Did the ampacity rules change in the 2020 NEC?

[ggunn]

For as long as I have been designing solar (~14 years) I have been using the John Wiles method for selecting 90 degree AC conductors, which is to pick a conductor that will a) carry 125% of the maximum inverter current at 75 degrees AND b) carry 100% of the inverter current after derating for conditions of use (ambient temperature and conduit fill) at 90 degrees. Calculation a) is to protect 75 degree terminals under continuous use and calculation b) is for 90 degree wiring insulation preservation.

I am reviewing a design done by someone else which shows AC conductors one size smaller than I would have chosen, and it is done under the 2020 NEC, so to exercise due diligence I have gone to 690.8(B), and I do not see anything about protecting 75 degree terminals. Did the rules change? Is everything I know wrong?

 

[ggunn]

I have been doing this so long the same way that I have forgotten what is going on behind the curtain. After consulting with another PE we have concluded that 110.14(C) is also in play along with the continuous use considerations, so my conductors are correct.

And yes, I know that I should be multiplying the 75 degree ampacity by 0.8 rather than Imax by 1.25, but the math works out to be the same. So sue me.

 

[pv_n00b]

Nothing has changed. The conductor sizing information was copied into 690 from other sections to remind PV installers how to do it but the basic conductor sizing requirements from the other sections still apply. The terminal temperature check never made it into 690.8 but it's still required. Just as a reminder for folks, there are three criteria for sizing conductors to the amperage of the circuit in PV systems, conductor size will be the largest conductor found:

1. 125% of the maximum current for the conductor with no conditions of use taken into account. Look up the base conductor ampacity using the conductor's rated insulation temperature, usually 90dC for PV, and make it equal to or greater than 125% of the max current.
2. 100% of the maximum current for the conductor with conditions of use taken into account. Look up the base conductor ampacity using the conductor's rated insulation temperature, usually 90dC for PV, correct for conditions of use, and make it equal to or greater than 100% of the max current.
3. 125% of the maximum current for the conductor with no conditions of use taken into account but using the terminal temperature rating in place of the conductor insulation rating. Look up the base conductor ampacity as if the conductor's rated insulation temperature was the terminal temperature rating of the termination, usually 75dC for PV, and make it equal to or greater than 125% of the max current.

Pick the worst case. In my experience #1 almost never give the largest conductor of the three. Usually, #2 is the largest, sometimes #3.
Then after all this, increase the conductor size to take voltage drop into account. This often requires a larger conductor than the bare minimum found above.

 

[pv_n00b]

I remember when I first started out I would do 125% with conditions of use corrections and way oversize conductors. I was not checking for terminal temperature, but since I was oversizing the conductor it was not an issue. I still see people doing this today.

 

[Designer101]

Nothing has changed. The conductor sizing information was copied into 690 from other sections to remind PV installers how to do it but the basic conductor sizing requirements from the other sections still apply. The terminal temperature check never made it into 690.8 but it's still required. Just as a reminder for folks, there are three criteria for sizing conductors to the amperage of the circuit in PV systems, conductor size will be the largest conductor found:

1. 125% of the maximum current for the conductor with no conditions of use taken into account. Look up the base conductor ampacity using the conductor's rated insulation temperature, usually 90dC for PV, and make it equal to or greater than 125% of the max current.
2. 100% of the maximum current for the conductor with conditions of use taken into account. Look up the base conductor ampacity using the conductor's rated insulation temperature, usually 90dC for PV, correct for conditions of use, and make it equal to or greater than 100% of the max current.
3. 125% of the maximum current for the conductor with no conditions of use taken into account but using the terminal temperature rating in place of the conductor insulation rating. Look up the base conductor ampacity as if the conductor's rated insulation temperature was the terminal temperature rating of the termination, usually 75dC for PV, and make it equal to or greater than 125% of the max current.

