly all cases of single-family residences, the ground current drops to zero when

the main is open. When that happens, then you know absolutely that there is not a super-nodesituation.

Rick, you are looking at this as an engineer and that is the problem.

An electrician walking into a house will have a few meters and is looking for a reason for theproblem, they ordinarily will not be given time to analyze every circuit so simplifying the taskand coming up with a reasonable / approximate answer is the best way to approach it andeliminating the L-L loads is a start to this conclusion

Now, if you are on call and available when the OP or our service man needs you, we can taketrouble shooting to a different level. ;)

Roger

roger 12-05-2009 04:33 PM

Quote:

Well for one thing the currents don't add up. 38-21 does not equal 26A.

erickench 12-05-2009 05:13 PM

Originally Posted by hardworkingstiff (Post 1134683)Why Eric, does that make so much sense?

I'm prone to brain farts, but here's the OP

Quote:

If every single load was L-N,( and wouldn't that be the worst possible situation for current on theneutral) it would not have that much current on the neutral.

realolman 12-05-2009 05:21 PM

I measured the amperage on a neutural that is part of a 3 wire 100 amp panel, 120/240volt. One hot carried 21 amps, second leg carried 38amps and the neutural carried 26amps. Does this sound normal for the n uetural to carry so much current?




11/26

If there were ANY L-L loads there would be less on the neutral yet.

So, I'm missing what the L-L loads have to do with it.It's 4 amps too much for the worst possible situation, with those magnitudes of current on eachphase..

Like I said, I often miss the obvious, but I can't think of a way that the neutral could NORMALLYcarry that much current under the load conditions he described.

I would say it is NOT normal for the OP's neutral to carry that much current.

Quote:

I agree and SG-1 has suggested a possible solution.

http://www.electrical- photos.com/data/500/Neutral2.gif

Roger

roger 12-05-2009 05:38 PM

Originally Posted by realolman (Post 1135041)I would say it is NOT normal for the OP's neutral to carry that much current.

Well now that I look at it that does seem like alot of amps on the neutral.38-21=17A. Where are the additional 9 amps coming from? Non-linear load perhaps?

erickench 12-05-2009 05:55 PM

Quote:

Stealing my material??? See post 3 of this thread.:grin:


Originally Posted by erickench (Post 1135066)38-21=17A.

I guess I just didn't see where the OP asked anything beyond whether there was too muchamperage on the neutral.

realolman 12-05-2009 06:08 PM




12/26

That's a nice, very understandable diagram, and quite possible. Maybe probable, even .:)

Quote:

No Roger, I am not looking at this as an engineer. It is basic circuit analysis that any electricianshould know, unless of course they stopped teaching electricity to electricians.

Quote:

So you're suggesting that 3 circuit paths is more simplified than 2 circuit paths? :-? Is this kindof like the new math?

In one breath (of the same sentence) you state that they don't have time to analyze everycircuit, and yet you suggest they subtract out the L-L currents from the measured feed currents.So how are they supposed to do that without analyzing all of those circuits? (I certainly hope youwouldn't suggest they start flipping breakers on active loads just to suit their curiosity.)

What you are not understanding is that figuring out the problem is EXACTLY the same regardlesswhether there are any L-L loads or not. It makes absolutely no difference to figuring it out. Bytrying to eliminate the L-L current, you are actually making the problem more complicated, notless complicated.

Rick Christopherson 12-05-2009 06:41 PM

Originally Posted by roger (Post 1135013)Rick, you are looking at this as an engineer and that is the problem.

Originally Posted by roger (Post 1135013)

...they ordinarily will not be given time to analyze every circuit so simplifying the taskand coming up with a reasonable / approximate answer is the best way to approach itand eliminating the L-L loads is a start to this conclusion

Rick, you've been putting that reply together for over an hour, I will have to conceed that I havemade this complicated. :grin:

Roger

roger 12-05-2009 06:46 PM

It is not so much that I supect an open neutral somewhere as much as I think the first stepshould be to determine where the extra neutral current is originating. The numbers do not matchsimple Single or 3-Phase calculations.

