Long distance from solar panels to batteries

[Chris1]

In my proposed system I have a 1800w solar array 300' from a electric boat that has a 48v 1200ah lead acid battery bank. I'm thinking that I want to wire the panels in series to get somewhere between 200v and 400v so that the wire size is 2awg or smaller, and size the controller accordingly. Is this basically correct?

[jflorey2]

In my proposed system I have a 1800w solar array 300' from a electric boat that has a 48v 1200ah lead acid battery bank. I'm thinking that I want to wire the panels in series to get somewhere between 200v and 400v so that the wire size is 2awg or smaller, and size the controller accordingly. Is this basically correct?

Well, 1800 watts at 400 volts is 4.5 amps, and real world will be closer to 4 amps. 2awg might be a bit of overkill.

If you want to keep losses to 1%, that's 4 volts of loss, which means max of 1 ohm. So you need 1 ohm over 600 feet of wire, which is 1.6 ohms per 1000 feet. You could do that with 10 gauge wire. Even 12 gauge is only 1.62 ohms per 1000 feet.

[SunEagle]

Well, 1800 watts at 400 volts is 4.5 amps, and real world will be closer to 4 amps. 2awg might be a bit of overkill.

If you want to keep losses to 1%, that's 4 volts of loss, which means max of 1 ohm. So you need 1 ohm over 600 feet of wire, which is 1.6 ohms per 1000 feet. You could do that with 10 gauge wire. Even 12 gauge is only 1.62 ohms per 1000 feet.

The wire size may be bigger then needed but I am more concerned with only 38 amps (1800w / 48v = 37.5 amp) to charge a 1200Ah battery system

[Sunking]

In my proposed system I have a 1800w solar array 300' from a electric boat that has a 48v 1200ah lead acid battery bank.

Just one tiny little problem. It takes 6000 watts to charge a 48 volt 1200 AH battery. Wire size at this point is the least of your concerns. Will not matter what the wire size is on a dead battery.

If you could get the panel voltage up to 200 volts @ 1800 watts results in 9 amps and at 600 foot loop length would require 6 AWG to slip in under 3%. At 400 volts or 4.5 amps requires 8 AWG to be at less than 3%

[jflorey2]

The wire size may be bigger then needed but I am more concerned with only 38 amps (1800w / 48v = 37.5 amp) to charge a 1200Ah battery system

Agreed there. I'd want to see at least 100 amps of charge current for that bank if it gets any kind of regular use. (About a 6000 watt array.) However, since this looks like a ground based array to charge a boat that is presumably some distance from the array, there may be other charging sources on the boat that he hasn't mentioned.

[Chris1]

Agreed there. I'd want to see at least 100 amps of charge current for that bank if it gets any kind of regular use. (About a 6000 watt array.) However, since this looks like a ground based array to charge a boat that is presumably some distance from the array, there may be other charging sources on the boat that he hasn't mentioned.

Oh shoot, my bad. I meant 1200ah worth of 12v batteries in series for 48v. So a 300ah 48v battery.

[jflorey2]

Oh shoot, my bad. I meant 1200ah worth of 12v batteries in series for 48v. So a 300ah 48v battery.

That's a lot more reasonable.

[SunEagle]

Oh shoot, my bad. I meant 1200ah worth of 12v batteries in series for 48v. So a 300ah 48v battery.

That is a better value for the battery system to charge with 1800 watts. Glad we cleared that up.

[Chris1]

If you could get the panel voltage up to 200 volts @ 1800 watts results in 9 amps and at 600 foot loop length would require 6 AWG to slip in under 3%. At 400 volts or 4.5 amps requires 8 AWG to be at less than 3%

Got it. I was using 1% as my acceptable loss. Maybe I will lose a little more and that's okay.

[Sunking]

Got it. I was using 1% as my acceptable loss. Maybe I will lose a little more and that's okay.

1% is just way to tight and not necessary. Once you go less than 3% is diminishing returns. Using your example 400 at 4.5 amps requires 10 AWG and gives you 2.9% loss at full power, vs 4 AWG at 0.98% loss. 600 feet of 10 AWG will cost you around $120, and 4 AWG is around $500.

Another mistake is assuming you will see and experience 4.5 amps, because you never will. At very best you might see 3.9 amps for a few minutes around solar noon. If you were to watch and log current for a whole day you would see something like this. When morning sun falls on the panels you would see around 0.2 amps. As the sun gets higher in the sky say around 10:00 pm you might see 2 amps. Current will continue to rise slowly as noon approaches and peaks at solar noon. Then a few minutes later the current starts to taper off and the process reverses itself. Your take away is as current less than 4.5 amps will be more efficient as current is a direct contributor with a weight of 50%. The other half is the voltage. So at say 2 amps on 10 AWG has a loss of 1.4%.

If you fail to understand that and use unrealistic maximum currents that only last a few minutes, is going to end up costing you big money. By using 4 AWG is costing you 4 times more than 10 AWG, and not buying you anything meaningful. It will work, but expensive. You crossed the line of diminishing returns.

[Chris1]

1% is just way to tight and not necessary. Once you go less than 3% is diminishing returns. Using your example 400 at 4.5 amps requires 10 AWG and gives you 2.9% loss at full power, vs 4 AWG at 0.98% loss. 600 feet of 10 AWG will cost you around $120, and 4 AWG is around $500.

Another mistake is assuming you will see and experience 4.5 amps, because you never will. At very best you might see 3.9 amps for a few minutes around solar noon. If you were to watch and log current for a whole day you would see something like this. When morning sun falls on the panels you would see around 0.2 amps. As the sun gets higher in the sky say around 10:00 pm you might see 2 amps. Current will continue to rise slowly as noon approaches and peaks at solar noon. Then a few minutes later the current starts to taper off and the process reverses itself. Your take away is as current less than 4.5 amps will be more efficient as current is a direct contributor with a weight of 50%. The other half is the voltage. So at say 2 amps on 10 AWG has a loss of 1.4%.

If you fail to understand that and use unrealistic maximum currents that only last a few minutes, is going to end up costing you big money. By using 4 AWG is costing you 4 times more than 10 AWG, and not buying you anything meaningful. It will work, but expensive. You crossed the line of diminishing returns.

That makes complete sense, thanks. So is the choice of voltage also one of economics? For instance a controller that can handle 400v might be much more expensive than one that can handle 200 but with 200v I'm paying more for wire.

[Sunking]

That makes complete sense, thanks.

You are welcome.

So is the choice of voltage also one of economics? For instance a controller that can handle 400v might be much more expensive than one that can handle 200 but with 200v I'm paying more for wire.

You are on the right track. Basically the higher the voltage is, the more efficient and at less cost. That is exactly why utilities run high voltages.