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The unit I will be using is 12V 0.5A

12V X 0.5A = 6 Watts 6W x 24hours = 144 Watts per day 144W / 12V = 12 amp hours

Now given a battery with 110 amp hours (http://www.defender.com/product.jsp?path=-1 |328|51495|306219&idQ496)

We should only draw the battery down 50% so we will actually only get

110 / 2 = 55 amp hours

12 amp hours needed / 55 amp hours available = 4.58 days so with this battery we have enough juice to potentially run our unit for 4.58 days w/out charging.

Now if the above is not correct then please explain. And the next question is how would I now figure out approx how large of a panel I would need to keep the battery charged. I am really looking for the formula and a understanding of how you got from A to B as oppose to "this panel should work".

Wrong. 6W x 24hours = 144 Watt-HOURS per day.

Watts are a unit of power. A watt-hour (like a kilowatt-hour) is a unit of energy.

Hope this helps.

This is really more a theoretical or academic answer than a practical one but here goes:

Assuming the battery is 12V (which it's not) and STAYS at 12V (which it won't), all you need to know is the draw (0.5A @ 12V) and the capacity of the battery (110AH).

Use your algebra and cross out the units.

110AH/0.5A = 220 hours to completely kill the battery or (approx) 110 hrs to kill the battery by 50%.

Of course, batteries don't discharge in a linear fashion, and ... and... and...

Yep. I think you have the basic idea, but in this stuff you have to be careful to dot your 'i's and cross your 't's. (this group is good at pointing out mix ups like this). It may seem rather 'anal' at times, but it***is*** important to get the right results ;-)

Well, you know that you need about 144 watt-hours per day on average. But how many hours a day does the sun shine in your area? That is the another factor needed to figure out the panel sizing.

If you're drawing 144 watt-hours***out*** of your battery, how many watt-hours
do you have to put in? More, unless you've found the perfect battery.
Assume 80% of the energy you put into the battery gets lost, so to get 144
watt-hours ***out***, you should plan on putting 144/0.80 = 180 watt-hours ***in***.

Now, perhaps the worst season would be winter, and perhaps in your area the sun only shines, 'on average' 3 full-sun hours a day in winter time. To get 180 watt-hours of energy to put***in*** your battery from a charge controller,
in just 3 hours, you would need a charge controller output of 180
watt-hours / 3 hours = 60 watts. (at 12V, that would be 5 amps).

Is this just for a year or two, or longer? The reason I ask, is that panel output degrades over time. Some manufacturers warrant for 80% after 20 years. I suspect you aren't planning some 20-year long plan here, so lets just***assume*** panel output is 90% of rating under the conditions you'll be
operating. So to be sure of 60 watts at the charge controller output, 60
watts / 0.90 = 67 watts panel output.

Of course, some days your sunshine will be more, and that will help recharge your battery. But if you run real close to just those 3 hours per day, then you're going to have a hard time recharging some weeks in the winter/cloudy season. Moral of this is, if you have another means to fully charge your battery sometimes, you might be okay. But if the panel is the sole charging, and no interruption of power can be tolerated, you'll need even more panels to manage recharging on those partly-cloudy days.

Hope this helps you along.

daestrom

You're right of course. I meant to say that perhaps only 80% of the energy you put into a battery is recoverable. So to get 144 watt-hours out, you would indeed need to put in 144/0.80 = 180 watt-hours in.

daestrom

#### Site Timeline

- posted on January 5, 2006, 9:24 pm

12V X 0.5A = 6 Watts 6W x 24hours = 144 Watts per day 144W / 12V = 12 amp hours

Now given a battery with 110 amp hours (http://www.defender.com/product.jsp?path=-1 |328|51495|306219&idQ496)

We should only draw the battery down 50% so we will actually only get

110 / 2 = 55 amp hours

12 amp hours needed / 55 amp hours available = 4.58 days so with this battery we have enough juice to potentially run our unit for 4.58 days w/out charging.

Now if the above is not correct then please explain. And the next question is how would I now figure out approx how large of a panel I would need to keep the battery charged. I am really looking for the formula and a understanding of how you got from A to B as oppose to "this panel should work".

--

Poor planning on your part does not constitute an emergency on my part.

Poor planning on your part does not constitute an emergency on my part.

