derating wire for conditions of use - value > OCPD?

[BBL Solar Design]

Hello,
I recently joined, so forgive me if this has been previously covered.. I read a similar topic last night, which did not 100% clarify it for me, so I will ask this again, using different words, at least in terms of how I am thinking about the subject:

My question is two parts:

• at what point (amperage level), in the Code are we allowed to round down for nearest breaker/fuse size to protect wire? (if ever)
• when we select and derate our wire for conditions of use, does the derated value (based on type of wire, temperature and number of conductors in conduit) need to be larger than the over current protection device protecting it? Or are we just trying to get the derated ampacity (corrected for conditions of use) to be simply greater than the continuous duty value (Full Load Amps x 1.25)?

The example I will use illustrates how the difference can be significant in terms of wire size. My main concern relative to the question is safety, and secondary interest is potential cost savings in wire size/quantity.

On a rooftop solar electric system, where the inverters (generators) are located on the rooftop in the desert. My question is specifically about the AC output side of inverters, which makes this a standard trade-related question (and not unique to the DC aspects of solar).

Ambient conditions are 44 deg C + I'm using a rooftop adder of 17 deg C (for 3.5" above roof) = 61 deg C @ conduit location.
In all cases, I'm assuming I will be running at max of 3 current carrying conductors in each conduit.

System specs:
(18) 27.7 kW inverters (480V, true 3 phase)
FLA of inverter = 36A
I'm using THWN-2, 90 deg C rated wire.
In 2011 NEC table 310.15(B)(2)(A) (formerly at the bottom of Table 310.16), the derate factor for 90 deg C column at this 61 deg C is 0.58.

18 x 36A = 648A actual current
648A x 1.25 = 810A continuous duty rated current


Method A:
Assuming that my derated wire value (for conditions of use) has to be greater than my breaker/fuse size, and assuming I have to round UP to get to my next breaker/fuse size (ignoring question #1 for now), I'm going with a 1000A breaker/fused switch. Then I would use the following equation:

1000 A fuse / 0.58 = 1724 A
Then I select parallel wire runs that exceed this number.
(4) sets of 500 kCMIL type THWN-2 (90 deg C rated), each set of 3 wires running in it's own conduit would result at 1720 A, which is just under this value, meaning that (4) sets of 500 kCMIL 90 deg rated conductors would be able to handle 997.6A (430A x 4 sets x 0.58) at 61 deg C in conduit in the hot desert sun. I wonder if this is sufficient. To get ABOVE this value one might call for (4) sets of 600 kCMIL. Is this sufficient/correct or overkill?


Method B:
I seem to recall from my notes and trainings over the years that we can size conductor supply feeds and over current devices based on NEC table 310.16 (now 310.15(B)(2)(A)) and use the greater of maximum inverter output current rated for continuous duty (1.25 factor) or maximum inverter output current corrected for maximum ambient temperature and multiple conductors - using whichever is greater.

If this is true, I would be size my wire according to the following equation:
648A actual / 0.58 = 1117 A
So (3) sets of 400 kCMIL THWN-2 (each set running in separate conduit) would provide 1140 A @ 30 deg C, and at a temperature of 61 deg C, a temperature corrected value of 661A (1140 x 0.58), which is > than the 648A of current running through it.
But my worry is that if we had a condition such as a ground fault where, say, 800A of current flows, the 1000A fuse does not blow, and the wire becomes exposed to a condition which exceeds its rating of 661A under heated conditions, and begins to melt.


Method C:
Hopefully not making this overly-complicated, a 3rd method which could be used is selecting wire so that the derated value of the wire (considering conditions of use) is greater than the FLA rated for continuous duty (x 1.25 factor) of the circuit. Here, the equation would be:

648A actual x 1.25 = 810A rated
810A rated / 0.58 = 1396 A
So (4) sets of 350 kCMIL THWN-2 (each set in separate conduit) would provide 1400 A @ 30 deg C, and at 61 deg C, it would provide 812 A (1400A x 0.56) which is > than the 810A continuous duty rating of the circuit. Again the potential is there, where if fuses are sized at 1000A (rounded up from 810A), that the wire could see, say, 900A of current (perhaps under fault condition) where the 1000A fuse is not clearing and yet the wire is subject to current and temperature beyond its conditional rating.


My main question is which method above (A, B or C) is the correct way to select my wire.

Question #1 is a little simpler -
Can I (ever) round my OCPD down to 800 from 810A?

Thanks for sticking with this, if you've read this far, and for your reply. I know I'm not alone here in figuring this out, and I definitely owe someone a beer for this one.

