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<DIV><FONT size=2 face=Arial>A battery is an open system. Heat is lost. 2nd
law of themodynamics. See <A
href="http://theory.uwinnipeg.ca/mod_tech/node70.html">http://theory.uwinnipeg.ca/mod_tech/node70.html</A></FONT></DIV>
<DIV> </DIV>
<BLOCKQUOTE
style="BORDER-LEFT: #000000 2px solid; PADDING-LEFT: 5px; PADDING-RIGHT: 0px; MARGIN-LEFT: 5px; MARGIN-RIGHT: 0px">
<DIV style="FONT: 10pt arial">----- Original Message ----- </DIV>
<DIV
style="FONT: 10pt arial; BACKGROUND: #e4e4e4; font-color: black"><B>From:</B>
<A title=hugh@scoraigwind.co.uk href="mailto:hugh@scoraigwind.co.uk">Hugh</A>
</DIV>
<DIV style="FONT: 10pt arial"><B>To:</B> <A
title=re-wrenches@lists.re-wrenches.org
href="mailto:re-wrenches@lists.re-wrenches.org">RE-wrenches</A> </DIV>
<DIV style="FONT: 10pt arial"><B>Sent:</B> Saturday, January 16, 2010 5:04
AM</DIV>
<DIV style="FONT: 10pt arial"><B>Subject:</B> Re: [RE-wrenches] discharging
Rolls batteries</DIV>
<DIV><BR></DIV>
<DIV>Hi Bruce,</DIV>
<DIV><BR></DIV>
<DIV>At 22:58 +1300 16/1/10, Bruce Geddes wrote:</DIV>
<BLOCKQUOTE cite="" type="cite"><FONT size=-1 face=Arial>in cold weather the
money counters operate more slowly and if the temperature rises they return
to normal speed. The money is still there,</FONT>....</BLOCKQUOTE>
<DIV><BR></DIV>
<DIV><BR></DIV>
<DIV>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.</DIV>
<DIV><BR></DIV>
<DIV>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.</DIV>
<DIV><BR></DIV>
<DIV>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.</DIV>
<DIV><BR></DIV>
<DIV>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?</DIV>
<DIV><BR></DIV>
<DIV>Thanks, Joel for the reading matter which I am working on! I hope
to become wiser in due course.</DIV>
<DIV><BR></DIV>
<DIV>best</DIV>
<DIV><BR></DIV>
<DIV>Hugh</DIV>
<DIV><BR></DIV>
<DIV><BR></DIV>
<DIV>At 17:45 -0800 15/1/10, Joel Davidson wrote:</DIV>
<BLOCKQUOTE cite="" type="cite"><BR></BLOCKQUOTE>
<BLOCKQUOTE cite="" type="cite"><FONT size=-1 face=Arial>Some charge energy
is lost in heat and some in coulombic efficiency.</FONT></BLOCKQUOTE>
<BLOCKQUOTE cite="" type="cite"><FONT size=-1
face=Arial>There are educational powerpoints, papers and other
information about batteries on the internet.</FONT></BLOCKQUOTE>
<BLOCKQUOTE cite="" type="cite"><FONT size=-1 face=Arial>See</FONT> <A
href="http://en.wikipedia.org/wiki/Faraday_efficiency"><FONT size=-1
face=Arial>http://en.wikipedia.org/wiki/Faraday_efficiency</FONT></A></BLOCKQUOTE>
<BLOCKQUOTE cite="" type="cite"><FONT size=-1 face=Arial>and</FONT> <A
href="http://www.mpoweruk.com/soc.htm"><FONT size=-1
face=Arial>http://www.mpoweruk.com/soc.htm</FONT></A></BLOCKQUOTE>
<BLOCKQUOTE cite="" type="cite"><FONT size=-1 face=Arial>and</FONT> <A
href="http://web.mit.edu/mit_energy/resources/iap/MatSciOfRenewEnergy_Lecture2_Batteries_2006.pdf"><SPAN></SPAN>http://web.mit.edu/mit_energy/resources/iap/MatSciOfRenewEnergy_Lect<SPAN></SPAN>ure2_Batteries_2006.pdf</A></BLOCKQUOTE>
