affordable 2000w inverter

BFL13 wrote:
1.2v with 24v for that 1000w is decent, so the wiring etc must be ok.

One of the reasons I went to 24V ...

brulaz wrote:

BFL13 wrote:
1.2v with 24v for that 1000w is decent, so the wiring etc must be ok.

One of the reasons I went to 24V ...

I get 0.6 with 12v, which is what I based my comment on. Many variables for wiring set-ups and battery types though.

"Good enough" is defined as being able to run the toaster the morning you are down to 50% with a recharge due later that day. No point doing more wiring upgrades if able to do that.

I would suspect that 24 volts may offer lower voltage drop than 12 volt. The amps needed for a given wattage are 50% of those on a 12 volt bank.

pianotuna wrote:
I would suspect that 24 volts may offer lower voltage drop than 12 volt. The amps needed for a given wattage are 50% of those on a 12 volt bank.

But the 24v guys use thinner wire, so wouldn't it be a wash? He is getting 1.2v with 24v and I am getting 0.6 with 12v, so that is the same percentage drop for a 1000w load. If his current is half mine then his R must be twice mine or is it four times mine? Brain just quit. :(

Unfortunately I don't have any data to compare 12 vs 24V V drops, as when we had 12V we only had a 1000W inverter and rarely loaded it up.

But according to my calcs, at 24V the wires should only account for 0.3-0.4VDC drop with an 1000W AC load. The remaining drop (0.8-0.9VDC) must come from the 4xGC2 in series at these loads (~50A at 24VDC).

Similarly I calculate that at 12V with the same wires there would be twice that V drop across the wires or 0.6-0.8VDC with a 1000W AC load.

IOW the wattage loss in the wires should be identical at 12 versus 24V. (I did use the same wires when changing from 12V to 24V).

So the question is how does the V drop of batteries behave at 12V versus 24V? If they behave like wire or resistors, then the wattage "loss" should be the same and V drop half? But I dunno ...

The load on the batteries is the same. i.e. 50 amps per cell. That's about double the draw that they are tested at. So voltage drop is pretty much the same whether you are using 12 or 24 volts.

Are the series connections the same or larger size than the feed wire to the inverter?

pianotuna wrote:
The load on the batteries is the same. i.e. 50 amps per cell. That's about double the draw that they are tested at. So voltage drop is the same whether you are using 12 or 24 volts.

Are the series connections the same or larger size than the feed wire to the inverter?

Battery cables are 1/0 I think, same as to the inverter.

The battery internal voltage drop is half, it as to be.

IMO (dubious) 1.2 = 0.6 for percentage drop 24/12. After that we play with V=IR where V is twice (but the same equivalent). We have I being half with 24 vs 12. So R must be way higher in absolute numbers, but it must come out even for same percentage V.

He used the same wire so its ampacity is the same (at 12v). But amps are half so you would expect less of a voltage drop but that would be in absolute numbers staying within the same voltage (12v)

It seems like something is wrong where if the wiring/batts can do 0.6 drop at 12v at 100 amps, it should do better (less than twice) at 50 amps 24v, but he gets twice, so something needs fixing--loose connections?

Or as mentioned is it peculiar to the battery bank being 6s in series instead of 12s in parallel or some such thing?

Google tells me that a battery can be treated as a simple resistor with respect to voltage drop. And so can all the wires and connectors. Apparently then Ohm's law applies.

Example:
A single 12V battery with 50A load (600W) has Vd voltage drop

Two 12V batteries in parallel, with 100A load (50A each, 1200W total), also has Vd voltage drop. Half the internal resistance but twice the current so Vd remains the same.

Two 12V batteries in series, with 50A load (@ 24V, 1200W total), has 2xVd voltage drop? Twice the internal resistance of a single battery but the same 50A current, so TWICE the voltage drop.

But, percentage-wise the Voltage drop of two batteries in parallel is the same as two in series. And the wattage (VA) drops is the same.

No idea if this happens in practice though.