SG-1
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- Ware Shoals, South Carolina
I should have said unaccounted for 10Amps, instead of the additional 10Amps.
amperage on a neutural
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I should have said unaccounted for 10Amps, instead of the additional 10Amps.
erickench
Senior Member
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- Brooklyn, NY
It has to be nonlinear loads. There is no other explanation.
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.
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Well I am confused. I noticed that this is a single phase being fed from a 3 phases system. 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 calculation since 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.
My question is wouldn't you calculate the neutral current based on a 3 phase calculation since 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.
boboelectric
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never saw him again.
From the OP:
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.
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You're correct. I read Zbang's post and I didn't realize his post was a question- I thought it was fact.
zbang said:
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 then carry the current of each half the element connected between phase and ground... 120 VAC divided by each half the element's resistance.
If both broken ends were to become grounded, it seems to me the circuit would behave similarly to a MWBC and the ECG current would be zero. If only one end of the element became grounded, and the other isolated, the EGC would carry the current of 120V divided by 1/2 the element's total resistance.
I started out to post that I thought the water heater element scenario would be plausible, but I think I have talked myself into thinking that if a water heater element broke somewhere along it's length you would likely get some sort of EGC current, but it would not cause the neutral current to rise to levels above normal operation, comparing phase currents to neutral current.
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SG-1
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- Ware Shoals, South Carolina
Suppose the water heater has a metal pipe ? Now the current can flow through the pipe to the GEC connection. Only it has to be the neighbors water heater.
Suppose there is another electrical system in the same facility and someone accidentally switched neutrals on a branch circuit between them ?
Suppose there is another electrical system in the same facility and someone accidentally switched neutrals on a branch circuit between them ?
Can you show the work for how you arrived at this answer?
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 decimal 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 21A
Ib 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 24.81A
Total 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.
I follow your math. I'm assuming that your math is referring to a 120/240V single phase service. The only question I have is weather or not the 40.75 deg angle you calculated from p.f. has to be added or subtracted from the Ib unity current which is at 180deg difference from Ia? Since the unity Ia current would be at 0deg wouldn't the unity Ib current be at -180deg and thus the Ib 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 phasor 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. Math assumes 120/240 single phase. Apologies if this doesn't help. I rounded the numbers to simplify.
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. Math assumes 120/240 single phase. Apologies if this doesn't help. I rounded the numbers to simplify.
Yes. 120-0-120.
It would. "Net Id in the neutral is thus (21-28.79) or -7.79A." accounts for that.
skeshesh
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- Los Angeles, Ca
To be honest I find the mathematical statement pretty easy to follow and more clear than the sinusiodal representation. I think that's a decent explanation. Of course I don't think Ia is simply at PF of unity, but a combination of power factor differences on Ia/Ib could account for it. I also see reason in the GEC path delivering current to the neutral. It's really difficult to rate either possibility as definitly more probable than the other since we really don't have that much information about the system and connected components. But I think ultimately to answer the OP 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 it needs to be checked out, so I wouldn't just disregard it.
Quite so.
A at 0.95 lead, B at 0.92 lag
A at 0.96 lag, B at 0.91 lead
A at 0.77 lag, B at unity.
I gave just one example of an infinite number of solutions.
This solution approach makes more sense to me since I am able to see the 40deg phase shift from Vb leading to an current angle of -220deg.
skeshesh
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- Los Angeles, Ca
Of course. I was just pointing it out for others who may be wondering. Your example demonstrated 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 one poster pointed out the first step would be to see if the neutral is bonded to the ground at the panel in question. While this possibility might be slim, the current path through the grounding conductor could energize metallic surfaces and be quite a hazard. I'm not an electrician but I would say it's a good idea to check to prevent possible liability.
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