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Author Topic: lead acid battery efficiency  (Read 746 times)
billi
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« on: February 05, 2012, 12:25:20 AM »

That is quite a good research


http://www.google.ie/url?sa=t&rct=j&q=leadacidbatterieefficiency&source=web&cd=4&ved=0CFIQFjAD&url=http%3A%2F%2Fwww.sandia.gov%2Fpv%2Fdocs%2FPDF%2Fbatpapsteve.pdf&ei=TsctT7mQHoSp0QXK2IWuCA&usg=AFQjCNEPUW8iKyzyB82BBrhufkmbhAjLfw&cad=rja


But still  i  do not understand efficiency ratings  of lead acid batteries  

Some reports rate them  as 60-90%  efficient , when i read this report it undermines my opinion  that, they are pretty efficient when not full or under 80 %  , but to charge them 100% full is a long way ............    

The question is ,  does it take  so much more Ah   to fully charge a battery    or is it just the time that it takes..........

If it is only longer time  , than why should efficiency drop so much , when AH pushed in seem the same ?

It is pretty obvious in my set-up , that AMPs from the PV  are reduced when battery gets close to full, but if i use these AMPS left over from the PV  for something else  , then these AMPS  should not be calculated in  .......

Or  am i totally wrong   and one needs much more AH to put in ,that one is able to take out

 wackoold


Billi

« Last Edit: February 05, 2012, 12:29:54 AM by billi » Logged

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camillitech
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« Reply #1 on: February 05, 2012, 06:54:53 AM »

Morning Billi,

not read the article yet, it's 6:50 and I'm off for a bath before going out to feed the pigs  Roll Eyes I know it's barmy but the water is boiling and I need to use it  Grin

Perhaps it's like filling an air cylinder with a compressor which has to work harder as the cylinder gets fuller  Huh

Cheers, Paul
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rogeriko
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« Reply #2 on: February 05, 2012, 06:57:14 AM »

Batteries are grossly inefficient one needs alot more ah in than you can ever take out. I use 12kwh per day (measured) but put in 12kwh of solar + the wind turbine probably another 8 to 12kwh per day and my batteries never seem to be fully charged. They rarely go to 28v. I do not have any kind of controller or dump load, just 2 refridgerators and 1 freezer + computers and 55" tv, I guess thats my dump load!
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camillitech
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« Reply #3 on: February 05, 2012, 07:22:47 AM »

Morning Roger,

'off topic' but do you have any experience with Studer inverter/chargers  Huh http://www.navitron.org.uk/forum/index.php/topic,16411.0.html I'm just about to go and have a look at it and know from previous that I'll not be able to make any sense of the manual  Cry

Cheers, Paul
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http://lifeattheendoftheroad.wordpress.com/

12kw Lister
11m turbine tower
10 hundred ah 48v battery bank
900' pennstock
8kw woodburner
7kw Lister
6 bladed Rutland
50w of solar
4 and a half Kw inverter
3kw Lister
2 hydro turbines
and a Proven in a pear tree :-)

Raasay, 57 27 537 N 06
rogeriko
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« Reply #4 on: February 05, 2012, 07:42:10 AM »

Good morning. To program the studer inverters you have to press the button aux for at least 2 seconds and then see which lights are flashing. the flashing lights are the ones that control the contacts. The 100% power led is the one that will start the generator on overload and the 4 green led's control the battery voltage.
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fred
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« Reply #5 on: February 05, 2012, 09:02:21 AM »

That is quite a good research


http://www.google.ie/url?sa=t&rct=j&q=leadacidbatterieefficiency&source=web&cd=4&ved=0CFIQFjAD&url=http%3A%2F%2Fwww.sandia.gov%2Fpv%2Fdocs%2FPDF%2Fbatpapsteve.pdf&ei=TsctT7mQHoSp0QXK2IWuCA&usg=AFQjCNEPUW8iKyzyB82BBrhufkmbhAjLfw&cad=rja


But still  i  do not understand efficiency ratings  of lead acid batteries  

Some reports rate them  as 60-90%  efficient , when i read this report it undermines my opinion  that, they are pretty efficient when not full or under 80 %  , but to charge them 100% full is a long way ............    

The question is ,  does it take  so much more Ah   to fully charge a battery    or is it just the time that it takes..........

If it is only longer time  , than why should efficiency drop so much , when AH pushed in seem the same ?

It is pretty obvious in my set-up , that AMPs from the PV  are reduced when battery gets close to full, but if i use these AMPS left over from the PV  for something else  , then these AMPS  should not be calculated in  .......

Or  am i totally wrong   and one needs much more AH to put in ,that one is able to take out

 wackoold


Billi



First law of entropy says you can't win, second law says you can't break even, it's a losing game as how you play it.

Those amp hours they print on batteries are like the mpg figures they give for cars, the conditions have to be identical to when they tested them. A 100 amp hour battery will give 1 amp for 100 hours, it won't give 2 amps for 50 hours though.

