PaulEngr wrote:

There are a lot of variables by the way. Just the basic radiative view factors turns into a "radiosity" problem. To get a flavor for the math, google "ray tracing" and "radiosity" until you can wrap your head around what's going on because radiation is the same whether it's thermal or visible light.

This will be what I dig into next. While I like the idea of using open source software to calculate an arc-in-a-box multiplying factor, I would feel more confident in my calculation if I could show the calculation done by hand. That would also help me document my calculations in the future, saying that I used software I found online can raise flags when my calculation is reviewed.

Paul, do you know of a paper or calculation that shows a PV DC arc flash calculation from start to finish using the Ammerman and Wilkins models?

This is my preliminary calculation methodology after reading your feedback, the Ammerman paper, the Wilkins paper and going through the example Jim Phillips explained in his "Know Your Arc: DC arc flash calculations" post:

**1. Collect module data:**--- V_MaxPower: I figured V_MaxPower would be more realistic than using V_OpenCircuit since current flow will not be zero.

--- I_ShortCircuit: This should be more conservative than I_MaxPower since the load has been bypassed with a lower resistance arc.

**2. Find fault values:**--- I_BoltedFault = TotalStringsInParallel x I_ShortCircuit

--- I feel like using a constant 125% multiplier here to account for higher Solar Insolation days is overkill. Looking at Solar Insolation graphs, 1060 w/m^2 is that maximum value for the Northern Hemisphere and that's at 12 noon on a bright clear summer solstice day.

--- I saw some calculations that used another 125% multiplier to comply with NEC's requirement for sizing output circuit conductors and OCPD for continuous loads. I don't see how that justifies the multiplier.

**3. Draw equivalent circuit**--- Circuit will consist of V_System, R_System, R_Arc in series.

--- V_System: Use V_MaxPower and the number of series modules in a string to find V_System (V_MaxPower x ModulesInSeries)

--- R_System = V_System/I_BoltedFault. I could use some feedback here. I opted to ignore cables to keep the calculation simpler. I see this as being more conservative as well.

--- In the examples I went through, I usually end up with an R_system of 0.3 to 0.5 ohms.

**4. Apply Stokes/Oppenlander equations (Ammerman Model)**--- Use I_arc = 50% of I_BoltedFault for the first iteration.

--- Obtain gap distance between positive and negative terminal in working location (in mm).

--- Find R_arc = (20+0.534*Zg)/(I_arc^0.88)

--- Use R_arc back in equivalent circuit and calculate new I_arc for next iteration I_arc = V_system/(R_system + R_arc)

--- Find R_arc for second iteration and so on until I_arc no longer changes significantly.

--- For the few examples that I have done, I ended up with I_arc = 0.75 * I_bolted

--- Find V_arc = (20 + 0.534*Zg)*I_arc^0.12 or simply I_arc * R_arc

**5. Calculate DC arc power**--- P_arc = I_arc * V_arc (watts)

**6. Calculate DC arc energy**--- E_arc = P_arc * FCT (Joules)

--- I am considering either using a 2 second fixed FCT value or looking up fuse TCCs.

--- I have yet to come across an example where the fuse burns in less than 2 seconds. Any experience here?

**7. Convert energy to calories:**--- E_arc / 4.18

**8. Find energy distributed across surface area of sphere: **--- E_sphere = E_arc/(4*PI*D^2)

--- D is typically 18" (45.7 cm )

**9. Find arc-in-a-box multiplying factor:**--- Still not sure here.

--- One option is to use the Wilkins equation with the 'k' and 'a' of a panel (per IEEE equipment types).

--- To me, a combiner box is much closer to a panel than LV switchgear because the depth is only 8" or so.

--- Looking at the Wilkins k values for IEEE equipment types, k seems to scale linearly with depth.

--- That would result in a MF of 1.5 from sphere to box.

--- Or, look into a better way to account for the box effect.

I'm still making the very conservative assumption that 100% of the electrical energy released will be heat energy but the final values are significantly lower than using the Maximum Power Transfer theorem and the 300% multiplier.

Any thoughts?