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#1
Posted to alt.solar.photovoltaic,alt.home.repair




Help with calculating and understanding
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=132851495306219&id=51496) 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. 
#2
Posted to alt.solar.photovoltaic,alt.home.repair




Help with calculating and understanding
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 (GelCell, 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: 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=132851495306219&id=51496) 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". 
#3
Posted to alt.solar.photovoltaic,alt.home.repair




Help with calculating and understanding
– Colonel – wrote: On 20060105 16:24:35 0500, "Bryan Martin" said: The unit I will be using is 12V 0.5A 12V X 0.5A = 6 Watts 6W x 24hours = 144 Watts per day Wrong. 6W x 24hours = 144 WattHOURS per day. That should be 6W x 24hr = 144 watthours. hvacrmedic Watts are a unit of power. A watthour (like a kilowatthour) is a unit of energy. Hope this helps. 
#4
Posted to alt.solar.photovoltaic,alt.home.repair




Help with calculating and understanding
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 watthours per day would that work out to? 144 / 5 hours of sunlight = 28.8 W panel? "Bennett Price" wrote in message . .. 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 (GelCell, 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: 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=132851495306219&id=51496) 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". 
#5
Posted to alt.solar.photovoltaic,alt.home.repair




Help with calculating and understanding
"Bryan Martin" wrote in message news Thanks for the replies. Unless I am missing something the formula is sound but not worded correctly? 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 ;) 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. Well, you know that you need about 144 watthours 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 watthours *out* of your battery, how many watthours 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 watthours *out*, you should plan on putting 144/0.80 = 180 watthours *in*. Now, perhaps the worst season would be winter, and perhaps in your area the sun only shines, 'on average' 3 fullsun hours a day in winter time. To get 180 watthours of energy to put *in* your battery from a charge controller, in just 3 hours, you would need a charge controller output of 180 watthours / 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 20year 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 partlycloudy days. Hope this helps you along. daestrom 
#6
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Help with calculating and understanding
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 
#7
Posted to alt.solar.photovoltaic,alt.home.repair




Help with calculating and understanding
Bryan Martin wrote:
The unit I will be using is 12V 0.5A 12V X 0.5A = 6 Watts 6W x 24hours = 144 Watthours per day 144W / 12V = 12 amp hours Now given a battery with 110 amp hours (http://www.defender.com/product.jsp?path=132851495306219&id=51496) Walmart has 115 ah deep cycle batteries for $55 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". 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 watthours daily.  Steve Spence Dir., Green Trust, http://www.greentrust.org Contributing Editor, http://www.offgrid.net http://www.rebelwolf.com/essn.html 
#8
Posted to alt.solar.photovoltaic,alt.home.repair




Help with calculating and understanding
wrote:
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 Using $85 T105's (6v * 225ah) at 50% discharge, he's good for 7+ years.  Steve Spence Dir., Green Trust, http://www.greentrust.org Contributing Editor, http://www.offgrid.net http://www.rebelwolf.com/essn.html 
#9
Posted to alt.solar.photovoltaic,alt.home.repair




Help with calculating and understanding
daestrom wrote: "Bryan Martin" wrote in message news Thanks for the replies. Unless I am missing something the formula is sound but not worded correctly? 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 ;) 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. Well, you know that you need about 144 watthours 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 watthours *out* of your battery, how many watthours 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, 20% efficiency for a battery is pretty poor, Daestrom. Perhaps you would like to rephrase this. so to get 144 watthours *out*, you should plan on putting 144/0.80 = 180 watthours *in*. Now, perhaps the worst season would be winter, and perhaps in your area the sun only shines, 'on average' 3 fullsun hours a day in winter time. To get 180 watthours of energy to put *in* your battery from a charge controller, in just 3 hours, you would need a charge controller output of 180 watthours / 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 20year 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 partlycloudy days. Hope this helps you along. daestrom 
#10
Posted to alt.solar.photovoltaic,alt.home.repair




