I am trying to understand how to calculate what my unbalanced load should be on my neutral. In one of my buildings which is fed with 3 phase 208/120, I have a panel that is 208 single phase with a neutral. I measured 34 one one phase, 27 one the other phase and 35 on the neutral. I would have thought the neutral load would have been much lower considering the loads were balanced close. Also, How does the phase conductor cancel each other out in the first place like on a 240/120 volt single phase system.
3 phase neutral load
- Thread starter JdoubleU
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see this link for an explanation of 3 phase vector theory
http://www.electrician2.com/electa1/electa3htm.htm
http://www.electrician2.com/electa1/electa3htm.htm
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mcclary's electrical
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√((Ia?+Ib?+Ib?)-(Ia*Ib)-(Ib*Ic)-(Ia*Ic))
Thank you for that information. I do have a few questions if you have time. If i understand it right. The Magnitude is 120v because that is what our voltage is and the vector degree is where each phase hits the x axsis. I don't quit understand the next step in connecting the lines.
dbuckley
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Firstly, I don't think you have a 208V single phase panel - you have a panel with two phases connected, with 208V between the two hot bars, and 120V either hot bar to neutral.
So what you actually have is a three phase panel, that is missing the third phase.
Looking at your ampmeter readings again, this means your three phase current loads are 34A, 27A and 0A. They dont look much to me like they are "balanced close"
Just 'cos you're only using two of the three phases, its still a three phase system.
There are two different vectors, one for voltage and one for current. When the load is purely resistive, the current vector assumes the same direction as the voltage vector. When there is a reactive load, the current vector is not aligned with the voltage vector. It will either lead or lag the voltage vector. To get the neutral current, you place the line current vectors tail to head. This is graphical vector addition. The neutral current vector is then from the last 'head' to the starting point.
If you have Word and Excel, you can download my Excel Neutral Calculator which permits you to enter the numbers and Excel does the calculating and plotting of the current vectors. You need Word because the Excel file is embedded in a Word file... because this site doesn't permit Excel uploads, but does permit Word file uploads.
http://forums.mikeholt.com/attachment.php?attachmentid=3369&d=1244942631
If you have Excel and not Word, send me a PM.
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Sounds more like he has a 120/208 1? 3W panel. Though it is usually supplied with two lines and a neutral conductor from a 208Y/120 3? 4W transformer-secondary supply, it is technically called a single phase panel per IEEE standards.
dbuckley
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- Canterbury, New Zealand
That's bizarre, and my oscilloscope begs to differ, but if some higher power has defined this name for that panel, then so be it. No wonder the one phase / two phase threads go on for hundreds of posts...
Doesn't alter the fact that the OP has a two phase load on a three phase system
hardworkingstiff
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- Wilmington, NC
That's bizarre, and my oscilloscope begs to differ, but if some higher power has defined this name for that panel, then so be it. No wonder the one phase / two phase threads go on for hundreds of posts...
I'm with you dbuckley!
When I ran my calculation (a few days ago but got too busy to post it), I used 120 degree separation of phase currents 34A,27A,0A. The resultant vector (the neutral) works out to be 31A. So your original measurements are in the right range. I think my calculation (just vector algebra) is only dead accurate with purely resistive loads but I need to think about that.
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Well go ahead and think... but I'll give you advanced confirmation
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