Solar Question

[has20]

There is a new 3phase 120/208 1200 amp service with 5 sets of 500mcm cables. The solar plans we received calls for piercing taps on one conductor for each phase. I guess my question is would this set up be a code violation? I see 310.4a in the 2008 NEC but not in any edition afterwards.

 

[jumper]

Look at 310.10(H)(2) and (3). 2014 NEC.

 

[pv_n00b]

You might want to ask this question over in the Photovoltaic forum. But basically, you can't tap one conductor in a parallel set of conductors because the NEC requires all parallel conductors have the same characteristics. What you have to do is bond all the conductors in a set together at a point and interconnect your single PV conductor to that bond. You can use something like a 6 terminal Polaris connector, 5 ports for the existing parallel conductors and one more for the PV conductor.

 

[ggunn]

I'm not advocating, only pointing out that it isn't explicitly addressed in the NEC and requires interpretation.

 

[Carultch]

There is a new 3phase 120/208 1200 amp service with 5 sets of 500mcm cables. The solar plans we received calls for piercing taps on one conductor for each phase. I guess my question is would this set up be a code violation? I see 310.4a in the 2008 NEC but not in any edition afterwards.

I've addressed this issue before. See Post #22 on this thread, for an example of why breaking the symmetry of the parallel sets is a bad idea, from the point of view of the physics of how current will be distributed in that circuit..
http://forums.mikeholt.com/showthread.php?t=186761&page=3

 

[pv_n00b]

I've addressed this issue before. See Post #22 on this thread, for an example of why breaking the symmetry of the parallel sets is a bad idea, from the point of view of the physics of how current will be distributed in that circuit..
http://forums.mikeholt.com/showthread.php?t=186761&page=3

But you let them off the hook by setting V2 to 480V and not V1, make them work it out and show their work.

The NEC is a little vague on the subject of parallel conductors. NEC 310.10(H)(2) lists the characteristics that parallel conductors need to have:

(1) Be the same length.​

(2) Consist of the same conductor material.​

(3) Be the same size in circular mil area.​

(4) Have the same insulation type.​

(5) Be terminated in the same manner.
​

But it fails to identify the actual goal we are trying to obtain and what it takes to reach that goal. Probably because there are many different ways to get there and they wanted to simplify it.

The goal is to have the current split evenly within some given tolerance, a tolerance that is not given. Since the current will split in a way that equalizes the voltage drop along the parallel runs anything that affects the voltage drop will affect the equal split. All the things in the list should affect voltage drop, although I'm not sure how #4 does. And injecting current into one of the parallel conductors will effect the overall voltage drop in that conductor, and require the current split to change to compensate.

The unanswered question in the code would be, how much of tolerance should be allowed in the current split? In any real-world installation, the parallel runs are going to be slightly different and the current split will not be equal. Is tolerance 0% and therefore not obtainable in the real world, 5%, 10%, or what? I would think a difference in current split up to the current carrying capacity of an individual leg was acceptable, but it's not written anywhere to allow that. If someone made up a set of parallel conductors where a perfect split was required to not exceed the conductor ampacity it would be code compliant but in operation, it could overload a conductor.

 

[ggunn]

But you let them off the hook by setting V2 to 480V and not V1, make them work it out and show their work.

The NEC is a little vague on the subject of parallel conductors. NEC 310.10(H)(2) lists the characteristics that parallel conductors need to have:

(1) Be the same length.​

(2) Consist of the same conductor material.​

(3) Be the same size in circular mil area.​

(4) Have the same insulation type.​

(5) Be terminated in the same manner.
​

But it fails to identify the actual goal we are trying to obtain and what it takes to reach that goal. Probably because there are many different ways to get there and they wanted to simplify it.

The goal is to have the current split evenly within some given tolerance, a tolerance that is not given. Since the current will split in a way that equalizes the voltage drop along the parallel runs anything that affects the voltage drop will affect the equal split. All the things in the list should affect voltage drop, although I'm not sure how #4 does. And injecting current into one of the parallel conductors will effect the overall voltage drop in that conductor, and require the current split to change to compensate.

The unanswered question in the code would be, how much of tolerance should be allowed in the current split? In any real-world installation, the parallel runs are going to be slightly different and the current split will not be equal. Is tolerance 0% and therefore not obtainable in the real world, 5%, 10%, or what? I would think a difference in current split up to the current carrying capacity of an individual leg was acceptable, but it's not written anywhere to allow that. If someone made up a set of parallel conductors where a perfect split was required to not exceed the conductor ampacity it would be code compliant but in operation, it could overload a conductor.

As I said, it is subject to (and requires) interpretation. It would be great if the NEC would address it explicitly.

 

[GoldDigger]

But you let them off the hook by setting V2 to 480V and not V1, make them work it out and show their work.

The NEC is a little vague on the subject of parallel conductors. NEC 310.10(H)(2) lists the characteristics that parallel conductors need to have:
...

