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batteries</title></head><body>
<div>Hi Bruce,</div>
<div><br></div>
<div>At 22:58 +1300 16/1/10, Bruce Geddes wrote:</div>
<blockquote type="cite" cite><font face="Arial" size="-1">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 type="cite" cite><br></blockquote>
<blockquote type="cite" cite><font face="Arial" size="-1">Some charge
energy is lost in heat and some in coulombic
efficiency.</font></blockquote>
<blockquote type="cite" cite><font face="Arial"
size="-1">There are educational powerpoints, papers and
other information about batteries on the internet.</font></blockquote>
<blockquote type="cite" cite><font face="Arial" size="-1">See</font> <a
href="http://en.wikipedia.org/wiki/Faraday_efficiency"><font
face="Arial"
size="-1">http://en.wikipedia.org/wiki/Faraday_efficiency</font></a
></blockquote>
<blockquote type="cite" cite><font face="Arial" size="-1">and</font> <a
href="http://www.mpoweruk.com/soc.htm"><font face="Arial"
size="-1">http://www.mpoweruk.com/soc.htm</font></a></blockquote>
<blockquote type="cite" cite><font face="Arial" size="-1">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 type="cite" cite><font face="Arial" size="-1">and</font> <a
href="http://ecee.colorado.edu/~ecen4517/materials/Battery.pdf"><font
face="Arial"
size="-1">http://ecee.colorado.edu/~ecen4517/materials/Battery.pdf</font
></a></blockquote>
<blockquote type="cite" cite><font face="Arial" size="-1">and</font> <a
href=
"http://users.ece.utexas.edu/~kwasinski/EE394V_DG_Fall2008_Week5%20part2.ppt#1"
><font face="Arial"
size="-1"
>http://users.ece.utexas.edu/~kwasinski/EE394V_DG_Fall2008_Week5%20pa<span
></span>rt2.ppt#1</font></a></blockquote>
<blockquote type="cite" cite><font face="Arial" size="-1">and for info
about long series strings of batteries see</font> <a
href="http://www.battcon.com/PapersFinal2004/SymonsPaper2004.pdf"><font
face="Arial"
size="-1">http://www.battcon.com/PapersFinal2004/SymonsPaper2004.pdf</font
></a></blockquote>
<blockquote type="cite" cite> </blockquote>
<blockquote type="cite" cite> <br>
</blockquote>
<div><br></div>
<div>Bruce:</div>
<blockquote type="cite" cite><font face="Arial" size="-1">Hi
Hugh.</font></blockquote>
<blockquote type="cite" cite> </blockquote>
<blockquote type="cite" cite><font face="Arial" size="-1">Positive
electrode: PbO</font><font face="Arial" size="-2">2</font><font
face="Arial" size="-1"> + 3H +HSO</font><font face="Arial"
size="-2">4</font><font face="Arial" size="-1"> + 2e</font><font
face="Arial" size="-2"> =</font><font face="Arial" size="-1">
PbSO</font><font face="Arial" size="-2">4 +</font><font face="Arial"
size="-1"> 2H</font><font face="Arial" size="-2">2</font><font
face="Arial" size="-1">O (e= electron)</font></blockquote>
<blockquote type="cite" cite> </blockquote>
<blockquote type="cite" cite><font face="Arial" size="-1">Negative
electrode: Pb + HSO</font><font face="Arial"
size="-2">4</font><font face="Arial" size="-1"> = PbSO</font><font
face="Arial" size="-2">4</font><font face="Arial" size="-1"> + H
+2e</font></blockquote>
<blockquote type="cite" cite> </blockquote>
<blockquote type="cite" cite><font face="Arial" size="-1">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 type="cite" cite> </blockquote>
<blockquote type="cite" cite><font face="Arial" size="-1">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 type="cite" cite> </blockquote>
<blockquote type="cite" cite><font face="Arial" size="-1">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 type="cite" cite> </blockquote>
<blockquote type="cite" cite><font face="Arial" size="-1">Bruce
Geddes</font></blockquote>
<blockquote type="cite" cite><font face="Arial"
size="-1">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 type="cite" cite><font face="Arial"
size="-1">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 type="cite" 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 type="cite" cite>
<blockquote type="cite" 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>
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<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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