# Simple question but I'm lost...

Registered Users Posts: 4
Hi I'm new to battery setups with solar but not solar in general. My issue is understanding how much battery I need. I'm a bit confused as to the AH and discharge state.. I have a simple set of yard fountains the I've been running off of a 45 watt 24v (OC) panel that I drop down to 8v with a buck converter, it works great and has for a year now. But I would like to have my fountain run at night also. So my question is this.... Let's say I bought a 100ah battery and only let it drop to 11.5v (say its 50%) does that mean I will only have the 50ah out of 100ah?

Some details.
45 watt
Custom charge controller pwm type (adjustable).
Amp draw for fountains is 1amp.
Northern California about 6 solid hours of sun at peek output.

Thanks for the help guys, I'm sure its simple.

• Solar Expert Posts: 232 ✭✭
You will want to convert all your devices to the universal unit of watts, then the calculations will be more obvious. And yea, half a discharge is the same as half capacity.
Welcome to the forum Kicker.

Amp*Volts = Watts (a rate, like gallons per hour)
Amps * Volts * Hours = Watt*Hour (an amount, like gallons of gasoline used)

Amp*Hours is very common when working with DC systems. If you have a 12 volt battery system, then you can use Amps and Amp*Hours for everythihng.

However, with off grid solar power, working with Amp*Hours gets confusing. Amp*Hours is missing a major component, Voltage. To give you an idea, look at using a 120 Watt light bulb.
• 120 Watts / 120 volts = 1 amp of current flow
• 1 amp * 5 hours = 5 Amp*Hours at 120 VAC
However, if you had a 120 Watt load on a 12 volt system:
• 120 Watts / 12 volts = 10 amps of current flow
• 10 amps * 5 hours = 50 Amp*Hours of current at 12 VDC
So--if you "forget" to include voltage, you can be off by a factor of 10x between a 120 VAC load and a 12 VDC load.

However, if you use Voltage in the equation, you get Watts and Watt*Hours, which is "voltage" independent:
• 120 Watts * 5 hours = 600 Watt*Hours of energy used
• 600 Watt*Hours / 120 VAC = 5 Amp*Hours
• 600 Watt*Hours / 12 VDC = 50 Amp*Hours
So you work in Watts and Watt*Hours, then convert where needed to the correct voltage.

Lets say you have a 20 Watt water pump that runs at 8 volts. Current would be:
• 20 watts / 8 volts = 2.5 amps @ 8 volts DC (just to show how math works).
Now, lets say you have a 100 AH @ 12 volt Deep Cycle battery (note: Car batteries and even Marine/deep cycle batteries will not last near as long as a true deep cycle battery).
• 100 AH * 12 volts * 0.50 maximum discharge = 600 Watt*Hours of "useful" storage
• 600 Watt*Hours * 0.85 buck converter efficiency * 1/20 Watt load = 25.5 Hours of operation (roughly, taking wild guesses)
And to recharge a 100 AH @ 12 volt battery, generally recommend 5% to 13% rate of charge:
• 100 AH * 14.5 volts charging * 1/0.77 panel+controller derating * 0.05 rate of charge = 94 Watt minimum solar array (weekend/seasonal usage)
• 100 AH * 14.5 volts charging * 1/0.77 panel+controller derating * 0.10 rate of charge = 188 Watt nominal solar array (daily usage (9+ months of the year)
• 100 AH * 14.5 volts charging * 1/0.77 panel+controller derating * 0.13 rate of charge = 244 Watt maximum "cost effective" array
And then there is sizing based on your power usage... Say you wan to use the pump 12 hours per day and you get a reasonable amount of sun, roughly 4 hours per day of sun for ~9 months of the year average (excluding bad weather/marine layer days):
• 20 Watts * 12 hours * 1/0.52 off grid system eff * 1/4 hours of sun per day = 115 Watt array minimum (based on my guesses)
Anyway, that is the basic math. Of course, you should have some tools to measure you actual power usage (at the very least, a DC Current/DMM like this is a great learning tool).

-Bill
Near San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset
• Registered Users Posts: 4
Wow that's a lot of info!. Thanks and I'm a little more confused but I think I get it. Let's see if I'm on the right path here. The battery question was would I only have 50% ah at 50% and that is clear to me now so thank you. On to the fountain draw. I measure amp with my meter going from my battery to the charge controller so that I can see total draw...is this a good way to get my math? If so here is what I get.

Amps out is only .63 at 12 so that's .63*12= 7.56 watts. Correct?

Or do I go by the buck converter volts? I don't think so because this is prior to it and that's the main draw.

So assuming I draw 7.56watts an hour and I wanted to run my pumps at night for 16 hours (to be on the safe side) that's 7.56*16=120.96 watts.

And 120.96/12=10.08ah that means I need to produce 12amps or 144 watts (added some for losses) to recharge my battery in the day. Now my panels are 45watt or 3.75ah but I've only seen it reach 2.6ah so 31.2watts, with this math let's see..

144/31.2=4.62 hours of full sun needed.

