Newbie simple (yea right) question for 3-phase

[Penwar]

Here I go, exposing my lack of history with 3-phase.....

for sizing the conductors and fuses for a disconnect:

1st time at this one, please be gentle.

208Y/3-PH/4-wire system. Connecting (6) inverters (solar) to an existing grid system. Output of each inverter is 25A. So a total of 150A*1.25 = 187.5. Utilizing a 3-ph panel with 2-pole breakers (figuring 35A breakers). Figuring that the current at each leg to be 62.5A. I am curious how I go about sizing the wires out of the panel and the final disconnect switch (200A). Do I size the wires based on the total load (187.5A), or do I size each of the 3-wires for the load from each leg off the panel (62.5A).

I am at a total loss on this. My gut tells me that I go for the entire load, which being in conduit, underground, 100' run would be 3/0 (with 1.37% drop). This is fine if correct, but why would it not be a #4 for each wire.

Along with this, for a 3-ph AC disconnect switch, would this be (3) 80A breakers, or (3) 200A breakers.

I believe I am missing something fundamentally.

 

[Smart $]

Here I go, exposing my lack of history with 3-phase.....

for sizing the conductors and fuses for a disconnect:

1st time at this one, please be gentle.

208Y/3-PH/4-wire system. Connecting (6) inverters (solar) to an existing grid system. Output of each inverter is 25A. So a total of 150A*1.25 = 187.5. Utilizing a 3-ph panel with 2-pole breakers (figuring 35A breakers). Figuring that the current at each leg to be 62.5A. I am curious how I go about sizing the wires out of the panel and the final disconnect switch (200A). Do I size the wires based on the total load (187.5A), or do I size each of the 3-wires for the load from each leg off the panel (62.5A).

I am at a total loss on this. My gut tells me that I go for the entire load, which being in conduit, underground, 100' run would be 3/0 (with 1.37% drop). This is fine if correct, but why would it not be a #4 for each wire.

Along with this, for a 3-ph AC disconnect switch, would this be (3) 80A breakers, or (3) 200A breakers.

I believe I am missing something fundamentally.

I assume your inverters are connected double delta, i.e. two per phase, line-to-line, 208V 1?. You are correct about each OCPD having to be a minimum 25A ? 125% = 31.25A... so 35A standard.

However, when combined the phase angles result in a total current on each leg to be less than the sum of their magnitudes, typically by a factor of .866, the square root of 3 divided by 2. If equal in magnitude, the result is the current on one phase times 1.732, the square root of 3.

In your case, simply figure the output in VA and proceed from there...

25A ? 6 ? 208V = 31,200VA

31,200VA ? (208V ? 1.732) = 86.7A

86.7A ? 125% = 108.3A

...for minimum conductor size to and OCPD rating of your PV System disconnecting means.

EDIT: The above is under 2014 Code. Under earlier editions, you have to use first OCPD rating of 35A as such...

35A ? 6 ? 208V = 43,680VA

43,680VA ? (208V ? 1.732) = 121.3A

 

[GoldDigger]

Here I go, exposing my lack of history with 3-phase.....

for sizing the conductors and fuses for a disconnect:

1st time at this one, please be gentle.

208Y/3-PH/4-wire system. Connecting (6) inverters (solar) to an existing grid system. Output of each inverter is 25A. So a total of 150A*1.25 = 187.5. Utilizing a 3-ph panel with 2-pole breakers (figuring 35A breakers). Figuring that the current at each leg to be 62.5A. I am curious how I go about sizing the wires out of the panel and the final disconnect switch (200A). Do I size the wires based on the total load (187.5A), or do I size each of the 3-wires for the load from each leg off the panel (62.5A).

I am at a total loss on this. My gut tells me that I go for the entire load, which being in conduit, underground, 100' run would be 3/0 (with 1.37% drop). This is fine if correct, but why would it not be a #4 for each wire.

Along with this, for a 3-ph AC disconnect switch, would this be (3) 80A breakers, or (3) 200A breakers.

I believe I am missing something fundamentally.

The first thing that you are missing is that if the inverters are, as is most common, connected line to line to 208V (in a 208Y/120 service), then the line current in each phase line will NOT be just the sum of the line-to-line currents connecting to that line. (It is a vector addition thing that happens in 3-ph.)

So, if the current across each phase is 50A, and you rate it as 62.5A, then the actual current in the line conductor that links four inverter connections (two going to each of the other two line conductors) will be 62.5 x 2 x (sqrt(3)/2), or 62.5 x 1.732 = 108.
So you will need individual two pole breakers for each inverter, giving you a total of six branch circuits, and each of those breakers needs to be rated for 31.25A or more. (Too bad it is over 30!) The wires in those six branch circuits need to be sized to be protected by that breaker.

Now once you have combined the six inverters onto your panel bus, the main disconnect needs to a three phase breaker rated for 108A or more per pole. (Ouch again on must missing a standard size.)

