How to calculate a battery capacity

"Simon Dice" wrote in news:1158603177.343083.179580 @i3g2000cwc.googlegroups.com:

Watts = Amps * Voltage

The Amp/hr rating of the battery (I think) says that if the device draws x amount of amps, the charge will last this many hours.

This should get you started.

Puckdropper

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============================= The amp hours is measured at a '10 hour rate' or a '20 hour rate', so a

120 AH battery is rated at 12A for 10hr... less hours if the current is greater. Also read up on Peukert Number on one of the Battery sites.

I think Lewin can respond well to this. But in the meantime, what latitude? Will it have exposure to the entire sky? Etc. Can you add more information here?

My vague memory says that in the US, over an entire year's use, you can get as little as 1 MWh m^-2 yr^-1 of insolation in the north. I think the figure is about double that in the south. With thin film efficiencies of about 6%, that's 60 kWh m^-2 yr^-1 in the north. At

12% for polycrystal, about 120 kWh m^-2 yr^-1. Again, about double that in the south. You are looking for an average of 80*365*24 or 701 kW h yr^-1. But if you are only planning on supplying battery support for night time, you will need to find the minimum average for the lowest energy day of the year and cover that. That figure will be less than the average by a fair amount, especially if you are including worst case conditions with precipitation and heavy cloud cover.

And if you are at the poles, you need to think about battery storage for 6 months -- so latitude has a lot to say about what you need to do.

I think you can already see several square meters in your future, even with relatively high efficiency panels.

I'm ignorant of all the details, but general theory tells me:

(1) Examine the site and latitude to find the better orientation for your panel. You may not be able to slant it optimally, or you may be able to include tracking the sun so that the panel efficiency is optimal all the time. I don't know. Keep also in mind that you will get increasing reflections as the sun moves and if your panel is fixed in position, losing usable light in the process. (2) Then estimate how much ground-level, visible wavelength light you are likely to get per meter^2 on the weakest day of the year with worst case cloudy conditions and precipitation -- include obstructions, trees, and varying angles as the sun crosses across the sky. You will need to consult gov't or industry figures for your area

-- perhaps contact the local power company about this. (3) Apply your estimated solar panel efficiency (6% for thin film, say 12% for polycrystal) and compute your average daily energy. (4) You will need 2kWh per day, so work out the number of square meters of area you will now require to achieve this average on the weakest day of the year. This number will be painful to see. (5) Decide how much time your batteries will have to support your 80W system -- probably should size this for at least 24 hours of use, or about 2 kWh.

I didn't include figures on how much efficiency you can expect in delivering the power to your load from the panel or from the battery, but there will be losses here. Also, there will be losses when you are charging batteries AND supplying power, during the day. Overall, those panels will just be getting larger and larger and your batteries bigger. It's going to hurt.

Jon

The amp-hour rating is usually based on a fairly "low" rate of discharge. The greater the discharge rate, the lower the battery's capacity. Lead-acid batteries are often rated at a stated discharge rate. Do a search on "peukert" for more information on this relationship.

You should also be able to find some sites with information on designing solar charging systems.

Chuck

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FYI, it's Amp-hrs (since the units are multiplied), not Amp/hr (which would be "amps per hour", which is different).

When you divide the amp-hour rating by the battery current (in amps), you will get how many hours the battery should last. Eg., a 10 Amp-hour battery lasts 10 hours at 1 amp, 5 hours at 2 amps, etc.

As others have said, though, for higher amperages, er, uh, I mean CURRENTS, the amp-hour value is actually less.

As for the solar panels ... I don't know your specific location, but a rough calculation gives:

Incident power of sunlight (about 1000 Watts per square meter), times the efficiency of solar panel (I'll use 12%), times the fraction of day that panel receives "full sun" (I'll use 6 hours out of 24), times the fraction of days that are sunny (I'll use 2/3) ***

equals:

(1000 W/m^2) x (0.12) x (6/24) x (2/3) = 20 W/m^2.

The actual, useful power is 20 Watts per square meter. So for an 80W,

24-hour-per-day system, you'd need 4 square meters of solar panel. That's just a ballpark figure, there are people who know the relevant numbers better than I do, and you'd need to factor in the efficiency of charging the battery from the solar panel too.

Hope this helps,

Mark

*** p.s. on the "sunny day" factor: use what is reasonable for your area. Also, if your area occasionally goes a few days to a week with rainy weather, you'll need a battery (or several batteries) that can run for a few days to a week without recharging.

Thanks, for all the answers It looks like reduce the power consuption is the best idea to follow...

Best Regards

PS location is Mexico north area...Monterrey city.

Thanks, for all the answers It looks like reduce the power consuption is the best idea to follow...

Best Regards

PS location is Mexico north area...Monterrey city.

Should be good!

"BobG" wrote in news: snipped-for-privacy@i42g2000cwa.googlegroups.com:

You mean I have to type that in? Ah man, that's hard. (That was sarcasm.)

For anyone looking on Wikipedia, it's apparently under Peukert's Law.

Puckdropper

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Wise is the man who attempts to answer his question before asking it.

To email me directly, send a message to puckdropper (at) fastmail.fm