Thank you kindly,
Brian

 

[GoldDigger]

Well, just for a subtle point: When you find that you have to change the conductor size in your conduit on the roof because of the combination of ambient and roof temperature adder, look at the length of the conduit run on the roof and compare it to the total length of that same circuit located inside the roof (assuming a roof penetration near the PV).

If the length located above the roof is less than 10 feet and also less than 10% of the total length of that circuit, then you may be able to make use of 310.15(A)(2) [2011] to avoid having to oversize the wire.

Now for your first question:

You are allowed to round up the breaker size in some cases. You are always allowed to use a smaller breaker than the one required to protect the wire, as long as the breaker is also able to carry the calculated load. Those are two separate conditions, one which sets a maximum breaker size and one which sets a minimum breaker size.
If the load requires a particular breaker size, then for the most part the wire must be large enough that it is also protected by that breaker (unless the "next size up" rule applies at that point.)

Note that the required sizes for neutrals and EGCs can be very different from the ungrounded circuit conductors, especially when an EGC is also serving as a DC ground electrode conductor for a solar array, for example.

I am sure other members will be addressing your detailed examples.

 

[BBL Solar Design]

Gold Digger, thank you for the point about wire length. Unfortunately in most cases here my wire runs are exposed to the hot desert sun for more than 10% of the run.

"You are allowed to round up the breaker size in some cases. "
I'm confused why the Code allows us to do this, it seems by nature this allows OCPD to be higher than, and therefore not protect the wire amacity rating. This seems (to me) contrary to the intent of circuit protection. However this is slightly off my topic.

"You are always allowed to use a smaller breaker than the one required to protect the wire, as long as the breaker is also able to carry the calculated load."
To clarify what you mean by "calculated load", would this be: FLA x 1.25 ?

Thanks, Brian
:dunce:

 

[david luchini]

Gold Digger, thank you for the point about wire length. Unfortunately in most cases here my wire runs are exposed to the hot desert sun for more than 10% of the run.

"You are allowed to round up the breaker size in some cases. "
I'm confused why the Code allows us to do this, it seems by nature this allows OCPD to be higher than, and therefore not protect the wire amacity rating. This seems (to me) contrary to the intent of circuit protection. However this is slightly off my topic.

"You are always allowed to use a smaller breaker than the one required to protect the wire, as long as the breaker is also able to carry the calculated load."
To clarify what you mean by "calculated load", would this be: FLA x 1.25 ?

Thanks, Brian
:dunce:

See NEC 240.4(B) and (C) for protecting wires at higher than their ampacity. Basically it is allowed for OCPDs 800A or less (and also for specific equipment such as motors.) In your example with a 1000A OCPD, you would not be allowed to use a conductor with an ampacity of less than 1000.

See 215.3 for OCPD sizing. The breaker must be sized for at least 125% of the continuous load plus 100% of the non-continuous load.

 

[bob]

Using your information, 648A actual x 1.25 = 810A rated and 810A rated / 0.58 = 1396 A total. 1396/4 runs = 349 amps per ckt. The minimum size conductor is 500 kcm at 380 amps. However if you use a 1000 amp breaker, the conductor must match the breaker rating as David indicated.

 

[ggunn]

Well, just for a subtle point: When you find that you have to change the conductor size in your conduit on the roof because of the combination of ambient and roof temperature adder, look at the length of the conduit run on the roof and compare it to the total length of that same circuit located inside the roof (assuming a roof penetration near the PV).

If the length located above the roof is less than 10 feet and also less than 10% of the total length of that circuit, then you may be able to make use of 310.15(A)(2) [2011] to avoid having to oversize the wire.

Another maybe-not-so-subtle point: OCP on inverter output conductors is not to protect the conductors from the inverter; it's to protect the conductors from current flowing from the service in the event of a fault. The sizing of the OCPD is set to open a current window wide enough for the maximum output of the inverter to reach the interconnect, but the protection is for current flowing the other way. The inverter is incapable of supplying enough current to endanger properly sized conductors.

BTW, this discussion should probably be in the Photovoltaics forum.

 

[BBL Solar Design]

Another maybe-not-so-subtle point: OCP on inverter output conductors is not to protect the conductors from the inverter; it's to protect the conductors from current flowing from the service in the event of a fault. The sizing of the OCPD is set to open a current window wide enough for the maximum output of the inverter to reach the interconnect, but the protection is for current flowing the other way. The inverter is incapable of supplying enough current to endanger properly sized conductors.

