Finally was able to watch the week 4 video, it made a lot of sense, but now I some questions
Could you please provide example using this 2000W inverter with 12v battery string?
First of all, I would like to let students know, that the information here will definitely not be on the NABCEP Entry Level exam. This is a question that is way beyond the scope of an entry level PV class, in fact, I heard that there were no wire sizing questions on the last NABCEP PV Installation professional exam. This is probably because wire sizing has many rules that are not clear and top experts in the field have disagreements on exactly how to do wire sizing.
Wire sizing is the most difficult part of my advanced PV class. There are many different factors that we have to take into consideration at the same time, such as number of conductors in conduit, is the conduit in sunlight on a rooftop, what is the ASHRAE 2% high ambient temperature, what is the temperature rating of the conductor, what is the temperature rating of the terminals and what is the rating of the overcurrent protection device.
Let me just give you an overview explanation of how an average electrician might approach this calculation.
If the battery is 12V nominal, then lets say that we are turning off the inverter when the battery voltage gets down to 11.8V. Let's also say that the inverter efficiency is 95%.
To determine the current on the battery to inverter circuit.
2000W output / .95 = 2105W inverter input
2105W / 11.8V = 178A
(that is a lot of current and higher voltage would be better)
Then we have to go to the 301.15(B) tables in the NEC
Let's say that we just have a single conductor that is in free air not in sunlight and not in conduit at an ambient temperature not over 30C to make it simple.
See link for screenshot:
Table shows that if we used a 90C rated copper conductor, then we can carry 190A on a 2AWG conductor. We will still have to check the terminal temperature column here and let's say that we are using 75C rated terminals. In this case, a 1AWG 75C rated wire/terminal can carry 195A. We will choose a 1AWG conductor at this stage.
Note: we would still use the 90C rated column when derating for conditions of use if we had the conductor in a hot place or had 4 or more current carrying conductors in a conduit or a raceway.
We then will check our overcurrent protection device size by multiplying maximum circuit current as described in 690.8(A)(4) here
by 125% and rounding up to the next common overcurrent protection device.
178A x 1.25 = 223A
round up to 225A overcurrent protection device
Here are the common overcurrent protection device sizes:
Will a 225A overcurrent protection device protect a 1AWG conductor that is rated for 195A in the 75C column (terminals)?
We will then round up the 195A conductor to a theoretical overcurrent protection device of 200A and this check shows us that a 195A can be protected by a 200A overcurrent protection device, but not a 225A overcurrent protection device.
We then go back to table 310.15(B)(17) and try the next larger size of wire, which is a 1/0 AWG wire rated for 230A.
Then we can round up 230A to 250A theoretical overcurrent protection device and this checks out, because a 1/0 AWG wire can be protected by a 250A overcurrent protection device. Therefore a 225A overcurrent protection device will protect better.
In this case we can use a 1/0 AWG wire!
If the wire was going any distance, we would also want to check for voltage drop, just for efficiency.
Also, if there were other derating factors that we mentioned above, we would have to take the other factors into consideration. When we derate for heat, such as ambient temperatures, 4 or more current carrying conductors in conduit or sunlight on a rooftop, we would use the 90C column and we will not have to apply the extra 125% of current.
I highly recommend that you take the advanced PV course if you want to learn more about wire sizing!