[RE-wrenches] Strings and series of batteries with reversereturn linkup

[James Surrette]

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_solar at 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 <boB at midnitesolar.com> wrote:

> From: boB Gudgel <boB at midnitesolar.com>
> Subject: Re: [RE-wrenches] Strings and series of batteries with reverse return linkup
> To: "RE-wrenches" <re-wrenches at 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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