I want to bring only the high voltage DC from the shed to the house and then invert it there. At the same place where the grid enters the house.
What implications are there for DC to traverse from the shed to the house?
Yes, high DC voltage (48v) .... distance shouldn't be the same kind of resistance drop as AC.
Well, you are right. The voltage drop between AC and DC is different. However, your voltage drop assumptions aren't correct. The voltage drop between AC and DC is basically the same (single phase, anyways).
So, if you are going to maintain AC devices in the house, it would be far better to invert it in the shed and feed it into the house. You'll loose far less voltage that way. And, you loose voltage, not amperage.
Using a voltage calculator:
Wire Size 1/0 Copper Wire
Voltage 120V
Single Conductor (1 Wire each way)
Distance from Source: 100m
Load current: 100A
Voltage drop: 6.45
Voltage drop percentage: 5.38%
Voltage at the end: 113.55
Wire Size 1/0 Copper Wire
Voltage 50V
Single Conductor (1 Wire each way)
Distance from Source: 100m
Load current: 100A
Voltage drop: 6.45
Voltage drop percentage: 12.90%
Voltage at the end: 43.55
So there'd be a larger voltage drop, 5.38% vs 12.90%. If you went to 240V, it'd be even better:
Voltage drop: 6.45
Voltage drop percentage: 2.69%
Voltage at the end: 233.55
Now, you
could raise your DC voltage to 120V or 240V to send to the house, but now you are getting into safety issues. I would must rather be hit with 120VAC than 120VDC. DC will create a nasty arc that will burn, best case, or lock you muscles up on the wires and cause you to fry in worse case. AC will at least let you have somewhat of control of pulling yourself off.