diy solar

diy solar

10.6KW bifacial off grid ground array powering multiple locations 500 feet away

Alright, today is the day I spend money on solar panels, inverter, and two charge controllers and 38KW of lithium. Hopefully I've learned enough from this board not to be making a huge mistake somewhere.

If interested in taking a peak at what I've decided and warning me of any iceberg I'm about to hit, please see this thread!

 
I'm reading with interest. I will eventually be doing something similar, to 3 new standing cabins, with shorter runs. 1 this year and the other 2 next year. Good luck Noob. I hope some take a peak and help you out.
 
I think you're letting the tail wag the dog. You might have a great location for all the gear, but the cheapest way to get the power 500 feet without a lot of buried welding cable is to connect your panels in longer series strings and run higher voltage DC to the cabin. You could buy a smallish container (or build a shed), hyper-insulate it, and heat it electrically to keep your gear at optimal temperatures and out of the house--nice for reducing the risk of fire. 10AWG wire would be fine for 600VDC at 8-10 amps over even 1000 feet of transmission. Probably $2K worth of wire. I'd worry about the yurt later--the needs of visitor structures are probably a fraction of what you need in the cabin. Probably a single run of 10AWG to the yurt powering much smaller equipment--an mppt controller, batteries and maybe a cheap inverter.

And since you're using bifacial panels you can boost your output a bit by simply adding reflectors. You can fiddle with reflector sun angle to compensate as much as possible for your lattitude.
 
So what's the "acceptable" power loss at these long runs? 5%? More? Because when sizing the wire you have to account that the panels don't run at 100% most of the time. So it's IMHO acceptable to have loss at full power and it's better to go for the middle and size the panel for let's say 2/3 or 4/3 at best.
Am I correct?
 
So what's the "acceptable" power loss at these long runs? 5%? More? Because when sizing the wire you have to account that the panels don't run at 100% most of the time. So it's IMHO acceptable to have loss at full power and it's better to go for the middle and size the panel for let's say 2/3 or 4/3 at best.
Am I correct?
This seems like gibberish to me? What is "2/3 or 4/3 at best" supposed to mean? Panels not running at 100% is referring to amperage output, not voltage. So, you can usually count on the amps being lower then expected, not the volts.

You commonly use 2% loss as your guideline, though the solar police are not going to show up at your door if your system has 3% loss. You can use this voltage drop calculator to determine your actual losses.

Remember though to carefully evaluate your total Voc with the coldest overnight temp your area has ever seen (50 to 100year low), not just your average low. I would keep my predicted max Voc at least 10% less than your unit's max Voc, just to sleep better at night.
 
Just to update the thread since there has been some activity, we are running the 500 foot distance with AC power using a 250kcml-250kcml-3/0 triplex aluminum wire. Will direct bury it. Voltage drop should be very small as a result.
 
Why not?

I have a PV array, and wouldn't I want a short run from it to the SCC and the Multiplus II and the battery bank? Where the Multiplus could then convert it to AC power and deliver it the long distances to the various structures that need the power?
Exactly
 
This seems like gibberish to me?
The correlation between energy loss and amps is clear. The higher the amps the higher the voltage drop. You can even check it in the calculator you've posted. But panels rarely run at rated power (at least in here). Most of the time they run at 50-80%. Hence why you IMHO don't need to size the wires for peak power but you can make them smaller (I mean to an extent, go too small and the insulation will melt).
 
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