Is the extra 10 Amps being generated by the premisis wiring & loads or is it part of somebody

SG-1 12-05-2009 10:47 PM




13/26

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elses load that is just passing through ?

If there were actual L-L loads the unbalance obtained by subtracting would be more than themeasured value of the neutral and not less. That's what's so strange about this problem.:-?

erickench 12-05-2009 11:14 PM

All times are GMT -4. The time now is 08:20 PM.


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Quote:

I should have said unaccounted for 10Amps, instead of the additional 10Amps.

SG-1 12-05-2009 11:45 PM

Originally Posted by erickench (Post 1135214)If there were actual L-L loads the unbalance obtained by subtracting would be more thanthe measured value of the neutral and not less. That's what's so strange about thisproblem.:-?

It has to be nonlinear loads. There is no other explanation.

erickench 12-06-2009 12:07 AM

Quote:

What would you get if a 2 pole electric water heater had one element broken close

enough to the grounded metal tank that current would flow from phase to egc ?

Maybe 9 amps ?

This is not an uncommon occurance.

benaround 12-06-2009 02:27 AM

Originally Posted by erickench (Post 1135228)It has to be nonlinear loads. There is no other explanation.

Quote:

Besoeker 12-06-2009 06:26 AM


1/14/2010http://forums.mikeholt.com/printthread.php?t=120522&pp=40&page=2


15/26

I agree.If one L-N load is unity power factor and the other about 0.76 lagging, the neutral current wouldbe 26A as measured.

Originally Posted by ga r (Post 1134829)091205-0953 EST

hardworkingstiff:

Your statement is correct for resistive loads in the center tapped supply circuit you areassuming.

When you make an ordinary current measurement there is no phase information. Thus, if one current is shifted in phase relative to the other there will be a different result thanthat of your equation.

In fact your equation with the minus sign and using the absolute values of the currentsimplies that the two currents flowing into the neutral are 180 deg out of phase.

Well I am confused. I noticed that this is a single phase being fed from a 3 phases system. Chrissays why isn't the neutral calculated as L1-L2= 17.

My question is wouldn't you calculate the neutral current based on a 3 phase calculation since itis a 3 phase power source. Thus we would use L1 at 38, L2 at 21 and L3 at 0. This would give us33 amps. Why the difference I don't know.

Dennis Alwon 12-06-2009 07:54 AM

Quote:

never saw him again.

boboelectric 12-06-2009 07:59 AM

Originally Posted by SPARKYNSJ (Post 1134581)I measured the amperage on a neutural that is part of a 3 wire 100 amp panel, 120/240volt. One hot carried 21 amps, second leg carried 38amps and the neutural carried 26amps. Does this sound normal for the nuetural to carry so much current?

Quote:

Besoeker 12-06-2009 08:11 AM

Originally Posted by Dennis Alwon (Post 1135275)




16/26

From the OP:Quote:

This I took to mean a centre tapped 120-0-120 supply giving L1, L2, and N as the three wires.I don't think there any mention of it being three phase.

Well I am confused. I noticed that this is a single phase being fed from a 3 phasessystem. Chris says why isn't the neutral calculated as L1-L2= 17.

My question is wouldn't you calculate the neutral current based on a 3 phase calculationsince it is a 3 phase power source. Thus we would use L1 at 38, L2 at 21 and L3 at 0.This would give us 33 amps. Why the difference I don't know.

3 wire 100 amp panel, 120/240 volt

Quote:

You're correct. I read Zbang's post and I didn't realize his post was a question- I thought it wasfact.

Quote:

Dennis Alwon 12-06-2009 08:18 AM

Originally Posted by Besoeker (Post 1135279)From the OP:

This I took to mean a centre tapped 120-0-120 supply giving L1, L2, and N as the threewires.I don't think there any mention of it being three phase.

Originally Posted by zbang Actually fed from two legs of 120/208 3-phase?

