Charging question

Dailypix

I have 4 6 volt batteries wired to give 24 volts. They are rated at 428AH each. Suppose I have no load whatsoever. I even turn off the inverter. I get 5 hours sun. 
How many watts solar panels do I need to completely charge the batteries in 1 day?
Also with the batteries wired this way is the capacity of the bank 428AH or is it 4 X 428AH.
i guess what I am trying to learn is how powerful I need the panels to be for a battery bank this size in order to charge it up fairly quickly. Assume I have good MPPT charge controller. I want to learn how to do the math minus the load.
regards

mcgivor

Send a message to BB,Bill or wait, I'm sure he will read your request sooner or later, he has a wealth of information I couldn't approach regarding charging, hours of sun, loss percentages and so on.

Your  battery capacity buy the way is 428Ah @ 24v, you would multiply if they were in parallel which would be 1712Ah @ 6v

Dailypix

Thank you

Dailypix

I suppose to simplify the question I am looking for the formula to calculate the watt hours for a 428 amp hour 24 volt system and will I have to multiply it times 4 because it is made up of 4 batteries.

BB.

I am here--Just had to get to laptop computer so I can type on a real keyboard and bigger screen....

For the batteries, when you add batteries in series, the voltage adds. When you add batteries in parallel, the Amp*Hours adds... For a couple of examples (as Migivor says):

• 4x 6 volt batteries in series @ 428 AH = 24 volts @ 428 AH
• 4x 6 volt batteries in parallel @ 428 AH = 6 volt @ 1,712
• 2series x 2parallel of 6 volt @ 428 AH = 12 volts @  856 AH

Note that the total energy storage is the same (in Watt*Hours):

• 24 volts * 428 AH = 6 volts * 1,712 AH = 12 volts * 856 AH = 10,272 Watt*Hours

Watts and Watt*Hours is less confusing when working with DC and AC off grid power systems... Watts and Watt*Hours are the total description of Power (W) and Energy (WH). If you use AH, you are "missing" the voltage part.

MPPT or PWM controller, it does not "matter" for this first set of math... The choices of MPPT/PWM, bus voltage, etc. depends on size of system, distances, cost issues, power needs, etc.... And we can choose that stuff later after we work out the basics.

I don't like to do a bunch of "what if" calculations--It tends to make things confusing. Since I do not know anything about your loads, power needs, where the system is, etc... I will start the design with your battery bank, and we can go from there (normally loads drive the sizing of the battery bank).

Just to give you an idea of what loads could be supported by your proposed battery bank (using our, relatively conservative, off grid rules of thumbs), for an AC inverter based system, with 2 days of storage (no sun), and 50% maximum battery discharge (longer battery life for Lead Acid batteries):

• 24 volts * 428 AH * 0.85 AC inverter eff * 1/2 days of storage * 0.50 maximum discharge = 2,183 Watt*Hours per day
  

And, a 428 AH @ 24 volt battery bank (flooded cell) can support a maximum AC inverter rating of:

• 428 AH * 1,000 Watts per 200 AH (at 24 volt battery bank) = 2,140 Watt maximum AC inverter (or AC loads) recommended (or smaller)

The above is also a suggested maximum solar array Watt rating (for various reasons).

Next, charging... We suggest 5% to 13% rate of charge--5% is good for a weekend system use, or seasonal (sunny summer usage). For a full time off grid system, suggest 10%+ rate of charge to keep battery happy and not agonize about energy usage every day.

• 428 AH * 29 volts charging * 1/0.77 panel+controller deratings * 0.05 rate of charge = 806 Watt array minimum
  
• 428 AH * 29 volts charging * 1/0.77 panel+controller deratings * 0.10 rate of charge = 1,612 Watt array nominal
  
• 428 AH * 29 volts charging * 1/0.77 panel+controller deratings * 0.13 rate of charge = 2,096 Watt array "cost effective" maximum

Now for charging deep cycle lead acid batteries, they really like ~10% to 13% rate of charge... Pick 10% for now, if you discharge to 50% state of charge, it will take a C/10 charge rate to fill a battery bank in ~5 hours + 3-5 hours for Absorb, or about 8-10 hours back to >90% state of charge (full enough for our needs).

However, with solar power, you can only charge when you have sun. So say you are near Brampton Ontario Canada:
http://www.solarelectricityhandbook.com/solar-irradiance.html

Brampton
Average Solar Insolation figures

Measured in kWh/m2/day onto a solar panel set at a 46° angle from vertical:
(For best year-round performance)

Jan	Feb	Mar	Apr	May	Jun
2.71	3.50	4.04	4.50	4.75	5.12
Jul	Aug	Sep	Oct	Nov	Dec
5.24	4.98	4.58	3.68	2.47	2.24

Lets say you are "doomed" to using a genset during the 3 winter months--That leaves February at 3.50 hours of sun per day as the "break even" month (note this are usually 20 year average sun based on weather conditions, etc.). And we use the daily load of 2,140 Watt*Hour per day nominal load (some days more, some days less--The "blessings of using the free sun power":

• 2,140 Watt*Hours of AC power * 1/0.52 AC off grid system eff * 1/3.50 hours of sun (Feb break even month) = 1,176 Watt array minimum (Feb break even).
  

Note that in February, you only get ~3.5 hours of sun per day (equivalent to 3.5 hours of noon time sun), so if you have a 10% array (1,612 Watts) and need ~5 hours of full sun and another 3-5 hours or so of partial sun (tapering charging current), you are looking at almost two day to fully recharge the battery bank (assuming no loads). This is a bit of "calculation hand waving", if you need more exact numbers, we can do that (trying to not make this a math assignment, but show how math helps us size the system).

Does this get you started/give you some ideas (I can give you the inverse of some equations, or you can do the algebra too).

Note that Solar power is accurate to within 10% or so (don't get down to the 0.4 watt numbers--I just used all the digits so you can repeat my math and not have round-off errors in further calculations, and you can follow where the numbers are used). A 1,612 Watt array is about the same as a 1,451 Watt array (of course, more array is more energy--But it is close enough for our needs).

And the above rules of thumbs are just to get us "close" to a workable system without going into a 5 page post with 5x more math and hand waving discussions. There is nothing "magic" about the above numbers and they can be moved around based on your actual power needs (day time vs night time loads, winter or summer usage vs full time, size of loads and surge currents--such as starting a well pump, etc.).

But the above should get you into the ball park.

-Bill