I know there are a few technical readers out there who might be interested in this stuff.  For the rest of you, browse through some old posts, some of them are quite entertaining.

The PV system sizing is based on a few ground facts, as we used to call them.  First, the current usage of the structure.  In my case, about 8,000 kWh per year.  Second, the average annual sun hours per day for the location (about 5), and third, the shading, if any and system losses.  Therefore, to calculate the system size, the following method is used:

1. Annual kWh ÷ 365 days = kWh per day
2. Percentage of electricity to offset (decimal)
3. kWh per day ÷ sun hours (about 5 hours in the Hudson Valley)
4. Figure in losses (temperature loss 88%, system derate 95%, inverter 95.5%)

Therefore, my system looks like this:

1. 8000 kWh ÷ 365 days = 21.9 kWh per day.
2. I want to offset 100 percent, so 21.9 kWh × 1.0 = 21.9 kWh
3. I have an average of 5 sun hours per day, so 21.9 kWh ÷ 5 hours = 4.38 kW
4. Calculate system temperature loss, 4.38 kW ÷ 0.88 = 4.98 kW
5. Calculate system derate, 4.98 kW ÷ 0.95 = 5.24 kW
6. Calculate inverter loss, 5.93 kW ÷ 0.955 = 5.26 kW

Therefore, according to this, I would need a 5.26 KW DC rated PV system.  Our system is 4.1 KW DC, which is a little bit lower than required.  I am waiting to see how the micro inverters do with the solar panels.  I will bet they are more efficient than large string inverters and thus, we will get close to the desired number.

Next, things like breaker sizes, wire sizes, voltage drop, temperature de-rate, conduit fill and grounding need to be addressed.  First, there is a three line diagram that shows how the array is wired:

3 line diagram, 4,100 watt PV system using Enphase M-210 inverters

There are two 240 volt 15 amp branch circuits, each one is connected to 10 Enphase microverters.  The inverters are connected in parallel on these circuits.  Each inverter is in turn connected to a single 205 watt Sanyo Hip-205N PV panel.  Therefore, each inverter is capable of 205watts / 240 volts = 0.85 amps.  Maximum branch circuit current is then 10 inverters x 0.85 amps or 8.5 amps.  The NEC states that breakers should normally run at 75% of there rated value, so 8.5 amps x 1.25 = 10.63 amps.  Therefore a 15 amp circuit breaker is satisfactory.

Next, wiring sizing.  A fifteen amp breaker calls for #14 AWG wire.  This will not be satisfactory, however, to deal with the voltage drop between the solar panel array and the service entrance panel.  The distance between them is 124 feet.  Since we paid so much money for the solar panels, I want to keep the voltage drop to 1% or so.  This will ensure that all of the power we generated at the solar array gets into our electrical system and will not be dissipated as heat.  Here is the calculations for voltage drop:

Vdrop = (I x 2 x d) / (1000Ft/Kft) x r

(It is a little hard to write this formula out on one line)

Vdrop – volts lost
I – current
d – distance
r – resistance of wire per 1000 ft (from NEC 2008, table 8, conductor properties)

Therefore, using 14 gauge (stranded) wire:

Vdrop = (8.5 amps x 2 x 124 ft) /1000 x 3.14Ω = 6.62 volts.

In a 240 volt circuit, each leg is 120 volts, therefore 6.62 volts / 120 volts = 0.0551 or 5.51% voltage drop.  Too high for our purposes.

Using 8 gauge (stranded) wire:

Vdrop =(8.5 amps x 2 x 124ft) / 1000 x 0.778Ω = 1.64 volts.

1.64 volts / 120 volts = 1.3%

Therefore, #8 AWG wire is appropriate for this application.

Next, temperature derate.  The wire itself is #8 THHN which is rated for 90°C.  This will be well within our specs, especially since we already accounted for voltage drop, above.  The wire will be in conduit.  In this case, we are derating the conductor for the maximum temperature that conductor is expected to experience.  Since our maximum temperature around here is about 100°F, according to NEC table 310.16, #8 AWG copper wire has a current caring capacity of 55 amps x 0.91 or 50 amps.  This is well above our maximum current of 10.63 amps so that is good.  This step is more critical on roof top installations where ambient temperatures can be very high.

Next is conduit fill.  One can’t just stuff as many wires as one can fit into a conduit.  Generally speaking, the NEC seems to expect about 50% conductor fill in any given conduit.  Thankfully, there are tables that give out this information.  For my purposes, I used 1 1/4 inch schedule 40 PVC conduit.  According to Table C.10 up to seven #8 AWG conductors can be placed in that conduit.  I have two 240 volt branch circuits, including neutrals and a ground wire.  That totals seven conductors.

Finally, grounding.  All non-current carrying metal parts, frames, etc. must be grounded. Fortunately, the 2008 NEC allows us to size our grounding conductor to the size of the over current device (circuit breaker).  In this cast that would be 15 amps, therefore 14 gauge.  Unfortunately, the Jurisdiction Having Authority (JHA) has not adopted the 2008 NEC yet, they are on 2002 which requires the grounding wire to be the same size as the current carrying wire, regardless of the sizing of the wire for voltage drop.  So, #8 AWG ground wire between the array and the service entrance panel is required.  All metal frames and mounting rails of the PV array need to be connected to this ground.  Any ground wire that is not protected e.g. run in conduit needs to be a #6 or bigger conductor.  Also, a separate grounding electrode needs to be installed at the PV array since it is not a part of the existing structure.

Instead of running a continuous ground wire to each module, inverter and mounting frame member, I was allowed to use a WEEB (Washer, Electrical Equipment Bond) type ground.  These little clips go between the modules and the frame, making the mounting frames the grounding conductor.  It saves time.

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