Quote:

realolman 12-06-2009 09:28 AM

Originally Posted by benaround (Post 1135240)What would you get if a 2 pole electric water heater had one element broken close

enough to the grounded metal tank that current would flow from phase to egc ?

Maybe 9 amps ?




17/26

The resistive heating element is surrounded by grounded conductive material, so that certainly

sounds to me like something that would be possible.

Assume a water heater element broke exactly in half and became grounded. The ECG would thencarry the current of each half the element connected between phase and ground... 120 VACdivided by each half the element's resistance.

If both broken ends were to become grounded, it seems to me the circuit would behave similarlyto a MWBC and the ECG current would be zero. If only one end of the element becamegrounded, and the other isolated, the EGC would carry the current of 120V divided by 1/2 theelement's total resistance.

I started out to post that I thought the water heater element scenario would be plausible, but Ithink I have talked myself into thinking that if a water heater element broke somewhere alongit's length you would likely get some sort of EGC current, but it would not cause the neutralcurrent to rise to levels above normal operation, comparing phase currents to neutral current.:)

This is not an uncommon occurance.

Suppose the water heater has a metal pipe ? Now the current can flow through the pipe to theGEC connection. Only it has to be the neighbors water heater.

Suppose there is another electrical system in the same facility and someone accidentallyswitched neutrals on a branch circuit between them ?

SG-1 12-06-2009 05:45 PM

Quote:

Can you show the work for how you arrived at this answer?

mull982 12-07-2009 09:30 AM

Originally Posted by Besoeker (Post 1135271)I agree.If one L-N load is unity power factor and the other about 0.76 lagging, the neutralcurrent would be 26A as measured.

Quote:

Besoeker 12-07-2009 02:11 PM

Originally Posted by mull982 (Post 1135568)




18/26

It's a bit difficult to give detail here.The reverse calculation is a little easier to show. Figures are rounded to three or four decimalplaces.

I took the 21A to be Ia at unity and the 38A to be Ib at 0.7575 pf lagging.This gives an angle of 0.7113 radians (or 40.75 deg).

Using Id and Iq to stand for in-phase and quadrature components:

Ia is at unity so has only an Id component of 21AIb has an Id component of 38*cos(0.7113) giving an Id value of 28.79A.Net Id in the neutral is thus (21-28.79) or -7.79A.

The Iq component of Ib is 38*sin(0.7113) giving an Iq value of 24.81ATotal neutral current is sqrt((-7.79)^2 + (24.81)^2)Which is 26A.

Sorry. That's much too turgid.I'll present it in graphical form shortly.

Can you show the work for how you arrived at this answer?

Quote:

I follow your math. I'm assuming that your math is referring to a 120/240V single phase service.

mull982 12-07-2009 02:29 PM

Originally Posted by Besoeker (Post 1135678)It's a bit difficult to give detail here.The reverse calculation is a little easier to show. Figures are rounded to three or fourdecimal places.

I took the 21A to be Ia at unity and the 38A to be Ib at 0.7575 pf lagging.This gives an angle of 0.7113 radians (or 40.75 deg).

Using Id and Iq to stand for in-phase and quadrature components:

Ia is at unity so has only an Id component of 21AIb has an Id component of 38*cos(0.7113) giving an Id value of 28.79A.Net Id in the neutral is thus (21-28.79) or -7.79A.

The Iq component of Ib is 38*sin(0.7113) giving an Iq value of 24.81ATotal neutral current is sqrt((-7.79)^2 + (24.81)^2)Which is 26A.

Sorry. That's much too turgid.I'll present it in graphical form shortly.




19/26

The only question I have is weather or not the 40.75 deg angle you calculated from p.f. has tobe added or subtracted from the Ib unity current which is at 180deg difference from Ia? Sincethe unity Ia current would be at 0deg wouldn't the unity Ib current be at -180deg and thus theIb current with the p.f. considered by 40deg +/- the unity Ib current?