I'm wondering if efficiency could be improved in the later stages of charging by supplying the charge current in the form of a sawtooth riding on the battery dc, letting a very steep leading edge kick the sulphur off the lead plates.
 
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bxman
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« Reply #6 on: February 05, 2012, 09:29:52 AM »

billi

I am afraid I cannot access the link.
Can you give us another route to the data.
Thanks.
« Last Edit: February 05, 2012, 10:06:05 AM by bxman » Logged
bxman
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« Reply #7 on: February 05, 2012, 10:03:24 AM »

http://evbatterymonitoring.com/WebHelp/Battery_Book.htm#Section_3.htm

Might be of interest particularly Fig 23 at the end of the document
other

sites
http://ecee.colorado.edu/~ecen2060/materials/lecture_notes/Battery3.pdf

http://pvcdrom.pveducation.org/BATTERY/charlead.htm

http://www.hagerbr.com.au/popup.php?scr=1024&id=84&action=faq

http://www.windsun.com/Batteries/Battery_FAQ.htm#Mini%20Factoids

All a little distressing,
But without having two lakes and 1000 difference in altitude
I suspect batteries are a viable means of storing energy.
« Last Edit: February 05, 2012, 10:06:54 AM by bxman » Logged
camillitech
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« Reply #8 on: February 05, 2012, 10:05:47 AM »

billi

I am afraid I cannot access the link.
Can you give us another route to the data.
Thanks.

Try and Google this, "sandia national laboratories photovoltaic applications download" BXman.

Good luck, Paul
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http://lifeattheendoftheroad.wordpress.com/

12kw Lister
11m turbine tower
10 hundred ah 48v battery bank
900' pennstock
8kw woodburner
7kw Lister
6 bladed Rutland
50w of solar
4 and a half Kw inverter
3kw Lister
2 hydro turbines
and a Proven in a pear tree :-)

Raasay, 57 27 537 N 06
bxman
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« Reply #9 on: February 05, 2012, 10:12:24 AM »

Thanks Paul
was it
http://www.localenergy.org/pdfs/Document%20Library/Fundamentals%20of%20batteries%20and%20charge%20control.pdf

that looks relevant.
cheers Patrick.
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billi
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« Reply #10 on: February 05, 2012, 02:31:59 PM »

no.......it was  this one

www.sandia.gov/pv/docs/PDF/batpapsteve.pdf
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1.6 kw and 2.4 kw   PV array  , Outback MX 60 and FM80 charge controller  ,24 volt 1600 AH Battery ,6 Kw Victron inverter charger, 1.1 kw high head hydro turbine as a back up generator , 5 kw woodburner, 36 solar tubes with 360 l water tank, 1.6 kw  windturbine
splyn
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« Reply #11 on: February 06, 2012, 02:59:11 PM »

Its even worse than you might think from a quick glance at the Sandia labs report - the 'less than 60% efficiency' when using only the top 20% of the battery capacity is only measuring the coulombic efficiency - Ah out/Ah in. When you account for the difference in voltage when charging and discharging, there's another 10 to 15% energy loss on top.

Factoring in the charger and inverter efficiencies and losses for periodic equalisation charges as well, the overall round trip efficiency could easily be less than 40%. So add the cost of the extra 150% electrical energy to the 15 to 20p per kWh for the battery replacement cost and whatever it costs to provide backup power (genny?) when needed and it don't come cheap!

Splyn
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billi
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« Reply #12 on: February 06, 2012, 03:26:18 PM »

I would say one can  produce  below  20 pence  a unit seen over the life(cycles ) time of the battery  ........  all  in   excl . backup

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« Reply #13 on: February 06, 2012, 05:06:59 PM »

I don't think this efficiency loss for the last 10% of charging makes much difference for off-grid PV systems.  Once you're charging that last 10% the controller will already be in absorb, so wouldn't it already be reducing the available power from the array?  Guess it only matters in winter afternoons when light is scarce and absorb still hasn't finished.
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Eleanor
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« Reply #14 on: February 06, 2012, 10:39:29 PM »

They don't really explain why the charge efficiency reduces as the battery becomes charged. One explanation could be that most of the "loss in efficiency" is down to gassing and heating of the electrolyte eg taking a theoretical 100Ah battery and charging it at 10A will give most of the 10A to the battery during the bulk phase until the gassing voltage of around 2.4V is reached. At this point the charger starts to reduce the current as the internal resistance of the battery increases. The electrolyte starts to gas and if, for example, 0.5A is going towards the gassing 9.5A will go into the battery giving an charge efficiency of 95%. When the current is 8A the battery will receive 7.5A so 94% efficiency. 5A gives 90%, 3A 83%, 2A 75%, 1A 50%. They are only made up figures but it shows how the charge efficiency would drop off quickly towards the end of the charge and why proportionally more has to be put in than can be retrieved. It's not that simple though as charging above the gassing voltage will help to improve the efficiency of the charging reaction by mixing the electrolyte and it also ignores any other losses.

If this is right we should see good results from the electrolyte circulation system we're getting with the new bank  stir
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