I could be reading into this too far, but are you saying Method A is correct? Where the derated value of the wire (wire ampacity x 0.58) needs to be greater than my OCPD?

BTW, this discussion should probably be in the Photovoltaics forum.

Respectfully, the issue of derating wire above OCPD vs. continuous duty rating of circuit vs. actual amps is somewhat irrelevant to PV. While it's peripherally involved, I chose to post here to steer away from DC related discussions.

Thanks

 

[BBL Solar Design]

Using your information, 648A actual x 1.25 = 810A rated and 810A rated / 0.58 = 1396 A total. 1396/4 runs = 349 amps per ckt. The minimum size conductor is 500 kcm at 380 amps. However if you use a 1000 amp breaker, the conductor must match the breaker rating as David indicated.

First, 500 kcmil at 380 amps is from the 75 deg C column. I'm using THWN-2 wire, which is 90 deg C rated. Shouldn't it be 430 A?

Second, having a conductor set that matches the 1000A breaker rating would look like Method A.
1000A / 0.58 = 1,724A
1,724A / 4 = 431 A

If I have to use the 75 deg C column, this would be (4) runs of 700 kcmil.
If I can use the 90 deg C column, (4) runs of 600 kcmil gets me there.

Third, is it acceptable to use an 1000A breaker with an adjustable trip setting and set the trip rating for 850A so I can use smaller wire?
I'd imagine (guessing) someone could come round and change the trip setting to the max rating and potentially expose the conductor run to higher than it's ampacity.
Also I am fairly ignorant of costs here, so perhaps if possible, it could be way more expensive. Any insights here are greatly appreciated.

:?

 

[Smart $]

First, 500 kcmil at 380 amps is from the 75 deg C column. I'm using THWN-2 wire, which is 90 deg C rated. Shouldn't it be 430 A?

Second, having a conductor set that matches the 1000A breaker rating would look like Method A.
1000A / 0.58 = 1,724A
1,724A / 4 = 431 A

If I have to use the 75 deg C column, this would be (4) runs of 700 kcmil.
If I can use the 90 deg C column, (4) runs of 600 kcmil gets me there.

Third, is it acceptable to use an 1000A breaker with an adjustable trip setting and set the trip rating for 850A so I can use smaller wire?
I'd imagine (guessing) someone could come round and change the trip setting to the max rating and potentially expose the conductor run to higher than it's ampacity.
Also I am fairly ignorant of costs here, so perhaps if possible, it could be way more expensive. Any insights here are greatly appreciated.

:?

Apparently you are not aware of 110.14(C) requirements... terminal temperature limitations. For over 100A or 1 AWG, see 110.14(C)(1)(b) in particular. What it amounts to (in almost, but not all cases) is that your conductor adjusted and corrected ampacity cannot exceed the 75?C equivalent of that size wire.

 

[Smart $]

...

Third, is it acceptable to use an 1000A breaker with an adjustable trip setting and set the trip rating for 850A so I can use smaller wire?
I'd imagine (guessing) someone could come round and change the trip setting to the max rating and potentially expose the conductor run to higher than it's ampacity.
Also I am fairly ignorant of costs here, so perhaps if possible, it could be way more expensive. Any insights here are greatly appreciated.

:?

Also, we need some clarification on your circuit wiring, more specifically how you are "combining" the inverters' outputs. It seems that to run 18 inverters rated 27+kW each directly to a 1000A breaker will require the reverse of "feeder taps" somewhere. Otherwise there should be disconnecting means and overcurrent protection between the inverters and the 1000A breaker.

 

[BBL Solar Design]

As far as "rounding down" to nearest breaker size, my understanding is a standard breaker/fuse must be greater than or equal to FLA x 125%.
In my example of (18) inverters at 36A FLA each, a 648A continuous load x 125% = 810A. The standard OCP would not be allowed to be smaller than 810A (unless it was a 100% Duty Rated breaker or fuse). So next standard size up is 1000A. And adjustable trip breaker could be used to limit current at 850A or 900A, if the adjustment is restricted.

For adjustable trip breakers, 240.6(B) defines if adjustment setting is external, I have to use the max possible setting of the breaker, and 240.6(C) says if adjustment is "restricted access" (meaning behind a sealable cover, bolted door, or locked), I can use the adjusted current setting of the breaker to protect the wire.

 

[BBL Solar Design]

stil unclear...

stil unclear...