<BLOCKQUOTE cite="" type="cite"><FONT size=-1 face=Arial>and</FONT> <A
href="http://ecee.colorado.edu/~ecen4517/materials/Battery.pdf"><FONT
size=-1
face=Arial>http://ecee.colorado.edu/~ecen4517/materials/Battery.pdf</FONT></A></BLOCKQUOTE>
<BLOCKQUOTE cite="" type="cite"><FONT size=-1 face=Arial>and</FONT> <A
href="http://users.ece.utexas.edu/~kwasinski/EE394V_DG_Fall2008_Week5%20part2.ppt#1"><FONT
size=-1
face=Arial>http://users.ece.utexas.edu/~kwasinski/EE394V_DG_Fall2008_Week5%20pa<SPAN></SPAN>rt2.ppt#1</FONT></A></BLOCKQUOTE>
<BLOCKQUOTE cite="" type="cite"><FONT size=-1 face=Arial>and for info about
long series strings of batteries see</FONT> <A
href="http://www.battcon.com/PapersFinal2004/SymonsPaper2004.pdf"><FONT
size=-1
face=Arial>http://www.battcon.com/PapersFinal2004/SymonsPaper2004.pdf</FONT></A></BLOCKQUOTE>
<BLOCKQUOTE cite="" type="cite"> </BLOCKQUOTE>
<BLOCKQUOTE cite="" type="cite"><BR> </BLOCKQUOTE>
<DIV><BR></DIV>
<DIV>Bruce:</DIV>
<BLOCKQUOTE cite="" type="cite"><FONT size=-1 face=Arial>Hi
Hugh.</FONT></BLOCKQUOTE>
<BLOCKQUOTE cite="" type="cite"> </BLOCKQUOTE>
<BLOCKQUOTE cite="" type="cite"><FONT size=-1 face=Arial>Positive
electrode: PbO</FONT><FONT size=-2 face=Arial>2</FONT><FONT size=-1
face=Arial> + 3H +HSO</FONT><FONT size=-2 face=Arial>4</FONT><FONT size=-1
face=Arial> + 2e</FONT><FONT size=-2 face=Arial> =</FONT><FONT size=-1
face=Arial> PbSO</FONT><FONT size=-2 face=Arial>4 +</FONT><FONT size=-1
face=Arial> 2H</FONT><FONT size=-2 face=Arial>2</FONT><FONT size=-1
face=Arial>O (e= electron)</FONT></BLOCKQUOTE>
<BLOCKQUOTE cite="" type="cite"> </BLOCKQUOTE>
<BLOCKQUOTE cite="" type="cite"><FONT size=-1 face=Arial>Negative
electrode: Pb + HSO</FONT><FONT size=-2 face=Arial>4</FONT><FONT
size=-1 face=Arial> = PbSO</FONT><FONT size=-2 face=Arial>4</FONT><FONT
size=-1 face=Arial> + H +2e</FONT></BLOCKQUOTE>
<BLOCKQUOTE cite="" type="cite"> </BLOCKQUOTE>
<BLOCKQUOTE cite="" type="cite"><FONT size=-1 face=Arial>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.</FONT></BLOCKQUOTE>
<BLOCKQUOTE cite="" type="cite"> </BLOCKQUOTE>
<BLOCKQUOTE cite="" type="cite"><FONT size=-1 face=Arial>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.</FONT></BLOCKQUOTE>
<BLOCKQUOTE cite="" type="cite"> </BLOCKQUOTE>
<BLOCKQUOTE cite="" type="cite"><FONT size=-1 face=Arial>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!"</FONT></BLOCKQUOTE>
<BLOCKQUOTE cite="" type="cite"> </BLOCKQUOTE>
<BLOCKQUOTE cite="" type="cite"><FONT size=-1 face=Arial>Bruce
Geddes</FONT></BLOCKQUOTE>
<BLOCKQUOTE cite="" type="cite"><FONT size=-1 face=Arial>PowerOn</FONT><BR>
<BLOCKQUOTE>----- Original Message -----</BLOCKQUOTE>
<BLOCKQUOTE><B>From:</B> <A
href="mailto:hugh@scoraigwind.co.uk">Hugh</A></BLOCKQUOTE>
<BLOCKQUOTE><B>To:</B> <A
href="mailto:re-wrenches@lists.re-wrenches.org">RE-wrenches</A></BLOCKQUOTE>
<BLOCKQUOTE><B>Sent:</B> Saturday, January 16, 2010 12:25 PM</BLOCKQUOTE>
<BLOCKQUOTE><B>Subject:</B> Re: [RE-wrenches] discharging Rolls
batteries</BLOCKQUOTE>
<BLOCKQUOTE><BR></BLOCKQUOTE>
<BLOCKQUOTE>hi</BLOCKQUOTE>
