<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">
<html>
<head>
<meta content="text/html;charset=ISO-8859-1" http-equiv="Content-Type">
<title></title>
</head>
<body bgcolor="#ffffff" text="#000000">
Hugh,<br>
<br>
If my understanding about this is right, the amphours aren't literally
lost when the battery is cold or discharge rate is high. You would
just have to operate the battery below 1.75 volts to get them. Not a
situation that is very useful. But warm them up or lower the discharge
rate and you'll get the full amount. <br>
<br>
Kent Osterberg<br>
Blue Mountain Solar, Inc.<br>
<br>
<br>
<br>
Hugh wrote:
<blockquote cite="mid:p06230942c776a734712a@%5B10.0.2.2%5D" type="cite">
<style type="text/css"><!--
blockquote, dl, ul, ol, li { padding-top: 0 ; padding-bottom: 0 }
--></style>
<title>Re: [RE-wrenches] discharging Rolls
batteries</title>
<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 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 -----</blockquote>
<blockquote><b>From:</b> <a moz-do-not-send="true"
href="mailto:hugh@scoraigwind.co.uk">Hugh</a></blockquote>
<blockquote><b>To:</b> <a moz-do-not-send="true"
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 type="cite" 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 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><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>
<a class="moz-txt-link-freetext" href="http://www.scoraigwind.co.uk">http://www.scoraigwind.co.uk</a><br>
</blockquote>
<blockquote>
<hr></blockquote>
</blockquote>
<br>
<div><br>
</div>
<x-sigsep></x-sigsep>
<pre>--
</pre>
<div>Hugh Piggott<br>
<br>
Scoraig Wind Electric<br>
Scotland<br>
<a class="moz-txt-link-freetext" href="http://www.scoraigwind.co.uk">http://www.scoraigwind.co.uk</a></div>
<br>
</blockquote>
</body>
</html>