Also, we need some clarification on your circuit wiring, more specifically how you are "combining" the inverters' outputs. It seems that to run 18 inverters rated 27+kW each directly to a 1000A breaker will require the reverse of "feeder taps" somewhere. Otherwise there should be disconnecting means and overcurrent protection between the inverters and the 1000A breaker.

re: circuit wiring: each inverter will have OCP in a panel board. Multiple panel boards are used and combine in a switchboard, which will also be mounted on the roof in the sun. It's the output of the switchboard to the point of connection where I'm concerned and have my question:

240.4(c) says for devices rated over 800A, the ampacity of conductors must be must be equal to or greater than rating of OCPD.

My question is:
Does the DE-RATED value of my wire need to be greater than OCPD used? Or is consideration of wire de-rating used with respect to max continuous load value only?

ggunn said, "The sizing of the OCPD is set to open a current window wide enough for the maximum output of the inverter to reach the interconnect, but the protection is for current flowing the other way. The inverter is incapable of supplying enough current to endanger properly sized conductors."
...which makes me think Method B, where I take max continuous current of 648A / 0.58 = 1117A. Where de-rated wire value under heated conditions is sufficient, relative to actual current flow in the circuit. Using (3) sets of 500 kCMIL THWN-2 conductors x 380A (from 75 deg column) = 1140 > 1000A OCPD. The conductors will be exposed to 61 deg C conditions, so I derate from the 90 deg column: 430A x 3 x 0.58 = 748A, which is > continuous current of 648A. That makes sense to me...but something tells me it's wrong...

on the other hand bob said, "648A actual x 1.25 = 810A rated and 810A rated / 0.58 = 1396 A total. 1396/4 runs = 349 amps per ckt. The minimum size conductor is 500 kcm at 380 amps. However if you use a 1000 amp breaker, the conductor must match the breaker rating as David indicated."
...which indicates Method A is correct, where if I'm using an 1000A breaker, I need to pick a wire where the de-rated value is greater than my OCPD. So if I'm using 1000A breaker or fuse, I would need to use (4) sets of 600 kCMIL: 4 x 475A (from 90 deg col) x 0.58 = 1102A > 1000A OCPD.

The difference between these two methods and resulting wire sizes is HUGE, which is why I ask. I'm sill hoping to get clarity around this particular point.
I can imagine, based on responses I see, there could be grey area here. I WAS hoping it would be black and white, though I suppose if it were easy, I might forget it

Thanks for all the input.

 

[david luchini]

240.4(c) says for devices rated over 800A, the ampacity of conductors must be must be equal to or greater than rating of OCPD.

My question is:
Does the DE-RATED value of my wire need to be greater than OCPD used?

Yes, the de-rated value of the wire needs to be greater or equal to your 1000A OCPD. The "de-rated" value is the ampacity of the conductor for the given conditions.

Or is consideration of wire de-rating used with respect to max continuous load value only?

Wire de-rating is used with respect to the actual load current. The de-rated ampacity must have an ampacity sufficient for the load to be served. BUT, the OCPD must also protect the wire at its ampacity (for over 800A) or at the next standard size up (for 800A or less.) The SIZE of the conductor also cannot be smaller than the conductor that has an ampacity that is at least 125% of the continuous load plus 100% of the non-continuous load BEFORE any adjustment or correction factors are applied.

 

[BBL Solar Design]

Yes, the de-rated value of the wire needs to be greater or equal to your 1000A OCPD. The "de-rated" value is the ampacity of the conductor for the given conditions.

Wire de-rating is used with respect to the actual load current. The de-rated ampacity must have an ampacity sufficient for the load to be served. BUT, the OCPD must also protect the wire at its ampacity (for over 800A) or at the next standard size up (for 800A or less.) The SIZE of the conductor also cannot be smaller than the conductor that has an ampacity that is at least 125% of the continuous load plus 100% of the non-continuous load BEFORE any adjustment or correction factors are applied.

Thank you David, that was helpful.

 

[kwired]

"You are allowed to round up the breaker size in some cases. "
I'm confused why the Code allows us to do this, it seems by nature this allows OCPD to be higher than, and therefore not protect the wire amacity rating. This seems (to me) contrary to the intent of circuit protection. However this is slightly off my topic.

I am not by any means an expert when it comes to installing PV equipment, but there are different rules in certain areas related to PV also.

Rounding up to the next higher overcurrent device to me only makes sense that it may not be allowed on PV in some cases. You are dealing with a source and not a load here. The source will supply whatever demand is there regardless of overcurrent device setting, as long as it is capable anyway, whereas the general rules that otherwise allow you to go to next higher device are more load oriented rather than source oriented and assume the load is fixed and will not increase under most general circumstances. There are cases where load could possibly increase, but next size up overcurrent device is usually not allowed in those cases.