I'll probably mess everything up hopping in here but and it is certainly easier looking at a phasordiagram but if you assume Va @ 0 degrees and Vb at -180 degrees, then Ia = 21 /_ 0 = 21 +j0in rectangular.

Assume Ib lags Vb by the ~40 degrees (draw in 2nd quadrant) then Ib = 38 /_ -220 = -28.8 + j24.8

Now you can draw resultant which is -7.8 + j 24.8 with a magnitude of ~ 26Amps. Mathassumes 120/240 single phase. Apologies if this doesn't help. I rounded the numbers to simplify.

dkarst 12-07-2009 03:11 PM

Quote:

Yes. 120-0-120.Quote:

It would. "Net Id in the neutral is thus (21-28.79) or -7.79A." accounts for that.

Besoeker 12-07-2009 03:16 PM

Originally Posted by mull982 (Post 1135684)I follow your math. I'm assuming that your math is referring to a 120/240V single phaseservice.

Originally Posted by mull982 (Post 1135684)Since the unity Ia current would be at 0deg wouldn't the unity Ib current be at -180deg

The graphical representation I promised in post #52.

http://i36.photobucket.com/albums/e3...ralcurrent.jpg

Besoeker 12-07-2009 03:18 PM

To be honest I find the mathematical statement pretty easy to follow and more clear than thesinusiodal representation. I think that's a decent explanation. Of course I don't think Ia is simplyat PF of unity, but a combination of power factor differences on Ia/Ib could account for it. I alsosee reason in the GEC path delivering current to the neutral. It's really difficult to rate eitherpossibility as definitly more probable than the other since we really don't have that much

skeshesh 12-07-2009 04:00 PM




20/26

information about the system and connected components. But I think ultimately to answer theOP question, this neutral current can possibly exist due to power factor issues, but I'd say,considering the magnitude as well as the numerous responses generated by this forum, that itneeds to be checked out, so I wouldn't just disregard it.

Quote:

Quite so.

A at 0.95 lead, B at 0.92 lag

A at 0.96 lag, B at 0.91 leadA at 0.77 lag, B at unity.

I gave just one example of an infinite number of solutions.

Besoeker 12-07-2009 04:15 PM

Originally Posted by skeshesh (Post 1135706)Of course I don't think Ia is simply at PF of unity, but a combination of power factordifferences on Ia/Ib could account for it.

Quote:

This solution approach makes more sense to me since I am able to see the 40deg phase shiftfrom Vb leading to an current angle of -220deg.

mull982 12-07-2009 04:29 PM

Originally Posted by dkarst (Post 1135693)I'll probably mess everything up hopping in here but and it is certainly easier looking at aphasor diagram but if you assume Va @ 0 degrees and Vb at -180 degrees, then Ia =21 /_ 0 = 21 +j0 in rectangular.

Assume Ib lags Vb by the ~40 degrees (draw in 2nd quadrant) then Ib = 38 /_ -220 = -28.8 + j24.8

Now you can draw resultant which is -7.8 + j 24.8 with a magnitude of ~ 26Amps. Mathassumes 120/240 single phase. Apologies if this doesn't help. I rounded the numbers tosimplify.

Quote:

skeshesh 12-07-2009 05:04 PM

Originally Posted by Besoeker (Post 1135713)




21/26

Of course. I was just pointing it out for others who may be wondering. Your exampledemonstrated your point quite clearly and I understand your choice of unity for Ia PF for ease of

calculation/demonstration.

I think it's probably necessary as some others suggested to check the ground path. As oneposter pointed out the first step would be to see if the neutral is bonded to the ground at thepanel in question. While this possibility might be slim, the current path through the groundingconductor could energize metallic surfaces and be quite a hazard. I'm not an electrician but Iwould say it's a good idea to check to prevent possible liability.

I gave just one example of an infinite number of solutions.