Help with calculating and understanding
daestrom wrote: 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 ;) I agree... If you're drawing 144 watthours *out* of your battery, how many watthours 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 watthours *out*, you should plan on putting 144/0.80 = 180 watthours *in*. 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" 
#11
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Help with calculating and understanding
Bryan Martin wrote:
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=132851495306219&id=51496) 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". Hi, 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 
#12
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Help with calculating and understanding
Hi Bryan
On Thu, 05 Jan 2006 21:46:19 0500, Steve Spence wrote: 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 watthours daily. 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 
#13
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Help with calculating and understanding
Thanks everyone for shedding such light on it. I think I have enough info
to get started "Bryan Martin" wrote in message ... 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=132851495306219&id=51496) 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. 
#14
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Help with calculating and understanding
"Tony Wesley" wrote in message ups.com... daestrom wrote: 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 ;) I agree... If you're drawing 144 watthours *out* of your battery, how many watthours 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 watthours *out*, you should plan on putting 144/0.80 = 180 watthours *in*. 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" 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 watthours out, you would indeed need to put in 144/0.80 = 180 watthours in. daestrom 
#15
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Help with calculating and understanding
On Thu, 05 Jan 2006 21:24:35 GMT, "Bryan Martin"
wrote: 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=132851495306219&id=51496) 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". 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) 
#16
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Help with calculating and understanding
Your calculations are a bit off, because batteries are rated at the 20 hour
rate, but you are assuming about a 120 hour discharge time, so the actual capacity will be somewhat more  around 20%. But that won't really affect the basic calculations  all it does really is give some safety factor. The panel size is a bit more complicated, but not rocket science. First you need to determine how important the load is  will it be a major event if whatever you are running goes offline. If so, then you need to factor in a pretty large safety factor  35% or more  for those rare occasions when you get much longer than usual cloudy periods. Basically, 100% uptime will cost you a lot more than 99% uptime. But ignore that for now  the basic calcuation is that you want your solar panel to supply what you use each day + around 20%. That is for the WORST case conditions  such as you might see seasonally with lots of clouds. Assuming you look at the insolation charts, and you get worst case in December of 4 hours full sun per day. You are drawing .5 amps over 24 hours, or 12 amphours. You need to replace that 12 AH in during the 4 hours of sun you get, so you need a panel that will supply 12/4 = 3 amps. Add 2025% or so safety factor, and you come up with around 3.5 amps. Most panels are rated at 17 volts or so, so 3.5 amps x 17 volts = about a 60 watt panel.  Free Solar Discussion Forum: http://www.windsun.com/smf/index.php  "Bryan Martin" wrote in message ... 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 
#17
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Help with calculating and understanding
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.  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 watthours daily.  Steve Spence Dir., Green Trust, http://www.greentrust.org Contributing Editor, http://www.offgrid.net http://www.rebelwolf.com/essn.html 
#18
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Help with calculating and understanding
Windsun wrote:
... Basically, 100% uptime will cost you a lot more than 99% uptime. If cloudy days were coinflips, storing enough energy for 1 day would make 50% uptime possible, 2 would make 75%, with 3 for 88%, 4 for 94%, 5 for 97%, 6 for 98%, and 7 for 99%... 100% would be impossible. Nick 
#19
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wrote in message ... Windsun wrote: ... Basically, 100% uptime will cost you a lot more than 99% uptime. If cloudy days were coinflips, But they are not, so the rest of your supposition is false. Anyone that watches the 6:00 o'clock local news and weather can tell that when you have one cloudy day, the chances of the next day also being cloudy are higher. Similarly, when one day is sunny, the probability of the next day also being sunny are higher than 50/50. (and yes, I have evidence for that 'article of faith'. Look on NOAA's web site for any given weather reporting station and do the stats on cloudy days for a year or two) storing enough energy for 1 day would make 50% uptime possible, 2 would make 75%, with 3 for 88%, 4 for 94%, 5 for 97%, 6 for 98%, and 7 for 99%... 100% would be impossible. daestrom 
