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3 phase nuetral question
#129365
02/04/05 05:48 PM

Joined: Dec 2004
Posts: 1,064
OP
Member

I think I have the wrong formula here, so help me out if you can. On three phase nuetral calculations is it:
A^+B^+C^  (AXB)+(BXC)+(AXC)
Sorry, wrong formula....
[This message has been edited by Dnkldorf (edited 02042005).]



Re: 3 phase nuetral question
#129366
02/04/05 06:32 PM

Joined: May 2004
Posts: 162
Member

The formular I use is solve the square root of A sq +B sq +C sq minus AB minus BC minus CA
Charlie



Re: 3 phase nuetral question
#129367
02/04/05 06:43 PM

Joined: Dec 2004
Posts: 1,064
OP
Member

Charlie, I am still missing something here. Her are my readings:
Aphase 22a Bphase 24a Cphase 23a
The nuetral has 30a on it. I keep coming up with 1.3 or 1.7 amps. Is the math wrong or do I have harmonics problems?



Re: 3 phase nuetral question
#129368
02/04/05 07:28 PM

Joined: May 2004
Posts: 162
Member

I get 1.7 amps on the neutral using the calculation I suggested. If your reading 30A with a clamp on meter you ,may have more than that for THD. 60Hz meters are not accurate at frequencies above 60Hz. and Harmonics are multiples of 60.
What are you using to read the neutral current and what is the nature of the load?
Charlie



Re: 3 phase nuetral question
#129369
02/04/05 08:27 PM

Joined: Dec 2004
Posts: 1,064
OP
Member

I was using typical clamp on ammeter, then swicthed to my Fluke 39 power anylyzer, got the same reading. I am not really familiar with the use of the 39, so I downloaded the manual from fluke itself. One of troubles I have with the manual, it doesn't state what normal readings should be in respect to THDR and THDF. I am not sure if I am rading it right but the THD was showing 58%. PF was at 87. But I don't trust the readings because of my umfamiliarity with testing harmonics at this time.
The place is a fiber optic component manufatcurer and testing lab. Current electric bills are out of the normal. High side. Most of the equipment in the place are computers, oscopes, signal generators, signal analyzers and such. Some (3) single phase 220V oven /chiller combo units.
Now their guy there tells me he thinks all the power supplies are of the linear type and not the "switching" that I am somewhat a custom too. They buy these type supplies because of potential noise problems with the "switching" type. Service incoming is 480/277v 3 phase 4 wire service with 480X208v 3 phase transformer "dry type" NOT a Kfactor transformer. Most of the loads are 120v.
These measurements were taken off the subpanel on the 208/120 side.
Lighting is all 277V. T8 fixtures
I have never seen the Nuetral current this far off before.
Any clues?
Dnk......



Re: 3 phase nuetral question
#129370
02/04/05 09:21 PM

Joined: May 2004
Posts: 162
Member




Re: 3 phase nuetral question
#129371
02/05/05 12:05 AM

Joined: Feb 2005
Posts: 19
Member

I have calculated the neutral imbalance to be the squre root of three or 1.73. The way that I got this is by taking the squre rtoot of A^ + B^ + C^ minus AB + AC + CB. Try plugging the numbers into this equation and you should come up with the square root of 3 or extended out it will be 1.73. hopefully this helps. Rick Buck
Buck



Re: 3 phase nuetral question
#129372
02/06/05 12:20 AM

Joined: Sep 2003
Posts: 649
Member

The math is simply not applicable in this case.
The equations for calculating neutral current given the phase currents make the assumption of pure sinusoidal current flow. If the current flow is _not_ a sinusoid, then the equations cannot be expected to work.
In a perfectly balanced system, the _fundamental_ current flow is exactly 120 degrees out of phase between each phase. There is no neutral current flow. The phase difference means that current flowing out of one phase goes directly into the others. In a real system, of course, the current flow probably isn't exactly 120 degrees out of phase, and the individual phase currents are not balanced, so you expect _some_ neutral current to flow.
In a perfectly balanced system with third harmonic current flow, the third harmonic component of the current flow is exactly 3*120 = 360 degrees out of phase between each phase...but 360 degrees == 0 degrees. In other words, in a perfectly balanced system with third harmonic, the currents are _in phase_ for all of the phases. Current flowing out of one phase will be matched by current flowing out of each of the other phases, and the only place for this current to go is down the neutral. In the perfectly balanced system, the third harmonic current down the neutral will be 3x the third harmonic current in any given phase. Of course, again, in the real world the currents will not be balanced, but the third harmonic current on the neutral will be pretty darn close to the _sum_ of the third harmonic currents on each of the phases.
If you can arrange it, put an oscilloscope with a current probe on the neutral in question. You'll probably see lots of 180 cycle and 360 cycle current flow.
Nonlinear loads in general introduce harmonic current flows into electrical systems. It could be a switching power supply or a linear power supply; somewhere there will be a diode rectifier on the input. This diode rectifier is a classic nonlinear load, only conducting in peaks for part of the AC cycle, providing a rich supply of harmonic current to the power system.
Jon




