Re: [RE-wrenches] discharging Rolls batteriesSome charge energy is lost in heat and some in coulombic efficiency. There are educational powerpoints, papers and other information about batteries on the internet. See http://en.wikipedia.org/wiki/Faraday_efficiency and http://www.mpoweruk.com/soc.htm and http://web.mit.edu/mit_energy/resources/iap/MatSciOfRenewEnergy_Lecture2_Batteries_2006.pdf and http://ecee.colorado.edu/~ecen4517/materials/Battery.pdf and http://users.ece.utexas.edu/~kwasinski/EE394V_DG_Fall2008_Week5%20part2.ppt#1 and for info about long series strings of batteries see http://www.battcon.com/PapersFinal2004/SymonsPaper2004.pdf
----- Original Message ----- From: Hugh To: RE-wrenches Sent: Friday, January 15, 2010 3:25 PM Subject: Re: [RE-wrenches] discharging Rolls batteries hi We know that batteries deliver less amphours at low temperature and at high currents. Volts drop quicker. That's my starting point. My question that I still do not hear an answer to is this: If the battery is a bank account and its harder to get the money out in cold weather and when you want to get your hands on a lot at once... Does this actually mean that some of the money gets lost? What happens to it? Is it perhaps available later when the bank warms up or the demand gets less hectic? Is there really less money in there or does it just seem like less due to the conditions? I notice that Ah capacity is actually defined as how much Amphours you can get out before the battery reaches a certain terminal voltage. I am wondering whether it is the ability to maintain voltage that is the limiting factor whereas the chemicals in there can still deliver amphours, given the right temperature and time later. You can certainly see recovery take place when a battery warms up and/or operates on lighter loads. One last time what happens to the chemicals (lead and lead oxide) that represent Amphours of charge in the battery plates? For me this is a little bit like current of 10 amps entering one end of a piece of wire and only 9 amps coming out the other end. I understand that the volts go down due to voltage drop (in this analogy) but loss of current is entirely a different matter. Thanks for any help with this rather obscure question.
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