[RE-wrenches] discharging Rolls batteries
Hugh
hugh at scoraigwind.co.uk
Sat Jan 16 05:04:31 PST 2010
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:hugh at scoraigwind.co.uk>Hugh
>To: <mailto:re-wrenches at 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:hugh at scoraigwind.co.uk>Hugh
>>
>>To: <mailto:re-wrenches at 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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>--
>Hugh Piggott
>
>Scoraig Wind Electric
>Scotland
>http://www.scoraigwind.co.uk
>
>
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Hugh Piggott
Scoraig Wind Electric
Scotland
http://www.scoraigwind.co.uk
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