Re: [RE-wrenches] discharging Rolls batteriesA battery is an open system. Heat 
is lost. 2nd law of themodynamics. See 
http://theory.uwinnipeg.ca/mod_tech/node70.html

  ----- Original Message ----- 
  From: Hugh 
  To: RE-wrenches 
  Sent: Saturday, January 16, 2010 5:04 AM
  Subject: Re: [RE-wrenches] discharging Rolls batteries


  Hi Bruce,


  At 22:58 +1300 16/1/10, Bruce Geddes wrote:
    in cold weather the money counters operate more slowly and if the 
temperature rises they return to normal speed.  The money is still there,....




  Yes but if that is the case then the actual capacity is not affected by 
temperature - just the ability to deliver.  I would like to think it's that 
simple, but in reality I suspect that there is some loss of amphours under 
these conditions.  I am not yet clear about the mechanism, but I suspect that 
there is more to it than just a 'volt-drop' style explanation such as you offer 
below.


  As far as I understand it, there are two losses: one of voltage due to 
internal resistance and chemical 'sluggishness' and another actual loss of 
capacity in amphours ( getting less amphours out of the battery than you put 
in).  I am trying to establish what happens to those missing amphours, and also 
to what extent they actually are missing and to what extent they are just 
rendered inaccessible by the decision to end the discharge at a certain voltage 
which in turn is affected by the previous 'volt-dop' issues.


  If it were really just a case of the bank tellers having cold fingers then it 
would seem reasonable to hammer the battery down to a much lower voltage in the 
confident knowledge that we are still only taking out 50% of the capacity as 
enshrined as 'good practice'.  however if some of the cash has actually got 
lost (where to?) then it is nt legitimate to hit the battery bank for more cash 
in this way.


  I haven't yet heard from any Wrench what actual voltages they would use to 
set the LBCO or the genstart on an Outback (or an SW), but one has told me 
off-list that it's a negotiation with the client.  Fair enough but what are the 
numbers used in the negotiation, and are they temperature dependant?


  Thanks, Joel for the reading matter which I am working on!  I hope to become 
wiser in due course.


  best


  Hugh




  At 17:45 -0800 15/1/10, Joel  Davidson wrote:


    Some 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





  Bruce:
    Hi Hugh.

    Positive electrode:  PbO2 + 3H +HSO4 + 2e = PbSO4 + 2H2O  (e= electron)

    Negative electrode:  Pb + HSO4 = PbSO4 + H +2e

    Sorry, I can't insert the superscript symbols to show electrical charge.  
If it is confusing let me know and I will repost this with the charge in 
brackets following the ion.

    In cold conditions the ion transfer rate slows so in effect the internal 
resistance of the cell rises.  It simply can't deliver the electrons under 
load.  If the electrons are taken out at a reduced rate then the Vdrop of the 
"internal resistance" is lowered and the cell terminal voltage stays up for 
longer.

    So, to use your analogy of the bank, in cold weather the money counters 
operate more slowly and if the temperature rises they return to normal speed.  
The money is still there, it is just the rate at which it comes out that 
varies.  In cold weather the counters simply won't deliver as much before they 
say "my fingers are too cold, that's all you get today!"

    Bruce Geddes
    PowerOn

      ----- Original Message -----
      From: Hugh
      To: RE-wrenches
      Sent: Saturday, January 16, 2010 12:25 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.


      Hugh


        A lead-acid battery is an electro-chemical processor (just like you and 
other living things). When you and your battery are cold or hot, performance 
changes because the chemical process is affected by temperature. Cold equals 
sluggish chemical reaction, reduces the capacity to perform work, and affects 
battery performance linearly. Battery chemistry is well understood. When I get 
some time, I'll google for temperature-based formulas and charts unless someone 
else posts the links first.

          ----- Original Message -----

          From: Hugh

          To: RE-wrenches

          Sent: Friday, January 15, 2010 12:02 AM

          Subject: Re: [RE-wrenches] discharging Rolls batteries



          Hi Jamie,





            Remember, as batteries cool actual capacity is reduced, so if 200AH 
is 50% @ 25C it is significantly more than 50% @ 5C.   Thus, you are 
discharging more deeply.



          But earlier you put it this way:



              Regarding temperature effects on capacity, earlier responses are 
spot on as the lower capacity is totally as a result of slower reaction times 
as a result of lower temperatures.  



          There is an issue here that I need to understand better.  You state 
that a battery has lower capacity in low temperatures.  Suppose you take a 
fully charged, 400 Ah battery and cool it down to -5 degrees C where according 
to our numbers it will only have 80% of its nominal capacity.  You then remove 
160 Ah (say 10 amps for 16 hours).  It will then be 50% discharged.  Now warm 
it up again to 20 degrees or whatever.  My question is: will you only have 200 
amphours left in it now?  And if so, what happened to the other 40 amphours?  
Does low temperature operation actually lose amphours, or is it just more 
sluggish?  What is the chemical explanation for the lost amphours?



          I understand batteries as a chemical process of converting amphours 
into chemical changes.  I assume that a given amount of electrical charge 
converts a given amount of lead into lead sulphate (and likewise) back again.  
I understand that cooling will make this process less efficient and thereby 
result in a rise in charging voltage and a drop in discharging voltage.  But 
does a low temperature actually mean that a given amount of lead being 
converted to sulphate actually give you less amphours electrically?



          (I have similar questions in relation to Peukert's equation where 
high discharge rates impact on the amphour capacity.  The capacity apparently 
'recovers' when the discharge rate is reduced.  To what extent is the capacity 
actually lost by using high discharge rates and to what extent is it just a 
voltage effect that impacts on the terminal voltage, rather than the actual 
chemical state of the battery?)




          I hope you can follow my descriptions.

          --

          Hugh Piggott

          Scoraig Wind Electric
          Scotland
          http://www.scoraigwind.co.uk


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      --
      Hugh Piggott

      Scoraig Wind Electric
      Scotland
      http://www.scoraigwind.co.uk


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-- 
Hugh Piggott

  Scoraig Wind Electric
  Scotland
  http://www.scoraigwind.co.uk


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