Help sorting this out!

moonbow · April 2018

Good Morning, I have a simple ( I think) question that I cannot quite get straight in my head. Here goes... In my RV, I have a nice solar system comprised of a roof panel, charge controller and: (2) - 6 volt 220 AH AGM Batteries wired in series to make a 12V 220AH bank. (1) - 3000W Pure sine Wave Inverter. I also have a generator to use when needed but my question is this. I have a 6 gallon water heater that draws 1550 Watts and 14 Amps. When I want to heat water, I bypass the solar system and turn on the generator for 30 to 40 min . For curiosity's sake, I decided to run the Water heater off the batteries / inverter. When I did this ( with a fully charged battery bank), my charge controller read that the battery bank voltage was dropping to 11.8v while the inverter was reading a 1560Watts of output. I wasn't sure what to make of this so, after about 30 sec, I stopped using the batteries for the WH . The voltage reading on my bank rebounded to 12.7 as soon as the inverter was turned off and the load removed. While not totally necessary (because I have a generator) I would think my fully charged battery bank would be able to handle a 14 Amp load for 30 to 40 min. Is it normal and OK to see the battery bank voltage drop like this during a heavy load draw? Thanks in advance

Photowhit · April 2018

First lets calculate the load, you have 1550 watts + your inverter is likely only 88% efficient. So the total load is more like (1550 ÷ .88=) 1761 watts. 1761 watts ÷ 12 volts = 146.75 amps... Perhaps I need not continue.

moonbow · April 2018

Ok, I suspected as much. Thank you for verifying that for me. Is it assumed that the above #'s are based on an hours use? Meaning for .5 hours my draw would be roughly 74amps? Either way, I will stick to using my generator to heat the water in the tank.

Estragon · April 2018

It would likely be worse than that. The battery voltage would continue to sag, so 146a at 12v would become 160a at 11v, and 176a at 10v. This would be compounded by higher wire losses etc with the higher currents, and the diminishing effective capacity of the bank.

The 220amp-hour capacity is likely a 20 or 24 hour rating. At a discharge rate approaching the 1 hour rate, the effective capacity is greatly reduced, so the bank may behave more like a 120ah capacity. My guess is by the .5hr point, the voltage would have sagged to the point the inverter would drop the load to protect itself.

You can get away with big loads for short times (eg microwave), but for longer periods you get into a sort of self-reinforcing death spiral.

Photowhit · April 2018

moonbow said:

Is it assumed that the above #'s are based on an hours use? Meaning for .5 hours my draw would be roughly 74amps?

Yeah, you've pretty much got it, just to expand your knowledge base. There is something else at work here. Batteries amp hour rating is usually based on a discharge over 20 hours. If you discharge faster than that they have effectively less capacity. If you discharge over a longer period, they have an effective larger capacity. As an example here is the amp hour capacity chart for Trojan T105 batteries;

BB. · April 2018

For flooded cell lead acid batteries, typically we recommend a maximum AC inverter capacity/power draw of ~500 Watts per 100 AH @ 12 volt battery bank. 220 AH @ 12 volts would be ~1,100 Watt recommended AC inverter maximum size (for various reasons on that size/type battery bank).

Also, remember that Power=Voltage*Current... You are pulling 14 amps @ 120 Volts for your electric water heater. At 12 volts, you are drawing from the battery bank:

Power = V*I = 120 volts * 14 amps = 1,680 Watts
I=P/V= 1,680 Watts / 12 volt battery bank = 140 Amps nominal from the battery bank

And for worst case electrical design:

1,680 Watts * 1/0.85 ac inverter efficiency * 1/10.5 volts AC inverter "cutoff voltage" = 188 Amps worst case

That is a lot of current to pull from a 220 AH capacity (20 hour rate) lead acid storage battery.

Realistically, an 8 hour discharge rate is about what I would suggest drawing from a standard deep cycle lead acid battery... In terms of AC Watts, that would be:

220 AH * 12 volt nominal * 1/8 hour discharge rate * 0.85 AC inverter efficiency = 280.5 Watt recommended continuous AC load

Now--You do have an AGM battery bank. AGM batteries are known for supporting very high current draw... 1 hour discharge or one brand even a 15 minute discharge (4x rated battery capacity).

HOWEVER--Drawing high current from AGM batteries can damage the batteries (UPS batteries are typical AGM or GEL batteries--And are replaced every 1-2 years or after a "significant" discharge/power outage or two).

Also, if you are going to draw such high currents from your lead acid (or AGM) batteries at such low voltages (12 volt battery bus), you need very heavy and short cables from the AC inverter to the battery bank. For example, 0.5 volt drop 3,000 Watt 12 volt AC inverter:

3,000 Watts * 1/0.85 AC inverter eff * 1/10.5 volt AC inverter shutdown voltage * 1.25 NEC wiring+breaker deratings = 420 amp rated wiring and branch circuit.

Using the NEC wiring table (relatively conservative):

https://lugsdirect.com/WireCurrentAmpacitiesNEC-Table-301-16.htm

You would need a pair of 4/0 cables in parallel for such high current draw (can do with smaller cables--This is a reasonably conservative recommendeation for DC bus cables in free air).

Then you want to have a maximum of 0.5 volt drop for your 12 volt wiring... Using a simple voltage drop accumulator and 2x 4/0 cable carrying ~4200 Amps, the maximum distance for the wiring would be:

5 feet (one way wire run--Some calculators would use 10 feet as round trip wire run)
Voltage drop: 0.52
Voltage drop percentage: 4.35%
Voltage at the end: 11.48

The short answer is I would not recommend a 1,680 Watt load on a 220 AH @ 12 volt battery bank.. It will support such a load for a relatively short time. However, in the long term it is very hard on the batteries and takes something like 1/2 of their capacity:

220 AH * 11.5 volts = 2,530 Watt*Hours of stored capacity (very roughly)
1,680 Watts * 1/0.85 AC inverter losses * 40 minutes run-time * 1/60 minutes per hour = 1,318 Watt*Hours used
1,318 WH / 2,530 WH capacity = 0.52 ~ 52% of battery capacity

I suggest that you don't take the battery below ~50% state of charge (for longer battery life). Also, when you discharge a battery with a very heavy discharge current (your discharge rate is getting close to 1 hour discharge rate)... Batteries are typically rated a 20 Hour Discharge Rate (for our needs). When you discharge at a 1 hour discharge rate, the apparent AH capacity of the battery will be less (say ~180-190 AH capacity or something...

-Bill

mcgivor · April 2018

Excellent information regarding the discharge, load versus capacity, I totally agree the battery would be overtaxed. Another consideration is what is withdrawn needed to be replenished, plus a little extra to account for battery, controller,wiring plus other losses. If the PV is unable to replace the amount withdrawn, the battery will likely suffer undercharging, which in addition to the high discharge rates would likely result in premature battery failure.

Help sorting this out!

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