All, Good afternoon. Please how are those doing AF studies for all the new EV charger installs modeling lithium ion battery energy? There doesn't seem to be good data/model equations for these. The mnfr data on these UL listed charger equipment are not forthcoming much with engineering data to do some first principles calcs? Please does NFPA 70E not cover such equipment? Talking to installers they have to commission these energized with 480VAC/3ph on the input side and 1000VDC on the output side. Appreciate any feedback. Thank you much.

Normal arc flash calculations and PPE on the AC side based on the source. The charger can only supply charging current. A connected battery can supply short circuit current according to its resistance, different from vehicle to vehicle. I would suggest not doing energized work when charging.

So the following is what I used for an Investigation to determine the Arc Flash Incident Energy w/n a Lithium-Ion Battery Bank for NASA:

Doan Method (Ref: NFPA 70E Informative Annex D), for Calculation Arc Power and Incident Energy:
Where, R1 (RBattery) = 2.67 mΩ; R2 (Rconductor)= 50 uΩ; R3 (Rarc) = TBD; Battery = 53.8 Vdc
Using the “Complete Guide to Arc Flash Hazard Calculation Studies,” by Jim Phillips; Ch.12 – “DC Arc Flash Calculations,” based on Daniel R. Doan’s Technical Paper, “Arc Flash Calculations for Exposures to DC Systems,” the following was determined:
Idc bolted = Vdc/Rdc, where Rdc = RBattery + Rconductor, and Rconductor was determined by the volume of the Cu Buss plate involved in this incident.
= 53.8V/ (2.67 mΩ +50 uΩ) = 53.8V/ (2.72 mΩ)
= 19.781X103 A
Then Idc arc = .5* Idc bolted, so:
Idc arc = .5*19.781X103 A = 9.89 X103 A
Calculating for DC Arc Resistance (Rarc):
1) G*.534, where G = gap in mm. So for a ¼” gap or G=6.35mm
G*6.35mm = 3.391mm
2) Step 1) + 20 = 23.391
3) Idc arc.88 = (9.89 X103).88 = 3,279.3
4) Rarc = Step2) / Step3) = 23.391/3,279.3 = 7.1328 mΩ
Then re-solving for Idc arc:
Idc arc = Vdc /(Rdc + Rarc) = 53.8V/ (2.72 mΩ + 7.1328 mΩ)
= 5.4604 X103 A
Note: Both Rarc & Idc arc, were re-solved for 5 more iterations, until these values did not change significantly from the previous values and converged to the final answers of:
Idc arc = 2.149 X103 A & 27.331 mΩ
Using the calculated Rarc & Idc arc, the Power and Energy in the Arc was determined as follows:
Parc = Idc arc2 * Rarc
= (2.149X103 )2 *(27.331X10-3)
= 126.22KW
Earc = Parc*tarc, again assuming tarc = .1s, then:
= (126.22KW)* (.1) = 12.622KJ
Incident Energy is determined by-
Ei = Earc/(4∏D2)
Ei, whereby D = 1”or 25.4mm, then: 12.622KJ /(4*∏*25.42) = 1.5577 J/mm2
For calories/cm2; take J/mm2*(23.39), then:

As you'll notice, I referenced Chapter 12, of Jim's “Complete Guide to Arc Flash Hazard Calculation Studies.” So, you may want to consider purchasing said doc if you don't have it already. I've realized it's well worth the price vs. beating your head against a wall trying to calculate such, especially w/ DC systems. Yes, I realize the math models (Arc flash S/W packages) are now realizing this arena, But....

So we must determine the volume of the CU buss plate for every make and model of electric vehicle on the market now and in the future?

Appreciate that feedback. LOL I think I actually did something similar utilizing the arcing resistance and iterative solver, appreciate good to know that first principles still apply. Will look into the book for additional reference to list in Bibliography. Thanks again.