<BLOCKQUOTE><BR></BLOCKQUOTE>
<BLOCKQUOTE>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:</BLOCKQUOTE>
<BLOCKQUOTE><BR></BLOCKQUOTE>
<BLOCKQUOTE>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?</BLOCKQUOTE>
<BLOCKQUOTE><BR></BLOCKQUOTE>
<BLOCKQUOTE>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.</BLOCKQUOTE>
<BLOCKQUOTE><BR></BLOCKQUOTE>
<BLOCKQUOTE>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.</BLOCKQUOTE>
<BLOCKQUOTE><BR></BLOCKQUOTE>
<BLOCKQUOTE>Thanks for any help with this rather obscure
question.</BLOCKQUOTE>
<BLOCKQUOTE><BR></BLOCKQUOTE>
<BLOCKQUOTE>Hugh</BLOCKQUOTE>
<BLOCKQUOTE><BR>
<BLOCKQUOTE cite="" type="cite"><FONT size=-1
face=Arial>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.</FONT><BR>
<BLOCKQUOTE>----- Original Message -----<BR></BLOCKQUOTE>
<BLOCKQUOTE><B>From:</B> <A
href="mailto:hugh@scoraigwind.co.uk">Hugh<BR></A></BLOCKQUOTE>
<BLOCKQUOTE><B>To:</B> <A
href="mailto:re-wrenches@lists.re-wrenches.org">RE-wrenches<BR></A></BLOCKQUOTE>
<BLOCKQUOTE><B>Sent:</B> Friday, January 15, 2010 12:02
AM<BR></BLOCKQUOTE>
<BLOCKQUOTE><B>Subject:</B> Re: [RE-wrenches] discharging Rolls
batteries<BR></BLOCKQUOTE>
<BLOCKQUOTE><BR></BLOCKQUOTE>
<BLOCKQUOTE>Hi Jamie,<BR></BLOCKQUOTE>
<BLOCKQUOTE><BR>
<BLOCKQUOTE cite=""
type="cite"><BR></BLOCKQUOTE></BLOCKQUOTE></BLOCKQUOTE></BLOCKQUOTE>
<BLOCKQUOTE>
<BLOCKQUOTE>
<BLOCKQUOTE>
<BLOCKQUOTE><FONT face="Lucida Grande">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.</FONT><BR></BLOCKQUOTE></BLOCKQUOTE>
<BLOCKQUOTE><BR></BLOCKQUOTE>
<BLOCKQUOTE>But earlier you put it this way:<BR></BLOCKQUOTE>
<BLOCKQUOTE><BR>
<BLOCKQUOTE cite="" type="cite">
<BLOCKQUOTE cite="" type="cite"><FONT
face="Lucida Grande">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.
</FONT><BR></BLOCKQUOTE></BLOCKQUOTE></BLOCKQUOTE></BLOCKQUOTE></BLOCKQUOTE>
<BLOCKQUOTE>
<BLOCKQUOTE>
<BLOCKQUOTE><BR></BLOCKQUOTE>
<BLOCKQUOTE>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?<BR></BLOCKQUOTE>
<BLOCKQUOTE><BR></BLOCKQUOTE>
<BLOCKQUOTE>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?<BR></BLOCKQUOTE>
<BLOCKQUOTE><BR></BLOCKQUOTE>
<BLOCKQUOTE>(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?)</BLOCKQUOTE>
<BLOCKQUOTE><BR></BLOCKQUOTE>
<BLOCKQUOTE><BR></BLOCKQUOTE>
<BLOCKQUOTE>I hope you can follow my descriptions.<BR></BLOCKQUOTE>
<BLOCKQUOTE><TT>--</TT><BR></BLOCKQUOTE>
<BLOCKQUOTE>Hugh Piggott<BR><BR>Scoraig Wind
Electric<BR>Scotland<BR>http://www.scoraigwind.co.uk<BR></BLOCKQUOTE>
<BLOCKQUOTE>
<HR>
</BLOCKQUOTE>
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<BLOCKQUOTE><BR></BLOCKQUOTE>
<BLOCKQUOTE><TT>--</TT></BLOCKQUOTE>
<BLOCKQUOTE>Hugh Piggott<BR><BR>Scoraig Wind
Electric<BR>Scotland<BR>http://www.scoraigwind.co.uk<BR></BLOCKQUOTE>
<BLOCKQUOTE>
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<DIV><BR></DIV><X-SIGSEP><PRE>--
</PRE></X-SIGSEP>
<DIV>Hugh Piggott<BR><BR>Scoraig Wind
Electric<BR>Scotland<BR>http://www.scoraigwind.co.uk</DIV>
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