Hi Darryl, 

To further on your comments (I hope); 

One thing we've noticed that is an often overlooked (potential) problem.  Even 
with a simple series sting (6, 12, 24 cells), the individual cell voltages are 
never identical -  meaning initial charge acceptance is never identical.  When 
you initially "activate" or "commission" a bank, assuming 48VDC you begin to 
charge at 57.5-60V.  The overall voltage is attained relatively quickly but the 
individual cell voltages can be shockingly different (2.2 - 2.7V).  The 
commissioning is complete when all cell voltages "balance" to the charging 
voltage, i.e. 2.4V.   As series voltages increase (120VDC+) or multiple banks 
are used, the initial commissioning becomes increasingly more important as 
overcoming imbalance within the bank become more difficult.  The most efficient 
commissioning occurs when low current is applied (2% of 20hr) until cells 
balance.  The point to the comment, you might get away without balancing / 
commissioning / activating a series string but is imperative if using multiple 
parallel banks. 

Regards, 

Jamie

>>> Darryl Thayer <daryl_so...@yahoo.com> 12/5/2009 9:31 PM >>>
This also assumes all the identical batteries are at identical temperature, and 
it one is slightly warmer or can not lose heat as fast, if being charged it 
will get warmer.  the warmer battery will do what? have a lower vltage, have 
higher current and receive more heat, and get warmer, this is significant.  I 
have never seen a battery set tha all batteries have the same thermal 
conductance to ambient
Darryl

--- On Sat, 12/5/09, boB Gudgel <b...@midnitesolar.com> wrote:

> From: boB Gudgel <b...@midnitesolar.com>
> Subject: Re: [RE-wrenches] Strings and series of batteries with reverse 
> return linkup
> To: "RE-wrenches" <re-wrenches@lists.re-wrenches.org>
> Date: Saturday, December 5, 2009, 3:45 PM
> Antony Tersol wrote:
> > Assume 3 parallel battery stings, with identical
> batteries of internal resistance r.
> >
> > Assume identical wire leads of resistance R.
> >
> > Then when the batteries have the same state of charge
> and their internal resistances are equal, one can solve for
> the current in each string.
> >
> > I1 = I3
> > I2 = I1 r / (r+R)
> >
> > where I2 is charging current thru the middle string.
> >
> > The amount that the middle current is reduced is a
> function of the relative sizes of the wire and battery
> resistances.
> >
> > For R << r, I2 --> I1.
> > For r << R, I2 --> 0.
> >
> > With a buss bar arrangement, I1 = I2 = I3 = V/(2R +
> r)
>
> One thing this does not take into account though, is the
> inductance of all the wires brought
> on by the wire lengths and their loop area, which in a
> usual inverter with ripple current, will also limit the
> current
> through the system...   That will also have
> some undesired consequences with the voltages
> seen at the inverter terminals due to resonances, etc.
>
> In ~can~ be significant.
>
> boB
>
>
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