- posted on January 5, 2006, 10:25 pm

Wrong. 6W x 24hours = 144 Watt-HOURS per day.

Watts are a unit of power. A watt-hour (like a kilowatt-hour) is a unit of energy.

Hope this helps.

- posted on January 5, 2006, 11:20 pm

– Colonel – wrote:

That should be 6W x 24hr = 144 watt-hours.

hvacrmedic

That should be 6W x 24hr = 144 watt-hours.

hvacrmedic

- posted on January 5, 2006, 10:35 pm

This is really more a theoretical or academic answer than a practical one but here goes:

Assuming the battery is 12V (which it's not) and STAYS at 12V (which it won't), all you need to know is the draw (0.5A @ 12V) and the capacity of the battery (110AH).

Use your algebra and cross out the units.

110AH/0.5A = 220 hours to completely kill the battery or (approx) 110 hrs to kill the battery by 50%.

Of course, batteries don't discharge in a linear fashion, and ... and... and...

- posted on January 6, 2006, 2:32 am

allow lots of money for battery replacements.

better to upsize battery bank and discharge less.

even deep cycle batteries have tough tme with frequent deep discharges

better to upsize battery bank and discharge less.

even deep cycle batteries have tough tme with frequent deep discharges

- posted on January 6, 2006, 3:03 am

snipped-for-privacy@aol.com wrote:

Using $85 T105's (6v * 225ah) at 50% discharge, he's good for 7+ years.

Using $85 T105's (6v * 225ah) at 50% discharge, he's good for 7+ years.

--

Steve Spence

Dir., Green Trust, http://www.green-trust.org

Steve Spence

Dir., Green Trust, http://www.green-trust.org

Click to see the full signature.

- posted on January 5, 2006, 10:39 pm

I'm not sure you realize it but some of your figures/formulae are
unnecessary.

More simply Battery 110 AH (/2) Load = .5 A 55/.5 = 110 hrs

Typically batteries are charged at .10 of their 1 hr rate (when fully discharged) but the type of battery (Gel-Cell, NiCad, NiMH, etc.), must be considered. Also, while .1C might be good for a discharged battery, the charge should diminish as the battery becomes fully charged (or some other mechanism, temperature rise, e.g.), must shut down the charger.)

See the battery mfr. for recommended charge rate and charger technology. Also, consider how fast a recharge you want. A 4 ampere charger will recharge the battery (but slowly) as will a 100 Amp charger

Bryan Martin wrote:

More simply Battery 110 AH (/2) Load = .5 A 55/.5 = 110 hrs

Typically batteries are charged at .10 of their 1 hr rate (when fully discharged) but the type of battery (Gel-Cell, NiCad, NiMH, etc.), must be considered. Also, while .1C might be good for a discharged battery, the charge should diminish as the battery becomes fully charged (or some other mechanism, temperature rise, e.g.), must shut down the charger.)

See the battery mfr. for recommended charge rate and charger technology. Also, consider how fast a recharge you want. A 4 ampere charger will recharge the battery (but slowly) as will a 100 Amp charger

Bryan Martin wrote:

- posted on January 5, 2006, 11:49 pm

Thanks for the replies. Unless I am missing something the formula is sound
but not worded correctly? I will try to find more info on this. The reason
behind the extended formula's is for better understanding and also the end
result and second part of this question is that I plan on hooking this up to
a solar panel if possible. I was trying to get an idea about what I would
be using per day/hour if run 24/7 and what size panel I would need for this.
Then depending on how that works out $$$ wise for the size panel I would
either stay at that or look at some kind of timer setup that would allow the
unit to run for X hours a day. I would rather have it on 24X7 but the
actual real world usage of the unit would be around 3 hours per day on a
weekday and posssibly more on the weekends.

So I am still needing some help with the panel size I will be needing to keep this thing up and running.

So given 144 watt-hours per day would that work out to?

144 / 5 hours of sunlight = 28.8 W panel?

So I am still needing some help with the panel size I will be needing to keep this thing up and running.

So given 144 watt-hours per day would that work out to?

144 / 5 hours of sunlight = 28.8 W panel?

- posted on January 6, 2006, 1:54 am

Yep. I think you have the basic idea, but in this stuff you have to be careful to dot your 'i's and cross your 't's. (this group is good at pointing out mix ups like this). It may seem rather 'anal' at times, but it

Well, you know that you need about 144 watt-hours per day on average. But how many hours a day does the sun shine in your area? That is the another factor needed to figure out the panel sizing.