SPARKYNSJ , what makes you think it is too much current thru the neutral? The neutral current

of the panel is the vector sum of the two-phase currents and I won't discount any value youmeasured in the neutral line.Let's say, of the VA at an angle 0 deg, VB at angle 120 deg and VC at angle -120 deg from a Ytransformer, phases VA and VC are connected to the panel in question. Let's say phase A load ispure resistive such that phase A current is IA at angle 0 and phase C load is pure inductive suchthat the phase C current is IC at angle 30 deg. Adding IA at angle 0 and IC at angle 30 willresult a substantial amount current flowing thru the panel neutral wire.

gmtt 12-08-2009 05:35 PM

FWIW, I received a PM from the OP stating that this is indeed 2 ungrounded and the grounded

conductors of a 208y/120 system.


Quote:

That makes more sense.

Roger

roger 12-08-2009 06:40 PM

Originally Posted by chris kennedy (Post 1136187)FWIW, I received a PM from the OP stating that this is indeed 2 ungrounded and thegrounded conductors of a 208y/120 system.

Quote:

Dennis Alwon 12-08-2009 06:45 PM




22/26

I guess I sensed it. :D


Quote:

Your my favorite superhero.


Originally Posted by Dennis Alwon (Post 1136219)I guess I sensed it. :D

i measured the amperage again. The panel is being fed from a 3 phase 4 wire 120/208 main.The subpanel is being fed by a 70 amp breaker w/ 2 2/0's hots and 1 2/0 neutural. one hot has22 amps the other 41, the neutral is carrying 30-35 amps. how is this possible? Always thoughtthat the neutural would carry the difference between both hots..

SPARKYNSJ 12-08-2009 07:59 PM

Quote:

A human non-contact voltage tester. :)

LarryFine 12-08-2009 08:05 PM

Originally Posted by Dennis Alwon (Post 1136219)I guess I sensed it. :D

Quote:

roger 12-08-2009 08:29 PM

Originally Posted by SPARKYNSJ (Post 1136261)i measured the amperage again. The panel is being fed from a 3 phase 4 wire 120/208main. The subpanel is being fed by a 70 amp breaker w/ 2 2/0's hots and 1 2/0 neutural.one hot has 22 amps the other 41, the neutral is carrying 30-35 amps. how is thispossible? Always thought that the neutural would carry the difference between bothhots..




23/26

You would be correct if you were dealing with a single phase 120/240 volt system but, since youare dealing with a wye you would have to figure it as follows.

SQRT IA + IB - (IA x IB)

SQRT 22 x 22 + 41 x 41 - (22 x 41)

SQRT 484 + 1681 - 902

SQRT 2165 - 902 = 1263

SQRT 1263 = 35.54 amps

Roger

091208-2116 EST

SPARKYNSJ:

If you draw a vector diagram and assume both phases only have resistive loads from the hotlines to neutral and the phases are 120 deg apart, then the result is:

Draw a vertical line P1 to P2 with a length of 41. Classify P1 as the south end and P2 as thenorth end. Next draw a line from P2 to the south-east with an included angle of 60 deg betweenthis new vector, called P2 to P3, and the vector P1-P2. Make P2-P3 with a length of 22. Nowmeasure the length from point P1 to point P3. Its value is 35.54871, and this corresponds toyour reading of 35 A. The phase angle of the current in the neutral is 90-57.6 = 33.4 deg fromthe voltage P2 to P1.

.

gar 12-08-2009 10:36 PM

Quote:

Electrically it carries the unbalanced current. When the current through the "hots" is in phasethrough a neutral-connected junction, the amount of current directed through the neutralconductor is exactly the arithmetic difference. The more the "hots" currents are out of phasethrough the neutral-connected junction, the less mutual current there is going through the "hot"conductors, so the neutral conductor has to make up for that.

In your case, the voltages are 120 out of phase with respect to the neutral. The respective

Smart $ 12-08-2009 10:56 PM

Originally Posted by SPARKYNSJ (Post 1136261)

... Always thought that the neutural would carry the difference between both hots..