(4) Have the same insulation type.​

...
All the things in the list should affect voltage drop, although I'm not sure how #4 does.

Different resistive heating based on different wire temperature while dissipating power? But the temperature coefficient of resistance for Cu or Al is quite small.
Even if there is no effect on voltage drop, #4 could affect the calculated ampacity of one of the wires relative to the others, making an even distribution of current the wrong goal.

 

[Carultch]

But you let them off the hook by setting V2 to 480V and not V1, make them work it out and show their work.

It was an arbitrary choice to set voltage that way. Prior to connecting the PV, the voltage V1 would be 493V, to make this equation work out. After connecting the PV, the voltage V1 would be 488V. This would be achieved by the 2.5% tap adjustment on the transformer. As you know, service voltage is a moving target, because it depends on the big picture of all the loads on the grid, and what the utility does in order to compensate and maintain it within their tolerance specification.

Here are the equations used to solve this:

Given data:
R1 = 0.0520 [Ohm]
R2 = 0.0001 [Ohm]
It = 200 [A]
IL = 500 [A]
V2 = 480 [V]

Kirchhoff's current law at each node:
Ia1 + Ib = Iu
Ia2 = Ia1 + It
IL = Ia2+Ib

Voltage drop in relation to resistances and currents, per the Ohm's law equation:
V1 - V2 = Ib*(R1 + R2)
Vt - V2 =Ia2*R2
V1 - Vt = Ia1*R1

6 equations, 6 unknowns (V1, Vt, Ia1, Ia2, Ib, Iu)

The solution when the 200A of PV are connected:
V1 = 487.8V
Vt = 480.035V
Iu = 300A
Ia1 = 149.8A
Ia2 = 349.8A
Ib = 150.2A

Had we divided current equally across both sets, we'd have 250A through both R2 resistors on both the A and B branches. But by breaking the symmetry and only tapping the A branch with the PV system output current, you get a serious amount of current in excess of what the wire can withstand, for the R2 segment of that branch. About 350A on a 255A rated conductor.

The current from the utility is 300A, and observe that it divides nearly equally across the two paths as 149.8A and 150.2A respectively. This is because the dominant resistance is in the 500 ft run, rather than in the 2 ft distance after the tap. That is the resistance that will govern how the utility current will divide, and it will divide nearly equally. But only one section of conductor, will receive the full current of the tapped in PV, which adds 200A onto the 150A. Had we tapped all conductors and not broken the symmetry, we'd instead add 100A onto each 150A, and stayed below the 255A limit.

This isn't like breaking the symmetry due to an extra couple of feet of conductor on one parallel set, because its conduit had the outside track around the bends. That's the kind of break in the symmetry that would probably go unnoticed, and if you check it with an ammeter, it is not going to be too serious of a cause of a disproportionate division of the current. Tapping only one set of conductors in parallel is a break in the symmetry of a much more serious amount of current, due to only adding the new source of current to one of the conductors.

 

[pv_n00b]

It was an arbitrary choice to set voltage that way. Prior to connecting the PV, the voltage V1 would be 493V, to make this equation work out. After connecting the PV, the voltage V1 would be 488V. This would be achieved by the 2.5% tap adjustment on the transformer. As you know, service voltage is a moving target, because it depends on the big picture of all the loads on the grid, and what the utility does in order to compensate and maintain it within their tolerance specification.

Here are the equations used to solve this:

Given data:
R1 = 0.0520 [Ohm]
R2 = 0.0001 [Ohm]
It = 200 [A]
IL = 500 [A]
V2 = 480 [V]

Nice analysis Carultch. It's interesting to look at it if R1 << R2 and the split would then be closer to 50/50 in the conductors, except in the small section represented by R1, which would be I1a-It or something approaching 50A as R1 approaches 0. The smaller R1/R2 the more it looks like "It" is connected directly to the common point at the top and the split goes back to 250A/conductor supplied by both the PV system and utility. But the NEC does not encourage users to do this kind of math. It wants to give nice conservative rules of thumb that don't require a lot of analysis to apply.

 

[Carultch]

Nice analysis Carultch. It's interesting to look at it if R1 << R2 and the split would then be closer to 50/50 in the conductors, except in the small section represented by R1, which would be I1a-It or something approaching 50A as R1 approaches 0. The smaller R1/R2 the more it looks like "It" is connected directly to the common point at the top and the split goes back to 250A/conductor supplied by both the PV system and utility. But the NEC does not encourage users to do this kind of math. It wants to give nice conservative rules of thumb that don't require a lot of analysis to apply.

Thank you.

Actually, the negligible length makes it worse. When R2 = R1, meaning you tap the one set at the midpoint of the main feeder, you get the "safest" outcome with Ia2 = 300A. In the limit as R2 approaches zero, Ia2 = 350A. If that 350A is going through any length of 250 kcmil wire at all, it is going through a chokepoint that isn't rated for the current, according to the NEC.

