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1. Assistance with battery selection

I thought I knew what I was doing but the information my research has uncovered is so varied and overwhelming on the subject that my head is about to explode. I am by no means a math whiz nor a solar system designer however, I have repaired existing systems before. My current conundrum is to simply select batteries for a new solar system. I made myself appear somewhat like an expert to some friends living on a Bahamian island and now have come up short. The specs on batteries on websites range from watt-hours, amp-hours, kw-hours, etc.. leaving me lost. Perhaps someone can shed some light on the subject for me.

Some of my math was correct when comparing it to the plethora of solar calculators on the web.

Here is the challenge.

Looking to operate 8 CFL for up to 12 hours a day (during the nighttime only - year round) with a new 45 watt array.
The wattage of the lamps is unknown so for calculation purposes I am using 23 watts (100w incan. equiv.) per lamp as the assumed load = 120v @ 184w / 1.5a - rounding up to 2 amps (24 amp-hours?)

Assuming a 20 percent loss after the "break in" period of the panels per the manufacturer specs
and per online calculators only 5.5 hours of average solar recharge time per day
using a 300 watt inverter.....
Not wanting to discharge the batteries less than 50 percent....

How many 12 volt batteries will it take to achieve this goal?
What should I be looking for?

Most of the specs on the marine batteries are rated in CCA which has confused me even more.

I also thought I could find a 12 volt timer or photo cell to cycle off the inverter during the day to save juice.

Thank you for looking at my problem and I would be most grateful for any response that may help me.

2. Wintertime you are collecting maybe 75 watts of power per day considering losses - you need to rethink the number of lamps or wattage of the lamps to start with

3. Originally Posted by Stageman
Looking to operate 8 CFL for up to 12 hours a day (during the nighttime only - year round) with a new 45 watt array.
OK, so:

8*12*23 watts / 85% inverter efficiency = 2.6 kwhr/day

45 watts * 80% (your derating) * 5.5 hours a day sunlight = .198 kwhr/day

So first step - you need way more solar. 900 watts is going to be a bare minimum.

Next step is battery size. You're going to want, at minimum, 3 days of operation with no sun. So 7kwhrs worth of battery. To limit discharge to 50% you'll need 14kwhrs worth of battery. A standard Trojan T-105 is 220AH at 6V, or 1.3kwhr/battery. That means 11 batteries. To get 12 volts you need 2 batteries so that means you'll need a bank of 12 batteries to meet your requirements.

I also thought I could find a 12 volt timer or photo cell to cycle off the inverter during the day to save juice.
Sure, these are easy to find. The SunLight SL-20L-12V 20A controller will do that. You don't need a separate photocell; just wire one of the panels through this controller to the battery and it will use the panel as a sun sensor. (You can't use it as the only charge controller since you'll be generating about 80 amps with your array.)

4. Originally Posted by billvon

45 watts * 80% (your derating) * 5.5 hours a day sunlight = .198 kwhr/day
Considering all consumption is from the batteries that is way too high.

5. Hardly seems worth the cost

Wow! It hardly seems worth the cost. 200 watts an hour, perhaps double the panels with a supplemental wind generator may be the way to go but then the investment needed to run these lights makes it cost prohibitive to do this project.

Perhaps an alternative would be LED without an inverter.

Originally Posted by billvon
OK, so:

8*12*23 watts / 85% inverter efficiency = 2.6 kwhr/day

45 watts * 80% (your derating) * 5.5 hours a day sunlight = .198 kwhr/day

So first step - you need way more solar. 900 watts is going to be a bare minimum.

Next step is battery size. You're going to want, at minimum, 3 days of operation with no sun. So 7kwhrs worth of battery. To limit discharge to 50% you'll need 14kwhrs worth of battery. A standard Trojan T-105 is 220AH at 6V, or 1.3kwhr/battery. That means 11 batteries. To get 12 volts you need 2 batteries so that means you'll need a bank of 12 batteries to meet your requirements.

Sure, these are easy to find. The SunLight SL-20L-12V 20A controller will do that. You don't need a separate photocell; just wire one of the panels through this controller to the battery and it will use the panel as a sun sensor. (You can't use it as the only charge controller since you'll be generating about 80 amps with your array.)

6. Any battery labled with CCA, is NOT going to work well as a deep cycle battery. Do you have a battery case size - like
Group 24,
Group 27 ? 24's are often about 85Ah, 27's 105Ah. (if they are DEEP CYCLE)

Measure and plan everything in watt hours, then convert to amps & volts later.

CFL's are seldom "23watts". That's the light output. throw in Power Factor, ballast losses, figure 35 watts consumed.

35 x 8 = 280watts round up to 300w (always round consumption UP, and harvest DOWN)

300 watts for 12 hours = 3600 watt hours consumed, plus another 20% for inverter losses = 4320 wh consumed

To provide for 25% discharge of the battery daily, you need 17,280 wh of storage. At 48 v, that's 360 amp hours
Closest battery is the L-16 400ah , 6V battery, 8 in series.
This cannot be done with a 12V system. Just forget 12V.

Now your harvest. You need to harvest 2x your daily consumption, as caculated on the shortest day. You have batteries to run 2 cloudy days, then you must start the generator, or damage your batteries with deep cycles.
PV array watts x hours of sun (from PV watts: http://www.nrel.gov/rredc/pvwatts/ ) = watt hours.

3600 wh consumed = 7200 wh harvest required.

enjoy

7. Originally Posted by Stageman
Perhaps an alternative would be LED without an inverter.
That would reduce load a little (20%?) but not by much. However, going to 1 watt LED's would be a big change and make the system much more reasonable. Depends on what you want to do. Do you want to light a classroom? You're going to need a lot of light. Do you want people to be able to find the bathroom at night? 1 watt is plenty.

8. Great information

Thanks for the great information and tips on the calculations.

Might be time for some formal instruction in alternate power calculation. I work with basic A/C electricity formulas (Ohms Law) at work where the power is abundant and seemingly never ending.

Will suggest LED to my friends. It really seems like a lot of effort and cost to run 8 lights.

9. Originally Posted by Mike90250
CFL's are seldom "23watts". That's the light output.
?? No, that's the power used. The light output is about 1/4 of that, since CFL's are about 25% efficient.

Also modern CFL's are electronically ballasted, and on modified sine inverters (most likely choice for a cheap system) their power factor is not an issue since peak rectification is not an issue on mod-sine. On a sine wave inverter system you will see peak rectification, and thus the peak power will be higher than the average power. However that's a wiring issue and does not significantly reduce efficiency.

On older magnetically ballasted bulbs power factor can be a big issue - indeed many mod-sine inverters won't work at all with magnetic ballasts, and/or will buzz like crazy. The simple way to tell them apart is to weigh them. The heavy bulbs have magnetic ballasts.

10. Originally Posted by billvon
?? No, that's the power used. The light output is about 1/4 of that, since CFL's are about 25% efficient.
Have a reference for the 25% efficient or better an efficacy number?

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