Educate me on batteries please...Links, charts?

[TheEquineFencer]

I'm not great at math, just electrical stuff.

Right now I have 8D 12V flooded batteries I'm wiring in to use on a "cobbled together system" so to speak to just get something going to play with. I'm working out some logistics to get a lot better system/panels in place.

I'm looking at some ReLion 6V REV L16-400 batteries for about $280 each. I need at least 4, but would like to get 8. They are 400Ah. My main question is about how much can I pull from these and for how long? If I get the panels I want, I should be able to charge them I'm pretty sure in a 5-8 hour day pretty easy. I'm waiting on another guy to get a quote on flooded batteries for solar use, but he's dragging his feet about it.

How do I calculate a load for them? I'd love to have an PSW inverter that puts out 120/240 but may have to settle for one that puts out 120 only. If I going into the inverter and out at 120/240 how do you calculate the amps load coming out of the battery and how long it sustain before I pull the batteries do to 70-80% ?

If this were a 12V system it'd be 12V in, 120V out, so not allowing for inverter loss if I put let's say a 40A load out at 120V I'd be pulling 400A out at 12V Right? So with a 400Ah battery it'd be dead in an hour? What would it be at 24V in then 200A from the batteries to the inverter?

Is there a chart or spreadsheet for calculating this? I'm trying to figure just what I can run off a 400Ah or a 800Ah battery bank and for how long.

I've got to do some testing here to see what loads my shop is actually pulling to see if/what I can run off this bank. I know I'm not going to be able to run my "big stuff" 7.5HP compressor, hot water heater or Mig welder off them, but the frig and lights I might be able to, maybe even the VFD's for the lathe and mill with limited usage.

Any insight would be helpful. Thanks in advance.

[frackers]

Take the capacity of the battery in amp-hrs and multiply by the voltage. If you have 8x 400ah batteries @ 12v then you have 38400watt-hours of storage. In the case of lead-acid you really don't want to use much more than half this if you want any decent life from the batteries so the useful storage is 20000watt-hours or 20kw-hrs.

Depending on the specs for the batteries (how much current they can supply), losses in cables and an inverter (assuming you are using one) then you would have a system that could supply 1kw for about 15 hours or 2kw for 7 hours (higher loads give higher losses). For a high capacity system you would want to series/parallel the batteries to provide 48volts so that you can keep the cable sizes reasonable at high loads, the calculations remain the same though.

Also keep in mind that an inverter will draw current even when you have no load on it - my home made OZInverter uses about 65w continuously due to the transformer not being as good as it could be - that adds up to 1.5kw-hr per day and if the sun doesn't come out then could be all I get from a 6kw solar array!!

[TheEquineFencer]

It's a 24V system, So it has 800Ah of battery storage.

[Bruce S]

It's a 24V system, So it has 800Ah of battery storage.

This depends on the voltage of the batteries if they are 6V then it will take 4 to make 24V , if they are all 400Ah then you will have 24V@400Ah
IF you have 8 of the 6V batteries, then you will assemble them into a series / parallel setup for 24Vdc@800Ah.

Hope this helps
Bruce S

[TheEquineFencer]

I understand series/paralleling of batteries as it relates to Ah, I'm trying to figure out the output ratio/time of what and how long I can run off them.

[XeonPony]

if lead acid then you have 50% usable so in 800AH configuration you have 400 usable.

So lets get to the watt of it: (400*24V = 9.6KwH) So you can do 9.6Kw for 1H, 4.8Kw for 2H, 2.4Kw for 4H, 1.2Kw for 6h, 600w for 8H (Starting to see a trend?)

Basically it all depends on the energy your consuming, this is where phantom loads add up dam quick, lets take a 50w continuous load: 50w* 24H = 1.2KwH of energy burned just sitting there. Coffe pot uses 1500w for 3-5 minutes, so in the end it uses barely any thing out of the battery bank.

So you need to figure out all your loads then multiply by run times then sum them all up to find your average power consumption, then add 25% safety factor, then multiply by the days of no charging to get a final battery capacity needed. If a generator is in the mix then you can just do the minimum battery so skip the days multiplier.


https://batteryuniversity.com/learn/

[TheEquineFencer]

if lead acid then you have 50% usable so in 800AH configuration you have 400 usable.

So lets get to the watt of it: (400*24V = 9.6KwH) So you can do 9.6Kw for 1H, 4.8Kw for 2H, 2.4Kw for 4H, 1.2Kw for 6h, 600w for 8H (Starting to see a trend?)

Basically it all depends on the energy your consuming, this is where phantom loads add up dam quick, lets take a 50w continuous load: 50w* 24H = 1.2KwH of energy burned just sitting there. Coffe pot uses 1500w for 3-5 minutes, so in the end it uses barely any thing out of the battery bank.