You certainly can tap at a spare position on a multilug device, such as an 800A breaker frame built with 3 terminations per phase while the main feeder is only occupying two of them. And that is the ideal solution, because now there effectively is a busbar instead of a conductor, eliminating this issue entirely.

 

[pv_n00b]

Actually, the negligible length makes it worse.

I'm not following how R1 approaching 0 will make it worse. If I interconnect the PV supply to the point before the parallel conductors so the total current at V1 is It+Iu then the current from both sources will split evenly in the parallel conductors.

 

[ggunn]

I'm not following how R1 approaching 0 will make it worse. If I interconnect the PV supply to the point before the parallel conductors so the total current at V1 is It+Iu then the current from both sources will split evenly in the parallel conductors.

I haven't verified the math for myself yet but a nontrivial consideration is that the PV inverter is a virtual current source while the grid is a nearly ideal voltage source.

 

[Carultch]

I'm not following how R1 approaching 0 will make it worse. If I interconnect the PV supply to the point before the parallel conductors so the total current at V1 is It+Iu then the current from both sources will split evenly in the parallel conductors.

If you do connect to a distribution block, busbar, or a third lug from a breaker that would otherwise be unused by the two main sets in parallel, then the tapped current isn't touching either of the two conductors in parallel, and nothing is causing the two conductors to carry dissimilar currents. This is the preferred way to do it, if practical, as it doesn't break the symmetry of the conductors in parallel.

By contrast, it is when you tap a tiny length away from the terminal, that you have a problem. It doesn't matter if that 250 kcmil conductor goes 6 inches or 6 feet, its terminations are only rated for 255A. By tapping just one of the parallel sets, a negligible length from its termination, but still on the wire, you put 350A through that chokepoint, which the termination isn't rated to withstand. Getting closer to the terminal increases this current by a marginal amount, approaching the limit of 350A. Tapping at the midpoint of the feeder, is where you get the least difference between the current on two conductors. 300A with the data I provided, reassigning the two resistances to be equal.

I haven't verified the math for myself yet but a nontrivial consideration is that the PV inverter is a virtual current source while the grid is a nearly ideal voltage source.

The way I've specified the problem, the PV will input 200A to the circuit, and set inverter output voltage to be whatever is necessary to get current from the inverter, through its output circuit, to the interconnection (which is what real inverters do). The utility at the service point is set up in this problem to have voltage be as large as necessary, in order to deliver the 480V at the position of the load in the circuit, and deliver the balance of current needed for the load to consume 500A. Suppose this is accomplished by a feedback system, possibly two people with walkie-talkies, one measuring voltage, one adjusting taps on the service transformer. Exactly how the two sources generate the conditions of the problem is a distraction from point I'm trying to make. The point is to show how current will flow among this network of nodes, given conductor resistances that represent a realistic situation, and cause an overload because the setup breaks the symmetry of conductors in parallel.

I'm also aware that I've simplified the problem by overlooking the 125% factor. Suppose that the relevant safety factor, is already accounted for, in the given currents.

 

[pv_n00b]

By contrast, it is when you tap a tiny length away from the terminal, that you have a problem. It doesn't matter if that 250 kcmil conductor goes 6 inches or 6 feet, its terminations are only rated for 255A. By tapping just one of the parallel sets, a negligible length from its termination, but still on the wire, you put 350A through that chokepoint, which the termination isn't rated to withstand. Getting closer to the terminal increases this current by a marginal amount, approaching the limit of 350A. Tapping at the midpoint of the feeder, is where you get the least difference between the current on two conductors. 300A with the data I provided, reassigning the two resistances to be equal.

Your assumption that Ia2 = Ia1 + It is only going to be a reasonable assumption for a large value of R1/R2, which is how you first described the circuit. When I said reducing R1 to approach zero this is no longer a reasonable assumption. This is a current divider circuit where It will split between the two paths based on the resistance of the paths.

The total resistance Rt is going to be 1/(1/(2*R1+R2)+1/R2) and It will split as It*Rt/(2*R1+R2) and It*Rt/R2. Let's say It splits into It1, going up, and It2, going down. So It1=It*Rt/(2*R1+R2) and It2 = It*Rt/R2

As R1/R2 increases Ia2 will approach Ia1 + It with less and less current taking the other path but as R1/R2 decreases It1 and It2 will approach 1/2It. So as I said, as R1 gets smaller compared to R2 it will reduce the current in a given section of the tapped conductor and reduce the current through the terminations on the two ends. At each end respectively the current would be Ia1-1/2It and Ia2 + 1/2 It and Iu would approach 300A. That leaves Ib = 1/2 Iu + 1/2 It = 250A and Ia2 = 1/2 Iu + 1/2 It = 250A and Ia1 = 1/2 Iu - 1/2 It = 150 - 100 = 50A at the point where R1<<R2.

I'll leave it as a problem for the reader to find the value of R1/R2 where the current through any point on the tapped conductor exceeds 255A.