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<DIV><FONT face=Arial size=2>Hi Hugh.</FONT></DIV>
<DIV><FONT face=Arial size=2></FONT> </DIV>
<DIV><FONT face=Arial size=2>Positive electrode: </FONT><FONT size=2><FONT
face=Arial>PbO<FONT size=1>2 </FONT><FONT size=2>+ 3H +HSO<FONT size=1>4</FONT>
</FONT><FONT size=1><FONT size=2>+ 2e</FONT> = <FONT size=2>PbSO</FONT>4 +
</FONT><FONT size=2>2H<FONT size=1>2</FONT>O (e=
electron)</FONT></FONT></FONT></DIV>
<DIV><FONT face=Arial size=2></FONT> </DIV>
<DIV><FONT face=Arial size=2>Negative electrode: Pb + HSO<FONT
size=1>4</FONT> = PbSO<FONT size=1>4</FONT> + H +2e</FONT></DIV>
<DIV><FONT face=Arial size=2></FONT> </DIV>
<DIV><FONT face=Arial size=2>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></DIV>
<DIV><FONT face=Arial size=2></FONT> </DIV>
<DIV><FONT face=Arial size=2>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></DIV>
<DIV><FONT face=Arial size=2></FONT> </DIV>
<DIV><FONT face=Arial size=2>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></DIV>
<DIV><FONT face=Arial size=2></FONT> </DIV>
<DIV><FONT face=Arial size=2>Bruce Geddes</FONT></DIV>
<DIV><FONT face=Arial size=2>PowerOn</FONT></DIV>
<BLOCKQUOTE
style="PADDING-RIGHT: 0px; PADDING-LEFT: 5px; MARGIN-LEFT: 5px; BORDER-LEFT: #000000 2px solid; MARGIN-RIGHT: 0px">
<DIV style="FONT: 10pt arial">----- Original Message ----- </DIV>
<DIV
style="BACKGROUND: #e4e4e4; FONT: 10pt arial; 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 12:25
PM</DIV>
<DIV style="FONT: 10pt arial"><B>Subject:</B> Re: [RE-wrenches] discharging
Rolls batteries</DIV>
<DIV><BR></DIV>
<DIV>hi</DIV>
<DIV><BR></DIV>
<DIV>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:</DIV>
<DIV><BR></DIV>
<DIV>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?</DIV>
<DIV><BR></DIV>
<DIV>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.</DIV>
<DIV><BR></DIV>
<DIV>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.</DIV>
<DIV><BR></DIV>
<DIV>Thanks for any help with this rather obscure question.</DIV>
<DIV><BR></DIV>
<DIV>Hugh</DIV>
<DIV><BR></DIV>
<BLOCKQUOTE cite="" type="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 -----</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> Friday, January 15, 2010 12:02 AM</BLOCKQUOTE>
<BLOCKQUOTE><B>Subject:</B> Re: [RE-wrenches] discharging Rolls
batteries</BLOCKQUOTE>
<BLOCKQUOTE><BR></BLOCKQUOTE>
<BLOCKQUOTE>Hi Jamie,</BLOCKQUOTE>
<BLOCKQUOTE><BR>
<BLOCKQUOTE cite="" type="cite"><BR></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:</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><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?</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?</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>I hope you can follow my descriptions.</BLOCKQUOTE>
<BLOCKQUOTE><TT>--</TT></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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<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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