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Thread ID: 120939	2011-10-03 22:25:00	Can anybody here do 3-phase current calculations?	Billy T (70)	PC World Chat

Post ID	Timestamp	Content	User
1235273	2011-10-03 22:25:00	My maths fell into disrepair about 40 years ago and I need to calculate the neutral current created by imbalanced phase loads of 110, 96 and 85 amps respectively . I found the following example for balanced loads (which illustrates that the neutral current will be zero) but the italicised line at the start throws me, because even I know that it is incorrect . Vector maths are involved, so maybe they meant the vector sum, but that is no help anyway because for me that level of maths is a closed book . I found an online calculator, but it didn't have NZ in the country list (see image) so I couldn't authenticate myself for a download, and then I found that it had hijacked Google and was offering itself up on every search no matter what the subject, so I fixed that and gave up the fight . The neutral current is the sum of the phase currents . :confused: Since R=1 I_{L1}=V_{L1}\,\! I_{L2}=V_{L2}\,\! I_{L3}=V_{L3}\,\! I_{N}=I_{L1}+I_{L2}+I_{L3}\,\! I_{N}=\sin x+\sin (x-\frac{2}{3} \pi)+\sin (x-\frac{4}{3} \pi) Using angle subtraction formulae I_{N}=\sin x+\sin x\cos(\frac{2}{3} \pi)-\cos x\sin(\frac{2}{3} \pi)+\sin x\cos(\frac{4}{3} \pi)-\cos x\sin(\frac{4}{3} \pi) I_{N}=\sin x-\frac{1}{2}\sin x-\frac{\sqrt{3}}{2}\cos x-\frac{1}{2}\sin x+\frac{\sqrt{3}}{2}\cos x I_{N}=0\,\! Any assistance would be appreciated . Normally I would just go the the site and meter it, but in this instance the site is remote and I have been given a neutral current value that I suspect is way too low for that level of imbalance, so I want to find out how much is escaping elsewhere via grounds etc . Cheers Billy 8-{)	Billy T (70)
1235274	2011-10-03 22:35:00	Geez too hard for me,electrical theory was more than 20 years ago. en.wikipedia.org seems clearer but still beyond me now. It also says the neutral current is the sum of the phase current, but I don't follow it.	dugimodo (138)
1235275	2011-10-03 22:41:00	Did a further search and found a simplified system (www.electrician2.com)that is good enough for my needs, and it also includes a calculator, but it is still beyond my understanding of the mathematical terminology and principles. Maybe this will be useful to somebody with less neanderthal mathematical skills than mine. :help: Cheers Billy 8-{)	Billy T (70)
1235276	2011-10-03 22:46:00	:) I read that page also, I think with a couple of hours I could probaly get the hang of it as it's really just trigonometry. The previous page gives a good way of approximating it with lines representing the phases. You could work it out with a compass and a ruler, but you probably got that already :)	dugimodo (138)
1235277	2011-10-03 23:11:00	:) You could work it out with a compass and a ruler, but you probably got that already :) LOL! :D I can barely spell trigonometry, let alone use it, and although I have a faint idea of how a ruler and compass might be helpful, the main problem is that I can't interpret the mathematical factors. The last recorded neutral current for very similar phase level differentials was just 11 amps, which my gut tells me is too low. I guess I'll just have to flag it as an unknown quantity and wait until the opportunity arises to check it. I would still be very appreciative if somebody else comes along and knows how to use the calculator in my previous post. Cheers Billy 8-{)	Billy T (70)
1235278	2011-10-03 23:15:00	Eeek that gave my physics knowledge a hard hit. Damn, study harder, exams just around the corner!	The Error Guy (14052)
1235279	2011-10-03 23:46:00	From here (forums.mikeholt.com) I find the following formula: SQRT I²A + I²B + I²C - (IA x IB) - (IB x IC) - (IC x IA) To clean it up (clarify it based on later posts), it's SQRT (I²A + I²B + I²C - (IA x IB) - (IB x IC) - (IC x IA)). IA = 110, IB = 96, IC = 85 = SQRT (12100 + 9216 + 7225 - 10560 - 8160 - 9350) = SQRT (471) = 21.7 amps According to my vague mental vector maths, that sounds about right. Hope this helps.	george12 (7)
1235280	2011-10-04 01:29:00	Thanks George, that fits nicely with my dumb-arse guesstimate, reached by squinting at the figures and adding half the length of my thumb! So, I'll write it up as approximately 22 amps, because the loads are dynamic and are constantly changing, therefore absolute precision is neither possible nor required . Cheers Billy 8-{) :thumbs: Now if that was a quadratic equation, I'd probably have to add twice the length of my thumb and use my extensible member to approximate the variable . :devil:	Billy T (70)
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