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>http://en.wikipedia.org/wiki/Faraday_efficiency
and <http://www.mpoweruk.com/soc.htm>http://www.mpoweruk.com/soc.htm
and
<http://web.mit.edu/mit_energy/resources/iap/MatSciOfRenewEnergy_Lecture2_Batteries_2006.pdf>http://web.mit.edu/mit_energy/resources/iap/MatSciOfRenewEnergy_Lecture2_Batteries_2006.pdf
and
<http://ecee.colorado.edu/~ecen4517/materials/Battery.pdf>http://ecee.colorado.edu/~ecen4517/materials/Battery.pdf
and
<http://users.ece.utexas.edu/~kwasinski/EE394V_DG_Fall2008_Week5%20part2.ppt#1>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>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: <mailto:h...@scoraigwind.co.uk>Hugh
To: <mailto:re-wrenches@lists.re-wrenches.org>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: <mailto:h...@scoraigwind.co.uk>Hugh
To: <mailto:re-wrenches@lists.re-wrenches.org>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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