If you're drawing 144 watt-hours

Now, perhaps the worst season would be winter, and perhaps in your area the sun only shines, 'on average' 3 full-sun hours a day in winter time. To get 180 watt-hours of energy to put

Is this just for a year or two, or longer? The reason I ask, is that panel output degrades over time. Some manufacturers warrant for 80% after 20 years. I suspect you aren't planning some 20-year long plan here, so lets just

Of course, some days your sunshine will be more, and that will help recharge your battery. But if you run real close to just those 3 hours per day, then you're going to have a hard time recharging some weeks in the winter/cloudy season. Moral of this is, if you have another means to fully charge your battery sometimes, you might be okay. But if the panel is the sole charging, and no interruption of power can be tolerated, you'll need even more panels to manage recharging on those partly-cloudy days.

Hope this helps you along.

daestrom

- posted on January 6, 2006, 4:40 am

daestrom wrote:

20% efficiency for a battery is pretty poor, Daestrom. Perhaps you would like to rephrase this.

so to get 144

20% efficiency for a battery is pretty poor, Daestrom. Perhaps you would like to rephrase this.

so to get 144

- posted on January 6, 2006, 4:49 am

daestrom wrote:

I agree...

I know what you're thinking and you'd get it right. But you worded it backwards.

Perhaps something like the following is what you meant: "Assume the battery storage is 80% efficient, 20% of the energy you put into the battery gets lost"

I agree...

I know what you're thinking and you'd get it right. But you worded it backwards.

Perhaps something like the following is what you meant: "Assume the battery storage is 80% efficient, 20% of the energy you put into the battery gets lost"

- posted on January 6, 2006, 1:24 pm

You're right of course. I meant to say that perhaps only 80% of the energy you put into a battery is recoverable. So to get 144 watt-hours out, you would indeed need to put in 144/0.80 = 180 watt-hours in.

daestrom

- posted on January 6, 2006, 2:46 am

Bryan Martin wrote:

Walmart has 115 ah deep cycle batteries for $55

Your math is correct. In NY, with about 3 full sun hours daily max, best case scenario I'd need a 50 watt panel. 50 watts * 3 hours = 150 watt-hours daily.

Walmart has 115 ah deep cycle batteries for $55

Your math is correct. In NY, with about 3 full sun hours daily max, best case scenario I'd need a 50 watt panel. 50 watts * 3 hours = 150 watt-hours daily.

--

Steve Spence

Dir., Green Trust, http://www.green-trust.org

Steve Spence

Dir., Green Trust, http://www.green-trust.org

Click to see the full signature.

- posted on January 6, 2006, 10:17 am

Hi Bryan

On Thu, 05 Jan 2006 21:46:19 -0500, Steve Spence

Steve and others have pointed out that your basic maths is correct, but there is a matter of losses to be considered. One that hasn't been mentioned I think in this thread is that a 60 watt panel does NOT put in 60 watts to your 12v battery unless you use a clever charge controller like an MPPT (which you wouldn't use for one panel). This is because panels are not rated at the voltage of the battery, but at some higher voltage called the maximum power voltage. One person commented that you should look at the CURRENT provided by the panel in your situation (rather than the wattage). A typical figure is for a panel to put about three quarters of its rated output into the battery (45 watts for a 60 watt panel). If you need about a 5 amp panel, this will be a panel rated at about 85 watts (= 5 amps X 17 volts). Personally, I would aim at about a 75 to 80 watt panel for what you want to do. The difference in price may turn out to be not all that great anyway as you pay more $ per watt for smaller panels usually.

Eric

On Thu, 05 Jan 2006 21:46:19 -0500, Steve Spence

Steve and others have pointed out that your basic maths is correct, but there is a matter of losses to be considered. One that hasn't been mentioned I think in this thread is that a 60 watt panel does NOT put in 60 watts to your 12v battery unless you use a clever charge controller like an MPPT (which you wouldn't use for one panel). This is because panels are not rated at the voltage of the battery, but at some higher voltage called the maximum power voltage. One person commented that you should look at the CURRENT provided by the panel in your situation (rather than the wattage). A typical figure is for a panel to put about three quarters of its rated output into the battery (45 watts for a 60 watt panel). If you need about a 5 amp panel, this will be a panel rated at about 85 watts (= 5 amps X 17 volts). Personally, I would aim at about a 75 to 80 watt panel for what you want to do. The difference in price may turn out to be not all that great anyway as you pay more $ per watt for smaller panels usually.