So would I be wise to add a second set of 45watt panels to catch morning sun? I have 16 sets to use if I need them (family member passed away and gave them to me.) These are those cheap harbor freight panels but they do pull the 2.4-2.6amps.

Thanks guys. Let me know if I'm off track here.
Amps out is only .63 at 12 so that's .63*12= 7.56 watts. Correct?

Probably--As I understand you can measure the current either in the 12 volt wiring or in the "8 volt" wiring for the pump... So, the Wattage is based on the voltage of the wiring where you measured the current.
Or do I go by the buck converter volts? I don't think so because this is prior to it and that's the main draw.

If you measured on the 12 volts side of the buck converter--You have both the power needed to run the pump, but also the power needed to run the buck converter too (typically around 85 to 95% efficient).

So a
ssuming I draw 7.56watts an hour and I wanted to run my pumps at night for 16 hours (to be on the safe side) that's 7.56*16=120.96 watts.

Just to be clear... Watts is a "rate" (like miles per hour). Watts is actually Joules per unit time. In physics it would be Joules per second, but for "large power systems", we use Hours as our unit of time, or everything gets 3,600x larger (3,600 seconds per hour).

So, just 7.56 Watts

And 7.56 Watts * 16 hours = 121 Watt*Hours
And 120.96/12=10.08ah that means I need to produce 12amps or 144 watts (added some for losses) to recharge my battery in the day. Now my panels are 45watt or 3.75ah but I've only seen it reach 2.6ah so 31.2watts, with this math let's see..

121 Watt*Hours / 12 Volts = 10 Amp*Hours

Your panels are (apparently) rated at 3.75 Amps (not Amp*Hours, amps is a rate too). If your panels are generally outputting 2.6 Amps, then you would need:

10 Amp*Hours / 2.6 amps (full sun) =

144/31.2=4.62 hours of full sun needed.

So would I be wise to add a second set of 45watt panels to catch morning sun? I have 16 sets to use if I need them (family member passed away and gave them to me.) These are those cheap harbor freight panels but they do pull the 2.4-2.6amps.

Thanks guys. Let me know if I'm off track here.[/QUOTE]

Near San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset
Amps out is only .63 at 12 so that's .63*12= 7.56 watts. Correct?

Probably--As I understand you can measure the current either in the 12 volt wiring or in the "8 volt" wiring for the pump... So, the Wattage is based on the voltage of the wiring where you measured the current.
Or do I go by the buck converter volts? I don't think so because this is prior to it and that's the main draw.

If you measured on the 12 volts side of the buck converter--You have both the power needed to run the pump, but also the power needed to run the buck converter too (typically around 85 to 95% efficient).

So a
ssuming I draw 7.56watts an hour and I wanted to run my pumps at night for 16 hours (to be on the safe side) that's 7.56*16=120.96 watts.

Just to be clear... Watts is a "rate" (like miles per hour). Watts is actually Joules per unit time. In physics it would be Joules per second, but for "large power systems", we use Hours as our unit of time, or everything gets 3,600x larger (3,600 seconds per hour).

So, just 7.56 Watts
• And 7.56 Watts * 16 hours = 121 Watt*Hours
And 120.96/12=10.08ah that means I need to produce 12amps or 144 watts (added some for losses) to recharge my battery in the day. Now my panels are 45watt or 3.75ah but I've only seen it reach 2.6ah so 31.2watts, with this math let's see..
• 121 Watt*Hours / 12 Volts = 10 Amp*Hours
Your panels are (apparently) rated at 3.75 Amps (not Amp*Hours, amps is a rate too). If your panels are generally outputting 2.6 Amps, then you would need:
• 10 Amp*Hours / 2.6 amps = 3.84 hours minimum of sun per day
Depending on where you live, in the sunny southwest you can see >4 hours of (noon time equivalent) sun per day for 9+ months a year. In many less sunny areas, you may see a long term average of 2-3 hours of sun per day in the 3 months of winter.
So would I be wise to add a second set of 45watt panels to catch morning sun? I have 16 sets to use if I need them (family member passed away and gave them to me.) These are those cheap harbor freight panels but they do pull the 2.4-2.6amps.

From a practical point of view, having 2x the number of solar panels (based on you loads) would be healthier for the battery bank (quickly recharge battery, especially if you have a few days of clouds).

If you have a 100 AH @ 12 Volt battery bank, see my above calculations for 5% to 13% rate of charge... You should have at least 5% minimum rate of charge (10%+ is better)... Of course, with more solar panels, and a good size battery bank, you can add some LED accent lighting and such.

We always try to design the system to support the loads. And lead acid batteries do not like to be under charged/over discharged--Solar panels have never been cheaper--So adding solar panels (plus charge controller capacity) generally will give you better battery life and help you through stormy weather (if you forget to turn off the pump with a week of rain--Do not want to over discharge the battery bank--it will kill it).