#20
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Help with calculating and understanding
For information on batteries the following site is very good:
http://www.batteryfaq.org "Bryan Martin" wrote in message ... Thanks everyone for shedding such light on it. I think I have enough info to get started "Bryan Martin" wrote in message ... 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=132851495306219&id=51496) 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. 
#21
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You are using a common math fallacy that proves nothing.
We have installed many many 100% uptime systems over the past 30 years or so. It is quite simple, it is just more expensive.  Northern Arizona Wind & Sun  Electricity From the Sun Free Solar Discussion Forum: http://www.windsun.com/smf/index.php  wrote in message ... Windsun wrote: ... Basically, 100% uptime will cost you a lot more than 99% uptime. If cloudy days were coinflips, storing enough energy for 1 day would make 50% uptime possible, 2 would make 75%, with 3 for 88%, 4 for 94%, 5 for 97%, 6 for 98%, and 7 for 99%... 100% would be impossible. Nick 
#22
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He did say "IF"
"Windsun" wrote in message news You are using a common math fallacy that proves nothing. We have installed many many 100% uptime systems over the past 30 years or so. It is quite simple, it is just more expensive.   Northern Arizona Wind & Sun  Electricity From the Sun Free Solar Discussion Forum: http://www.windsun.com/smf/index.php   wrote in message ... Windsun wrote: ... Basically, 100% uptime will cost you a lot more than 99% uptime. If cloudy days were coinflips, storing enough energy for 1 day would make 50% uptime possible, 2 would make 75%, with 3 for 88%, 4 for 94%, 5 for 97%, 6 for 98%, and 7 for 99%... 100% would be impossible. Nick 
#23
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Windsun wrote:
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. 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.greentrust.org Contributing Editor, http://www.offgrid.net http://www.rebelwolf.com/essn.html 
#24
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In article , "Windsun" wrote:
You are using a common math fallacy that proves nothing. We have installed many many 100% uptime systems over the past 30 years or so. It is quite simple, it is just more expensive. Hogwash. *Nothing* works *every* time.  Regards, Doug Miller (alphageek at milmac dot com) It's time to throw all their damned tea in the harbor again. 
#25
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Try stopping breathing for 15 minutes. It works
***EVERYTIME*** Wash your pigs someplace else. "Doug Miller" wrote in message . com... Hogwash. *Nothing* works *every* time.  Regards, Doug Miller (alphageek at milmac dot com) It's time to throw all their damned tea in the harbor again. 
#26
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Doug Miller wrote:
We have installed many many 100% uptime systems over the past 30 years or so. It is quite simple, it is just more expensive. Hogwash. *Nothing* works *every* time. The sad truth is that solar weather is probabilistic, like 100year floods or coin flips. No matter how much energy storage we provide, if we wait long enough, there will always be a long enough cloudy day sequence to exhaust our energy store and leave the house cold or the lights out, unless we have some other backup system, eg a woodstove or a generator. Nick 
#27
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#28
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It has been one dark dreary depressing year end season
eh? We saw some sun for an hour this am then about 30 minutes too late at 4pm today. Yuk! My PV panels are running dry and shrivelling up from lack of use. Steve(Crocket)Spence must be on the parafin candles by now...LOL "Tony Wesley" wrote in message ps.com... wrote: Windsun wrote: ... Basically, 100% uptime will cost you a lot more than 99% uptime. If cloudy days were coinflips, storing enough energy for 1 day would make 50% uptime possible, 2 would make 75%, with 3 for 88%, 4 for 94%, 5 for 97%, 6 for 98%, and 7 for 99%... 100% would be impossible. Just an FYI, in the Detroit area, we just went from Dec 20 through Jan 5 (inclusive) with 14 minutes total sunshine. 
#30
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Help with calculating and understanding
Jim Baber
Tony Wesley wrote: wrote: Windsun wrote: .. Basically, 100% uptime will cost you a lot more than 99% uptime. If cloudy days were coinflips, storing enough energy for 1 day would make 50% uptime possible, 2 would make 75%, with 3 for 88%, 4 for 94%, 5 for 97%, 6 for 98%, and 7 for 99%... 100% would be impossible. Just an FYI, in the Detroit area, we just went from Dec 20 through Jan 5 (inclusive) with 14 minutes total sunshine. Another FYI, 'Total sunshine' is not required to generate appreciable power. Today it has been cloudy for the 9 hours (so far) of daylight. My 10 kW system has been far from it's most productive for this time of year today. It has only produced 21 kWh today Feb. 4, 2006. On on the other side of the coin, on Feb. 1, 2006 it made 44 kWh on a beautiful Fresno winter day. So far this year the least I have produced was 0.568 kWh (truly terrible) on one cold rainy dark day in January. (Which is why I'm on the grid.) :) 
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