[RE-wrenches] discharging Rolls batteries

Sat Jan 16 09:46:19 PST 2010

Hi Hugh I am copying you and wrenches as some of my posts have not shown up.
The way I look at it is:  A charged battery has some number of molecules of Ions? each carry 2 electrons.  That is you store of Amp-hours, 6.25x10^18 electrons make 1 amp-sec.  Now at rest and 25 C the electrons want to excape at 2.1 volts per cell the voltage times the charge is the energy.  and if you charge slowly and discharge slowly one amp hour in should equal 1 amp hour out.  If you charge fast the ions cannot line up fastenough to collect their electrons and the voltage rises and some of the electrons make H2 and O2, causing water loss. You are charging at higher voltage than you will ever get out, and wasting both amp-hrs and voltage ie wasting energy. (some electrons make H2SO4 and some make H2 and O2) 

Now for Discharge: The way you get amp-hrs is the ions give up theie electrons. If the battery is cold the the Ions can not move very fast, and you have to force them, by using part of their desire to rid themselves of their electrons and the voltage drops.  (Now it also depends upon the distance the Ions have to move, so when you first start discharge the voltage is higher and after a while the voltage falls do to a lack of Ions near the plates called hysterises) The amp-hrs are there but the watt-hrs are not.  The watt-hrs you can get out depends upon the voltage and amp-hrs.  The voltage as we see depends upon the discharge rate, it also depends upon hysterisis. 

In my solar class I have the students measure hystersis at 25 C and slow rates.  

As to your important question can the LVCO be set and what should it be set at?. It is to complicated and I think the SOC method should be best.   

--- On Sat, 1/16/10, Hugh <hugh at scoraigwind.co.uk> wrote:

> From: Hugh <hugh at scoraigwind.co.uk>
> Subject: Re: [RE-wrenches] discharging Rolls batteries
> To: "RE-wrenches" <re-wrenches at lists.re-wrenches.org>
> Date: Saturday, January 16, 2010, 7:04 AM
> 
> Re: [RE-wrenches] discharging Rolls
> batteries 
> 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
> 
> and http://www.mpoweruk.com/soc.htm
> and 
> http://web.mit.edu/mit_energy/resources/iap/MatSciOfRenewEnergy_Lecture2_Batteries_2006.pdf
> and http://ecee.colorado.edu/~ecen4517/materials/Battery.pdf
> 
> and 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
> 
>  
>  
> 
> 
> 
> 
> 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: Hugh
> To: 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: Hugh
> 
> 
> To: 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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