 

[ggunn]

I am not by any means an expert when it comes to installing PV equipment, but there are different rules in certain areas related to PV also.

Rounding up to the next higher overcurrent device to me only makes sense that it may not be allowed on PV in some cases. You are dealing with a source and not a load here. The source will supply whatever demand is there regardless of overcurrent device setting...

Not exactly. Grid tied PV inverters behave as current sources and the current delivered by them is not determined by demand. As long as they are connected to the grid they deliver whatever current they can produce from the sunlight striking their arrays up to their maximum rated output. The output conductors from an inverter are sized to handle 125% of the maximum rated current of that inverter, which means that the inverter is incapable of endangering them. Even if an inverter were capable of feeding a dead short (it isn't) the maximum available fault current from the inverter would still be less than the ampacity of the conductors.

 

[kwired]

Not exactly. Grid tied PV inverters behave as current sources and the current delivered by them is not determined by demand. As long as they are connected to the grid they deliver whatever current they can produce from the sunlight striking their arrays up to their maximum rated output. The output conductors from an inverter are sized to handle 125% of the maximum rated current of that inverter, which means that the inverter is incapable of endangering them. Even if an inverter were capable of feeding a dead short (it isn't) the maximum available fault current from the inverter would still be less than the ampacity of the conductors.

What is the rated current of an inverter? The amount it can continuously deliver without overheating, or the absolute maximum amount it can deliver period? Most other sources can deliver more current than their rated current but will result in increased heating, maybe will also start experiencing changes in voltage, frequency, etc. when overloaded.

Like I said, I don't know a lot about PV. I could also see the PV panel itself having a rating, yet being capable of delivering more than that rating, but with consequences.

 

[ggunn]

What is the rated current of an inverter? The amount it can continuously deliver without overheating, or the absolute maximum amount it can deliver period? Most other sources can deliver more current than their rated current but will result in increased heating, maybe will also start experiencing changes in voltage, frequency, etc. when overloaded.

Like I said, I don't know a lot about PV. I could also see the PV panel itself having a rating, yet being capable of delivering more than that rating, but with consequences.

It's the maximum amount it can deliver, period, at the lowest line voltage at which it can remain operational. It's a current source, so loading is not a factor, and voltage and frequency are determined by line conditions. Most other sources are voltage sources.

PV modules (panels) can indeed deliver more power than their rated output under certain conditions, but an inverter can only accept what it is rated for and it will clip the rest.

 

[bob]

648A actual x 1.25 = 810A rated and [/COLOR]810A rated / 0.58 = 1396 A total. 1396/4 runs = 349 amps per ckt. The minimum size conductor is 500 kcm at 380 amps. However if you use a 1000 amp breaker, the conductor must match the breaker rating as David indicated."
...which indicates Method A is correct, where if I'm using an 1000A breaker, I need to pick a wire where the de-rated value is greater than my OCPD. So if I'm using 1000A breaker or fuse, I would need to use (4) sets of 600 kCMIL: 4 x 475A (from 90 deg col) x 0.58 = 1102A > 1000A OCPD.

If you used the method I proposed, "648A actual x 1.25 = 810A rated and 810A rated / 0.58 = 1396 A total ampacity required. 1396/4 runs = 349 amps per ckt. The minimum size conductor is 500 kcm at 380 amps is adequate. " and compare it to this method "rated load = 810 amps. Using 90C rating 500 kcm 430 amps x .58 = 249.4 amps x 4 = 997.6 the results are the same as far as conductor size required. However if you use the 1000 amp breaker the 500 kcm is not correct. I think the 2nd calculation is the way it is usually done but would the 1st calculation be incorrect? Any comments?

 

[kwired]

It's the maximum amount it can deliver, period, at the lowest line voltage at which it can remain operational. It's a current source, so loading is not a factor, and voltage and frequency are determined by line conditions. Most other sources are voltage sources.

PV modules (panels) can indeed deliver more power than their rated output under certain conditions, but an inverter can only accept what it is rated for and it will clip the rest.

So when inverter experiences an overload or short circuit condition does it limit the current to it's max output or does it trigger some kind of overcurrent protection and shut down the output? Or is either scenario a possibility?

I am starting to see this is a source that doesn't exactly follow the same rules as other sources often do.

most of my inverter experience is with small units used to derive AC power from a 12 volt DC source, or with VFD's.