Okay, let's break down this situation and clarify the NEC requirements. You're on the right track, but the PE's comment highlights a crucial point of interpretation and application of NEC 215.2(A)(1).

Understanding the PE's Concern

The PE's comment is essentially saying this: After all derating factors are applied, the final ampacity of your conductor must still be equal to or greater than 125% of the inverter output current. They're not wrong, but the way they stated it might be slightly misleading. It's not just that the derated value needs to be higher than 360.8A, but that the entire conductor ampacity calculation must yield a final ampacity equal to or greater than 360.8A.Code Sections at Play NEC 215.2(A)(1): Feeders - Minimum Size and Rating

This is the heart of the matter. It states: "(a) Feeder Conductors. Feeder conductors shall have an ampacity not less than the sum of the noncontinuous load plus 125 percent of the continuous load." "(b) Conductor Ampacity before Adjustment or Correction. The ampacity of conductors before the application of any adjustment or correction factors shall not be less than the noncontinuous load plus 125 percent of the continuous load."This is where the confusion lies. Both (a) and (b) need to be satisfied.

NEC 690.8(A): Circuit Sizing and Current

This section, specifically 690.8(A)(1) applies to PV source and output circuits. It generally requires that the maximum PV system current be multiplied by 125%. While this section doesn't directly apply to your feeder (which is after the inverters and is considered a load), it's the source of that continuous load, and informs how you must size it.

NEC 310.15: Ampacities of Conductors

This section deals with the ampacity of conductors based on insulation type and temperature ratings, and also covers adjustment factors (like those for ambient temperature and number of conductors in a raceway).

Where Your Calculation Needs Adjustment

Your calculation shows that the derated ampacity of the #250 kcmil Al conductors is 334.9A. This is less than the 360.8A required by NEC 215.2(A)(1). You have two options:1. Increase Conductor Size

As the PE suggested, increasing the conductor size to #4/0 Cu will likely resolve the issue. With 4/0 CU at 90C being 260A and parallel you would have 520A. Using the same deratings, 520A x 0.91 x 0.8 = 378.56A which is more than 360.8A.
2.

Reduce Derating Factors

If possible, modify the installation to reduce the temperature derating or the number of conductors in the raceway. This might involve moving the raceway to a cooler location, increasing the size of the conduit, or separating the conductors. This is often impractical or impossible.

Here's a revised approach to the calculation to be sure you cover everything1. Calculate Inverter Output Current

You correctly identified this as 288.6A (2 x 144.3A).
2.

Apply 125% Factor for Continuous Load

288.6A 1.25 = 360.8A. This is your minimum required ampacity for the feeder

after all derating factors are applied.3. Determine Conductor Ampacity Before Derating

You need to select a conductor size whose ampacity before applying any derating factors is high enough so that after you apply the derating factors, the final ampacity is at least 360.8A.
4.

Apply Derating Factors



Ambient Temperature Correction

You've used a derating factor of 0.91. Verify that this is the correct value for the actual maximum ambient temperature expected at the location of the feeder. Consult NEC Table 310.15(B)(2)(a).

Number of Conductors in Raceway

You've used a derating factor of 0.8 for 6 current-carrying conductors. Verify that this is correct. NEC Table 310.15(C)(1) provides the adjustment factors for the number of current-carrying conductors in a raceway. Remember that neutral conductors in a 3-phase system are only counted as current-carrying if they carry significant harmonic currents (NEC 310.15(C)(1) Note 5).5.

Calculate Final Ampacity

Multiply the conductor ampacity (from step 3) by all applicable derating factors. The result must be greater than or equal to 360.8A.
6.

Select Overcurrent Protection

The overcurrent protection device (breaker) must be sized according to NEC 240.4 and other relevant sections. You've correctly chosen a 400A breaker, which is suitable for a 288.6A continuous load. Ensure the breaker's interrupting rating is adequate for the available fault current.

In Summary

The issue isn't about checking the load at one point in the middle of the run. It is about following 215.2.A.
1.a and b in that order. You calculated the ampacity and breaker correctly at the point of use where you consider the continuous load. But the second half of 215.2.A.1 requires that the final ampacity after deration is more than the continuous load ampacity required by 215.
2.A.1.a, in this case 360.8A.Think of it this way: The derated ampacity represents the actual, usable current-carrying capacity of the conductor in the specific conditions where it is installed. That usable capacity must be sufficient to handle the 125% continuous load.You were close, and your thoroughness in questioning the PE's comment is commendable. Always double-check and clarify any concerns about code interpretation. Flag for review