Pick the worst case. In my experience #1 almost never give the largest conductor of the three. Usually, #2 is the largest, sometimes #3.
Then after all this, increase the conductor size to take voltage drop into account. This often requires a larger conductor than the bare minimum found above.

means example

one 11.4 kw inverter
max out put current 47.5A
OCPD WILL BE 60A
but wire size (100 percent if used) second case
for 2 hot one neutral and one ground

using 90 c column
#8 AWG
55x0.87 ( cou)= 47.85A
47.85A> 47.5A (100 PERCENT) that means #8 wires would be enough for 11.4 kw inverter??

IF WE DO THE SAME CALCS BY 125 PERCENT first case
47.5AX1.25=59.81 A

90 C COLUM FOR #8 WIRES IS RATED ONLY FOR 55 Amps ( NO CONDITIONS OF USE)
which doesn't support for # 8 wires so have to se #6

Which is correct
??

 

[Carultch]

means example

one 11.4 kw inverter
max out put current 47.5A
OCPD WILL BE 60A
but wire size (100 percent if used) second case
for 2 hot one neutral and one ground

using 90 c column
#8 AWG
55x0.87 ( cou)= 47.85A
47.85A> 47.5A (100 PERCENT) that means #8 wires would be enough for 11.4 kw inverter??

IF WE DO THE SAME CALCS BY 125 PERCENT first case
47.5AX1.25=59.81 A

90 C COLUM FOR #8 WIRES IS RATED ONLY FOR 55 Amps ( NO CONDITIONS OF USE)
which doesn't support for # 8 wires so have to se #6

Which is correct
??

#6 Cu is correct in this case.

47.85A * 1.25 governs the OCPD being 60A, and governs the minimum wire termination rating prior to applying conditions of use factors (formerly known as derate factors). Since the industry norm is that terminations are rated for 75C, this means you need #6 Cu wire to meet this criteria, which has a 75C termination rating of 65A. You do have a burden of proof to use the 75C rating for equipment 100A and less, although the vast majority of equipment does meet this burden of proof. The default if not listed otherwise is that 100A and less is rated 60C, which is a rare case today.

Once we establish that #6 Cu is required for termination ampacity, we check whether the wire temperature rating is good for the conditions of use. 75A * 0.87 = 65.25A, which is sufficient for a 47.85A load. You don't need a 1.25 factor in this part of the calculation.

A further check that is required is that the OCPD where required, needs to protect the wire according to article 240. The section 240.4(B) is commonly a section that helps you, as it means your wire and termination ampacities need to "round up" to the OCPD you are using. So this means you'd need at least 50.01A worth of wire ampacity to use on a 60A breaker. 50A on the dot is insufficient. Both the termination rating at 75C and the derated 90C ampacity need to exceed the previous standard OCPD rating, to qualify for 240.4(B). It is counter-intuitive to me that we are allowed to use this rule, and I don't understand the physical reason for it, but it is something that can work in our favor.

Once your circuit is over 800A of OCPD, this rule no longer applies. Read 240.4(B) in its entirety to see this. Because it specifically refers to standard sizes, by omission, it also doesn't apply if you use exotic ratings of fuses and breakers, like a 63A breaker (common in rest of the world, maybe something you inherit as a built-in breaker in your equipment) or a 330A fuse (maybe your equipment comes with this exotic size for a reason not entirely clear), as these non-standard OCPD ratings would require the wire and termination ampacities to both meet or exceed the trip rating.

 

[kwired]

I have been doing this so long the same way that I have forgotten what is going on behind the curtain. After consulting with another PE we have concluded that 110.14(C) is also in play along with the continuous use considerations, so my conductors are correct.

And yes, I know that I should be multiplying the 75 degree ampacity by 0.8 rather than Imax by 1.25, but the math works out to be the same. So sue me.

Multiplying Imax by 1.25 saves time. you calculating minimum ampacity needed then look at 310.16 to find a conductor that is same or higher and that is what you need.

If multiplying the conductor ampacity by .8 results in too small of a conductor, then you need to do it all over again with another conductor size.

Can do same thing when ambient temp or adjustments are needed, - figure out minimum ampacity or derate a conductor only to find out it isn't large enough, then start process all over with a larger conductor.

Sure after doing this enough times you maybe already know what to select off top of your head for many commonly used sizes and situations.