Eric

- posted on January 7, 2006, 2:28 pm

It does not work that way because panels are rated at 17 volts, not 12. You
have to work with the panels amps not watts.
---------------------

- posted on January 8, 2006, 1:30 am

Windsun wrote:

unless one has a mppt controller.

without, a 50 watt panel might produce 3a x 3h or 9ah.

9ah x 12v = 108 wh (not 150wh) into the battery, of which 86wh (7ah) might be available for use (20% conversion inefficiency). That's less than his needed 12ah, so a 75 watt panel would be more appropriate in NY, but I do not know where the OP is located.

unless one has a mppt controller.

without, a 50 watt panel might produce 3a x 3h or 9ah.

9ah x 12v = 108 wh (not 150wh) into the battery, of which 86wh (7ah) might be available for use (20% conversion inefficiency). That's less than his needed 12ah, so a 75 watt panel would be more appropriate in NY, but I do not know where the OP is located.

--

Steve Spence

Dir., Green Trust, http://www.green-trust.org

Steve Spence

Dir., Green Trust, http://www.green-trust.org

Click to see the full signature.

- posted on January 6, 2006, 5:48 am

Bryan Martin wrote:

In the case of float charging, your charger has to generate higher output than constant current draw rate plus some + tolerance to compensate loss. Otherwise the battery will run down eventually. Also think about winter, summer situation. Tony

In the case of float charging, your charger has to generate higher output than constant current draw rate plus some + tolerance to compensate loss. Otherwise the battery will run down eventually. Also think about winter, summer situation. Tony

- posted on January 6, 2006, 12:53 pm

Thanks everyone for shedding such light on it. I think I have enough info
to get started :)

- posted on January 7, 2006, 11:27 pm

For information on batteries the following site is very good:

http://www.batteryfaq.org

http://www.batteryfaq.org

- posted on January 6, 2006, 2:54 pm

wrote:

I would approach it a bit differently. And I think your error in your calculations is an assumption that the battery charge/discharge is 100% efficient. It's closer to 80%.

1. Daily Load is 12AH (@ 12V) as you calculated. But with DC loads you can do your calculations directly in Amps. So 0.5A * 24 hrs = 12AH (@12V).

2. Add 20% for system losses and a safety factor --> 14.40 AH. You don't get 100% efficiency in charging/discharging batteries.

3. Calculate your PV requirement based on the number of effective sun hours at your site during the worst month. Not knowing where you live it's tough to say, but, for example, the San Francisco area that might be 4 hrs.

That gives you a required PV array current of 3.60 amperes (14.40/4). Now you can go look for an appropriate PV panel (or two, if need be). A BP365 panel is rated at 3.69A at PMax, as an example.

4. Now decide on your DOA (days of autonomy) to obtain required battery size. 5 DOA with a 50% maximum discharge --> 144AH battery.

Or, working backwards, your 110AH battery would give you 3.8 DOA.

-- ron (off the grid in Downeast Maine)

I would approach it a bit differently. And I think your error in your calculations is an assumption that the battery charge/discharge is 100% efficient. It's closer to 80%.

1. Daily Load is 12AH (@ 12V) as you calculated. But with DC loads you can do your calculations directly in Amps. So 0.5A * 24 hrs = 12AH (@12V).

2. Add 20% for system losses and a safety factor --> 14.40 AH. You don't get 100% efficiency in charging/discharging batteries.

3. Calculate your PV requirement based on the number of effective sun hours at your site during the worst month. Not knowing where you live it's tough to say, but, for example, the San Francisco area that might be 4 hrs.

That gives you a required PV array current of 3.60 amperes (14.40/4). Now you can go look for an appropriate PV panel (or two, if need be). A BP365 panel is rated at 3.69A at PMax, as an example.

4. Now decide on your DOA (days of autonomy) to obtain required battery size. 5 DOA with a 50% maximum discharge --> 144AH battery.

Or, working backwards, your 110AH battery would give you 3.8 DOA.

-- ron (off the grid in Downeast Maine)

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