Vines41
Member
- Location
- Silicon Valley CA
- Occupation
- PV Designer
Hi all, I have been installing and designing circuits for a few years but a recent comment by the stamping PE confused me.
A few years ago, I passed my test and got my Jcard for California, so I have some knowledge. I am willing to learn more and be wrong. If I am missing something, I want to be sure to know exactly what and follow the code to understand the issue. I also am the answer man when teammates come with hard questions, so if I have been wrong I have to own that too.
In this case, I am sizing a feeder for 2x Solar Edge 120K inverters landing in a 400A subpanel. That subpanel feeds a 400A Disconnect and on to 2000A 480Y/277VAC service. The feeder from the 400A rooftop subpanel to the point of interconnection is the question of the day. I thought I knew how to size this, but a recent comment by the stamping PE made me revisit my textbooks and this forum for some advice. Here is my feeder calculation:
AC2.2 CONDUCTORS FROM 400A SUBPANEL TO AC DISCONNECT
SOLAR EDGE S120KUS MAX OUTPUT CURRENT = 144.3A COMBINE (2) IN 400A SUBPANEL = 288.6A
MINIMUM CIRCUIT BREAKER = 288.6A X 1.25 = 360.8A -->USE 400A BREAKER AND 2 PARALLEL 250 KCMIL AL = 410A --> OK
2 PARALLEL #250 kcmil AL THWN-2 90°C AMPACITY PER NEC 310.16 = 460A
TEMP RANGE CALCULATION: USE DERATING FACTOR OF 0.91 (OUTDOOR)
(6) MAX CURRENT CARRYING CONDUCTORS IN CONDUIT: 0.8 FILL FACTOR DERATING
THEREFORE 2 PARALLEL #250 KCMIL AL AMPACITY IS (460A X 0.91 X 0.8)=334.9A
OK -->(MORE THAN 288.6A PV OUTPUT)
The part in bold got the following quoted comment from the PE team: "Hello, this derated value needs to be equal to or higher than the inverter output current x 1.25 (360.8A). Suggest changing to #4/0 Cu"
I hope to understand where I went wrong, or that they are wrong, in the opinion of smart folks here. Please help with any relevant code sections.
I had thought this feeder calculation was controlled by 215.2.A.1.a and b. In my mind, part 215.2.A.1.b is the part in bold and indicates the check that happened in the middle of the run, where fill deration and temperature deration apply, but the load was only checked at 100% of the maximum load, so 288.6A in this case.
The first 3 lines of the calc are intended to calculate the part in 215.2.A.1.a, where we are checking the ampacity at the equipment where the wire is landed. In part .a we are required to use 125% of the continuous load plus 100% of the non-continuous load. So since there are just 2 inverters in the panel, the continuous load * 1.25 is 360.8A. This is taken without temperature or fill deration factors.
A few years ago, I passed my test and got my Jcard for California, so I have some knowledge. I am willing to learn more and be wrong. If I am missing something, I want to be sure to know exactly what and follow the code to understand the issue. I also am the answer man when teammates come with hard questions, so if I have been wrong I have to own that too.
In this case, I am sizing a feeder for 2x Solar Edge 120K inverters landing in a 400A subpanel. That subpanel feeds a 400A Disconnect and on to 2000A 480Y/277VAC service. The feeder from the 400A rooftop subpanel to the point of interconnection is the question of the day. I thought I knew how to size this, but a recent comment by the stamping PE made me revisit my textbooks and this forum for some advice. Here is my feeder calculation:
AC2.2 CONDUCTORS FROM 400A SUBPANEL TO AC DISCONNECT
SOLAR EDGE S120KUS MAX OUTPUT CURRENT = 144.3A COMBINE (2) IN 400A SUBPANEL = 288.6A
MINIMUM CIRCUIT BREAKER = 288.6A X 1.25 = 360.8A -->USE 400A BREAKER AND 2 PARALLEL 250 KCMIL AL = 410A --> OK
2 PARALLEL #250 kcmil AL THWN-2 90°C AMPACITY PER NEC 310.16 = 460A
TEMP RANGE CALCULATION: USE DERATING FACTOR OF 0.91 (OUTDOOR)
(6) MAX CURRENT CARRYING CONDUCTORS IN CONDUIT: 0.8 FILL FACTOR DERATING
THEREFORE 2 PARALLEL #250 KCMIL AL AMPACITY IS (460A X 0.91 X 0.8)=334.9A
OK -->(MORE THAN 288.6A PV OUTPUT)
The part in bold got the following quoted comment from the PE team: "Hello, this derated value needs to be equal to or higher than the inverter output current x 1.25 (360.8A). Suggest changing to #4/0 Cu"
I hope to understand where I went wrong, or that they are wrong, in the opinion of smart folks here. Please help with any relevant code sections.
I had thought this feeder calculation was controlled by 215.2.A.1.a and b. In my mind, part 215.2.A.1.b is the part in bold and indicates the check that happened in the middle of the run, where fill deration and temperature deration apply, but the load was only checked at 100% of the maximum load, so 288.6A in this case.
The first 3 lines of the calc are intended to calculate the part in 215.2.A.1.a, where we are checking the ampacity at the equipment where the wire is landed. In part .a we are required to use 125% of the continuous load plus 100% of the non-continuous load. So since there are just 2 inverters in the panel, the continuous load * 1.25 is 360.8A. This is taken without temperature or fill deration factors.