XFMR Secondary conductors

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slipknotbmg

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Recently I had an inspection in which the secondary conductors from my xfmr to my distribution panel were parallel feeds of 500mcm. Total length of 1 set was 6' and the other set was around 9'. Due to space limitations this is how it had to be done. Now the AHJ wanted me to shorten the 9' set to match the 6' set citing difference of percentage. This was accomplished as I flipped the main breaker and was able to shorten the 1 set and satisfy the AHJ. My question is why did I have to do this? What did he mean by percentage (length?) And how would the minimal difference in length affect anything? As the AHJ could give me no clear answer except that "the percentage had to be equal" lol. And that "it would not be an issue if the runs were say 200' and 205'".
 
Re: XFMR Secondary conductors

310.4
The paralled conductors in each phase, neutral, or grounded circuit conductor shall
(1) be the same length
(2)...
 
Re: XFMR Secondary conductors

The problem with 310.4(1) is that it leaves wide open the question of how close is close enough. Judging by percentage alone does not make sense to me, any more than limiting the difference in length to a fixed number (e.g., within 2 feet, regardless of the total length of the run).

The problem that you are trying to avoid is having one conductor carry so much more of the total current than the other conductor that it exceeds its ampacity limits. The AHJ is, of course, within his rights to make the call that you described. But my inclination is to say that it was unnecessary.
 
Re: XFMR Secondary conductors

Charlie,
I think that the percentage is a very good way. The resistance of the conductor is directly related to its length. In this case there was a very good chance that the 6' conductors would be overloaded. With two sets of conductors the current will divide in inverse proportion to their length. In this case I would expect that the 6' set would be carrying about 60% of the total current and the 9' set about 40%.
Don
 
Re: XFMR Secondary conductors

This is actually a variation of the old maximum power transfer problem.

I suspect the load impedance would so outweigh the conductor impedance that the current division imbalance between the paralleled conductors would be negligible for such short conductor runs.
 
Re: XFMR Secondary conductors

I put some numbers to this issue. I assumed a 750A load and used an effective Z of 0.048 ohms per 1000' of 500kcmil per Table 9. That gives us a Z of 0.00288 ohms for the 6' conductor and 0.00432 for the 9' one. The Z in parallel for these two lengths is 0.0001728 ohms. Then I solved for the voltage drop of this circuit using 750A and 0.0001728 ohms. The voltage drop is 0.1296 volts. If you use I =E/R you can solve for the current in each of the conductors. In this case it shows 450A on the 6' conductor and 300A on the 9' conductor. This exactly matches the 60% and 40% obtained using the percentage method.
Don
 
Re: XFMR Secondary conductors

Bob,
I really don't see how the impedance of the load enters into this. If the current is flowing in the circuit it will devide based on the impedances of the available paths.
Don

spelling correction

[ March 15, 2005, 12:10 PM: Message edited by: don_resqcapt19 ]
 
Re: XFMR Secondary conductors

You're right Don. A 50% difference in length would make that much difference in a current divider.
 
Re: XFMR Secondary conductors

Don is correct, based on basic circuit analysis for parallel circuits. Instead of comparing the difference of only 3ft, it should be compared as a percentage, 9 ft is 50% longer and have 50% more impedance than 6ft of comparitive conductors. The comparison of the 2 lengths can be be "60 ft to 90 ft" or 600 ft to 900 ft", because the percentage relationship is the same.
 
Re: XFMR Secondary conductors

Heh, I've been dealing so much with transformer loadings, my brain went dead. :eek:
 
Re: XFMR Secondary conductors

Thank you all for your input, as I now have a much better understanding of the AHJ's intentions. You wouldn't think 3' would be much of a difference until it is explained well, as it always seems to be here. Once again thanks guys.
 
Re: XFMR Secondary conductors

I agree with Don and Bob, with regard to the mathematics of current division, as they pertain to this situation. I would have come up with the same answer. True, but irrelevant. What I had said is that I though judging by percentages ALONE does not make sense to me. First of all, beyond what percent difference in length would you be in violation of the NEC? Secondly, below what percentage would you be in compliance? There is, of course, no answer to either question. So what I mean is that if the AHJ picks any arbitrary number and uses that as the only criterion, then I would call that a poor inspection technique.

For example, suppose you have a pair of 3/0 copper in parallel. Suppose there is a difference of 50% in the length of the two sets of conductors. We know (from the discussion above) that the current will split 60/40. Now suppose that the calculated load was 330 amps. That means that one set of conductors will draw 198 amps and the other will draw 132 amps. This is not an ideal situation, of course. But please note that the conductors drawing 198 amps are still within their ampacity of 200 amps. So there is no danger of failure.

What I would prefer to see is some intelligence brought into play. Look at each specific situation, and see what makes sense for that situation.
 
Re: XFMR Secondary conductors

Originally posted by charlie b:
So what I mean is that if the AHJ picks any arbitrary number and uses that as the only criterion, then I would call that a poor inspection technique.
Click to expand...
I would call it a poor installation technique. As an inspector, I am tired of being the heavy for making a person fix their mistakes. When I was in the feild, I measured each of my parrallel raceway with a true tape, and bought the longest length times X sets. I then laid the conductors out and cut them to be sure they were the same length. When I terminated, I cut the shortest one first. If that meant (because of bending and turning radii) that I had to fold 10' of 500 Kcmil in the enclosure to keep them the same length, thats what I did. The fact that an installer is too lazy to keep them the same length is not my fault

[ April 15, 2005, 06:13 PM: Message edited by: ryan_618 ]
 
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