1000' run from good solar site to house/shop - Questions

[timselectric]

Looking at either SOK 48v 100Ah Racks, or I have a connection with KiloVault. Just getting started looking at their solutions. Either way, they will be LiFePo4.

All of Northern California == High Fire Hazard.

Oh
Keeping fire out.
That makes sense, now. lol

 

[timselectric]

My power room is the size of a walk in closet, attached to the room I sleep in. My equipment enjoys the same (goldilocks) environment that I do.

 

[jameshowison]

For sure. When I say "shed", I mean it only as an outbuilding. It will most likely be cinderblock (high fire hazard area) and clean inside. Only use will be for these electronic components.

You mentioned 115F top temps. You might want to check the likely temps inside your shed, I think that Victron derates aggressively from around 100F (other brands might not but they likely suffer for it long term).

Perhaps cross ventilation and sufficient insulation to knock back solar gain will keep it all good, but something to consider.

Perhaps the Aussies in the thread have some guidance, maybe not an issue? It was for me in an RV because cross ventilation wasn’t possible (security and humidity issues). So I set the AC at 90F and it’s all good.

 

[schmism]

another HI voltage DC vote. 600v DC is a thing

 

[Pyrofx]

My run is 190 feet one way. I went 480 volt dc to the power house. This allowed me to run number 12 copper and almost no loss.
17kw panels , 90kw lifepo4 batteries.
2 sunny boy 7.7 , 1 sb6000us, 2 sunny island 6048.
Powers the world

 

[Arron]

Hi
Just seen this thread and thought I would put a few thoughts on it.
Remember your loss through the cables is simply [current * current * resistance] IIR
Therefore:
DC from panels to shed with 5000W (500V) and a cable size of 6mm or 10 gauge (approx 1.1 ohm per 1000ft)
Double the wire resistance because you have two wire connecting between source and load (in series)
This would give 10A of current, so you would have (10A * 10A * 2.2) 220 Watts of loss in the cable.
There would be a bit more from joints etc, but around 250W
If you are sending 10000W at 500V the current would now be 20A, which will have (20*20*2.2) 880Watts loss in the same cables.
Plus a bit for other losses, call it nearly 1KW

As you can see keeping current to a minimum is very beneficial, if you double the current you increase the loss 4 times because of the square law.
Do the math for each scenario and work out your losses.
If possible try to keep your currents to around 10A or less.