So you need to figure out all your loads then multiply by run times then sum them all up to find your average power consumption, then add 25% safety factor, then multiply by the days of no charging to get a final battery capacity needed. If a generator is in the mix then you can just do the minimum battery so skip the days multiplier.


https://batteryuniversity.com/learn/

OK, I sort of what you're saying about the watts and/or amps on the battery side....to take my 120V usage on the inverter side and then calculate how much that translates to usage out of the Battery side...

Is wattage or Watt/Hours on the inverter side equal to what's on the battery side? Less the power lost in the conversion from DC to A/C of course?

[XeonPony]

yup.

Watts are watts they remain constant

you just multiply the efficiency

So lets say you have 1600w on the AC side: 1600*(100-95% = 1.05) = 1680 divide by battery voltage (1680/24v = 70Amps)

So with a 95% efficient inverter loaded to 1600w will draw 70amps from the battery

Now we can bring time in and simply multiply it by the time to get Ah

70+ 1h = 70Ah
70 * 4H = 280AH

To change back to Watts: 280 * 24V = 6720Wh. This is why it is easier to do it in Watt Hours as it ignores the voltage, and you can derive the volts or amps as needed on either the AC or DC side

[TheEquineFencer]

I guess I was looking at it this way....I have several circuits that are rated 20A 120V, that's 2400W...

I guess I was looking at it if I was pulling 20A@120V...then assuming I had the batteries wired 12V not 24V it'd be a 10-1 ratio at 12V so that would be 200A and at 24V ...I was assuming it'd be half that and put it at 100A out of the batteries plus what I'd be loosing through the conversion loss...

But I'm seeing how you're doing it with wattage...it makes a bit more sense now....I just think in Amps and Voltage I guess....

Thanks.

[XeonPony]

Rating are for circuit breakers and max continuous loads, so other wise meaningless.

To size things you need to know the actual loads, and to size batteries you need to know the watt hours the whole system needs to produce in a 24H period

Then take that to size the panels. Too much battery Vs panels = wrecked battery bank
Too much solar Vs average load = wasted money if you'll never expand (but all so means long panel life and plenty head room)

To get a good match you really need to crunch the numbers.

It all ways comes down to the watts in the end, even for figuring what gauge of wire and the needed fuse/Circuit breaker size!

[TheEquineFencer]

Good advice...

Just crunching some rough numbers so far...I'm sort of pissing in the wind other than learning right now. The batteries I already have are way too short on Ah for what I'd truly like to run long term. The ones I want to use are a big investment for now, The charge controller...well it's pretty much free along with a couple other smaller MPPTs I have. I may just wire up the three 100W Mono panels I have for now and run them through a SS16 MPPT I have and dump it into a pair for 8D batteries for now. If I can come up with a bunch more of the $30.00 8D take outs that would work for a while, but I'd rather use the L16-6V's the guy quoted me for the big system.

Thanks for the help, you put things in perspective!

[XeonPony]

8D is a starting battery, so they have massive current output but weak on sustained power. The L16s will be fine.

To give you an idea, I have 6 golf cart batteries that are very old now. they are 6v at 230AH, Series parallel configuration gives me 12v at 690 AH

that's 8,280Wh with 4,140 available, this runs my house, Computer at 150w for up to 6h a day, fridge, lights, and at their current age will run things through the night on an 1800w nominal 2.5Kw peak inverter.

Solar array is 540w, that is undersized, I really need 3Kw to have excess power, but the battery bank is good enough (I'll add one more set for a total of 920Ah)

Because again all it needs to do is run the house over night which is fridge, and phantom loads. For prolonged bad weather I have a genset, and I am making a wind turbine to supplement them.

Summer my 540w panels make an average of 5 to 3 KwH, Winter 1 to 2.5KwH of total energy, this needs to run the house and charge the batteries (Which brings us back to not enough panel and why my batteries are in such bad shape)

[TheEquineFencer]

You'd make out good with some of the 250W poly panels I found for about $100 each...or the 300W mono for around $150 each...If I had the $$$ I'd order a pallet of the 250's...29 of them....$3357.50 delivered to my place...

[Simen]

When you start loading the batteries continuously with higher loads than the C-rating, you should take Peukert's Law into consideration.

[XeonPony]

When you start loading the batteries continuously with higher loads than the C-rating, you should take Peukert's Law into consideration.

This only applies to NiFe/NiCd wet cells and Lead acids

[Simen]

When you start loading the batteries continuously with higher loads than the C-rating, you should take Peukert's Law into consideration.

This only applies to NiFe/NiCd wet cells and Lead acids

This topic hasn't been talking about anything else than Lead Acid batteries i believe, so my advice does apply.

[XeonPony]

I was specifying for his knowledge.