# How to calculate a battery capacity

S

#### Simon Dice

Jan 1, 1970
0
I'm working on a project in wich I need to have a system running 24hrs
a day , power supply will be solar panels feeding a battery. Can any
one share with me formulas on how to calculate de Amp/hr for the
battery and the power of the solar panel?
Systems will consume average 80watts and need to keep running on a few
clowdy days as well.
Thanks for the assistance!

P

#### Puckdropper

Jan 1, 1970
0
I'm working on a project in wich I need to have a system running 24hrs
a day , power supply will be solar panels feeding a battery. Can any
one share with me formulas on how to calculate de Amp/hr for the
battery and the power of the solar panel?
Systems will consume average 80watts and need to keep running on a few
clowdy days as well.
Thanks for the assistance!

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

B

#### BobG

Jan 1, 1970
0
Puckdropper said:
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.
=============================
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.

J

#### Jonathan Kirwan

Jan 1, 1970
0
I'm working on a project in wich I need to have a system running 24hrs
a day , power supply will be solar panels feeding a battery. Can any
one share with me formulas on how to calculate de Amp/hr for the
battery and the power of the solar panel?
Systems will consume average 80watts and need to keep running on a few
clowdy days as well.
Thanks for the assistance!

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

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
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
(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

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

C

#### chuck

Jan 1, 1970
0
Puckdropper said:
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

The amp-hour rating is usually based on
a fairly "low" rate of discharge. The
greater the discharge rate, the lower
batteries are often rated at a stated
discharge rate. Do a search on "peukert"

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

Chuck

R

#### redbelly

Jan 1, 1970
0
Simon said:
I'm working on a project in wich I need to have a system running 24hrs
a day , power supply will be solar panels feeding a battery. Can any
one share with me formulas on how to calculate de Amp/hr for the
battery and the power of the solar panel?
Systems will consume average 80watts and need to keep running on a few
clowdy days as well.
Thanks for the assistance!

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.

S

#### Simon Dice

Jan 1, 1970
0
It looks like reduce the power consuption is the best idea to follow...

Best Regards

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

S

#### Simon Dice

Jan 1, 1970
0
It looks like reduce the power consuption is the best idea to follow...

Best Regards

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

P

#### Puckdropper

Jan 1, 1970
0
=============================
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.

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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