-Bill

Near San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset
• Registered Users Posts: 4
Ok seems i'm on the right path now, Thanks for all the info on this. This is for testing now and getting my feet wet for some day moving to larger setups. I'm looking at just running these fountains right now. I was looking at a simple battery on the site http://www.solar-electric.com/batteries-meters-accessories/batteries/unba/unba35amseag.html. I understand that if I get a rainy day i'm more then likely not going to get 2 nights on it because it's only 17.5 usable AH or 210 watts correct?. I'm ok if it doesn't last more then one night as this is just for my fountains and testing. I'm just looking for something that can last the 16 hours (121 watts) i need for 1 night. Would this little battery cover me assuming I can charge it in one day?

I think for faster charging i'll toss up 1 more 45 watts giving me a total of 90 watts at peek. That should based on my amps i see with the one... i'm not getting full 45 watts because 2.6(amps)*12v=31.2 watts so 31.2*2=62.4 watts

So to get my needed 120 watts i'd need 120/62=1.94 hours.

Is this the smart option to being able to use my pump throughout the summer months?

Like I said i'm not looking for the weather coverage so if it's cloudy I understand it won't work and i'm fine with that. I just want to make sure my math is correct so that I can plan for bigger projects down the road. Awesome Forum here by the way. Very knowledgeable people willing to help us slower users :P
Sorry to be correcting your units... But once you get the hang of them, it makes things much easier to understand and model.
Ok seems i'm on the right path now, Thanks for all the info on this. This is for testing now and getting my feet wet for some day moving to larger setups. I'm looking at just running these fountains right now. I was looking at a simple battery on the site http://www.solar-electric.com/batteries-meters-accessories/batteries/unba/unba35amseag.html. I understand that if I get a rainy day i'm more then likely not going to get 2 nights on it because it's only 17.5 usable AH or 210 watts*Hours correct?. I'm ok if it doesn't last more then one night as this is just for my fountains and testing. I'm just looking for something that can last the 16 hours (121 watts*Hours) i need for 1 night. Would this little battery cover me assuming I can charge it in one day?

Remember there are losses too... An AGM battery is roughly 90% (or better) efficient (using Watts/Watt*Hours). And a flooded cell battery is around 80% or so efficient (again based on power).

The battery losses are from charging at ~14.7 volts and discharging at ~12.0 volts. (AGM batteries have a smaller voltage swing between charging and discharging voltages).

AGMs are "better" lead acid batteries (cleaner, don't need watering, lower internal resistance). However, because you cannot measure the specific gravity of the electrolyte, they are a bit more difficult to monitor for state of charge. Also, because they are sealed, they are easier to damage by overcharging.

http://www.windsun.com/Batteries/Battery_FAQ.htm
http://www.batteryfaq.org/
http://batteryuniversity.com/

Lead acid batteries are near 100% efficient for Amp and Amp*Hours efficiency (another reason using Amp*Hours for capacity is common).
I think for faster charging i'll toss up 1 more 45 watts giving me a total of 90 watts at peek. That should based on my amps i see with the one... i'm not getting full 45 watts because 2.6(amps)*12v=31.2 watts so 31.2*2=62.4 watts

Yes--But remember that your batteries need to charge about 14.4 volts for ~2-6 hours to fully recharge. So your charging wattage is roughly:

2.6 amps * 14.4 volts = 37.4 Watts
So to get my needed 120 watts*Hours i'd need 120 Watt*Hours / 62 Watts =1.94 hours.

It does get a bit confusing with all of the conversions and keeping track of losses... But that is a good estimate.
Is this the smart option to being able to use my pump throughout the summer months?

As you say, you are experimenting and learning. About the only thing you can damage is the Lead Acid Battery--By over charging (if you don't have a charge controller, or it is set wrong). Or by over discharging (basically taking the battery below ~12.0 to 11.5 volts).

Otherwise, as long as you don't connect the solar panels "backwards" (connecting solar panels backwards directly to the battery bank can smoke the panels very easily), or short stuff out--It should be pretty forgiving.
Like I said i'm not looking for the weather coverage so if it's cloudy I understand it won't work and i'm fine with that. I just want to make sure my math is correct so that I can plan for bigger projects down the road. Awesome Forum here by the way. Very knowledgeable people willing to help us slower users :P

You are very welcome... Let us know how it all works out for you. We all started at the same place. And we learned too (hear, school, playing around at home, etc.).

-Bill
Near San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset
• Registered Users Posts: 4
Yeah i've read a lot into the AGM batteries already. My charge controller is from a site i came across that I thought was good for my little project for the price and the easily repairable parts (i'm a nut about being able to repair/rebuilt my things). This is the charge controller i'm using for my battery setup i've been playing with... http://www.opend.co.za/hardware/200ds232/index.html I also like the fact i can adjust it to different settings. Take note that i'm not using the calibration of 10.5v - 13.8 instead i'm set to 11.5v-14.4v.

I'm starting to see that I was over thinking this project as far as converting everything to amps and going off that...watts just makes more since now. Thanks again and i'll be reporting back once i can get some time to gather a new battery.