Since you have line-to-line loads (sources), you should use a single three pole main breaker with common trip. The feeder wires need to sized to carry at least 108A and to be protected by whatever size breaker you have at the upstream panel (if there is one.)

Depending on your code cycle, the inspector may try to apply the 120% rule to the panel bus, which will force you to use a panel rated at 225A or more. (200 x 1.20 = 240, but 160 (4x40 per phase) + 125 (main) (depending on what size main breaker you are able to find) is greater than 240.
With a 225A bus, the limit is 270, which could still be a problem. (The proper way of applying the sum of breakers math to a three phase panel is open to interpretation!)
You may also be told to size the feeder wires for 280A, which would be a real shame.

 

[Penwar]

so...

so...

I had forgotten that 1.732! Doh!!!!

Going forward, whether my load is 108 or 121.3A (as proved by goldigger and Smart$), would I size the individual legs based on those numbers, or would it be based on my the total output of 187.5A or 210A as was noted based on a previous code cycle? I can see doing this with 1/0 wire and 125A fuses in the main disconnect. Note that the summation panel will not have a main CB.

 

[GoldDigger]

I had forgotten that 1.732! Doh!!!!

Going forward, whether my load is 108 or 121.3A (as proved by goldigger and Smart$), would I size the individual legs based on those numbers, or would it be based on my the total output of 187.5A or 210A as was noted based on a previous code cycle? I can see doing this with 1/0 wire and 125A fuses in the main disconnect. Note that the summation panel will not have a main CB.

The wire sizes are based on the amps that will be flowing through each individual conductor, not the sum of the current in all three wires.

 

[Penwar]

see...

see...

That is what we are thinking, but it just seems wrong to some extent. Granted, one benefit of 3-phase is smaller wire sizes (sometimes).

Thanks for the help.

 

[jaggedben]

...(The proper way of applying the sum of breakers math to a three phase panel is open to interpretation!) ...

I'd disagree with that. 705 is pretty clear about connections to a 'busbar or conductor'. You count the breakers connected to each busbar individually regardless if it's split-phase or three-phase.

The math in this case works out to 35Ax4, plus the rating of the 3-pole utility breaker, for each busbar. A 225A panel with a 125A main breaker works out just barely with the sum of breakers at 265A and the 120% rule allowing 270. All breakers are sized correctly for the output as properly calculated with the three phase math.

It's important that the breakers are equally spread around each busbar and not installed willy-nilly, and probably those 'inverter output, do not relocate..' stickers should be used.

Let's just be glad that this system is balanced. Otherwise we'd have panels and breakers oversized for some of the output, and the math would be over my head.

 

[Penwar]

Let's add a little spice to this one

Let's add a little spice to this one

OK, now let's tweak this a little to see if we are all in agreement, or maybe better still, to further confuse the who knows what out of me.....

What if we had a mix of inverters hitting that same summation panel. Say we had six inverters; two of them at 25A and 4 of them at 35A. Is the same logic applied? We would have an unbalanced load to an extent, but for the sake of sizing the output conductors and subsequent disconnect breakers, would this still apply:

(25+25+35+35+35+35)= 39520VA
39520VA / (208V * 1.732) = 109.67A
109.70A * 125% = 137.12A, for which we use as a basis for sizing the conductors and OCPD (150A) at the D/S

Or do we base it on the size of the breakers in the panel, in this case being 35,35, 50, 50, 50, 50, which would give us:

(35+35+50+50+50+50)= 56160VA
56160VA / (208V * 1.732) = 155.89A
155.89A * 125% = 194.86A, for which we use as a basis for sizing the conductors and OCPD (200A) at the D/S

 

[ggunn]

You know that here is a photovoltaic forum, right?

 

[Penwar]

I do, but wanted more of a basic electrical forum as a sounding board. Perhaps a mod can move this over to the PV forum.....if it is decided that this is the wrong forum...

 

[jaggedben]

OK, now let's tweak this a little to see if we are all in agreement, or maybe better still, to further confuse the who knows what out of me.....

What if we had a mix of inverters hitting that same summation panel. Say we had six inverters; two of them at 25A and 4 of them at 35A. Is the same logic applied? We would have an unbalanced load to an extent, but for the sake of sizing the output conductors and subsequent disconnect breakers, would this still apply:

(25+25+35+35+35+35)= 39520VA
39520VA / (208V * 1.732) = 109.67A
109.70A * 125% = 137.12A, for which we use as a basis for sizing the conductors and OCPD (150A) at the D/S

Or do we base it on the size of the breakers in the panel, in this case being 35,35, 50, 50, 50, 50, which would give us:

(35+35+50+50+50+50)= 56160VA
56160VA / (208V * 1.732) = 155.89A
155.89A * 125% = 194.86A, for which we use as a basis for sizing the conductors and OCPD (200A) at the D/S

The EE's on this thread may be able to help you understand it better, but I believe that vector math makes it not as simple as either of those. What I know about that stuff is from trying to understand this article. Give it a shot if you like. The max current on one or two phases is going to be larger than the other(s) so averaging them out like you did is going to result in too small a number for the conductors and breaker.