24/26

"hot" currents will likely be out-of-phase accordingly, handling the same type of load (i.e. havingnearly the same power factor). This is why you get the sum for neutral current if both "hots" areconnected to the same phase. Being 120 out-of-phase yields a neutral current somewhere inbetween the arithmetic sum and difference.

Roger has given an example of the basic math solution... but such method does not account fordiffering power factors.

Quote:

So we have about 60 posts based on the OP saying "3 wire 100 amp panel, 120/240 volt" andnow he PMs you to say it is completely different?

I say we let him figure it out himself.

ghrist 12-09-2009 12:00 AM


Neutural

Thanks for the answers guys.. I have to study them to fully understand the math. Thanks again.

SPARKYNSJ 12-09-2009 09:33 AM

Quote:

mull982 12-09-2009 02:41 PM

Originally Posted by ga r (Post 1136356)091208-2116 EST

SPARKYNSJ:

If you draw a vector diagram and assume both phases only have resistive loads from thehot lines to neutral and the phases are 120 deg apart, then the result is:

Draw a vertical line P1 to P2 with a length of 41. Classify P1 as the south end and P2 asthe north end. Next draw a line from P2 to the south-east with an included angle of 60deg between this new vector, called P2 to P3, and the vector P1-P2. Make P2-P3 with alength of 22. Now measure the length from point P1 to point P3. Its value is 35.54871,and this corresponds to your reading of 35 A. The phase angle of the current in theneutral is 90-57.6 = 33.4 deg from the voltage P2 to P1.

.




25/26

Shouldn't the included angle between these two vectors be 120deg instead of 60deg due to thefact that they are L-N loads?

Quote:

120 when tail to tail... but when doing vector addition graphically, they are arranged tail tohead and thus a 60 included angle.

Smart $ 12-09-2009 03:28 PM

Originally Posted by mull982 (Post 1136623)Shouldn't the included angle between these two vectors be 120deg instead of 60deg dueto the fact that they are L-N loads?

Quote:

Yup you are right that is my mistake.

mull982 12-09-2009 03:47 PM

Originally Posted by Smart $ (Post 1136658)120 when tail to tail... but when doing vector addition graphically, they are arranged tailto head and thus a 60 included angle.

091209-2110 EST

mull982:

Try to visualize it this way:

Two sine waves are being added together.

Call the angle of the vector P1-P2 0 deg and indicate phasing by putting an arrow at the P2 end.

Next create a second vector P2'-P3 with its arrow at P3. This vector is at an angle of -120 degrelative to the first vector. To add these two move P2'-P3 so that P2' is coincident with P2.

The resultant current vector is P1-P3.

You can turn these three vectors into two right triangles. From this you can solve for themagnitude and angle of P1-P3 relative to P1-P2. Or you can use a CAD program for a graphical

gar 12-09-2009 10:11 PM




26/26

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solution.

Consider the case where the 120 deg is changed to 180 and you will see that the second vectoris parallel to the first and in the opposite direction. Thus, the magnitude of the second subtractsfrom the first.

Another way:Go to http://en.wikipedia.org/wiki/List_of...ric_identities A little more than half way down the long page, and just above "Other sums of trigonometricfunctions" find the sentence "More generally, for an arbitrary phase shift, we have", and you willsee the equation for the sum of two sine waves of different amplitude and phase shift. Note: thevector diagram is easier.

.

091210-2250 EST

A question for you to consider.

Why is it possible to use vector math to solve problems about the relationship of sine waves of the same frequency but possibly of different amplitudes and phases?

To answer this consider what is the result of of adding sine waves of the same frequency. Myprevious post may give a hint.

.

gar 12-10-2009 11:55 PM

Check for harmonics

You didn't say what your 3rd leg measured. It sound liek your legs are not balanced.

I have measured this sort of imbalance with computer equipment, particularly with olderswitching power supplies. Usually, it's a phenomenon measurable only in the 3rd orderharmonics using something like a Fluke Power Quality meter.

paulcbrowne 01-14-2010 11:03 AM

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