Another New Guy wiring question

srhillsrhill Registered Users Posts: 8
Hi,
What I would like to do is: run a DC line from a ground mount pv array(8.5kw) 200 feet to my garage, install the SB8000 inverter there, then run AC 200 feet to the house.
My problem is what do I do about combining and fusing the 3 strings at the array since the SMA DC disconnect / fuse box is intended to mount on the inverter and not as a stand alone unit ? Am I going to need to locate the inverter (& DC disconnect) waaaay out at the array ? Or is there another low cost option...
thanks so much !

Comments

  • ggunnggunn Solar Expert Posts: 1,973 ✭✭✭
    Re: Another New Guy wiring question
    srhill wrote: »
    Hi,
    What I would like to do is: run a DC line from a ground mount pv array(8.5kw) 200 feet to my garage, install the SB8000 inverter there, then run AC 200 feet to the house.
    My problem is what do I do about combining and fusing the 3 strings at the array since the SMA DC disconnect / fuse box is intended to mount on the inverter and not as a stand alone unit ? Am I going to need to locate the inverter (& DC disconnect) waaaay out at the array ? Or is there another low cost option...
    thanks so much !
    You can bypass the SB's integrated combiner with a single input, but you'll need to get a combiner box to put out at the array to combine the strings onto one set of conductors, otherwise you'll need to run the home runs from the strings all the way to the inverter. You'll want to look at the wiring cost both ways to decide which is more cost effective; when you combine the strings the conductors will have 3X the current at the same voltage so be sure to look at voltage drop when you size them. Also. look at voltage drop on the AC side; it could be more cost effective to run the DC all the way to the house and put the inverter there, or as you suggest, put the inverter out at the array. Consider all your alternatives; copper is expensive.
  • prasetopraseto Registered Users Posts: 22
    Re: Another New Guy wiring question

    200 ft is a lot of $ in copper. Why you reconsider mounting the inverter on a side of the house closer to the array. The SB8000 is suitable for outdoor use.
  • solarvicsolarvic Solar Expert Posts: 1,060 ✭✭✭✭
    Re: Another New Guy wiring question

    Is it 200 ft. to garage and another 200 ft to garage meaning your house is 400 ft from panels? Or is ithe house 200 ft. from panels? I was going to install an inverter at the garage as you proposed. It was closer to my house from the panels to the house as my house was between the panels and garage. So I put the inverter in the house in my utility room. That way I only needed a 250 ft dc run from my panels to inverter and about 12 ft of 240 vac to service box from inverter. Like fellow in post # 2 recomended you need combiner box at panels. On mine I used the safetouch fuse system in the combiner box and run positive wire out of combiner box to 600 vdc shutoff at panels. My inverter is a fronius and I never hear it running. If I would have run to garage I would have had about a 400 ft run of dc wire and aproximatly 100 ft of 240 ac wireing. Solarvic
  • srhillsrhill Registered Users Posts: 8
    Re: Another New Guy wiring question

    Thanks everyone for the advise. To clarify things, the distance from my house to the array is 400 feet with the garage right in the middle (@200 feet)
  • ggunnggunn Solar Expert Posts: 1,973 ✭✭✭
    Re: Another New Guy wiring question
    srhill wrote: »
    Thanks everyone for the advise. To clarify things, the distance from my house to the array is 400 feet with the garage right in the middle (@200 feet)
    That's the way I was reading it. 400' is a long way, and some power loss to voltage drop is inescapable. In general, it usually makes sense to make your highest voltage run the longest one. If, for example, your PV array voltage were 500VDC and your inverter output voltage were 120VAC, the way to go would probably be to run the DC all the way to the house and make the AC run as short as possible. From there it's a balancing act between the cost of the copper and the power loss in the conductors. P = VI = (I^2)(R) = (V^2)/(R); get R from Tables 8 and 9 in the NEC, and remember that the current makes a loop, so if the array to the house is 400' it's really 800' of wire in your calculations.
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