What I can tell you is this:
1) If you had different inverters but the system remained balance (e.g. 3 at 25A and 3 at 35A) you can still use simple math like you did above.
2) For the unbalanced case, you would use the largest sum of breakers any one busbar had connected to calculate the 120% rule. You'd probably end up over sizing the utility side feeder conductor and breaker for at least one phase just because you won't find any mixed ampacity breakers for sale.

 

[Smart $]

I'd disagree with that. 705 is pretty clear about connections to a 'busbar or conductor'. You count the breakers connected to each busbar individually regardless if it's split-phase or three-phase.

...

... For the unbalanced case, you would use the largest sum of breakers any one busbar had connected to calculate the 120% rule. ...

I agree with that for 2011 and earlier NEC compliance.

However, 705 {2014} says to use 125% of inverter rated output current for ampacity calculations... and that infers we can determine such using standard 3? current formulas. JMO/YMMV

 

[Smart $]

I do, but wanted more of a basic electrical forum as a sounding board. Perhaps a mod can move this over to the PV forum.....if it is decided that this is the wrong forum...

As such, can you limit your application to a specific NEC edition?

 

[GoldDigger]

However, 705 {2014} says to use 125% of inverter rated output current for ampacity calculations... and that infers we can determine such using standard 3? current formulas. JMO/YMMV

Using the rated output of the inverters (times 1.25) does not imply, to me anyway, that you can use the vector sum rather than the sum of the magnitudes to determine bus contribution. The number you are working with may be smaller, because of standard breaker sizes, but there is still no recognition of the differences caused by the 3-phase configuration.

 

[Smart $]

Using the rated output of the inverters (times 1.25) does not imply, to me anyway, that you can use the vector sum rather than the sum of the magnitudes to determine bus contribution. The number you are working with may be smaller, because of standard breaker sizes, but there is still no recognition of the differences caused by the 3-phase configuration.

But the removal of breaker rating sum previously stipulated opens calculation up to using standard 3? calculation of current... IMO

If you had a resistive fault that drew enough current to saturate inverter outputs and utility source without tripping any OCPD, how much current would the inverters contribute to the fault? ...compared to breaker ratings sum of the inverters?

 

[ggunn]

If you had a resistive fault that drew enough current to saturate inverter outputs and utility source without tripping any OCPD, how much current would the inverters contribute to the fault? ...compared to breaker ratings sum of the inverters?

It seems to me that in such a scenario it would be extremely unlikely that the inverters would keep running, so I guess my answer would be zero. What I think I would do if I had to connect three mismatched inverters phase to phase on a three phase system would be to size the OCPD and conductors on their combined output as if they were all the size of the largest inverter. But first I would try to avoid having to do that.

 

[Smart $]

It seems to me that in such a scenario it would be extremely unlikely that the inverters would keep running, so I guess my answer would be zero. What I think I would do if I had to connect three mismatched inverters phase to phase on a three phase system would be to size the OCPD and conductors on their combined output as if they were all the size of the largest inverter. But first I would try to avoid having to do that.

While yes they would likely shut down, I said saturate inverter output... as in max current output... and I meant equally sized across phases... say, rated 40A, providing 40A, 125% at 50A and 50A breaker, each.

Would max bus/conductor current after combined be:

A) 80A
B) 69A
C) 100A
D) 87A
E) __A?​

 

[ggunn]

While yes they would likely shut down, I said saturate inverter output... as in max current output... and I meant equally sized across phases... say, rated 40A, providing 40A, 125% at 50A and 50A breaker, each.

Would max bus/conductor current after combined be:

A) 80A
B) 69A
C) 100A
D) 87A
E) __A?​

As I said, if I had, say, two 6kW inverters and one 7kW inverter all phase to phase around a three phase interconnection, I would size conductors and OCPD appropriately for each inverter going into the AC combiner and then size conductors and OCPD from the combiner to the interconnection as if the inverters were all 7kW.

Your question is IMO meaningless; if the inverters would somehow keep running the current between the inverters and the fault would be whatever the inverters were producing at the time, and the current between the interconnection and the fault would be whatever the OCPD at the interconnection allowed onto the conductors.

The likelihood of such a delicately balanced sustainable fault occurring and the inverters not shutting down is, IMO, less than 1% of a snowball's chance at noon on a Texas day in August.

 

[Smart $]

... and then size conductors and OCPD from the combiner to the interconnection as if the inverters were all 7kW.

And this last part is what is in question. Under 2014 Code, what size would you use? Please elaborate on your calculation of the minimum required ampacity.

 

[Smart $]

...

Your question is IMO meaningless; if the inverters would somehow keep running the current between the inverters and the fault would be whatever the inverters were producing at the time, and the current between the interconnection and the fault would be whatever the OCPD at the interconnection allowed onto the conductors.

Well then, forget about the fault part of the question.

What is the maximum current after combined of 2 inverters connected and operating at 60? to each other on the connected end (phase-wise they'd be at 120?)?